Paper 1 –Set A Solutions

Regn No: _________________

Name: ___________________
(To be written by the candidate)

9th NATIONAL CERTIFICATION EXAMINATION – December, 2009

FOR
ENERGY MANAGERS AND ENERGY AUDITORS

PAPER – 1: General Aspects of Energy Management & Energy Audit

Date: 19.12.2009 Timings: 0930-1230 HRS Duration: 3 HRS Max. Marks: 150

General instructions:
o Please check that this question paper contains 12 printed pages
o Please check that this question paper contains 64 questions
o The question paper is divided into three sections
o All questions in all three sections are compulsory
o All parts of a question should be answered at one place

Section – I: OBJECTIVE TYPE Marks: 50 x 1 = 50

(i) Answer all 50 questions.

(ii) Each question carries one mark.
(iii) Please hatch the appropriate oval in the OMR answer sheet with Black Pen or HB
pencil, as per instructions

1 To maximize the combustion efficiency, which of the following in the flue gas needs to be
done?

a) maximize O2 b) maximize CO2 c) minimize CO2 d) maximize CO

2 Which of the following is not a primary energy source?

a) electricity b) coal c) wood d) natural gas

3 Which one of the following is not an example of air pollution from furnace oil fired boilers
and furnaces?

a) sulphur dioxide (SO2) b) chloro-fluro carbons (CFC)

c) nitrous oxide(NOX) d) carbon monoxide (CO)
4 Which industry among the following is not a designated consumer as per EC Act-2001?

a) fertilisers b) chlor alkali c) cement d) nuclear power stations

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5 Propane is an example of

a) nuclear energy b) radiant energy

c) chemical energy d) thermal energy
6 The density of a fuel oil is 0.86. Its specific gravity will be

a) 0.75 b) 0.86 c) 1.75 d) 0.0086

7 An indication of sensible heat content in air-water vapour mixture is

a) wet bulb temperature b) dew point temperature

c) density of air d) dry bulb temperature
8 The total mechanical energy of a body free falling in a vacuum

a) increases b) decreases
c) remains the same d) depends on the shape of the body
9 How much carbon emission will be reduced per year by replacing 60 Watt incandescent
lamp with 15 Watt CFL Lamp, if emission per unit is 1 kg CO2 per kWh and annual burning
is 3000 hours?

a) 45 ton b) 3 ton c) 0.135 ton d) 183 ton

10 Which of the following is false?

a) electricity is high-grade energy

b) high grade forms of energy are highly ordered and compact
c) low grade energy is better used for applications like melting of metals rather than
heating water for bath
d) the molecules of low grade energy are more randomly distributed than the molecules of
carbon in coal
11 The present value of equipment is Rs. 10,000 and discount rate is 10%. The future value
of the cash flow at the end of 2 years is:

a) Rs. 10000 b) Rs. 12,100 c) Rs. 8100 d) Rs. 8264

12 What will be the energy saving if one 1500 watts, 25 liter water heater , which is normally
put on for 20 minutes per day and 250 days in an year, is replaced with one 100 liter
capacity solar water heater?

a) 581units b) 750 units C) 125 units d) 169 units

13 An energy audit as defined in the Energy Conservation Act 2001 does not include

a) action plan to reduce energy consumption

b) verification, monitoring and analysis of use of energy
c) submission of technical report with recommendations
d) implementation of all the recommendations of energy audit
14 The difference between GCV and NCV of coal is

a) the heat of vaporization of the moisture and hydrogen in coal

b) the difference in calorific value with theoretical air and with allowable excess air
c) difference in accounting the un-burnt content in the ash
d) none of the above

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15 A public expression of organization’s commitment to energy conservation would be

a) reduce contract demand b) energy audit

c) energy policy d) improve power factor
16 For calculating plant energy performance which of the following data is not required

a) current year’s production b) reference year’s production

c) reference year energy use d) capacity utilization
17 Which of the following is not applicable to liquid fuels?

a) the viscosity of a liquid fuel is a measure of its internal resistance to flow.

b) the viscosity of all liquid fuels decreases with increase in its temperature
c) higher the viscosity of liquid fuels, higher will be its heating value
d) viscous fuels need heat tracing
18 Non contact speed measurements can be carried out by

a) odometer b) tachometer c) stroboscope d) oscilloscope

19 In the first two months the cumulative sum is 4 and 12 respectively. In each of the next
two months Ecalculated is more than Eactual by 3. The energy savings at end of the fourth
month would be

a) -6 b) 0 c) 6 d) none of the above

20 The statement not applicable in the case of Sankey diagram

a) useful tool to represent an entire input and output energy flow

b) represents visually various outputs and losses
c) depicts rejection and wastage of material flow
d) helps energy manager to focus on finding improvements in a prioritized manner
21 The Designated National Agency (DNA) of India for Clean Development Mechanism
(CDM) is

a) Ministry of Environment and Forests (MoEF)

b) Bureau of Energy Efficiency (BEE)
c) Central Electricity Authority (CEA)
d) Central Electricity Regulatory Commission (CERC)
22 Which among the following can be best implemented through an ESCO (Energy Service
Company) route:

a) coal procurement contract for captive power plant

b) energy efficient design of a municipal lighting system
c) large Waste Heat Recovery System in a large process plant, where external
financing is sought
d) energy and mass balance study of a Steel Plant
23 A CUSUM graph follows a random fluctuation trend and oscillates around

a) 50% b) 100% c) 0 d) mean value

24 Statement not applicable to TOD (Time of the Day) in electricity tariff structure?

a) higher energy charges during peak period

b) it is an incentive to maximize off- peak consumption
c) it is an incentive to minimize peak time power draw from the grid by consumers

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d) it is a disincentive for Distribution Company
25 Which of the following will be true of load factor for a continuous process

a) higher than batch process plants

b) comparable to that of a five star hotel with 60% occupancy
c) less than that of an energy efficient municipal lighting system
d) closer to the regional grid load factor
26 In a heat treatment furnace the material is heated up to 800 °C from ambient temperature
of 30 °C. Considering the specific heat of material as 0.13 kCal / kg °C, what is the energy
content in one kg of material after heating?

a) 700 kCal b) 250 kCal c) 350 kCal d) 100 kCal

27 In an industry the average electricity consumption is 4.6 lakh kWh for the period, the
average production is 40000 tons with specific electricity consumption of 10 kWh /ton for
the same period. The fixed electricity consumption for the plant is :

a) 60000 kWh b) 46000 kWh c) 20000 kWh d) none of the above

28 The process by which Annex 1 countries can invest in the GHG mitigation projects in
developing countries is called:

a) green trading b) clean development mechanism

c) conference of parties d) certified emission reduction
29 All the activities falling in the critical path of the PERT network will have

a) ES = LS and EF = LF b) LF = LS and ES=EF

c) only ES=LS d) only EF = LF
30 The cost of replacement of inefficient chiller with an energy efficient chiller in a plant was
Rs. 10 lakh .The net annual cash flow is Rs 2.50 lakh .The return on investment is:

a) 18% b) 20% c) 15 % d) none of the above

31 Condensation of saturated steam releases

a) sensible heat b) super heat c) latent heat d) none of the above

32 Calorific Value of coal is measured by a device called

a) bomb calorimeter b) calorifier

c) infrared thermometer d) none of these
33 The first vital step in an energy management program is

a) measurement b) setting goals

c) energy audit d) top management commitment
34 The calorific value of coal is 5000 kCal /kg .Find out the oil equivalent of 200 kg of coal if
the calorific value of oil is 10000 kCal/kg.

a) 100 kg b) 108 kg c) 105 kg d) none of the above

35 Which gas has the least impact on global warming?

a) carbon dioxide b) methane

c) ozone d) carbon monoxide

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36 Which one is not an energy consumption benchmark parameter?

a) kCal/kWh of electricity generated b) kg/ deg C.

c) kW/ton of refrigeration d) kWh/kg of yarn
o
37 Calculate the amount of electricity required to heat 100 litres of hot water from 20 to 60 C

a) 4.65 kWh b) 0.465 kWh c) 465 kWh d) 2 kWh

38 A person can do the following with wind energy

a) destroy it b) convert it c) create it d) burn it

39 In a force field analysis in energy action planning, high price of energy acts as

a) positive force b) negative force

c) neutral forces d) none of the above
40 The four pillars of successful energy management are technical ability, monitoring system,
top management support and ______

a) strategy plan b) energy audit plan c) quality plan d) financial plan

41 The 80/20 Rule in management means

a) few (20%) are vital and many (80%) are trivial

b) many (80%) are vital and few (20%) are trivial
c) 80% of work is outsourced
d) 20% of work is outsourced
42 The contractor provides the financing and is paid an agreed fraction of actual savings
achieved. This payment is used to pay down the debt costs of equipment and/or services.
This is known as

a) traditional contract b) extended technical guarantee/service

c) performance Contract d) shared savings performance contract
43 In project financing ,sensitivity analysis is applied because

a) almost all the cash flow methods involve uncertainty

b) of the need to assess how sensitive the project to changes in input parameters
c) what if one or more factors are different from what is predicted
d) all the above situation
44 One thousand liters of fuel oil cost Rs 20,000. How much does one kg of fuel oil cost if
density is 0.98

a) 20.40 b) 20.0 c) 19.02 d) none of the above

45 Installed capacity of nuclear power plants in India as a % of total installed capacity is

a) 10% b) 25% c) 3% d) 55%

46 Name the sector which is the biggest consumer of commercial energy

a) industry b) agriculture c) transport d) residential

47 To calculate internal rate of return , the net present value is set to

a) 1 b) 0 c) 10 d) 100

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48 Project management technique which uses three time estimates.

a) PERT b) CUSUM c) CPM d) none of the above

49 One Certified Emission Reduction (CER) in equivalent of CO2 emission is

a) 1 ton of CO2 b) 1 kg of CO2 c) 10 kg of CO2 d) 10 ton of CO2

50 A system uses 100 kg of raw material A, 200 kg of raw material B and 220 kg of raw
o
material C. The mix is heated to 220 C. The air is blown to cool which carries away on an
average 60% of raw material A and 30% of raw material B through the chimney. The
output product would be

a) 520 kg b) 400 kg c) 312 kg d) 208 kg

……. End of Section – I …….

Section – II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40

(i) Answer all Eight questions

(ii) Each question carries Five marks

S-1 Explain the difference between contract demand and maximum demand.

Ans

Contract Demand
Contract demand is the amount of electric power that a customer demands from
utility in a specified interval. Unit used is kVA or kW. It is the amount of electric
power that the consumer agreed upon with the utility. This would mean that utility
has to plan for the specified capacity.
(2.5 Marks)
Maximum Demand
Maximum demand is the highest average kVA recorded during any one-demand
interval within the month. The demand interval is normally 30 minutes, but may
vary from utility to utility from 15 minutes to 60 minutes. The demand is measured
using a tri-vector meter / digital energy meter.
(2.5 Marks)
S-2 An induction motor draws 8 kW with a lagging reactive power of 4 kVAr.
Calculate the operating power factor of the motor.

kW
Ans. Power factor =
(kW ) 2  (kVAr ) 2
(2 Marks)
8
=
(8) 2  ( 4) 2
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 Paper 1 –Set A Solutions
= 0.905
(3 Marks)

S-3 What are the environmental impacts of combustion of fossil fuels?

Ans Sulphur dioxide (SO2), nitrous oxide (NOX) and carbon monoxide (CO) and
Carbon Dioxide are major emissions from combustion of fossil fuels. SO2
and NOx lead to acid rain. CO is unstable and toxic. CO2 is a green house
gas, which leads to global warming.
(5 Marks)

S-4 Fuel substitution may not always result in energy savings. Explain with an
example.

Ans Let us say we are replacing an oil fired boiler by a wood fired boiler for
economic reasons. The losses in wood fired boiler may be more and hence
to compensate for that we need to supply more energy for the same output.
Thus it will increase the energy consumption.
(5 Marks)

S-5 What parameters are measured with the following instruments?

a) Pitot tube
b) Stroboscope
c) Fyrite
d) Lux meter
e) Power analyser

Ans

a. Pitot tube - static, dynamic and total pressure

b. Stroboscope - speed, RPM
c. Fyrite - CO2 or O2
d. Lux meter - Light levels in Lux (lumens/m2)
e. Power analyser - kW, kVA, Power factor, V, A etc
(1 Mark each)

S-6 The initial temperature of 150g of ethanol was 22oC. What will be the final
temperature of the ethanol if 3240 J was needed to raise the temperature of the
ethanol? (Specific heat capacity of ethanol is 2.44 J/goC).

Ans q = m x Cg x (Tf - Ti)

q = 3240 J
m = 150g
Cg = 2.44 J/goC
Ti = 22oC
3240 = 150 x 2.44 x (Tf - 22)
3240 = 366 (Tf - 22)
8.85 = Tf - 22
Tf = 30.9oC(5 Marks)
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S-7 Explain the importance of TOD (time of the day) tariff and its advantage to
utilities and to users. List some of the Industries where TOD benefits can be
fully utilised.

Ans Many electrical utilities like to have flat demand curve to achieve high
plant efficiency. They encourage user to draw more power during off-peak
hours (say during night time) and less power during peak hours. As per
their plan, they offer TOD Tariff, which may be incentives or disincentives.
Energy meter will record peak and non-peak consumption separately by
timer control. TOD tariff gives opportunity for the user to reduce their
billing, as off peak hour tariff charged are quite low in comparison to peak
hour tariff.
(3 Marks)
a. Operation of additional Cement Mills in Night periods
b. Operation of Melting furnaces in night shift
c. Heat treatment furnace operation in Night shift
d. Storage pump operation in large complex in night shift
(2 Marks)

S-8 Explain why a project with a high IRR is not necessarily more attractive than a
project with lower IRR?

Ans The IRR can not be distinguished between lending and borrowing and
hence a high IRR is not necessarily a desirable feature.
Also the NPV is the increase in shareholder or companies wealth through
the project. It therefore can happen that a project A has a very high IRR
but low NPV; while another Project B has low IRR but High NPV .In this
case B should be selected
(5 Marks)

……. End of Section – II …….

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60

(i) Answer all Six questions

(ii) Each question carries Ten marks

L-1 Steam flow from a water tube boiler is measured by measuring the feed water
tank levels. In two hours there is a drop of 200 cm in level. The cross sectional
area of the tank is 9 m2. Assuming no blowdown, calculate the steam flow rate.
The enthalpy of steam is 662 kCal/kg and feed water temperature is 80oC.
The calorific value of fuel used in the above boiler was measured by a
continuous flow calorimeter. The following data were obtained.

Mass of fuel : 2.25 kg

Inlet water temp : 30oC
Quantity of water : 360 litres
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Outlet water temp : 84oC
% heat transferred to water in bomb calorimeter : 85 %
(a) Calculate the calorific value of the sample (kCal/kg).
(b) If the efficiency of the boiler is 85% find out the fuel firing rate (kg/hr).

Ans To calculate the calorific value of the fuel

(a) Heat carried by water = 360 x 1x 54

= 19440 kcal

Calorie released by fuel = 19440/0.85

= 22871 kcal
= 10,164 kcal/kg
(5 Marks)
(b) Volume of flow in 2 hours = 9 x 2m
= 18 m3
Steam flow rate = 18/2 = 9 TPH
Fuel firing rate = 9000 kg/hr x (662-80)/(0.85 x 10164)
= 606.29 kg/hr

(5 Marks)

L-2 Investment for a set of interrelated energy efficiency projects identified in a

medium size process plant woks out to Rs.12.00 lakh. Annual savings for the
first four consecutive years are Rs. 300,000, Rs. 400,000, Rs. 400,000 and Rs.
450,000, respectively. The cost of capital is 12% p.a.
What is the net present value (NPV)? And as per NPV, suggest weather the
plant can go ahead with the projects.

Ans. NPV= (-) 12,000,00/(1.12)0+3,50,000/(1.12)1+4,00,000/(1.12)2+

4,00,000/(1.12)3+ 4,50,000/(1.12)4

= (-)12,00,000+12,02073 = Rs. 2073

(8 Marks)

Hence the decision rule associated with the net present value criterion is:
“Accept the project if the net present value is positive and rejects the
project if the net present value is negative”.
Therefore the project is acceptable.
(2 Marks)

L-3 What is Energy Security? Mention few strategies countries adopt to ensure this?

Ans. The basic aim of energy security for a nation is to reduce its dependence
on the imported energy sources for its economic growth.
(2 marks)

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Some of the strategies for energy security are:

 Building Stock Piles

 Diversification of energy supplies sources.
 Increased capacity of fuel switching
 Demand restraint.
 Development of renewable energy sources.
 Energy Efficiency.
 Sustainable development
(8 marks, 2 marks for each for any four points)
L-4 (a) Construct a PERT Chart diagram for the data given below
(b) Identify the critical path and compute the project duration. Also compute the
earliest start, earliest finish, latest start & latest finish of all activities

Activity Precedent Time, weeks

A 4
B A 3
C A 2
D B 5
E B 3
F C,D 4
G E,F 3

Ans (i)
E
B3
3
A B
5 G
1 2 3 D F 5 6
3
4 C
4 4
2
E

[3 marks]
D1 and D2 are dummy activities

ii) Critical Path

A- B- D- F- G

Total time on critical path: 19 weeks

[1 mark]

Early start (ES), Early Finish (EF), Latest start (LS), Latest finish (LF)

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S.no Activity Duration ES EF LS LF

1 A 4 0 4 0 4
2 B 3 4 7 4 7
3 C 2 4 6 10 12
4 D 5 7 12 7 12
5 E 3 7 12 13 16
6 F 4 12 16 12 16
7 G 3 16 19 16 19

(6 marks)
L-5 A 500 MW coal plant based on conventional pulverized fuel has a gross
efficiency of 38%. The Gross calorific value of the coal used is 4000 kCal/kg
with 40% total carbon. A supercritical unit of 500 MW replaces the plant with a
gross efficiency of 40% using the same characteristic coal. Calculate the
following

(a) Specific coal consumption after replacement

(b) Amount of coal and carbon di-oxide saved during a year if the plant works
for 8000 hours.
Ans Case (a) Before replacement
Heat rate = 860 = 2263 kcal/kwh
0.38

Specific coal consumption = heat rate / GCV = 2263/4000 = 0.565

kg/kwh
[3 marks]
Case (b) After replacement

Heat rate of the plant = 860 = 2150 kcal/kWh

0.40

Specific coal consumption = heat rate / GCV

= 2150/4000 = 0.537 kg/kWh

Saving in Coal consumption = 0.565 – 0.537 = 0.028 kg/kWh

Amount of power generated in kWh = 500 x 1000 x8000 = 4 x 10 9

Amount of coal saved in a year = 0.028kg/kWh * 4 x 10 9

= 112000 tonnes

Amount of CO2 saved = 44/12 * 112000* 0.40 = 164267 tonnes

[7 marks]

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L-6 (a) When the same quantity of heat is added to the same mass of Iron and
copper pieces, the temperature of Iron piece rises by 15 OC. Calculate the
rise in temperature of Copper piece, if the specific heat of Iron is 470 J/kgoC
and that of Copper is 390 J / kg oC?

(b) The input to a textile dryer is 60 kg of wet cloth per hour with 55% moisture.
If it is dried to 10% in a dryer, estimate the moisture removed per hour.

Ans Mass of Iron x Sp. Heat Iron x 15 OC = Mass of Copper x Sp. Heat Copper
(a) x (Rise in Temp of Copper OC)
Since mass of Iron = Mass of Copper
Sp. Heat Iron x 15 OC = Sp. Heat Copper x (Rise in Temp of Copper OC
Sp. Heat of Iron = 470 J / kg OC
Sp. Heat of Copper = 390 J / kg OC
Hence, Rise in Copper Temp. = (470 x 15 ) / 390
= 18.08 OC
[5 marks]

Ans 60 kg of wet cloth contains, 60 x 0.55 kg water = 33 kg moisture,

(b) and 60 x (1-0.55) = 27 kg bone dry cloth.
As the final product contains 10% moisture,
If m is the moisture in kg, [m/(27+m) ]= 0.1
m =2.7+0.1m
0.9m = 2.7
m = 2.7/0.9
the moisture in the product is 27/9 = 3 kg
And so Moisture removed / hr = 33 - 3 = 30 kg/hr
[5 marks]

……. End of Section – III …….

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 Paper-1 Set A Key

Regn No: ____ _____________

Name: ___________________
(To be written by the candidates)

8th NATIONAL CERTIFICATION EXAMINATION – May, 2009

FOR
ENERGY MANAGERS & ENERGY AUDITORS

PAPER – 1: General Aspects of Energy Management & Energy Audit

Date: 23.05.2009 Timings: 0930-1230 HRS Duration: 3 HRS Max. Marks: 150

General instructions:
o Please check that this question paper contains 7 printed pages
o Please check that this question paper contains 64 questions
o The question paper is divided into three sections
o All questions in all three sections are compulsory
o All parts of a question should be answered at one place

Section – I: OBJECTIVE TYPE Marks: 50 x 1 = 50

(i) Answer all 50 questions

(ii) Each question carries one mark
(iii) Please hatch the appropriate oval in the OMR answer sheet with Black Pen or
HB pencil, as per instructions

1. India’s proven oil reserves is about __% of total world reserves

a) 0.1 b) 2 c) 0.4 d) 4
2. The average operating Station Heat Rate (SHR) of a Thermal Power Plants is 3000
kCal/kWh; corresponding to this the station thermal efficiency in percentage is

a) 28.7% b) 33.3% c) 40.1% d) 15.5%

3. Which of the following may not be a suitable energy security option for India ?

a) improving Energy Efficiency b) increasing Jatropha Cultivation

c) increasing Renewable Energy use d) increasing oil fired thermal power stations
4. The Sector which is not a Designated consumer as per the latest Gazette is

a) Railways b) Iron & Steel c) Thermal Power plants d) Commercial Buildings

5. As per Energy Conservation Act, 2001 appointment of BEE Certified Energy Manger is
mandatory for

a) Designated Consumers b) Urban Local Bodies

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c) All Captive Power Plants d) Electrical Distribution Licensees

6. Which of the following could be a better index of Building Energy Performance?

a) solar Heat Gain factor by wall and roof

b) ratio of air-conditioned to total carpet area
c) kWh per unit area of the building per year(kWh /Sq. meter/year)
d) ratio of Captive generation to grid power consumed
7. Calculate the amount of electricity required to heat 1 bucket (20 lit) of hot water from 30-
0
50 C

a) 465 kWh b) 1 kWh c) 0.465 kWh d) 2 kWh

8. Which of the following energy storage methods has the highest energy densities (kWh per
kg)?

a) hydrogen b) compressed air c) gasoline d) lead acid battery

9. The rate of energy transfer is measured in

a) Watts b) kCal/hr c) BTU/hr d) all of the above

10. A three phase induction motor is drawing 15 Ampere at 440 Volts. If the operating power
factor of the motor is 0.80 and the motor efficiency is 90%, then the power drawn by the
motor is

a) 9.14 kW b) 8.2 kW c) 10.16 kW d) 5.86 kW

11. Nameplate kW or HP rating of a motor indicates

a) input kW to the motor b) output kW of the motor

c) minimum input kW to the motor d) maximum input kW to the motor
12. Water flow in pipes is measured by

a) pressure gauge b) infrared imaging c) ultrasonic meters d) anemometer

13. The generally accepted energy performance parameter for thermal power plants would be

a) heat rate (kCal/kWh) b) % aux. power consumption

c) specific coal consumption d) all the above
14. The type of energy possessed by the charged capacitor is

a) kinetic energy b) electrostatic c) potential d) magnetic

15. Matching energy usage to requirement means providing

a) just theoretical energy needed b) just the design needs

c) energy with minimum losses d) less than what is needed
16. In general Specific Energy Consumption of a process plant for a given product mix and
raw material mix

a) increases with increasing capacity utilization

b) decreases with increasing capacity utilization
c) remain almost constant irrespective of capacity utilization
d) has no relation with capacity utilization

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17. Which of the following will not motivate employees to conserve energy?

a) training b) awareness c) enforcing targets d) incentives

18. Which of the following best defines the role of an energy manager ?

a) energy auditor and in charge of the finance department of the plant

b) intermediate player between top management, energy and cost centers of the plant
c) in charge of the captive power plant and mediator between plant and Electricity boards
d) mediator between production manager and project manager
19. Ten barrels of crude oil is equivalent to

a) 1600 liters b) 2050 litres c) 2000 litres d) 10 liters

20. If feed of 100 tonnes with 40% moisture, is dried to 20% moisture, the amount of water
vapour evaporated would be

a) 20 b) 28 c) 30 d) 40
21. The force field analysis in energy action planning considers

a) positive forces only b) negative forces only

c) both negative and positive forces d) no forces
22. An oil fired boiler operates at an excess air of 20%. If the stoichiometric air to fuel ratio is
14 and oil consumption is 200 kg per hour then the mass of flue gas leaving the boiler
chimney in kg/ hour would be

a) 3560 b) 3360 c) 3460 d) 3660

23. Which of the following is false?

a) 1 calorie = 4.187 kJ b) 1 calorie = 4.187J

c) 1000 kWh = 1 MWh d) 860 kCal = 1 kWh
24. The contract in which the ESCO provides financing and is paid an agreed fraction of
actual savings as they are achieved is called

a) traditional contract b) extended financing terms

c) shared savings performance contract d) energy service contract
25. Which of the following equation is used to calculate the future value of the cash flow?
n n n n
a) NPV (1 – i) b) NPV + (1 – i) c) NPV (1 + i) d) NPV/ (1 + i)
26. The cost of installation of a variable speed drive in a plant is Rs 2 lakh. The annual
savings are Rs 0.5 lakh. The maintenance cost is Rs. 5,000/year. The return on
investment is

a) 25% b) 22.5% c) 24% d) 27.5%

27. A sensitivity analysis is an assessment of

a) cash flows b) risks due to assumptions

c) capital investment d) best financing source
28. For an investment which has a fluctuating savings over its project life which of these
analysis would be the best option

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a) Payback b) NPV c) ROI d) IRR

29. ROI must always be ___ than interest rate

a) lower b) higher c) equal d) no relation

30. A sum of Rs 10,000 is deposited in a bank at the beginning of a year. The bank pays 10%
interest annually. How much money is in the bank account at the end of the fifth year, if no
money is withdrawn?

a) 16105 b) 15000 c) 15500 d) 16000

31. A contract in which the costs are paid from all or part of the energy cost savings is called

a) performance contract
b) traditional contract
c) extended technical guarantee contract
d) guaranteed savings performance contract
32. The annual electricity bill for a plant is Rs 10 lakhs and accounts for 38% of the total
energy bill. Furthermore the total energy bill increases by 5% each year. How high is the
plant’s annual energy bill at the end of the third year?

a) Rs 30.46 lakhs b) Rs 26.32 lakhs c) Rs 38.42 lakhs d) none of the above

33. In Project Management the critical path in the network is

a) the quickest path b) the shortest path

c) the path from start to finish d) the path where no activities have slack
34. The technique used for scheduling the tasks and tracking of the progress of energy
management projects is called

a) CPM b) Gantt chart c) CUSUM d) PERT

35. If air consists of 77% by weight of nitrogen and 23% by weight of oxygen, the mean
molecular weight of air is,

a) 11.9 b) 28.92 c) 17.7 d) none of the above

36. The calorific value of oil is 10000 kCal/ kg. Find out the coal equivalent to replace 1 kg of
oil (Coal GCV = 16000 KJ/kg) assuming same efficiency

a) 0.625 kg b) 2.6 kg c) 2 kg d) none of the above

O O
37. In a heat exchanger, inlet and outlet temperatures of cooling water are 28 C & 33 C. The
cooling water circulation is 200 litres /hr. The process fluid enters the heat exchangers at
O O
60 C and leaves at 45 C. Find out the flow rate of the process fluid?
(Cp of process fluid =0.95)

a) 70 b) 631 c) 63 d) 570
38. Which of these is not a duty of an energy manager

a) Report to BEE
b) Provide support to accredited energy auditing firm
c) Prepare a scheme for efficient use of energy
d) Sign an energy policy
39. Large scattering on production versus energy consumption trend line indicates

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a) poor process control b) many forced outages

c) poor process monitoring d) none of the above
40. In a cumulative sum chart, if the graph is horizontal, then

a) nothing can be said

b) energy consumption is reduced
c) specific energy consumption is increasing
d) actual and calculated energy consumption are the same
41. Due to high oil prices, a firm goes for substitution of oil and with biomass in a boiler. The
following scenario is most likely

a) higher fuel cost b) better boiler efficiency

c) higher energy consumption d) less smoke in the stack
42. Which of the following Financial Analysis Techniques fails to consider the time value of
money

a) Simple Payback Period (SPB) b) Internal Rate of Return (IRR)

c) Net Present Value (NPV) d) none of the above
43. Assume project A has an IRR of 85% and NPV of Rs 15,000 and project B has an IRR of
25% and NPV of 200,000. Which project would you implement first if financing is available
and project technical life is the same?

a) B b) A c) cannot be decided d) question does not make sense

44. The fixed energy consumption for the company is 1000 kWh. The slope in the energy –
production chart is found to be 0.3. Find out the actual energy consumption if the
company is to produced 80,000 Tons of product

a) 25,000 b) 24,000 c) 26,000 d) 38,000

45. Which of the following is not an environmental issue of global significance

a) Ozone layer depletion b) Global Warning

c) Loss of Biodiversity d) Suspended particulate Matter
46. The commitment period for emission reductions of annex-I countries as per Kyoto protocol
is

a) since 2001 b) 2008 – 2012 c) there is none d) 2012 -2022

47. The Global Warming Potential (GWP) of Nitrous Oxide gas is

a) 1 b) 21 c) 270 d) 100
48. The national inventory of greenhouse gases in India indicates that about __ % of the total
CO2 emissions come from energy sector

a) 55% b) 15% c) 80% d) Negligible

49. Ozone layer is found in the stratosphere between

a) 5 to 10 km above the ground b) 10 to 50 km above the ground

c) 50 to 100 km above the ground d) 100 to 200 km above the ground
50. In CDM terminology CER means

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a) Carbon Emission Reduction b) Clean Environment Rating

c) Certified Emission Reduction d) Carbon Emission Rating

…….……. End of Section – I ………..….

Section - II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40

(i) Answer all Eight questions

(ii) Each question carries Five marks

S-1 What is a Sankey diagram and what are its uses ? Explain with an example.

ANS The Sankey diagram is a very useful tool to proportionally represent an entire input
and output energy flow in any energy equipment or system such as boilers, fired heaters,
furnaces etc. after carrying out an energy balance calculation, This diagram represents
visually various outputs and losses so that energy managers can focus on finding
improvements in prioritized manner.

Example:

S-2 Calculate the net present value over a period of 3 years for a project with the
following data. The discount rate is 15%.

Year Investment (Rs) Savings (Rs)

0 1,00,000
1 50,000 75,000
2 75,000
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3 75,000

ANS NPV = -100,000 + (75000-50000)/(1+0.15) + 75000/(1+0.15)2 + 75000/(1+0.15)3

= -100,000 + 21,739 + 56,710 + 49, 313

= Rs. 27,762

S-3 In a compressed air Dryer, electrical heater is used for regeneration of silica gel.
The present Electrical energy consumption is 100 kWh/day. The management
intends to replace the electrical heater by steam coil.
a) How much steam is need per day?
b) Calculate cost savings/year. Cost of power is Rs.4/kWh and cost of steam is
Rs. 500/ton (Assuming only latent heat of steam used. Latent heat of steam is
540 kCal/kg. Efficiency of steam heating is 70%, operating days = 300)

ANS Latent heat of steam = 540 kCal/kg

a) Amount of steam required = 100 * 860/(540*0.7)
= 227.5 kg/day

b) Cost of power = 100 * 4 = Rs.400/day

Cost of steam = 0.2275 * 500
= Rs. 113.75/day
Annual cost savings = (400 – 113.75) * 300
= 286.25*300
= Rs. 85,875/-

S-4 Renovation and Modernization (R&M) of a 210 MW coal fired thermal power plant
was carried out to enhance the operating efficiency from 28% to 32%. The specific
coal consumption was 0.7 kg/kWh before R&M. For 8000 hours of operation per
year, and assuming the coal quality remains the same, calculate
a) the coal savings per year and
b) the expected avoidance of CO2 into the atmosphere in Tons/year if the
emission factor is 1.53 kg CO2/kg coal

ANS Specific coal consumption of conventional power plant = 0.7 kg/kWh

Specific coal consumption of power plant after R&M= 0.7 x (0.28/0.32)
= 0.6125 kg/kWh
Coal savings per annum = (0.7-0.6125) x 210 x 1000 x 8000
= 1,47,000 Tonnes per annum
CO2 avoidance = 1,47,000 x 1.53

= 224910 tonnes per annum

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S-5 Explain Time of Day (TOD) Tariff and how it is beneficial for the power system
and consumers?

ANS i. In Time of the Day Tariff (TOD) structure incentives for power drawl
during off-peak hours and disincentives for power drawl during peak
hours are built in.
ii. Many electrical utilities like to have flat demand curve to achieve high
plant efficiency.
iii. ToD tariff encourage user to draw more power during off-peak hours
(say during 11pm to 5 am, night time) and less power during peak
hours. Energy meter will record peak and off-peak consumption and
normal period separately.
iv. TOD tariff gives opportunity for the user to reduce their billing, as off
peak hour tariff is quite low in comparison to peak hour tariff.
v. This also helps the power system to minimize in line congestion, in
turn higher line losses and peak load incident and utilities power
procurement charges by reduced demand

S-6 An energy meter connected to a 3 phase, 18.75 kW pump shows 108 units
consumption for six hours of operation. The load on the motor was steady. The
consumer doubted the energy meter reading and electrical parameter such as
current, voltage and power factor were measured. The measured values were
430 V line volts, 25 amps line current and 0.80 Power Factor. Find out if the
energy meter reading is correct.

ANS Energy consumption = √ 3 x 0.430 (kV) x 25(A) x 0.80(PF) x 6(hours)

= 89. 37 kWh
The consumption shown by energy meter is very high; higher by 21%

S-7 What are the three flexible mechanisms available under Kyoto protocol for
achieving GHG reduction targets. Explain briefly the mechanism applicable to
India.

ANS 1. Joint implementation

2. Emissions trading
3. Clean Development Mechanism (CDM)

CDM is applicable to India. It is a mechanism by which the developed countries

finance/fund projects intended to reduce GHG emissions in return for emission reduction
credits. This will help them to achieve the emission reduction targets. Alternatively the
emission reductions achieved by developing countries can also be purchased,

S-8 What is meant by the following terms ?

a) Normalising of data
b) Benchmarking

ANS a) Normalising of data

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The energy use of facilities varies greatly, partly due to factors beyond the energy
efficiency of the equipment and operations. These factors may include weather or certain
operating characteristics. Normalizing is the process of removing the impact of various
factors on energy use so that energy performance of facilities and operations can be
compared.

a) Benchmarking
Comparison of energy performance to peers and competitors to establish a relative
understanding of where our performance ranks.

…….……. End of Section – II ………..….

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60

(i) Answer all Five questions

(ii) Each question carries Ten marks

L-1 An agency is implementing energy efficiency measures in municipal water pumping

under ESCO route. The investment is Rs. 5 crores. Present annual electricity bill is
Rs. 4 Crores. The expected savings are 20%. (Cost of electricity = Rs. 4/ kWh,
Annual maintenance cost -10% of investment) The expected CDM revenues would
be Rs. 50 Lakhs/ year. Calculate IRR for this project over a 10 year period, after
including the CDM benefit.

ANS year outflow inflow Net flow

Rs.
Lakhs Savings CDM
0 -500 0 0 -500
1 -50 80 50 80
2 -50 80 50 80
3 -50 80 50 80
4 -50 80 50 80
5 -50 80 50 80
6 -50 80 50 80
7 -50 80 50 80
8 -50 80 50 80
9 -50 80 50 80
10 -50 80 50 80
IRR 10%

L-2 Energy saving measures was implemented in a process plant prior to Jan-2008.
Use CUSUM technique and calculate energy savings for 6 months period of 2008.
The company produced consistently 3000 T/month in each of the six months.
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Refer the graph given in table below.

The predicted Specific energy consumption for 3000 MT production is 260 kWh/MT.
It may be noted that retrofits were not functioning during March & May 2008.

Actual Specific Energy Consumption Profile

300
290 290
Consumption,kWh/MT

280 280
Specific Energy

270
260
250
240
240 239
236 235
230
220
210
200
Jan Feb Mar Apr May Jun

Month

ANS

Actual - Predicted Difference=

2008- SEC, SEC- Actual-
Month kWh/MT kWh/MT Predicted CUSUM
Jan 240 260 - 20 -20
Feb 236 260 - 24 -44
Mar 280 260 + 20 -24
Apr 235 260 - 25 -49
May 290 260 + 30 -19
Jun 239 260 - 21 -40

= 40 kWh/MT x 3000 MT
Energy Savings for six months = 1.20 lakh kWh

L-3 Production rate from a paper machine is 340 tonnes per day (TPD). Inlet and outlet
dryness to paper machine is 50% and 95% respectively. Evaporated moisture
temperature is 80 °C. To evaporate moisture, the steam is supplied at 3.5 kg /cm 2.
Latent heat of steam at 3.5 kg /cm2 is 513 kCal/kg. Assume 24 hours/day operation.
i) Estimate the quantity of moisture to be evaporated.
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ii) Input steam quantity required for evaporation (per hour).

Note: Consider enthalpy of evaporated moisture as 632 kCal/kg.
ANS Production rate from a paper machine = 340 TPD
= 14.16 TPH (tonnes per hour)

Inlet dryness to paper machine = 50%

Outlet dryness from paper machine = 95%

i) Estimation of moisture to be evaporated:

Paper weight in final product = 14.16 x 0.95
= 13.45 TPH

Weight of moisture after dryer = 0.708 TPH

Paper weight before dryer on dry basis = 13.45 TPH

Weight of moisture before dryer = ((13.45 x 100)/50) – 13.45 = 13.45 TPH

Evaporated moisture quantity : 13.45 - 0.708 = 12.742 TPH

ii) Input steam quantity required for evaporation

Evaporated moisture temperature = 80 °C
Enthalpy of evaporated moisture = 632 kCal/kg
Heat available in moisture (sensible & latent) = 632 x 12742
= 8052944 kCal /h

For evaporation minimum equivalent heat available should be supplied from

steam

Latent Heat available in supply steam (at 3.5 kg/cm2 (g)) = 513 kCal/kg

Quantity of steam required = 8052944/ 513

= 15697.75 Kg/hr
= 15.698 MT/hour

L-4 a) Draw PERT Chart for the following for the task, duration and dependency given
below.
b) Find out:
i. Critical Path
ii. expected project duration

Task Predecessors
Tasks
(Dependencies) Time (Weeks)
A - 3
B - 5
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C - 7
D A 8
E B 5
F C 5
G E 4
H F 5
I D 6
J G-H 4
ANS

 The critical path is through activities C, F, H, J

 The expected project duration is 21 weeks (7+5+5+4)

L-5 a) Draw an energy balance for the DG set with following data.
A Diesel Generator trial gives Set 3.5 kWh per Liter of diesel. The cooling water
loss and exhaust flue gas loss as percentage of fuel input are 28% and 32%
respectively. Assume calorific value of diesel as 10200 kCal/kg. The Specific
gravity of Diesel is 0.85. Calculate unaccounted loss as percentage of input
energy.
b) Explain the following terms in heat transfer with examples.
i) Conduction ii) Convection

ANS a) Power Output = 3.5 kWh x 860 kCal/kWh = 3010 kCal/litre

Power Output as % of fuel input = (3010/(0.85x10200)x100 ~ 34.7%
Losses as % of fuel input:
Cooling Water Loss = 32%
Exhaust Flue Gas Loss = 28%
Therefore, unaccounted Loss = 100-(34.7+32+28) = 5.3%

b)
i) Conduction
The conduction of heat takes place, when two bodies are in contact with one

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another. If one body is at a higher temperature than the other, the motion of the
molecules in the hotter body will vibrate the molecules at the point of contact in the
cooler body and consequently result in increase in temperature.

ii) Convection The transfer of heat by convection involves the movement of a fluid
such as a gas or liquid from the hot to the cold portion. There are two types of
convection: natural and forced. In case of natural convection, the fluid in contact
with or adjacent to a high temperature body is heated by conduction. As it is
heated, it expands, becomes less dense and consequently rises. This begins a fluid
motion process in which a circulating current of fluid moves past the heated body,
continuously transferring heat away from it. In the case of forced convection, the
movement of the fluid is forced by a fan, pump or other external means. A
centralized hot air heating system is a good example of forced convection.

L-6 a) Explain the functioning of an ESCO in performance contracting.

b) Name two macro and micro factors considered in the sensitivity analysis of
major energy saving projects.
Ans a)

ESCOs are usually companies that provide a complete energy project service, from
assessment to design to construction or installation, along with engineering and
project management services, and financing.
In performance contracting, an end-user (such as an industry, institution, or utility),
seeking to improve its energy efficiency, contracts with ESCO for energy efficiency
services and financing. An agreed portion of the savings is shared with the ESCO.
The ESCO gets back the invested money plus profit over a period of time

b)

Micro factors

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• Operating expenses (various expenses items)

• Capital structure
• Costs of debt, equity
• Changing of the forms of finance e.g. leasing
• Changing the project duration

Macro factors Macro economic variables are the variable that affects the operation
of the industry of which the firm operates. They cannot be changed by the firm’s
management. Macro economic variables, which affect projects, include among
others:
• Changes in interest rates
• Changes in the tax rates
• Changes in the accounting standards e.g. methods of calculating depreciation
• Changes in depreciation rates
• Extension of various government subsidized projects e.g. rural electrification
• General employment trends e.g. if the government changes the salary scales
• Imposition of regulations on environmental and safety issues in the industry
• Energy Price change
• Technology changes

…….……. End of Section – III ………..….

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Paper -1 Set A Solutions

13th NATIONAL CERTIFICATION EXAMINATION

FOR
ENERGY MANAGERS & ENERGY AUDITORS – September, 2012

PAPER – 1: General Aspects of Energy Management & Energy Audit

Date: 15.09.2012 Timings: 09:30-12:30 Hrs Duration: 3 Hrs Max. Marks: 150

Section - I: OBJECTIVE TYPE Marks: 50 x 1 = 50

(i) Answer all 50 questions

(ii) Each question carries one mark
(iii) Please hatch the appropriate oval in the OMR answer sheet with Black Pen or HB
pencil, as per instructions

1. The primary energy content of fuels is generally expressed in terms of ton of oil equivalent
(toe) and is based on the following conversion factor

a) 1 toe=10x106 kCal b) 1 toe=11630 kWh

c) 1 toe=41870 MJ d) all the above
2. Largest share of global primary energy consumption is from which of the following fuels:

a) oil and natural gas b) coal and oil c) oil and nuclear d) coal and nuclear

3. 1 kg of wood contains 15% moisture and 7% hydrogen by weight. How much water is
evaporated from wood during complete combustion of 1 kg of wood ?

a) 0.78 kg b) 0.22 kg c) 0.15 kg d) 0.63 kg

4. A power utility distributed 1 million 15 Watt CFLs for Rs 15 million, replacing 60

Watt incandescent lamps under Bachat Lamp Yojna. What will be the drop in power
in the evening on the demand side, if 80% of the lights are on at that time, assuming
similar numbers of incandescent lamps were switched on during the same period?
a) 360 kW b) 12 MW c) 36 MW d) 60
MW

5. Which of the following with respect to fossil fuels is true?

a) Reserve / Production (R/P) ratio is a constant once established

b) R/P ratio varies every year with only changes in production
c) R/P ratio varies every year with only changes in reserves
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d) R/P ratio varies every year with changes in both production and reserves
6.
From rated V, A and PF given in the name-plate of a motor , one can calculate:
a) rated input Power b) rated output Power c) both a & b d) none of
these

7.
Air velocity in the ducts can be measured by using ___________ and manometer
a) orifice meter b) Bourden gauge c) Pitot tube d) anemometer

8. One certified emission reduction (CER) is equivalent to:

a) one kg of carbon b) one kg of carbon dioxide
c) one ton of carbon d) one ton of carbon dioxide

9. Speed measurement (RPM) of an electric motor is measured with a

a) stroboscope b) ultrasonic meter c) lux meter d) rotameter

10. How much power generation potential is available in a run of river mini hydropower plant
for a flow of 40 liters/second with a head of 24 metres. Assume system efficiency of 60% ?

a) 5.6 kW b) 2.4 kW c) 4.0 kW d) 2.8 kW

11. A process requires 10 kg of fuel with a calorific value of 5000 kCal/kg. The system
efficiency is 80% The losses then will be

a) 10000 kCal b) 45000 kCal c) 40000 kCal d) 20000 kCal

12. Ratio of average load (kW) to maximum load (kW) is termed as

a) load factor b) demand factor c) form factor d) utilization factor

13. The ISO standard for Energy Management System is

a) ISO 9001 b) ISO 50001 c) ISO 14001 d) none of the above

14. Material and energy balance is used to quantify

a) material and energy losses b) profit

c) cost of production d) all of the above
15. Which is not a part of “ Energy Audit” defined as per the Energy Conservation Act, 2001

a) monitoring and analysis of energy use

b) ensuring implementations of recommended measures followed by review
c) submission of technical report with recommendations
d) verification of energy use

16. Which of the following statements regarding ECBC are correct?

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i) ECBC defines the norms of energy requirements per sq. metre of area taking into
account climatic region where building is located
ii) ECBC does not encourage retrofit of Energy conservation measures
iii) ECBC prescribes energy efficiency standards for design and construction of
commercial and industrial buildings
iv) One of the key objectives of ECBC is to minimize life cycle costs (construction and
operating energy costs)

a) i & ii b) i & iii c) ii & iii d) i & iv

17. Which of the following statements regarding BLY (Bachat Lamp Yojana) are correct ?

i) BLY aims at large scale replacement of all fluorescent lamps of poor lumen intensity
with CFL of high lumen intensity
ii) CDM is used as a tool to recover market price difference between lower cost replaced
incandescent lamps of 60 W and higher cost CFLs of 11 W
iii) BLY involves public, private partnership and DISCOM partnerships
iv) DSM is used as a tool to recover market price difference between lower cost replaced
incandescent lamps of 60 W and higher cost CFLs of 11 W

a) i & ii b) i & iii c) ii & iii d) i & iv

18. The average gross efficiency of thermal power generation on all India bases is about

a) 30 – 34% b) 36 – 38% c) 39 - 41% d) 25 - 28%

19. Which of the following is not the activity related to restructured APDRP?

a) separate feeders for agricultural pumps

b) energy auditing at distribution transformer level
c) GIS mapping of the network and consumers
d) establishing targets for reducing power consumption

20. Assuming total conversion of electrical energy to heat energy, how much heat is produced
by a 200 W heater in 5 minutes?

a) 200 kJ b) 40 kJ c) 1000 kJ d) 60 kJ
21. Which of the following statements regarding DSM is incorrect?

a) potential areas for DSM thrust activity are agriculture, domestic and municipalities
b) savings accrued through DSM can be treated as new power addition on supply side
c) under DSM, demand can be shifted from off-peak to peak hours thereby avoiding
imported power during off peak hours
d) DSM programs may result in demand as well as energy reduction

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22. A motor with 10 kW rating in its name plate, will draw Input power of____

a) 10 kW at full load b) more than 10 kW at full load

c) less than 10 kW at full load d) 10 kW at 110% of full load

23. Which of the following statements is not true regarding Maximum Demand Control?

a) Maximum demand control offers a way of ‘shaving’ the peaks and ‘filling’ the valleys in
the consumer load diagram
b) Maximum demand control is carried out by concerned utility at customer
premises
c) Maximum demand control focuses on critical load for management
d) All of the above

24. Which of the following statements is false?

a) reactive current is necessary to build up the flux for the magnetic field of inductive
devices
b) some portion of reactive current is converted into useful work
c) Cosine of the angle between kVA and kW vector is called power factor
d) power factor is unity in a pure resistive circuit

25. Steam leak reduction program can be best achieved through

a) Small Group Activities b) Autonomous Maintenance

c) TPM d) All of the above

26. Consider two competitive projects A and B each entailing investment of Rs.85,000/- .
Project A returns Rs.50,000 at the end of each year, but Project B returns Rs.115,000 at
the end of Year 2. Which project is superior?

a) project A since it starts earning by end of first year itself and recovers cost before end
of two years
b) project B since it offers higher return before end of two years
c) both projects are equal in rank
d) insufficient information to assess the superiority
27. Which of the following statements regarding Internal Rate of Return (IRR) is correct?

a) IRR distinguishes between lending and borrowing

b) Internal rate of return is the discount rate at which net present value is equal to
zero
c) if the IRR is higher than current interest rate, the investment is not attractive
d) between two alternative projects, the project with lower internal rate of return would be
considered more attractive

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28. Global warming will not result in

a) melting of the ice caps

b) increasing sea levels
c) increasing the size of the hole in the ozone layer
d) unpredictable climate patterns

29. The ozone layer found in the stratosphere:

a) protects against the sun’s harmful UV rays

b) can react with atmospheric pollutants to form smog
c) is toxic to plants
d) is capable of disintegrating fabric and rubber on earth

30. The process of capturing CO2 from point sources and storing them is called
____________

a) carbon capture and sequestration b) carbon sink

c) carbon capture d) carbon absorption

31. Which of the following statements is false regarding wind turbine?

a) wind power does not vary as the cross-sectional area of the rotor
b) wind power varies as cube of wind velocity
c) cut-in wind speed is always less than rated wind speed
d) theoretical maximum amount of energy in the wind that can be collected by wind turbine
rotor is about 95%

32. What percentage of the sun’s energy falling on a silicon solar panel gets converted into
electricity?

a) 25% b) 15% c) 75% d) 50%

33. Which of the following fuels is non-renewable?

a) LPG b) lignite c) nuclear d) all of the above

34. India’s share of world oil reserves is _________

a) 5% b) 2% c) 0.5 % d) 3%

35. Nuclear power development in India is constrained by

a) low % of Uranium in the ore b) inadequate supply of Uranium

c) constraints in import of Uranium d) all of the above

36. In a contract when all or part of the savings are guaranteed by contractor, and all or part of
the costs of equipment and/or services are paid out of savings as they are achieved, is
termed as
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a) traditional contract b) guaranteed saving performance contract

c) shared saving performance contract d) extended technical guarantee contract

37. A fuel cell is

a) an electromagnetic cell b) a magnetic cell
c) an electrochemical device d) none of the above

38.
If 3350 kJ of heat is supplied to 20 kg of ice at 0o C, how many kg of ice will melt into
water at 0o C (latent heat of melting of ice is 335 kJ/kg)
a) 1 kg b) 4.18 kg c) 10 kg d) 29
kg

39.
If oxygen rich combustion air (25% vol oxygen) is supplied to a furnace instead of normal
air (21% vol oxygen), the % CO2 in flue gases will
a) reduce b) increase c) remain same d) will become zero

40.
The return on investment (ROI), is expressed as
a) annual cost / capital cost b) (first cost / first year benefits) x 100
c) NPV / IRR d) (annual net cash flow x 100) /
capital cost

41.
The time between its earliest and latest start time, or between its earliest and latest finish
time of an activity is
a) delay time b) slack time c) critical path d) start time

42. In project management work breakdown structure defines

a) temporary endeavour undertaken to create unique product or service

b) the activities to be completed in the projects
c) how realistic were the assumptions underlying the project
d) none of the above

43.
The empirical relationship used to plot Production Vs Energy consumption is………………
( where Y= energy consumed for the period; C = fixed energy consumption; M = energy
consumption directly related to production; X= production for same period).

a) X=Y+MC b) Y=MX+C c) M=CX+Y d) Y= MX-C

44.
The main constituent of greenhouse gases (GHG) in atmosphere is

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a) CO2 b) SOx c) nitrogen d) water vapor

45. The Global Warming Potential (GWP) of nitrous oxide (N2O) is

a) 1 b) 23 c) 300 d) 5700
46. In project management, the critical path in the network is

a) the path where activates have slack b) the shortest path

c) the path where no activities have slack d) none of the above

47. The cost of a new heat exchanger is Rs. 1.0 lakh. The simple payback period in years
considering annual savings of Rs 60,000 and annual operating cost of Rs. 10,000 is

a) 0.50 b) 1.66 c) 2.00 d)

6.00
48. Of the total natural gas used in India, the largest share goes to__________sector.

a) petrochemicals b) fertilizers c) power d) domestic

49. What is the future value of Rs. 1000/- after 3 years if the interest rate is 10% ?

a) 1331 b) 3000 c) 3300 d) 2420

50. If the asset depreciation is considered, then net operating cash inflow would be

a) higher b) lower c) no effect d) none of these

Section – II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40

(i) Answer all Eight questions

(ii) Each question carries Five marks

S-1 Explain the concept of Bachat Lamp Yojana (BLY)?

ANS BLY aims at the large scale replacement of inefficient incandescent bulbs in
households by Compact Fluorescent Lamps (CFLs).

It seeks to provide CFLs to households at the same price to that of incandescent

bulbs. Clean Development Mechanism (CDM) is used to recover the cost difference
between the market price of the CFLs and the price at which they are sold to
households.

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BLY involves public-private partnership between the Government of India, private

sector CFL suppliers and State level Electricity Distribution Companies (DISCOMs).

Briefly explain ‘Renewable Purchase Obligation (RPO)’ and means by which this
S-2
requirement can be met ?
ANS RPO is Renewable Purchase Obligation requires each retail seller of electricity to
include in its resource portfolio a certain proportion of power is from renewable
sources such as wind, solar, small hydro or various forms of biomass energy.

The retailer can meet this requirement by owning a renewable energy facility and
producing power or purchasing power from another renewable energy facility.

Give a short description about Availability Based Tariff (ABT).

S-3
Introduction of availability based tariff(ABT) and scheduled interchange charges for
ANS
power introduced in 2003 for interstate sale of power , have reduced voltage and
frequency fluctuation (any three)
• It is performance-based tariff system for the supply of electricity by
generators owned and controlled by the central government.
• It is also a new system of scheduling and despatch, which requires both
generators and beneficiaries to commit to day ahead schedule.
• It is a system of rewards and penalties seeking to enforce day ahead pre-
committed schedules, though variations are permitted if notified one and a
half hours in advance.
• The order emphasis prompt payment of dues , non-payment of prescribed
charges will be liable for appropriate action

Any other relevant points as appropriate may also be given marks

In a process plant , an evaporator concentrates a liquor containing solids of 6% w/w

S-4
(weight by weight) to produce an output containing 30% solids w/w. Calculate the
evaporation of water per 400 kgs of feed to the evaporator.

Inlet solid contents = 6 %

ANS
Output solid contents=30%
Feed=400kgs
Solid contents in kg in feed =400 x 0.06 = 24 Kg
Outlet Solid contents in kg =24 kg

quantity of water evaporated=[400 – {(100) x 24}] = 320

30
Also right = 24x400/30 = 320

In the management of financial aspects, state what are micro and macro factors and
S-5
list three factors in each, which influence sensitivity analysis?
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ANS
Micro factors are variables related to the project being implemented and can be
influenced

Any three in the following list

ü Operating expenses (various expenses items)

ü Capital structure
ü Costs of debt, equity
ü Changing of the forms of finance e.g. leasing
ü Changing the project life

Macro factors are variables that affect the operation of industry of which the
company operates and cannot be changed by the company management.

Any three in the following list

ü Changes in interest rates

ü Changes in the tax rates
ü Changes in the accounting standards e.g. methods of calculating depreciation
ü Changes in depreciation rates
ü Extension of various government subsidized projects e.g. rural electrification
ü General employment trends e.g. if the government changes the salary scales
ü Imposition of regulations on environmental and safety issues in the industry
ü Energy Price change

A plant is using 6 ton/day of coal in a boiler to generate steam at 72% boiler

S-6
efficiency. The gross calorific value of coal is 3300 kCal/kg. The cost of coal is Rs.
4200/ton . The plant substitutes coal with agro-residue , as a boiler fuel, which has a
gross calorific value of 3100 kCal /kg and costs Rs. 1800/ton. The boiler efficiency
reduces to 68%. Calculate the annual cost savings for 300 days of operation with
agro residue as fuel.

ANS
Useful energy to generate steam by 6 tonnes of coal per day
= 6000 x 3300 x 0.72 = 14256000 kcal/day

To deliver 14256000kcal/day , daily amount of agro residue required

= 14256000
3100 x 0.68
= 6763 kg/day

Daily saving = 6000 x 4200 - 6763 x 1800

1000 1000
= 25200-12173
= Rs 13027/-

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Annual saving =13027 x 300

=Rs 39,08,100/-

Based on local pollution control department norms the maximum limit of dust in the
S-7
gas stream leaving the industry should not exceed one ton per day. A bag filter was
installed to reduce the pollution from the exhaust gas stream. Find out the dust
collected in ton per day if the gas stream to the dust collector was entering at the rate
of 130,000 m3 per hour containing 6 g/m3 and leaving at the rate of 150,000 m3 per
hour, inclusive of ingress of air) containing 300 mg/m3. Also find out whether, the
industry met the pollution norms if the plant operates for 24 hours a day at same
capacity.

ANS
Amount of Dust in the inlet stream = 130,000 x 6
= 780,000 grams/hour

Amount of Dust in the outlet stream = 150,000 x 0.30 g/hr

= 45,000 grams/hour

Amount of Dust in the Bag Filter = 780,000 - 45,000

= 735,000 grams/hour x 24/106

= 17.64 Tonnes/day

Amount of dust leaving the Industry = 45,000 grams / hour x 24/106

= 1.08 Tonnes / day

Since it is more than 1 tonne/ day, the industry does not meet the pollution norms

In an industry the existing winding of a motor has burnt out. Calculate the annual
S-8
energy savings and simple payback for replacing the burnt out motor with an energy
efficient motor of the same capacity instead of rewinding.

The data given are

• Efficiency after rewinding of burnt out motor - 86%
• Cost of rewinding - Rs 7500
• Efficiency of energy efficient motor - 94%
• Cost of new energy efficient motor - Rs 42000
• Operating hours/year - 6900 hours
• % loading of motor - 82%
• Energy cost - Rs 5.2/kWh
• Name plate rating of motor - 22 kW

Energy cost saving (Rs / year)

ANS

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=[(kW) x (% loading) x [(1/efficiency of rewound standard motor)

- (1/ Efficiency of energy efficient motor)] x (HRS/annum) x (Rs/kwh)

=22 x 0.82 x 6900 x 5.2 [(100/86) – (100/94)]

=Rs 64055

Simple payback period

= [(42000-7500)/64055]
=6.5 months

------- End of Section - II ---------

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60
(i) Answer all Six questions
(ii) Each question carries Ten marks

L-1 An oil fired reheating furnace heats steel billets from 40oC to 1220oC at a furnace efficiency of
28%. The furnace operates for 4700 hours per annum. The GCV of furnace oil is10,000 kCal
/kg and density is 0.94kg/litre. The cost of furnace oil is Rs.45 /liter. The specific heat of billets
is 0.12 kCal/kgoC.
a. Calculate the amount of energy necessary to heat 12 tons
of steel billets per hour
b. Calculate liters of furnace oil fired per tons of steel billets.
c. If the efficiency of the furnace is improved from 28% to
30% by adopting ceramic fibre insulation, calculate the hourly furnace oil cost
saving
d. What is the simple payback period if the investment is Rs.
20 lakhs ?
e. How large could be the investment to improve the
efficiency at an internal rate of 16% and per year over 6 years.
ANS : a) Amount of energy necessary to heat 12 tons of steel billets
= m x cp x t
= 12000 Kgs x 0.12 x (1220- 40) kCals/hr
= 16,99,200 Kcals/hr

B) Litres of furnace oil fired per ton of steel billet

= [( 1699200/12)]
= 141600 Kcal/ tonne of billet
Input energy per ton of billet = 141160/0.28
= 505714 kcal/ tonne of billet
Furnace oil required in kg = 505714
10,000
= 50.57 Kg/ tonne of billet
Furnace oil required in litres = 50.57
0.94
= 53.79 litres/ tonne of billet

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c). Hourly furnace oil cost savings/ton

= 53.79 x [ 1 – (0.28/0.30)] * Rs 45
= Rs 161.37/tonne
Hourly furnace oil cost savings for 12 tons
= Rs 161.37 *12
=Rs 1936/hr

d). Simple payback period @ 4700 hrs of operation

=20,00,000/1936*4700
= 0.352 yrs or 4.2 months

e) Net cash inflow per annum= 1936*4700

= Rs 91.0 lakhs
Investment = 91.0 [ 1 + 1 + 1 + 1 + 1 + 1 ]
1.16 (1.16)2 ( 1.16)3 (1.16)4 (1.16)5 (1.16)6
= 91.0 [0.862+ 0.743+0.641+0.552+0.476+0.410]
= Rs. 3.35 crores

L-2 The Energy- production data (for Jan-June, 2011) of an industry follows a relationship :
Calculated energy consumption = 0.5 P +220.

A Waste heat recovery system was installed at end of June 2011 and further data was
gathered up to December 2011.

Using CUSUM technique, calculate energy savings in terms of ton of oil equivalent (toe) and
the reduction in specific energy consumption achieved with the installation of waste heat
recovery system.
The plant data is given in the table below.

Actual Energy Actual production,

2011-Month Consumption, toe/month ton/month
Jan 620 760
Feb 690 960
Mar 635 790
Apr 628 830
May 545 610
Jun 540 670
July 590 760
Aug 605 820
Sep 670 940
Oct 582 750
Nov 512 610
Dec 540 670
ANS
The table below gives values of actual energy consumption Vs. calculated (predicted)

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energy consumption from July –Dec. 2011.

Specific energy consumption monitored Vs predicted for each month. The variations are
calculated and the Cumulative sum of differences is calculated from Jan-June-2011.

2011- Ecal
Month Eact. 0.5P+220 Eact - Ecal CUSUM
July 590 600 -10 -10
Aug 605 630 -25 -35
Sept 670 690 -20 -55
Oct. 582 595 -13 -68
Nov. 512 525 -13 -81
Dec. 540 555 -15 -96
Energy savings achieved = 96 toe

Reduction in specific energy consumption = 96/4550 = 0.021 toe/tonne of production

(Production for 6 months = 760+820+940+750+610+670 = 4550 tonnes)

L-3 A proposed energy efficiency improvement project requires an initial investment of

Rs.5,00,000 and generates cash flow as given below:
Savings in year Cash flow (Rs.)
1 120000
2 115500
3 130000
4 116500
5 117250
6 200000

Calculate IRR of the project by interpolation method by taking initial discount rate as 11%.
ANS a) NPV at 11% = -500000 + 120000/(1+0.11)1 + 115500/(1+0.11)2 +130000/(1+0.11)3 +
116500/(1+0.11)4 +117250/(1+0.11)5 +200000/(1+0.11)6
= 50157.88

NPV at 16% = -500000 + 120000/(1+0.16)1 + 115500/(1+0.16)2 +130000/(1+0.16)3 +

116500/(1+0.16)4 +117250/(1+0.16)5 +200000/(1+0.16)6
= -25176.29

NPV is between 11% and 16%

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IRR with Linear Interpolation

IL = 11%
NPVL = 50157.88
IU= 16%
NPVU = -25176.29

IRR = IL + (IU -IL) x NPVL

(NPVL - NPVU)

= 0.11 + (0.16-0.11) (50157.88) / [(50157.88 – (- 25176.20)]

= 0.11 + (0.05 x 50157.88) / 75334.17
= 0.11 + 2507.894 / 75334.17
= 0.11 + 0.0333
= 0.1433
= 14.33%

L-4 Write short notes on any two of the following

a)BEE’s Standards and Labeling programme for equipment and appliances
b)Role of ESCOs
c)Evacuated tube collector for solar energy applications
ANS a) Standards & Labeling

Standards and Labeling would ensure that only energy efficient equipment and appliances
would be made available to consumers.
Main provisions of S&L are:
• Evolve minimum energy consumption and performance standards for notified equipment
and appliances
• Prevent manufacture, sale and import of equipment which do not meet the standards
• Introduce a mandatory labeling scheme for notified equipment and appliances to enable
consumers to make informed choices
• Spread information on benefits to consumers
For establishing standards, agreed testing procedures are defined and values of energy
performance are measured.

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Energy labels are the best way to implement the standards. They are information affixed to
manufactured products to describe the product’s energy performance usually in form of
energy use or efficiency. These give data to consumers to make informed purchases.

b) Role of ESCOs
Conduct of Investment grade energy audit
Arranging finance
Purchase, installation and maintenance of installed energy efficient equipment
Operation & Maintenance training
Monitoring of operations and energy savings

c) Evacuated Tube Collector

Evacuated tube collector comprises of two concentric glass tubes fused in the ends. The air is
evacuated from the gap between the tubes. The evacuated double-walled glass tube provides
thermal insulation similar to that of thermally insulated "Thermos" bottle. The outer glass tube
is clear, and the surface of the inner glass tube is coated with a special heat material that
absorbs the sun's energy.

Sun rays penetrate the outer clear glass and heat energy is absorbed by the inner coated
glass. The vacuum permits the heat radiation to enter the outer tube. The absorbent coating
on the inner tube converts short wave radiation to long wave radiation thus preventing re-
radiation to atmosphere. Since conduction cannot take place in vacuum, heat loss due to
conduction back to atmosphere is also prevented. Because of this principle, more heat is
trapped compared to a flat plate collector. The heat loss in Evacuated tube collector is less
than 10% compared with 40% for a flat plate collector. Water flows in through a third,
innermost concentric feeder tube and hot water flows out in the annulus outside the feeder
tube in contact with the absorber tube surface. This type of solar collector can reach high
temperatures upto 150°C.

L-5 a) Explain in brief the process of gasification of biomass.

b) Find out the conversion efficiency of a gasifier, if 20 kg of wood (having a calorific value of
3200 kCal / kg) produces 46 m3 of producer gas having an average calorific value of 1000
kCal / Nm3.

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a) Gasification of Biomass
ANS
Biomass contains Carbon, Hydrogen and Oxygen molecules. Complete combustion of
Biomass would produce Carbon Dioxide and water vapour, whereas combustion under
controlled conditions ie partial combustion produces Carbon Monoxide and Hydrogen, which
are combustible gases. The biogas produced through gasification is called as Producer Gas.

Gasification is a partial oxidation of biomass and takes place at temperature of about 1000oC.
Partial oxidation is facilitated by supplying air less than the stoichiometric requirements. The
products of combustion are gases like Carbon Monoxide, Hydrogen and traces of Methane
and non- useful products like tar & dust. The production of these gases is by reaction of water
vapour and Carbon Dioxide through a glowing layer of charcoal.

A gasification system consists of 4 main steps:

§ Feeding of Feedstock
§ Gasifier reactions where gasification takes place.
§ Cleaning of resultant gas
§ Utilization of cleaned gas

Biomass Gasifier is a thermo-chemical convertor / reactor where various physical and

chemical reactions take place. Biomass is passed through following zones before being
converted to high quality producer gas:

• Drying Zone
• Distillation Zone
• Pyrolysis zone
• Combustion Zone
• Reduction Zone

The following reactions take place:

C + O2 = CO2

H2 + ½ O2 = H2O

C + O2 = 2CO

C + H2O = CO + H2

CO2 + H2 = CO + H2O

C +2 H2 = CH4

The Producer gas has relatively a low calorific value ranging from 1000 to 1200 kCal/Nm3. The
conversion efficiency of Gasifier is in the range of 60 – 70%. It can be used for combustion in

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a reciprocating engine.

b) Heat Input in the Gasifier = 20 x 3200

= 64000 kCal

Heat Output as Producer gas = 46 x 1000

= 46000 kCal

Conversion efficiency of Gasifier = Heat Output / Heat Input

= 46000 x 100 / 64000
= 71.88 % Ans

L-6 a) Construct a PERT/CPM network diagram for a project for which the data is given below

b) Compute the earliest start, earliest finish, latest start, latest finish and slack for all the
activities

c) Also compute the project duration, identify critical activities and the critical path(s)

Activity Predecessor Time in weeks

A - 3
B - 5
C A 4
D A 6
E C 5
F D 3
G B 2
H E,F 1
I G,H 2

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ANS (a) PERT/CPM Network Diagram

b) Early start (ES), Early Finish (EF), Latest start (LS), Latest finish (LF)

Activity Duration ES EF LS LF Slack

(weeks)
A 3 0 3 0 3 0
B 5 0 5 6 11 6
C 4 3 7 3 7 0
D 6 3 9 3 9 0
E 5 7 12 7 12 0
F 3 9 12 9 12 0
G 2 5 7 11 13 6
H 1 12 13 12 13 0
I 2 13 15 13 15 0

c) Critical Paths

Two critical paths exist. They are

A-C-E-H-I & A-D-F-H-I

Project duration : 15 weeks

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14th NATIONAL CERTIFICATION EXAMINATION

FOR
ENERGY MANAGERS & ENERGY AUDITORS – August, 2013

PAPER – 1: General Aspects of Energy Management & Energy Audit

Date: 24.08.2013 Timings: 09:30-12:30 HRS Duration: 3 HRS Max. Marks: 150

Section - I: OBJECTIVE TYPE Marks: 50 x 1 = 50

a) Answer all 50 questions

b) Each question carries one mark
c) Please hatch the appropriate oval in the OMR answer sheet with Black Pen or HB
pencil, as per instructions

As per Energy Conservation Act, 2001, a BEE Certified Energy Manger is required to be
appointed/designated by the

a) state designated agencies b) all industrial consumers

c) designated consumers d) electrical distribution licensees
The type of energy possessed by a charged capacitor is

a) kinetic energy b) electrostatic c) potential d) magnetic

The process of capturing CO2 from point sources and storing them is called __________.

a) carbon sequestration b) carbon sink

c) carbon capture d) carbon adsorption
What is the heat content of 200 liters of water at 5 oC in terms of the basic unit of energy in
kilojoules ?

a) 3000 b) 2388 c) 1000 d) 4187

Nameplate kW rating of a motor indicates

a) input to the motor b) rated output of the motor

c) no-load input to the motor d) rated input to the motor
Which of the following has the highest specific heat?

a) lead b) mercury c) water d) alcohol

What is the average conversion efficiency of a solar photo voltaic cell?

a) 22% b)15% c) 98% d)50%

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An indication of sensible heat content in air-water vapour mixture is

a) wet bulb temperature b) dew point temperature

c) density of air d) dry bulb temperature
Which of the following statements with respect to Reserve / Production (R/P) ratio is true?

a) is a constant once established

b) varies every year with changes in production
c) varies every year with changes in reserves
d) varies every year with changes in production and reserves
Which issue is not addressed by Integrated Energy Policy of India?

a) consistency in pricing of energy

b) scope for improving supply of energy from varied sources
c) energy conservation, research and development
d) removal of subsidies for energy across all sectors
In inductive and resistive combination circuit, the resultant power factor under AC supply
will be

a) less than unity b) more than unity c) zero d) unity

Which of the following statement is not true regarding energy security?

a) impaired energy security can even reduce agricultural output

b) energy security is strengthened by minimising dependence on imported energy
c) diversifying energy supply from different countries weaken energy security
d) increasing exploration to find oil and gas reserves improves energy security
An energy policy at the plant level is to be preferably signed by

a) chief executive b) energy Manager

c) energy auditor d) chief executive with approval of state designated agency
The energy benchmarking parameter for air conditioning equipment is

a) kW/Ton of Refrigeration b) kW/ kg of refrigerant handled

c) kW/m3 of chilled water d) kW/EER
How much carbon dioxide emission will be reduced annually by replacing 60 Watt
incandescent lamp with a 15 Watt CFL Lamp, if emission per unit is 1 kg CO2 per kWh and
annual burning is 3000 hours?

a) 45 ton b) 3 ton c) 0.135 ton d) 183 ton

Which of the following statement is not correct regarding Demand Side Management
(DSM)?

a) agriculture and municipalities are potential areas for DSM activities

b) savings accrued through DSM cannot be treated as avoided capacity on supply
side

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 Paper 1 –Set A Solutions

c) under DSM, demand can be shifted from peak to off peak hours thereby avoiding
imported power during peak hours
d) DSM programs may result in demand as well as energy reduction

_________ considers impact of cash flow even after payback period

a) net present value b) return on investment

b) sensitivity analysis d) simple payback period
Consider two competitive projects entailing investment of Rs.85,000/- . Project A returns
Rs.50,000 at the end of each year, but Project B returns Rs.115,000 at the end of year 2.
Which project is superior?

a) Project A since it starts earning by end of first year itself and recovers cost before end
of two years
b) Project B since it offers higher return in two years
c) both projects are equal in rank
d) insufficient information
__________ determines the project viability in response to changes in input parameters.

a) Life cycle analysis b) Financial analysis

c) Sensitivity analysis d) Payback analysis

Which of the following is false?

a) 1 calorie = 4.187 kJ b) 1 calorie = 4.187J

c) 1000 kWh = 1 MWh d) 860 kcal = 1 kWh
The annual electricity bill for a plant is Rs 110 lakhs and accounts for 38% of the total
energy bill. Furthermore the total energy bill increases by 5% each year. The plant’s
annual energy bill at the end of the third year will be about ________

a) Rs 335 lakhs b) Rs 268 lakhs c) Rs 386 lakhs d) Rs 418 lakhs

The retrofitting of a variable speed drive in a plant costs Rs 2 lakh. The annual savings is
Rs 0.5 lakh. The maintenance cost is Rs. 5,000/year. The return on investment is

a) 25% b) 22.5% c) 24% d) 27.5%

For a project to be financially attractive, ROI must always be ___ than interest rate.

a) lower b) higher c) equal d) no relation

A sum of Rs 100,000 is deposited in a bank at the beginning of a year. The bank pays
10% interest annually. How much money will be in the bank account at the end of the fifth
year, if no money is withdrawn?

a) 161050 b) 150000 c) 155000 d) 160000

The technique used for scheduling the tasks and tracking of the progress of energy
management projects through a bar chart is called

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 Paper 1 –Set A Solutions

a) CPM b) Gantt chart c) CUSUM d) PERT

____________is a statistical technique which determines and quantifies the relationship
between variables and enables standard equations to be established for energy
consumption.

a) linear regression analysis b) time-dependent energy analysis

c) moving annual total d) CUSUM
Which of the following is not an environmental issue of global significance?

a) ozone layer depletion b) global Warning

c) loss of Biodiversity d) suspended particulate Matter
The power generation potential in mini hydro power plant for a water flow of 3 m3/sec with
a head of 14 meters with system efficiency of 55% is

a) 226.6 kW b) 76.4 kW c) 23.1 kW d) none of the above

If the wind speed doubles, energy output from a wind turbine will be:

a) 2 times higher b) 4 times higher c) 6 times higher d) 8 times higher

Which of the following two statements are true regarding application of Kaizen for energy
conservation?

i) Kaizen events are structured for reduction of only energy wastes

ii) Kaizen events engage workers in such a way so that they get involved in energy
conservation efforts
iii) Implementation of kaizen events takes place after review and approval of top
management
iv) In a Kaizen event, it may happen that small change in one area may result in
significant savings in overall energy use

a) ii & iv b) i & iv c) iii & iv d) i & iv

The electrical power unit Giga Watt (GW) may be written as

a) 1,000,000 MW b) 1,000 MW c) 1,000 kW d) 1,000,000 W

Which of the following statements regarding TOD tariff is true?

a) an incentive to induce user to draw more power during peak period

b) discourages user from drawing more power during off peak period
c) both a and b are true
d) encourages user to shift load from peak period to off peak period
The producer gas basically consists of

a) Only CH4 b) CO & CH4 c) CO, H2 & CH4 d) Only CO & H2

The ozone layer in the stratosphere acts as an efficient filter for ____

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 Paper 1 –Set A Solutions

a) UV- B rays b) X-rays c) Gamma rays d) beta rays

Which of the following macro factors is used in the sensitivity analysis of project finance?

a) Change in tax rates b) Changes in maintenance cost

c) Changes in debt: equity ratio d) Change in forms of financing

Which among the following has the lowest Global Warming Potential?

a) Perflurocarbon b) chloroflurocarbons c) methane d) nitrous oxide

Which of the following statements is correct regarding ‘float’ for an activity?

a) Time between its earliest start time and earliest finish time
b) Time between its latest start time and latest finish time
c) Time between latest start time and earliest finish time
d) Time between earliest finish time and latest finish time

In a cumulative sum (CUSUM) chart, if the graph is going up, then

a) nothing can be said b) actual and calculated energy consumption are the same
c) energy consumption is reduced d) specific energy consumption is going up
CO2 measurement in a Fyrite kit is based on
a) Weight basis (dry) b) Volume basis (dry)
c) Weight basis (wet) d) Volume basis (wet)
The depletion of Ozone layer is caused mainly by _________

a) nitrous oxide b) carbon dioxide c) choloroflourocarbons d) methane gas

Portable combustion analyzers may have in-built chemical cells for measurement of stack
gas components. Which combination of chemical cells for measurement of stack gas
components is not possible?

a) CO, SOx, O2 b) CO2, O2 c) O2, NOr, SOx, CO d) O2, CO

The Energy Conservation Act,2001 requires that all designated consumers should get
energy audits conducted periodically by

a) certified energy manager b) certified energy auditor

c) accredited energy auditor d) state Designated Agencies
The term missing in the following equation (kVA) 2 = (kVA cos phi) 2
+ ( ? ) 2 is

a) cos phi b) sin phi c) kVA sin phi d) kVArh

The weight (kg) of the water vapour in each kg of dry air(kg/kg) is termed as :

a) Specific Humidity b) relative humidity c) humidity d) saturation ratio

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Which of the following tool is made use of to assess the input, conversion efficiency,
output, losses, quantification of all material, energy and waste streams in a process or
system?

a) material balance b) energy balance

c) material and energy balance d) Sankey diagram
If feed of 100 tonnes per hour at 10% concentration is fed to an evaporator, the product
obtained at 25% concentration is equal to ____ tonnes per hour.

a) 25 b) 40 c) 50 d) 62.5
2000 kJ of heat is supplied to 500 kg of ice at 0 oC. If the latent heat of fusion of ice is 335
kJ/kg then the amount of ice in kg melted will be

a) 1.49 b) 83.75 c) 5.97 d) None of the above

An electric heater draws 5 kW of power for continuous hot water generation in an industry.
How much quantity of water in litres per min can be heated from 30 oC to 85oC ignoring
losses?.

a) 1.3 b) 78.18 c) 275 d) none of the above

The fixed energy consumption of a company is 2000 kWh per month. The line slope of the
energy (y) versus production (x) chart is 0.3. The energy consumed in kWh per month for
a production level of 80,000 tons/month is

a) 24,000 kWh b) 24,200 kWh c) 26,000 kWh d) 38,000 kWh

The major constituent of natural gas is

a) Methane b) Ethane c) Propane d) Hydrogen

……. End of Section – I …….

Section – II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40

(i) Answer all Eight questions

(ii) Each question carries Five marks

S-1 The rating of a single phase electric geyser is 2000 Watts, at 230 Volt.
Calculate:
a) Rated current
b) Resistance of the geyser in Ohms
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c) Actual power drawn when the measured supply voltage is 210 Volts

Answer:
a)Rated Current of the Geyser, I = P/V= 2000/230 = 8.7 Ampere

b)Resistance Value , R = V/I = 230/8.7= 26.4 Ohms

c)Actual Power drawn at 210 Volts = (V/R)*V = (210/26.4) x210

= 1670 Watts
OR (210/230)2 x 2000 = 1667 Watts

S-2 A Diesel Generator performance trial gives specific generation of 3.5 kWh per
liter of diesel. The cooling water loss and exhaust flue gas loss as percentage of
fuel input are 28% and 32% respectively. The calorific value of diesel is 10,200
kcal/kg. The specific gravity of Diesel is 0.85. Calculate unaccounted loss as
percentage of input energy.

Ans CV of Diesel = 10,200 kcal/kg

Heat in input diesel = 10,200 x 0.85 = 8670 kcal/litre

Heat in kWh energy output = 3.5 x 860 = 3010 kcal/litre

% of heat used for kWh output = 3010/8670 = 34.72 %

Unaccounted loss = 100 – (34.72+28+32) = 5.28 %

A renovation and modernization (R&M) program of a 110 MW coal-fired thermal

S-3
power plant was carried out to enhance the operating efficiency from 28% to
32%. The specific coal consumption was 0.7 kg/kWh before R&M. For 7000 hours
of operation per year and assuming the coal quality remains the same, calculate

a) the coal savings per year and

b) the expected avoidance of CO2 into the atmosphere in Tons/year if the

emission factor is 1.53 kg CO2/kg coal

a) Specific coal consumption after modernization = 28 x 0.7/32 = 0.6125 kg/kwh

Ans
Annual savings = (0.7 – 0.6125) x110 x 1000 x 7000/1000 = 67,375 Tonnes per year

b) CO2 emission reduction =67,375 x 1.53 = 103083.75 Tonnes of per year

S-4 Briefly compare NPV and IRR method of financial analysis.
In NPV method, NPV is determined by assuming the discount rate (cost of capital). In
Ans
IRR calculations, we set the NPV as zero and determine the discount rate (internal rate
of return) which satisfies this condition.
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The net present value method calculates the present value of all the yearly cash flows
(i.e. capital costs and net savings) incurred or accrued throughout the life of a project
and summates them. Costs are represented as negative value and savings as a
positive value. The higher the net present value, the more attractive the proposed
project.

The calculation procedure for determining IRR is tedious (iterative) and usually requires
a computer spreadsheet. The exact internal rate of return can be found by interpolation
or plotting the net present value on a graph. If this discount rate is greater than current
interest rate, the investment is sound.

NPV is essentially a tool which enables number of different projects to be compared

while IRR method is designed to assess whether a single project will achieve a target
rate of return.

The project is accepted if the net present value is positive and rejected if the net
present value is negative. A negative net present value indicates that the project is not
achieving the return standard and thus will cause an economic loss if implemented. A
zero NPV is value neutral.

In IRR, the criterion for selection among alternatives is to choose the investment with
the highest rate of return. The internal rate of return figure cannot distinguish between
lending and borrowing and hence a high internal rate of return need not necessarily be
a desirable feature.

Both the NPV and IRR takes into account the time value of money and it considers the
cash flow stream in entire project life.

When the same quantity of heat is added to equal masses of iron and copper
S-5
pieces, the temperature of iron piece rises by 15 OC. Calculate the rise in
temperature of copper piece, if the specific heat of iron is 470 J / kg / OC and that
of copper is 390 J / kg / oC.
Mass of Iron x Sp. Heat Iron x 15 OC = Mass of Copper x Sp. Heat Copper x
Ans
(Rise in Temp of Copper OC)

Since mass of Iron = Mass of Copper

Sp. Heat Iron x 15 OC = Sp. Heat Copper x (Rise in Temp of Copper OC)
Sp. Heat of Iron = 470 J / kg / OC
Sp. Heat of Copper = 390 J / kg / OC

Hence, Rise in Temp. of Copper piece = (470 x 15 ) / 390

= 18.08 OC

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Write down the parameters, which can be measured by following instruments:

S-6
a) Stroboscope
b) Sling Psychrometer
c) Fyrite
d) Tachometer
e) Pitot tube

Ans
a) Stroboscope : Speed (Non Contact)
b) Sling Psychrometer : Dry & Wet Bulb Temperatures
c) Fyrite : O2 or CO2 in Flue Gases
d) Tachometer : Speed (Contact type)
e) Pitot tube : Velocity pressure of moving gases

An industrial plant is consuming 400 kW of power with a maximum demand of

S-7
520 kVA. The demand charge is Rs. 150/-per kVA. Determine the savings
possible by improving power factor to 0.95 and payback period if investment on
capacitor bank is Rs 1,50,000/-.

Ans
Present Power Factor : 400 / 520 = 0.77
Present Demand Charges Rs. : 520 * 150 = 78000/-
Future Demand with higher PF : 400 / 0.95 =421 kVA
Modified Demand Charges : 421 * 150 = Rs. 63150/-
Savings =78000 – 63150
= Rs. 14850/- per Month

Capacitor Investment =Rs. 1.50,000/-

Simple Payback Period =1,50,000 / 14850

=10.1 Months

S-8 A 100 tonnes per day capacity chlor-alkali plant produced 30,000 tonnes per
annum (TPA) of caustic soda with annual energy consumption of
90 million kWh in the reference year 2009-10. During the year 2011-
12, the annual production was 25,000 TPA, with an annual energy
consumption of 80 million kWh. Calculate the Plant Energy
Performance.

Ans
Production Factor = 25000 / 30000
= 0.833

Reference year energy equivalent = Reference year energy use x Production factor

= 90 x 0.833 = 75 million kWh

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Excess Energy Consumption in 2011-2012 = 80 – 75 = 5 million kWh

Plant Energy Performance (PEP) = [ (75 - 80) / 75 ] x 100

= (-) 6.67 %

The performance in the year 2011 – 2012 is poor as compared to the reference
year

------- End of Section - II ---------

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60

(i) Answer all Six questions

(ii) Each question carries Ten marks

L1 A bag house is being used to remove dust from an air exhaust stream flowing at
100 m3/min. The dirty air contains 15 g/m3 of particles, while the cleaned air from
the bag house contains 0.02 g/m3. The industry's operating permit allows the
exhaust stream to contain as much as 0.9 g/m3.

For various operating reasons, the industry wishes to bypass some of the dirty air
around the bag house and blend it back into the cleaned air so that the total
exhaust stream meets the permissible limit. Assume no air leakage and negligible
change in pressure or temperature of the air throughout the process.

Draw a schematic diagram and calculate the flow rate of air through the bag
house and the mass of dust collected per day in kg.

ANS Draw a flow diagram of the process as shown in Figure 1.

In this problem two balances can be made, namely, flow rate of dust in g/m 3 and flow
rate of air in m3/min. Balancing of flow rate of air in m 3/min is possible because the
temperature and pressure of air remains constant in the system.

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Write a balance for dust around the total system:

Input = Output from bag house +Output in the mixed exhaust

Or

Dust removed from bag house (Z) = 100m3/min. x 15 g/m3 —100m3/min. x 0.90 g/m3
=1410 g/min.

Or

Daily dust Output = 1410g/min x 24h/1d x 60min/1h x 1kg/1000g= 2030 kg

Write a balance for airflow

100 = X+Y, where X and Yare bypass stream and flow through bag house,
respectively.

Write a balance for dust around B:

15X + 0.02Y = 0.9x100

Solving the last two equations

X, the bypass stream = 5.9 m3/ min.
Y, the flow through bag house = 94.1 m3/min.

L2 a. Explain the difference between GCV and NCV.

b. A gas fired water heater heats water flowing at the rate of 1.2 M 3 / hour from
20oC to 65oC. If the GCV of the gas is 4x10 7 J/kg, what is the rate of combustion
gas in kg/hr. The efficiency of water heater as 80%,

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a. The calorific value is the measurement of heat or energy produced, and is measured
either as gross calorific value or net calorific value.
The difference being the latent heat of condensation of the water vapour
produced during the combustion process.
Gross calorific value (GCV) assumes all vapour produced during the combustion process is
fully condensed. Net calorific value (NCV) assumes the water leaves with the combustion
products without fully being condensed.

b)
Mass of water heated = 1.20 M3 /hr
= 1.2 x 1000 / 60 = 20 kg/min
Heat required by Water = m x Cp x (t2 – t1)
= 20 kg/min x 4.187 x 103 J/kg/0C x (65-20) oC
= 3.77 x 106 J/min
Mass of Gas kg/min = 3.77 x 106 / 0.8 /(4 x 107 )
Mass 0f Gas Required = 0.1178 kg / min
= 7.068 kg / Hr
L-3 Answer any two of the following

a) Benefits of Monitoring and Targeting system

b) Duties and responsibilities of energy manager

c) Energy substitution need not save energy: Explain with an example

ANS a) Benefits of Monitoring and Targeting system

The ultimate goal is to reduce energy costs through improved energy efficiency and
management control measures. Other benefits include

 Identify and explain an increase or decrease in energy use

 Draw energy consumption trends (weekly, seasonal, operational)
 Improve energy budgeting corresponding to production plans
 Observe how the organization reacted to changes in the past
 Determine future energy use when planning changes in operations
 Diagnose specific areas of wasted energy
 Develop performance targets for energy management programs / energy action
plans
 Manage energy consumption rather than accept it as a fixed cost that cannot be
controlled.

b) Duties and responsibilities of energy manager

Responsibilities:
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1. Prepare an annual activity plan and present to management concerning

financially attractive investments to reduce energy costs.
2. Establish an energy conservation cell within the firm and agree with
management about the mandate and task of the cell.
3. Initiate activities to improve monitoring and process control to reduce energy
costs.
4. Analyze equipment performance with respect to energy efficiency.
5. Ensure proper functioning and calibration of instrumentation required to assess
level of energy consumption directly or indirectly.
6. Prepare information material and conduct internal workshops about the topic for
other staff.
7. Improve disseminating of energy consumption data down to shop level or profit
center of a firm.
8. Establish a methodology to accurately calculate the specific energy consumption
of various products/services or activity of the firm.
9. Develop and manage training programme for energy efficiency at operating
levels.
10. Co-ordinate nomination of management personnel to external programs.
11. Create knowledge bank on sectorial, national and international development on
energy efficiency technology and management system and information
denomination.
12. Develop integrated system of energy efficiency and environmental up gradation.
13. Wide internal and external networking
14. Co-ordinate implementation of energy audit/efficiency improvement projects
through external agencies.
15. Establish and / or participate in information exchange with other energy
managers of the same sector through association.

Duties of Energy Manager:

1. Report to BEE and State level Designated Agency once a year. The information
with regard to energy consumed and action taken in the recommendation of the
accredited energy auditor, as per BEE – Format.
2. Establish an improved data recording, collection and analysis system to keep
track of energy consumption.
3. Provide support to Accredited Energy Audit Firm retained by the company for the
conduct of energy audit.
4. Provide information to BEE as demanded in the Act, and with respect to the
tasks given by the mandate, and the job description.
5. Prepare a scheme for efficient use of energy and its conservation and implement
such scheme keeping in view the economic stability of the investment in such
firm and manner as may be provided in the regulations of the Energy
Conservation Act.

c) Energy substitution need not save energy: Explain with an example

The objective of energy substitution may be to reduce the use of costlier energy
source to maximize the profit and to improve the efficiency of the process.
Example: Replacement of conventional energy by renewable energy.

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The efficiency of fuel oil fired systems will be higher than rice husk fired systems.
Hence the total energy input to the systems will increase for rice husk fired systems.
Even though material handling cost, specific fuel (energy) consumption increases for
rice husk, this is a cheap locally available renewable form of energy which will bring
down the fuel cost and transport cost compared to fuel oil.

Hence Energy substitution need not save energy.

Any other similar example.

L-4 For the following tasks, durations, and predecessor relationships in the following
activity table,

Activity Immediate Optimistic Most Likely Pessimistic

Description Predecessor(s) (Weeks) (Weeks) (Weeks)
A --- 4 7 10
B A 2 8 20
C A 8 12 16
D B 1 2 3
E D, C 6 8 22
F C 2 3 4
G F 2 2 2
H E, G 4 8 12
I H 1 2 3

a) Draw the network

b) Calculate expected time for all tasks
c) Calculate variance for all tasks
d) Determine all possible paths and their estimated durations
e) Identify the critical path

Network diagram is shown below:

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Formulas used for filling the above table:

Te = (To + 4 Tm + Tp)/6

= (Tp –To)/6

V = ((Tp –To)/6)2

Activity Immediate Optimistic Most Likely Pessimistic Te Variance

Description Predecessor(s) (Weeks) (Weeks) (Weeks)
A --- 4 7 10 7 1.00
B A 2 8 20 9 9.00
C A 8 12 16 12 1.78
D B 1 2 3 2 0.11
E D, C 6 8 22 10 7.11
F C 2 3 4 3 0.11
G F 2 2 2 2 0.00
H E, G 4 8 12 8 1.78
I H 1 2 3 2 0.11

The critical path is A – C – E – H – I

Duration of critical path is 39 days
.

A - B – D – E – H –I 7+9+2+10+8+2 = 38
A–C–E–H–I 7+12+10+8+2 = 39
A–C–F–G–H–I 7+12+3+2+8+2 = 34

The critical path is A – C – E – H – I

Duration of critical path is 39 weeks.

L-5 Write short notes on any two of the following:

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a) Advantages of Demand Side Management (DSM) for end user and utility
b) ISO 50001 Energy Management System
c) Distinction between energy conservation and energy efficiency
ANS a) Advantages of DSM
End user:
End use demand can be shifted from peak to off peak hours thereby reducing the need
for buying expensive energy during peak hours
Helps better manage the load curve and thus reduce the demand improve the
profitability
Utility:
Energy saving through DSM is treated same as new additions in supply side
Can reduce the capital needs for power capacity expansion
Improved loading of utility power plants and hence improved efficiency and profitability

b) ISO 50001 features

ISO 5001involves the following features:
Goal outlined in Energy policy
Objectives to achieve the goal
Targets which are more specific than objectives which outlines actual energy
conservation measures to be implemented. An objective may have one or more
targets.
Action plans to implement the targets which outline actions, time frame, responsibility
and resources for implementation.
All the above with other related documents are audited during internal and external
audits.

c) Energy conservation and Energy efficiency

Energy conservation is achieved when energy consumption is reduced in physical
terms as a result of productivity increase or technology change. On the other hand,
energy efficiency is achieved when energy intensity is reduced in a specific product,
process or area of production without affecting the output, consumption or comfort
levels. Energy efficiency means using less energy to perform the same function.
Energy efficiency promotion will contribute to energy conservation and therefore a part
of energy conservation policies.

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L-6 It is proposed to install a heat recovery device in a process industry. The capital
cost of installing the device is Rs.2,00,000 and after 5 years its salvage value is
envisaged at Rs.15,000. The savings accrued by the heat recovery device are as
shown below. Determine the net present value after 5 years for a discount rate of
8%.

Year 1 2 3 4 5
Savings (Rs.) 70,000 60,000 60,000 50,000 50,000

Ans
Year Discount factor Capital Net savings Present value (Rs.)
for 8% Investment (Rs.) (Rs.)
0 1.00 -200000 -200000
1 0.926 70000 +64820
2 0.857 60000 +51420
3 0.794 60000 +47640
4 0.735 50000 +36750
5 0.681 50000 +15000 +44265
NPV=+44895
It is evident that over a 5-year life-span the net present value of the project is 44895.

-------- End of Section - III ---------

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15th NATIONAL CERTIFICATION EXAMINATION

FOR
ENERGY MANAGERS & ENERGY AUDITORS – August, 2014

PAPER – 1: General Aspects of Energy Management & Energy Audit

Date: 23.08.2014 Timings: 09:30-12:30 HRS Duration: 3 HRS Max. Marks: 150

Section - I: OBJECTIVE TYPE Marks: 50 x 1 = 50

a) Answer all 50 questions

b) Each question carries one mark
c) Please hatch the appropriate oval in the OMR answer sheet with Black Pen or HB pencil

1. A geothermal field may yield

a) dry steam b) wet steam c) hot air d) all of these

2. Which of the following is not a greenhouse gas ?

a) CFCs b) SO2 c) PFC d) SF6

3. Bio-gas generated through anaerobic process mainly consists of

a) only methane b) methane and carbon dioxide c) only ethane d) none of these
4. Which of the following statements are true?

i) bagasse is a source of secondary energy

ii) beneficiated coal belongs to primary energy
iii) electricity is basically a convenient form of primary energy
iv) steam is a convenient form of secondary energy

a) (ii) & (iii) b) (i) & (iii) c) (ii) & (iv) d) (ii) & (i)

5. Natural gas contains

a) methane, ethane and propane in equal proportions

b) only butane and propane in equal proportions
c) methane, propane and pentane in equal proportions
d) mostly methane and minor amounts of other gases
6. Which issue is not addressed by Integrated Energy Policy?

a) consistency in pricing of energy

b) scope for improving supply of energy from varied sources
c) energy conservation Research and Development

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d) reducing price of energy

7. Which of the following statement is not true regarding energy security?

a) impaired energy security will not affect agricultural output

b) energy security is strengthened by avoiding dependence upon imported energy
c) diversifying energy supply from different countries strengthen energy security
d) strengthening energy security requires increasing exploration to find oil and gas
reserves
8. In a boiler substitution of coal with rice husk leads to

a) energy conservation b) energy efficiency

c) both energy conservation and energy efficiency d) carbon emission reduction
9. A building intended to be used for commercial purpose will be required to follow Energy
conservation building code under Energy Conservation Act, 2001 provided its

a) connected load is 120 kW and above

b) contract demand is 100 kVA and above
c) connected load is100 kW and above or contract demand is 120 kVA and above
d) connected load is 500 kW and contract demand is 600 kVA
10. Which of the following is not a part of energy audit as per the Energy Conservation Act,
2001?

a) monitoring and analysis of energy use

b) verification of energy use
c) submission of technical report with recommendations
d) ensuring implementation of recommended measures followed by review
11. Which of the following criteria is a responsibility of Designated Consumers?

a) designate or appoint an accredited Energy Auditor

b) adhere to stipulated energy consumption norms and standards as notified
c) submit the status of energy consumption information every three years
d) conduct energy audit through a certified energy auditor periodically
12. Which of the following is an energy security measure?

a) fully exploiting domestic energy resources

b) diversifying energy supply source
c) substitution of imported fuels for domestic fuels to the extent possible
d) all of the above
13. An induction motor with 11 kW rating and efficiency of 90% in its name plate means

a) it will draw 12.22 kW at full load

b) it will always draw 11 kW at full load

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c) it will draw 9.9 kW at full load

d) nothing can be said about how much power it will draw as motor power factor is not
given
14. Which of the following statement is true regarding maximum demand controller?

a) maximum demand controller enables a way of ‘shaving’ the peaks in the consumer
load profile
b) maximum demand controller enables a way of improving the system power factor
c) enables a way for using more electrical energy at lower total cost of energy without
investment in expansion of power supply
d) maximum demand controller is installed by concerned utility at customer premises
15. Which of the following statements are true?

i) reactive current is necessary to build up the flux for the magnetic field of inductive
devices
ii) some portion of reactive current is converted into work
iii) the cosine of angle between kVA and kVAr vector is called power factor
iv) the cosine of angle between kW and kVA vector is called power factor

a) i & iv b) ii & iii c) i & iii d) iii & iv

16. Which of the following is a standard for Energy Management System?

a) ISO 14001 b) ISO 9001 c) ISO 18001 d) ISO 50001

17. Which of the following statements are true regarding simple payback period?

a) considers impact of cash flow even after payback period

b) takes into account the time value of money
c) considers cash flow throughout the project life cycle
d) determines how quickly invested money is recovered
18. Which of the following statements are true regarding CPM?

i) work breakdown structure are used to list the activities in the project as a first step
in CPM
ii) CPM takes into account variation in the completion time and average time is used
for any activity
iii) if the project is to finish earlier, it is necessary to focus on activities other than
critical path
iv) critical path is the longest path in the network.

a) i & iv b) i & iii c) ii & iv d) iii & iv

19. Which of the following statements is not correct?
Global warming will result in:

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a) melting of the ice caps b) increasing sea levels

c) severe damage to ozone layer in stratosphere d) unpredictable climate patterns

20. The process of capturing CO2 from point sources and storing them is called

a) carbon sequestration b) carbon sink c) carbon Capture d) carbon absorption

21. Which of the following statements are true regarding wind turbine?

i) wind power varies as cube of rotor size

ii) wind power varies as cube of wind velocity
iii) wind speed has more influence on wind power than turbine area
iv) practical maximum amount of energy in the wind that can be collected by wind
turbine rotor is about 79%

a) i & ii b) ii & iii c) iii & iv d) ii & iv

22. Which of the following statements regarding evacuated tube collectors (ETC) are true?

i) ETC is used for high temperatures upto 150oC

ii) because of use of vacuum between two concentric glass tube, higher amount of
heat is retained in ETC
iii) heat loss due to conduction back to atmosphere from ETC is high
iv) performance of evacuated tube is highly dependent upon the ambient temperature

a) i & iii b) ii & iii c) i & iv d) i & ii

23. What percentage of the sun’s energy can silicon solar panels convert into electricity?

a) 30% b) 15% c) 75% d) 50%

24. How much theoretical power you would expect to generate from a river-based mini
hydropower with flow of 20 litres/second and head of 12 metres

a) 2.35kW b) 2.44MW c) 1.67kW d) none of the above

25. Which among the following has the highest flue gas loss on combustion due to
Hydrogen in the fuel ?
a) natural gas b) furnace oil c) coal d) light diesel oil

26. Energy in one Tonne of Oil Equivalent (toe) corresponds to

a) 4.187 GJ b) 1.162 MWh c) 1 Million kcal d) none of the above

27. Tonnes of Oil Equivalent energy consumption / GDP in Million US $ is termed as

a) energy intensity b) per capita oil consumption
c) per capita energy consumption d) energy performance

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28. Assume CO2 equivalent emissions by the use of a 60 W incandescent lamp are of the
order of 60 g/hr. If it is replaced by a 5 W LED lamp then the equivalent CO2 emissions
will be
a) nil b) 5 g/hr
c) 12 g/hr d) 300 g/hr

29. Under the Energy Conservation Act, the designated consumer is required to get the
mandatory energy audit conducted by
a) certified energy manager b) certified energy auditor
c) accredited energy auditor d) in-house engineer

30. If the relative humidity of air is 100%, then which of the following statements is correct
a) only dew point & wet bulb temp. are same
b) only dew point & dry bulb temp. are same
c) only wet bulb & dry bulb temp. are same
d) all dew point , wet bulb & dry bulb temp. are same

31. Among which of the following fuel is the difference between the GCV and NCV
maximum?
a) coal b) furnace oil c) natural gas d) rice husk

32. Non-contact speed measurements can be carried out by

a) tachometer b) stroboscope c) oscilloscope d) speedometer

33. Which of the following instrument is used for assessing combustion efficiency ?
a) lux Meter b) pitot tube & manometer c) ultrasonic flow meter d) fyrite

34. The benchmarking parameter for a vapour compression refrigeration system is

a) kW / kg of refrigerant used b) kcal / m3 of chilled water
c) BTU / Ton of Refrigeration d) kW / Ton of Refrigeration

35. If 800 kcal of heat is supplied to 20 kg of ice at 0o C, how many kg of ice will melt into
water at 0oC. (Latent heat of fusion of ice is 80 kcal/kg)
a) 1 kg b) 4 kg c) 10 kg d) 20 kg
36. If feed of 100 tonnes per hour at 5% concentration is fed to a crystallizer, the product
obtained at 25% concentration is equal to ____ tonnes per hour.

a) 15 b) 20 c) 35 d) 40

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37. The return on investment (ROI), is expressed as

a) annual cost / capital cost b) (first cost / first year benefits) x 100
c) NPV / IRR d) (annual net cash flow x 100) / capital cost

38. The rate of energy transfer from a higher temperature to a lower temperature is
measured in

a) kCal b) Watt c) Watts per second d) none of the above.

39. Cost of a new heat exchanger is Rs. 1.5 lakh. The simple payback period (SPP) in years
considering annual savings of Rs 60,000 and annual maintenance cost of Rs 10,000 is
a) 0.4 b) 2.5 c) 3 d) 6

40. Energy sources which are inexhaustible are known as

a) commercial energy b) primary energy c) renewable energy d) secondary energy
41. 1 kg of wood contains 15% moisture and 5% hydrogen by weight. How much water is
evaporated during complete combustion of 1kg of wood?
a) 0.6 kg b) 200 g c) 0.15 kg d) none of the above

42. In an industry the average electricity consumption is 5.8 lakh kwh for a given period. The
average production is 50000 tons with a specific electricity of 11 kwh/ton for the same
period. The fixed electricity consumption for the plant is
a) 58000 kWh b) 30000kWh c) 80000kWh d) none of the above

43. The cost of replacement of inefficient compressor with an energy efficient compressor in
a plant costs Rs. 8 lakhs. The net annual cash flow is Rs. 2 lakhs. The return on
investment
a) 18% b) 20% c) 15% d) none of the above

44. The amount of electricity required to heat 100 litres of water from 30oC to 70 oC through
resistance heating is
a) 0.465 kWh b) 4.65 kWh c) 465 kWh d) 2 kWh

45. In project financing, sensitivity analysis is applied because

a) almost all the cash flows involve uncertainly
b) it evaluates how sensitive the project is to change in the input parameters
c) it assesses the impact of ‘what if one or more factors are different from what is
predicted’
d) it is applicable to all the above situations

46. A process requires 120 kg of fuel with a calorific value of 4800 kcal/kg for heating with a
system efficiency of 82 %. The loss would be
a) 576000 kcal b) 472320 kcal c) 103680 kcal d) 480000 kcal

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47. The internal rate of return is the discount rate for which the NPV is
a) positive b) zero c) negative d) less than 1

48. Having energy policy

a) satisfies regulations b) shows top management commitment
c) indicates energy audit skills d) adds to the list of number other policies

49. The producer gas is basically

a) CO, H2 and CH4 b) only CH4 c) only CO and CH4 d) only CO and H2

The time between its earliest and latest start time, or between its earliest and latest
50.
finish time of an activity is

a) delay time b) slack time c) critical path d) start time

……. End of Section – I …….

Section – II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40

(i) Answer all Eight questions

(ii) Each question carries Five marks

Calculate Net Present Value over a period of 3 years for a project with an investment of
S-1
Rs 70,000 at the beginning of the first year and second investment of Rs 70,000 at the
beginning of the second year and fuel cost saving of Rs 95,000 in second and third
year. The discount rate is 14%

Ans
NPV = –70,000 – (70000/1.14) + [95000/(1.14x1.14)] + [95000/(1.14x1.14x1.14)]
= –70000 – 61404 +73099 + 64122
= –131404 +137221
= Rs 5817/-
A water pumping station fills a reservoir at a fixed rate. The head and flow rate are
S-2
constant and hence the power drawn by the pump is always same. The pump operates
at 100 m head and delivers 250 litres per second. The power consumption was
measured as 300 kW.

Calculate energy consumption to pump 13,500 kL of water to the reservoir.

Ans
Time taken to pump water in hours = 13,500 × 103 L
250 L/s x 3600 sec/hr
= 15 hours
Power required to pump water = 300 kW

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Energy consumption = 300 x 15 =4500 kWh

S-3 A conveyor delivers coal with a width of 1 m and coal bed height of 0.25 m at a speed
of 0.5 m/s. Determine coal delivery in tons per hour considering coal density of 1.1
ton/m3.

Ans Volume of coal delivered per hour = area x length travelled per second
=1 m x 0.25 m x 0.5 m/s
= 0.125 m3/s = 450 m3/hr
Coal delivery rate = 450 m3/hr x 1.1 t/m3
= 495 t/hr

S-4 In a process industry, 12,000 kg/hr water is currently being heated from 18oCto 80oC by
indirect heating of steam. An opportunity has been identified which would preheat the
inlet water to 45oC to reduce the steam required.
Estimate the reduction in steam in kg/hr considering latent heat of steam as 520 kcal/kg
in both the cases.

Ans Without heat recovery

Heating required (Q1) = mCp T
= 12,000 x 1 x (80-18)
= 744,000 kcal/hr
Steam required = 744,000 / 520
= 1431 kg/hr
After heat recovery
Heating required (Q2) = 12,000 x 1 x (80 – 45)
= 420,000 kcal/hr
Steam required = 420,000/520
= 808 kg/hr
Reduction in steam required = 1431 - 808 = 623 kg/hr
Write short notes on any two of the following:
S-5
a) Building envelope
b) Standards and Labeling
c) Demand Side Management (DSM)
(a) Building envelope includes all components of building exposed to outside
Ans
environment such as outside doors, windows, roofs etc. Its main purpose is to
protect employees from outside environment.

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(b) Standards and Labeling (Page -35): There is a wide variation in energy
consumption of similar products by various manufacturers. Also information on
energy consumption is often not easily available. Standards and Labeling (S&L)
has been identified as a key activity for energy efficiency improvement. The S&L
program, when in place, would ensure that only energy efficient equipment and
appliances would be made available to the consumers.

Standards: Energy-efficiency standards are procedures and regulations prescribing

the energy performance of energy-consuming products. The sale of products that
are less energy efficient than minimum standards, often called Minimum Energy
Performance Standards (MEPS) may be prohibited. For establishing the standards,
agreed testing protocols (test procedures) are defined and value of energy
performance is determined.

Labels: Energy-efficiency labels are informative labels affixed to manufactured

products to describe the product’s energy performance (usually in the form of
energy use, efficiency). These labels give consumers the data necessary to make
informed purchases. Star rating is a system initiated by BEE to determine energy
efficiency of an appliance. Label indicates the energy efficiency levels through the
number of stars highlighted in colour on the label. It is being applied to many
products such as refrigerators, TVs, ACs and so on.

(c) Demand Side Management (Page -36) : Demand Side Management (DSM) means
managing of the demand for power, by utilities / Distribution companies, among
some or all its customers to meet current or future needs. DSM programs result in
energy and / or demand reduction. For example, under this process, the demand
can be shifted from peak to off peak hours thereby reducing the need for buying
expensive imported power during peak hours. DSM also enables end-users to
better manage their load curve and thus improves the profitability. Potential energy
saving through DSM is treated same as new additions on the supply side in MWs.
DSM can reduce the capital needs for power capacity expansion.

a) Briefly explain why combustion of biomass fuels is considered as carbon neutral?

S6

b) Name five energy intensive industries having annual energy consumption of 30,000
metric tonne of oil equivalent and above, notified as designated consumers under
the EC Act 2001

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Ans a) The CO2 emitted by combustion of biomass fuels is largely balanced by the
absorption/capture of carbon dioxide during its growth.

b)
1. Thermal Power Stations
2. Fertilizer
3. Cement
4. Iron & Steel
5. Pulp & Paper

S7 Briefly explain the difference between flat plate collector and evacuated tube collector.

Ans. (Page 260-261)

Flat plate collector:

 The most common collector is called a flat-
plate collector.
 Heat the circulating fluid to a temperature of
about 40-60°C.
 Usually comprises of copper tubes welded to
copper sheets (both coated with a highly
absorbing black coatings) with toughened
glass sheet on top for cover and insulating
material at the bottom. The entire assembly is Solar Flat Plate Collector
placed in a flat box.

Evacuated tube collector:

 Used For higher temperatures.
 Evacuated tube collector is less dependent upon
ambient temperature unlike flat plate collector
and its efficiency does not drop with ambient
temperature.
 Evacuated glass tubes are used instead of
copper in which case a separate cover sheet
and insulating box are not required. Evacuated Tube Collector
 Can reach high temperatures upto 150°C

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A sample of coal being used in a boiler is found to contain 60% carbon and 23% ash.
S8
The refuse obtained after combustion is analysed and found to contain 7% carbon &
the rest is ash.

Compute the percentage of the original carbon in coal which remains as unburnt in the
refuse.
Ans
Let the quantity of Refuse sample =100 kg
Amount of unburnt Carbon in Refuse = 7 kg
Amount of Ash in the Refuse = 93 kg

Total ash in the coal that has come into the Refuse = 23% of coal

93 kg of Ash corresponds to 23% ash in the coal

Therefore, quantity of total raw coal = 93 / 0.23

= 404.35 kg
Quantity of original Carbon in the coal = 0.60 x 404.35
= 242.61 kg
Quantity of unburnt coal in Refuse = 7 kg

%age of the original carbon unburnt in the refuse = (7 / 242.61) x 100

= 2.89%

------- End of Section - II ---------

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60

(i) Answer all Six questions

(ii) Each question carries Ten marks

L1 In pre-treatment process of a plating section of an engineering industry, LPG was being

used indirectly to heat 6000 litres/hr of water by 100C. The industry is planning to convert
from LPG to electrical heating.

Other data:
Annual operating hours = 3000 hours
Efficiency of indirect heating with LPG = 85%
Calorific value of LPG = 11000 kcal/kg,
Landed cost of LPG = Rs.75/kg
Cost of electricity = Rs.6/kwh.

a) If LPG is replaced with electrical heating with an investment is Rs.1.5 lakhs, compute
simple payback period.
b) Calculate the CO2 emissions in both the cases. Consider emission factors for LPG as 3
tons of CO2/Ton of LPG and Electricity as 0.81 tons of CO2/MWh

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Ans Water flow rate 6000 Litres/hr

Temperature rise 10oC
Heat provided with LPG (6000 x 10)=60000 kCal/hr
LPG consumption 60000/(11000 x 0.85) =6.4 kg/hr

Annual LPG consumption 6.4 x 3000 = 19.2 Tons/yr

Annual CO2 emission with LPG heating 19.2 x 3 =57.6 t CO2
Cost of heating with LPG 19.2 x 1000 x 75
= Rs.14.4 lakhs/annum

Electricity equivalent of LPG (60000/860) = 70 kW

70 x 3000 = 210 MWh

Average cost of electricity Rs. 6/kWh

Cost of electrical heating (70 x 6) = Rs. 420/hr
420 x 3000 = Rs.12.6 lakhs

Annual CO2 emission with electrical heating 210 x 0.81 =170 t CO2

Annual cost savings (14.4-12.6)

Potential annual savings Rs.1.8 lakhs

Investment for electrical heating Rs.1.5 lakhs

Payback period 1.5/1.8
0.83 years (< 10 months)

L2 A project has the following activities, precedence relationships, and time estimates in
weeks:

Activity Immediate Optimistic Most Likely Pessimistic

Predecessors Time Time Time
A - 15 20 25
B - 8 10 12
C A 25 30 40
D B 15 15 15
E B 22 25 27
F E 15 20 22
G D 20 20 22

a) Draw the network diagram (expected time may be rounded to the nearest whole number)
b) Identify the critical path and
c) Determine the project duration.

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ANS (a) Expected Time calculation:

Activity Immediate To Tm Tp Expected Time
Predecessors Te
A - 15 20 25 20
B - 8 10 12 10
C A 25 30 40 30.8
D B 15 15 15 15
E B 22 25 23 24.8
F E 15 20 22 19.5
G D 20 20 22 20.3

a)Network PERT Diagram

C (31)
A (20) (20, 51)
(0, 20)
2
(24, 55)
(4, 24) G (20) 6
1 (25, 45)
B (10) (35, 55)
(0, 10) D (15) 4 F (20)
(10, 25)
(0, 10) (35, 55)
(20, 35) E (25) (35, 55)
3 (10, 35)
(10, 35) 5
b)Critical Path : B-E-F

c)Project Duration : 55 weeks

L3 A company has got following two investment options:

Option A:. Investment envisaged Rs. 40 lakhs with an annual return of Rs. 8
lakhs; Life of the project is 10 years

Option B: Investment envisaged Rs. 24 lakhs; Annual return Rs. 5 lakhs;

Life of the project is 8 years

Calculate IRR of both the options and suggest which option the company should select.

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Option A:
Investment = Rs. 40 lakh
Annual Return = Rs. 8 lakh
Life of project = 10 years

0 = [(-) 40 x 105 ] + [(8 x 105) / (1 + 0.15)1] + [(8 x 105) / (1 + 0.15)2 ] + ---------- + [(8 x 105)
/ (1 + 0.15)9 ] + [ ( 8 x 105) / (1 + 0.15)10 ]

= 15.12 %

Option B:
Investment = Rs. 24 lakh
Annual Return = Rs. 5 lakh
Life of project = 8 years

0 = [(-) 24 x 105 ] + [ (5 x 105) / (1 + 0.13)1 ] + [(5 x 105) / (1 + 0.13)2] + ---------- + [(5 x 105) /
(1 + 0.13)7 ] + [ (5 x 105) / (1 + 0.13)8 ]
= 13.04 %

Based on IRR, the Option A has higher IRR value and the company may opt for Option A.

L4 Write short note on any two of the following

a) Energy Service Companies (ESCOs)

b) Sensitivity analysis for financing of energy conservation projects
c) Sankey diagram

Ans a) Energy Service Companies (ESCOs) : page -171

ESCOs are usually companies that provide a complete energy project service, from
assessment to design to construction or installation, along with engineering and project
management services, and financing.

Depending on the company’s capability to manage the risks (equipment performance,

financing, etc.) the company will delegate some of these responsibilities to the ESCO. In
general, the amount of risk assigned to the ESCO is directly related to the percent savings
that must be shared with the ESCO.

For example, a lighting retrofit has a high probability of producing the expected cash flows,
whereas a completely new process does not have the same “time tested” reliability. If the
in-house energy management team cannot manage this risk, performance contracting may
be an attractive alternative.

The ESCO will usually offer the following common types of contracts:
Fixed fee
Shared savings
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Guaranteed savings

Services offered by an ESCO usually include:

 An investment grade energy audit to identify energy and operational savings

opportunities, assess risks, determine risk management/mitigating strategies, and
calculate cost-effectiveness of proposed measures over time.
 Financing from its own resources or through arrangements with banks or other
financing sources.
 The purchase, installation and maintenance of the installed energy efficient
equipment; possibly maintenance on all energy-consuming equipment.
 New equipment training of operations and maintenance (O&M) personnel.
 Training of O&M personnel in energy-efficient practices.
 Monitoring of the operations and energy savings, so reduced energy consumption
and operation costs persist.
 Measurement and savings verification; and
 A guarantee of the energy savings to be achieved.

b) Sensitivity analysis for financing of energy conservation projects : (page168 -169)

Many of the cash flows in the project are based on assumptions that have an element of
uncertainty. The cash flows such as capital cost, energy cost savings, maintenance costs
can usually be estimated fairly accurately. Even though these costs can be predicted with
some certainty, it should always be remembered that they are only estimates. Cash flows in
future years normally contain inflation components and project life itself can vary
significantly.

Sensitivity analysis is an assessment of risk. Because of the uncertainty in assigning values

to the analysis, it is recommended that a sensitivity analysis be carried out - particularly on
projects where the feasibility is marginal. How sensitive is the project's feasibility to
changes in the input parameters? What if one or more of the factors in the analysis is not
as favourable as predicted? How much would it have to vary before the project becomes
unviable? What is the probability of this happening?

Sensitivity analysis is undertaken to identify those parameters that are both uncertain and
for which the project decision taken through the NPV or IRR is sensitive. The effect of
switching values of key variables required for the project decision (from acceptance to
rejection) can be compared with the post evaluation results of similar projects. Sensitivity
and risk analysis should lead to improved project design, with mitigation actions against
major sources of uncertainty involved.

The various micro and macro factors / variables that are considered for the sensitivity
analysis are listed below.

Micro factors:
 Operating expenses (various expenses items)
 Capital structure
 Costs of debt, equity
 Changing of the forms of finance e.g. leasing
 Changing the project life
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Macro factors: Macro economic variables are the variable that affects the operation of the
industry of which the company operates. They cannot be changed by the firm’s
management. Macro economic variables, which affect projects, include among others:

 Changes in interest rates

 Changes in the tax rates
 Changes in the accounting standards e.g. methods of calculating depreciation
 Changes in depreciation rates
 Extension of various government subsidized projects e.g. rural electrification
 General employment trends e.g. if the government changes the salary scales
 Imposition of regulations on environmental and safety issues in the industry
 Energy price change
 Technology changes

c) Sankey diagram: (page - 127)

The Sankey diagram is very useful tool to represent an entire input and output energy flow
in any energy equipment or system such as boiler generation, fired heaters, furnaces after
carrying out energy balance calculation. Usually the flows are represented by arrows. The
width of the arrows is proportional to the size of the actual flow. Better than numbers, tables
or descriptions, this diagram represents visually various outputs (benefits) and losses so that
energy managers can focus on finding improvements in a prioritized manner.

The Figure shows a Sankey diagram for

an internal combustion engine. From the
Figure, it is clear that exhaust flue gas
losses are a key area for priority
attention. Since the engines operate at
high temperatures, the exhaust gases
leave at high temperatures resulting in
poor system efficiency. Hence a heat
recovery device such as a waste heat
boiler has to be necessarily part of the
system. The lower the exhaust
temperature, higher is the system
efficiency. Sankey Diagram for an Internal Combustion Engine

L5 a) Calculate the annual energy savings and simple payback from replacing standard
existing motor with energy efficient motor versus rewinding the existing motor.

The data given:

Average cost of rewinding =Rs 6500
Cost of new high efficient motor =Rs 37000
Efficiency after rewound of standard motor =87%
Efficiency of energy efficient motor = 94%
Operating hours = 7200 hrs
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% loading of motor = 82%

Power cost = Rs 5.2 / kWh
Name plate rating of motor = 20 kW

b) During an air pollution monitoring study, the inlet gas stream to a bag filter was 200,000
m3 per hour. The outlet gas stream from the bag filter was little bit higher at 220,000m 3 per
hour. The dust load at the inlet was 5 g/m3 and at the outlet 0.2 g/m3.

How much dust in kg/hour was collected in the bag filter bin?
a)
a) Solution: Energy cost savings (Rs/year)
=[(KW)*(% loading)*{(100/efficiency of rewound standard motor)-(100/efficiency of energy
efficient motor)}*(Hrs/annum)*(Rs/kwh)]
= 20*0.82*7200*[(100/87)-(100/94)]*5.2
=118080*[1.1494-1.0638]*5.2
=52560/-

Simple payback period = [(Rs 37000-Rs 6500)/52560]

= 7 months

b) Dust (gas in) = dust (in gas out) + dust (in bin)
200000 x 5 = 220000 x 0.2 + X
X = 1000000 – 44000
= 956000 gm/hr
= 956 Kg/hr

L6 Write short note on any two of the following.

a) 5S
b) KAIZEN
c) ISO 50001
d) TPM

Ans a) 5S: (page-145)

5S, abbreviated from the Japanese words Seiri, Seiton, Seiso, Seiketsu, and Shitsuke,
are simple but effective methods to organize the workplace.

The 5S, translated into English are: housekeeping, workplace organization, cleanup,
maintain cleanliness, and discipline. They can be defined as follows:

Housekeeping. Separate needed items from unneeded items. Keep only what is

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immediately necessary item on the shop floor.

Workplace Organization. Organize the workplace so that needed items can be
easily and quickly accessed. A place for everything and everything in its place.
Cleanup. Sweeping, washing, and cleaning everything around working area
immediately.
Cleanliness. Keep everything clean in a constant state of readiness.
Discipline. Everyone understands, obeys, and practices the rules when in the plant.

Implementing 5S methods in the plant would help the company to reduce waste hidden in
the plant, improve the levels of quality and safety, reduce the lead time and cost, and thus
increase Company’s profit.

b) KAIZEN: (page -147)

“KAIZEN”, is a practice developed by Japanese for increasing productivity. KAIZEN is the

Japanese word made up of two components.

KAI – Change
ZEN – Good (for the better)
KAIZEN – Change for the better or continuous improvement.

KAIZEN means continuous improvement involving everyone – Managers and Workers

alike. It emphasizes improvement on working standard through small, gradual
improvement. Its philosophy assumes that our way of working life, our social life or our
human life – deserves to be constantly improved.

Kaizen events focuses on reducing various forms of wastes and often energy reduction
result from projects that focus on an area or a process. For example, a project that was
implemented to reduce colouring chemicals in a plastic industry resulted in small changes
in plant layout and material flow to its use. This ended up in big reduction in amount of
forklift travelled and fuel used in the forklift.

Implementation:

Step-1: Identification of a problem, i.e. waste, defect or something not working. The
operator writes and describes the problem
Step-2: Operator later develops an improvement idea and goes to immediate supervisor
Step-3: Supervisor / Kaizen team members review it and encourage immediate action and
fills up the Kaizen form
Step-4: The idea is implemented & checked.
Step-5: The operator is rewarded.

c) ISO 50001: (page -151)

ISO 50001 will establish a framework for industrial plants, commercial facilities or entire
organizations to manage energy. The document is based on the common elements found
in all of ISO’s management system standards, assuring a high level of compatibility with
ISO 9001 (quality management) and ISO 14001 (environmental management).

Energy Management System enables an organization to take a systematic approach in

order to achieve continual improvement of energy performance, energy efficiency and
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energy conservation.

An energy management system addresses:

- Energy supply;
- Measurement;
- Documentation and reporting of energy use; and
- Procurement & design practices for energy-using equipment, systems and
processes.

To simply put it, ISO 50001 is “saying what you do and doing what you say”.

d) TPM : (page -148)

Total productive maintenance (TPM) is the method that focuses on optimizing the
effectiveness of manufacturing equipment. TPM builds upon established equipment-
management approaches and focuses on team-based maintenance that involves
employees at every level and function.

The goal of TPM is to build a robust organisation by maximizing production system

efficiency (overall effectiveness).

TPM addresses the entire production system lifecycle and builds a concrete, shop floor-
based system to prevent all losses. It aims to eliminate all accidents, defects, and
breakdowns.
TPM involves all departments from production to development, sales, and
administration.
Everyone participates in TPM, from the top executive to shop floor employees.
TPM achieves zero losses through overlapping team activities.

-------- End of Section - III ---------

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REGULAR Paper 1 –Set A

MODEL SOLUTIONS

16th NATIONAL CERTIFICATION EXAMINATION

FOR
ENERGY MANAGERS & ENERGY AUDITORS – September, 2015

PAPER – 1: GENERAL ASPECTS OF ENERGY MANAGEMENT & ENERGY

AUDIT

Date: 19.09.2015 Timings: 0930-1230 HRS Duration: 3 HRS

Max. Marks: 150

Section – I: OBJECTIVE TYPE Marks: 50 x 1 =

50

(i) Answer all 50 questions

(ii) Each question carries one mark
(iii) Please hatch the appropriate oval in the OMR answer sheet with
Black Pen, as per instructions

1. Which one is not a consequence of global warming

a) rise in global temperature b) rise in sea level

c) food shortage and hunger d) fall in global temperature
2. Which of the following will not be a major component of mass balance

a) steam b) water c) raw materials d) lubricating oil

3. Which of the following terms does not refer to specific energy consumption

a) kWh/ton b) kCal/ton c) kJ/kg d) kg/kCal

4. Which of the following GHGs has the longest atmospheric life time

a) CO2 b) Sulfur Hexafluoride (SF6)

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REGULAR Paper 1 –Set A

c) CFC d) Per FluoroCarbon (PFC)

5. Which of the following comes under mandatory labeling programme

a) diesel Generators b) induction motors

c) tubular Fluorescent Lamps d) LED lamps
6. Transit time method is used in which of the instrument

a) lux meter b) ultrasonic flow meter c) pitot tube d) fyrite

7. To improve the boiler efficiency, which of the following needs to be done

a) maximize O2 in flue gas b) maximize CO2 in flue gas

c) minimize CO2 in flue gas d) maximize CO in flue gas
8. The simplest technique for scheduling of tasks and tracking the progress of
energy management projects is called

a) Gantt chart b) CPM c) PERT d) WBS

9. The ratio of wind power in the wind actually converted into mechanical
power and the power available in the wind is about

a) 75% b) 59% d) 44% e)10%

10. The quantity of heat required to raise the temperature of 1 kg of water by 1
O
C is termed as

a) latent heat b) one kilojoule c) one kilo calorie d) none of the above
11. The present value of Rs. 1,000 in 10 years’ time at an interest rate of 10% is

a) Rs. 2,594 b) Rs. 386 c) Rs. 349 d) Rs. 10,000

12. The number of moles of water contained in 54 kg of water is ------------

a) 2 b) 3 c) 4 d) 5
13. The monthly electricity bill for a plant is Rs. 100 lakhs which accounts for 45%
of the total monthly energy bill. How much is the plant’s monthly energy bill

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REGULAR Paper 1 –Set A

a) Rs 222.22 lakhs b) Rs 45 lakhs c) Rs 138 lakhs d) None of above

14. The major share of energy loss in a thermal power plant is in the

a) generator b) boiler c) condenser d) turbine

15. The ISO standard for Energy Management System is

a) ISO 9001 b) ISO 50001 c) ISO 140001 d) None of the

above
16. The indicator of energy performance in a thermal power plant is

a) heat rate (kCal/kWh) b) % aux. power consumption

c) specific coal consumption d) all the above
17. The fixed energy consumption for the company is 1,000 kWh. The slope in
the energy –production chart is found to be 0.3. Find out the actual energy
consumption if the production is 80,000 Tons

a) 25,000 b) 24,000 c) 26,000 d) 23,000

18. The cost of replacement of inefficient compressor with an energy efficient
compressor in a plant was Rs 50 lakhs. The net annual cash flow is Rs 12.5
lakhs. The return on investment is

a) 15% b) 20% c) 25% d) 19.35%

19. The contractor provides the financing and is paid an agreed fraction of actual
savings achieved. This payment is used to pay down the debt costs of
equipment and/or services. This is known as

a)traditional contract b) extended technical guarantee/service

c) performance contract d) shared savings performance
contract
20. PERT/CPM provides which of the following benefits

a) predicts the time required to complete the project

b) shows activities which are critical to maintaining the schedule
c) graphical view of the project

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REGULAR Paper 1 –Set A

d) all the above

21. Input fuel of fuel cell

a) petrol b) hydrogen c) nitrogen d) natural gas

22. In India power sectors consumes about_______% of the coal produced
a) 75% b) 50% c) 25% d) 90%

23. In an industry the average electricity consumption is 5.8 lakhs kWh for the
period, the average production is 50,000 tons with a specific electricity of 11
kWh/ton for the same period. The fixed electricity consumption for the plant
is

a) 58000 kWh b) 30000 kWh c) 80000 kWh d) none of the

above
24. In a drying process, moisture is reduced from 60% to 30%. Initial weight of
the material is 200 kg. Calculate the weight of the product

a) 104 b) 266.6 c) 130 d) 114.3

25. In a DG set, the generator is consuming 400 litres per hour diesel oil. If the
specific fuel consumption of this DG set in 0.30 litres/kWh at that load then
what is the kVA loading of the set at 0.6 power factor

a) 1200 KVA b) 2222 KVA c) 600 KVA d)1600 KVA

26. In a 50 Hz AC cycle, the current reverses directions ________ times per
second

a) 50 times b) 100 times c) Two times d) 25 times

27. If we heat the air without changing absolute humidity, % relative humidity
will

a) increase b) decrease c) no Change d) can’t say

28. If the pressure of water is 0.7 kg/cm2 then boiling point will be approximately

a) 100 b) 73 c) 114 d) Can’t say

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REGULAR Paper 1 –Set A

29. If heat rate of power plant is 860 kcal/kWh then the cycle efficiency of power
plant will be

a) 41% b) 55% c) 100% d) 86%

30. If air consists of 77% by weight of nitrogen and 23% by weight of oxygen, the
mean molecular weight of air is

a) 11.9 b) 28.8 c) 17.7 d) insufficient data

31. How much power generation potential is available in a run of river mini
hydropower plant for a flow of 40 liters/second with a head of 24 metres.
Assume system efficiency of 60%

a) 5.6 kW b) 9.4 kW c) 4.0 kW d) 2.8 kW

32. Fuel cell using methanol as anode and oxygen as cathode is

a) proton exchange membrane fuel cell b) phosphoric acid fuel cell

b) alkaline fuel cell d) direct methanol fuel cell
33. For expressing the primary energy content of a fuel in tonnes of oil equivalent
(toe) which of the following conversion factors is appropriate

a) toe=1x106 kcal b) toe=116300 kwh c) toe=41.870 GJ d) all the above

34. For calculating plant energy performance which of the following data is not
required

a) current year’s production b) reference year’s production

c) reference year energy use d) capacity utilization
35. ESCerts cannot be

a) bought b) sold c) banked for next cycle d) traded directly

between DCs
36. Energy intensity is the ratio of

a) fuel consumption / GDP b) GDP/fuel consumption

c) GDP/ energy consumption d) energy consumption / GDP

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REGULAR Paper 1 –Set A

37. Costs associated with the design, planning, installation and commissioning of
a project are

a) variable costs b) capital costs c) salvage value d) none of the

above
38. At standard atmospheric pressure, specific enthalpy of saturated water,
having temperature of 50 OC will be _________ kcal/kg

a) 1 b) 50 c) 100 d) Can't say

39. AT & C losses means

a) administration transmission and commercial

b) aggregate technical and commercial
c) average technical and commercial
d) none of the above
40. As per primary commercial energy consumption mix in India, the fuel
dominating the energy production mix in India is
a) natural gas b) oil c) coal d) nuclear energy

An oil-fired boiler operates at an excess air of 6 %. If the stoichiometric air

41. fuel ratio is 14 then for an oil consumption of 100 kg per hour, the flue gas
liberated in kg/hr would be

a)1484 b) 1584 c) 106 d) 114

42. An activity has an optimistic time of 15 days, a most likely time of 18 days and
a pessimistic time of 27 days. What is the expected time

a) 60 days b) 20 days c) 19 days d) 18 days

43. Among which of the following fuels, the difference between the GCV and
NCV is maximum

a) coal b) furnace Oil c) natural gas d) rice husk

44. A waste heat recovery system costs Rs. 54 lakhs and Rs. 2 lakhs per year to

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REGULAR Paper 1 –Set A

operate and maintain. If the annual savings is Rs. 20 lakhs, the payback
period will be

a) 8 years b) 2.7 years c) 3 years d) 10 years

45. A process requires 10 Kg of fuel with a calorific value of 5000 kcal/kg. The
system efficiency is 80% and the losses will be

a) 10000 kcal b) 45000 kcal c) 500 kcal d) 2000 kcal

46. A centrifugal pump draws 12 m3/hr. Due to leakages from the body of the
pump a continuous flow of 2 m3/hr is lost. The efficiency of the pump is 55%.
The flow at the discharge side would be

a) 12 m3/hr b) 10 m3/hr c) 5.5 m3/hr d) 6.6 m3/hr

47. A 400W lamp was switched on for 10 hours per day. The supply volt is 230V
(current= 2 amps & PF= 0.8). What is the energy consumption per day

a) 3.68 kWh b) 6.37 kWh c) 0.37 kWh d) 4.0 kWh

48. 20 m3 of water is mixed with 30 m3 of another liquid with a specific gravity of
0.9. The volume of the mixture would be

a) 47 m3 b) 48 m3 c) 50 m3 d) 53 m3
49. 100 tons of coal with a GCV of 4200 kcal/kg can be expressed in ‘tonnes of oil
equivalent’ as

a) 42 b) 50 c) 420 d) 125
50. 1 kg of wood contains 15% moisture and 7% hydrogen by weight. How much
water is evaporated during complete combustion of 1 kg of wood

a) 0.78 kg b) 220 grams c) 0.15 kg d) 0.63 kg

…….……. End of Section – I ………..….

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 REGULAR Paper 1 –Set A

Section – II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 =

40

(i) Answer all Eight questions

(ii) Each question carries Five marks

S-1 A gas fired water heater heats water flowing at a rate of 20 litres per minute
from 250 C to 85oC. If the GCV of the gas is 9200 kcal/kg, what is the rate of
combustion of gas in kg/min (assume efficiency of water heater as 82%)
Solution:

Volume of water heated = 20 liters/min

Mass of water heated = 20 Kg/min
Heat supplied by gas * efficiency = Heat required by water.
Mass of gas Kg/min * 9200 * 0.82 = 20 Kg/min * 1 kcal/Kg/oC)* (85-
25)oC

Mass of gas Kg/min = (20*1*60)/ (9200*0.82)

= 0.159 Kg/ min.

S-2 Calculate the net present value over a period of 3 years for a project with
the following data. The discount rate is 12%.

Year Investment (Rs) Savings (Rs)

0 75,000
1 25,000
2 75,000
3 50,000 75,000
4 35,000

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 REGULAR Paper 1 –Set A

Ans:
NPV = - 75,000 + 25,000/(1+0.12) + 75,000/(1+0.12)2 + (75,000 –
50,000)/(1+0.12)3

= -75,000 + 22,321 + 59,789 + 17, 794

= 24,904 Rs.

S-3 In a process plant, an evaporator concentrates a liquor containing solids of

6% by w/w (weight by weight) to produce an output containing 30% solids
w/w. calculate the evaporation of water per 500 kgs of feed to the
evaporator.
Solution :
Inlet solid contents = 6 %
Output solid contents = 30%
Feed = 500 kgs
Inlet solid content in kg in feed = 500 x 0.06 = 30 kg
Outlet solid content in kg = 30 kg

Quantity of water evaporated = [500 – {(30 / 30) x 100}] = 400 kg.

S-4 List down at least five schemes of BEE under the Energy Conservation Act –
2001
Ans:
Schemes of BEE under the Energy Conservation Act - 2001 are as
follows:
 Energy conservation building codes (ECBC)
 Standards and labeling (S&L)
 Demand side management (DSM)
 Bachat Lamp Yojana (BLY)
 Promoting energy efficiency in small and medium enterprises
(SME’s)
 Designated consumers
 Certification of energy auditors and energy managers

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 REGULAR Paper 1 –Set A

S-5 What parameters are measured with the following instruments?

a) Pitot tube
b) Stroboscope
c) Fyrite
d) Psychrometer
e) Anemometer
Ans:

a. Pitot tube Static, Dynamic and Total Pressure of Gas

b. Stroboscope Speed, RPM
c. Fyrite CO2 % or O2 %
d. Psychrometer Dry Bulb Temperature and Wet Bulb
Temperature
e. Anemometer Air or wind velocity

S-6 What are ESCerts and explain the basis for their issue and trading under PAT
scheme ?
PAT scheme provides the option for industries who achieve superior
savings to receive energy savings certificates for this excess savings, and to
trade the additional certified energy savings certificates with other
designated consumers (energy intensive industries notified as
Designated Consumers under the Energy Conservation Act and included
under PAT Scheme) who can utilize these certificates to comply with
their specific energy consumption reduction targets. Energy Savings
Certificates (ESCerts) so issued will be tradable at Power Exchanges. The
scheme also allows units which gain ESCerts to bank them for the next cycle
of PAT, following the cycle in which they have been issued.

S–7 Pressure of a nitrogen gas supplied to an oil tank for purging is measured as
100 mm of water gauge when barometer reads 756 mm of mercury.

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 REGULAR Paper 1 –Set A

Determine the volume of 1.5 kg of this gas if it’s temperature is 25 0C.

Specific gravity of mercury: 13.6. Take R = 8.3143 kJ/(kMol x K)
Ans:
Nitrogen pressure = 100 mm of Water Gauge = 100 / 13.6 = 7.353 mm
of Hg

Absolute Temperature, T = 25 + 273 = 298 K,

Mass = 1.5 kg & Barometric pressure = 756 mm of Hg.

Absolute pressure = 756 + 7.353 = 763.353 mm of Hg

Pressure, P = Density, (kg/m3) x Gravity, g (m/s2) x Mtr of Liquid, h

(Mtr) / 1000
= (13,600 x 9.81 x 0.763)/1000
= 101.79 kPa

Molar mass of Nitrogen = 28 kg/kMol.

Number of kMol, n = Mass / Molar Mass = 1.5/ 28 = 0.0536 kMol

Using the ideal gas equation and putting the above values;

PV = nRT
101.79 x V = 0.0536 x 8.3143 x 298
V = 1.395 m3

S–8 Distinguish between designated agency and designated consumer as per

energy conservation act 2001
Ans:
Designated Agency: Designated agency means an agency which
coordinates, regulates and enforces of Energy Conservation Act
2001within a state.

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REGULAR Paper 1 –Set A

Designated Consumer: Designated consumer means any users or

class of users of energy in the “energy intensive industries and other
establishments” specified in Schedule as designated consumer.

…….……. End of Section – II ………..….

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 REGULAR Paper 1 –Set A

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 =

60

(i) Answer all Six questions

(ii) Each question carries Ten marks

L-1 a) A furnace heating steel ingots is fired with oil having a calorific value of 10,500
kCal/kg and efficiency of 75%. Calculate the oil consumption per hour when the
throughput of the furnace is 50 TPH and the temperature of the finished product is
600 oC. Take ambient temperature as 30 oC and Specific Heat of Steel as 0.12
kCal/kg oC

b) In Steel industry, different types of gases are generated during steel making
process.

Volumetric Flow rate and Calorific Values of each gases are:

Type of Gas Flow (SM3/hr) CV (kCal/SM3)

Coke Oven Gas 75,000 4,000
COREX Gas 50,000 2,000
BOF Gas 55,000 1,500
Blast Furnace Gas 80,000 700

All these gases are mixed in the gas mixer before combustion. Find out the
Calorific Value (in kCal/SM3) of mix gas.

Ans:
a) Oil Consumption / hr

50 (TPH) x 0.12 (kCal/kg oC) x (600 – 30) (oC)

= ------------------------------------------------------------------
0.75 (%) x 10,500 (kCal/kg)

= 0.43 TPH

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 REGULAR Paper 1 –Set A

b) Total flow of Mix Gas = 75,000 + 50,000 + 55,000 + 80,000 = 2,60,000 SM3/hr

CV of Mix Gas =
[(75,000 x 4,000) + (50,000 x 2,000) + (55,000 x 1,500) + (80,000 x 700)] /
2,60,000

= 2,071 kCal/SM3

L–2 A) Briefly explain the following terms with respect to energy management?

I. Normalizing
II. Benchmarking

B) Explain the meaning of Fuel and Energy substitution with examples.

Ans:
A) I) Normalizing:

The energy use of facilities varies greatly, partly due to factors beyond the energy
efficiency of the equipment and operations. These factors may include weather or
certain operating characteristics. Normalizing is the process of removing the
impact of various factors on energy use so that energy performance of facilities
and operations can be compared.

II) Benchmarking:

Comparison of energy performance to peers and competitors to establish a

relative understanding of where our performance ranks.

B) Fuel and Energy substitution with examples:

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 REGULAR Paper 1 –Set A

Substituting existing fossil fuels/energy with more efficient and / or less cost/less
polluting fuel.

Few examples of fuel substitution

 Natural gas is increasingly the fuel of choice as fuel and feedstock in the
fertilizer, petrochemicals, power and sponge iron industries.
 Replacement of coal by coconut shells, rice husk etc.
 Replacement of LDO by LSHS

Few examples of energy substitution

 Replacement of electric heaters by steam heaters.
 Replacement of steam based hot water by solar systems.

L-3 The details of activities for a pump replacement project is given below:

a) Draw a PERT chart

b) Find out the duration of the project
c) Identify the critical path.

Activity Immediate Time

Predecesso (days)
rs
A - 1
B A 2
C B 4
D C 6
E C 3
F C 5
G D, E, F 8
H G 7
Ans:

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 REGULAR Paper 1 –Set A

Duration = 28 days

Critical Path: A-B-C-D-G-H

L–4 The production capacity of a paper drying machine is 500 TPD and is currently
operating at an output of 480 TPD. To find out the steam requirement for drying,
the Energy Manager measures the dryness of the paper both at inlet and outlet of
the paper drying machine which found to be 60% and 95% respectively.

The steam is supplied at 4 kg/cm2, having a latent heat of 510 kCal/kg. The
evaporated moisture temperature is around 100 0C having enthalpy of 640
kCal/kg. Plant operates 24 hours per day. Assume only latent heat of steam is
being used for drying the paper and neglect the enthalpy of the moisture
in the wet paper.

i) Estimate the quantity of moisture to be evaporated per hr.

ii) Input steam quantity required for evaporation per hr.

Ans:
Output of the drying machine = 480 TPD with 95% dryness.

Bone dry mass of paper at the output = 480 x 0.95 = 456 TPD

Since the dryness at the inlet is 60%,

Total mass of wet paper at the inlet = (456 x 100) / 60 = 760 TPD

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 REGULAR Paper 1 –Set A

Moisture evaporated per hour = (760 – 480)/ 24 = 11.67TPH

Mass of Steam, m = (11.67 x 640)/ 510 = 14.6 TPH

L-5 Use CUSUM technique to develop a table and to calculate energy savings for 8
months period. For calculating total energy saving, average production can be
taken as 6,000 MT per month. Refer to field data given in the table below.

Month Actual SEC, Predicted SEC,

kWh/MT kWh/MT
May 1311 1335
June 1308 1335
July 1368 1335
Aug 1334 1335
Sept 1338 1335
Oct 1351 1335
Nov 1322 1335
Dec 1320 1335

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 REGULAR Paper 1 –Set A

Ans

Actual Diff = ( Act - Pred

Predicted CUSUM
Month SEC, ) (-=
SEC, kWh/MT ( - = Saving )
kWh/MT Saving )
May 1311 1335 -24 -24
June 1308 1335 -27 -51
July 1368 1335 33 -18
Aug 1334 1335 -1 -19
Sept 1338 1335 3 -16
Oct 1351 1335 16 0
Nov 1322 1335 -13 -13
Dec 1320 1335 -15 -28

Savings in energy consumption over a period of eight months are 28 x 6000

=1,68,000 kWh

L-6 Write short notes on?

1. Time of the day tariff

2. Comparative label
3. Endorsement label
4. Benefits of ISO 50001

Solution:

1) In Time of the Day Tariff (TOD) structure incentives for power drawl during
off-peak hours and disincentives for power drawl during peak hours are built
in.
 Many electrical utilities like to have flat demand curve to achieve high
plant efficiency.
 ToD tariff encourage user to draw more power during off-peak hours

18
REGULAR Paper 1 –Set A

(say during 11pm to 5 am, night time) and less power during peak hours.
Energy meter will record peak and off-peak consumption and normal
period separately.
 ToD tariff gives opportunity for the user to reduce their billing, as off
peak hour tariff is quite low in comparison to peak hour tariff.
 This also helps the power system to minimize in line congestion, in turn
higher line losses and peak load incident and utilities power
procurement charges by reduced demand

2) Comparative label: allow consumers to compare efficiency of all the

models of a product in order to make an informed choice. It shows the
relative energy use of a product compared to other models available in
the market.

3) Endorsement label: define a group of products as efficient when they

meet minimum energy performance criteria specified in the respective
product schedule/regulation/statutory order.

4) ISO 50001 will provide the following benefits

 A framework for integrating energy efficiency into management

practices

 Making better use of existing energy-consuming assets

 Benchmarking, measuring, documenting, and reporting energy intensity

improvements and their projected impact on reductions in greenhouse
gas (GHG) emissions

 Transparency and communication on the management of energy

resources

 Energy management best practices and good energy management

19
REGULAR Paper 1 –Set A

behaviours

 Evaluating and prioritizing the implementation of new energy-efficient

technologies

 A framework for promoting energy efficiency throughout the supply

chain

 Energy management improvements in the context of GHG emission

reduction projects.

…….……. End of Section – III ………..….

20
 Paper 1 –Set A Reg with Solutions

17th NATIONAL CERTIFICATION EXAMINATION

FOR
ENERGY MANAGERS & ENERGY AUDITORS – September, 2016

PAPER – 1: GENERAL ASPECTS OF ENERGY MANAGEMENT & ENERGY

AUDIT

Date: 24.09.2016 Timings: 0930-1230 HRS Duration: 3 HRS

General instructions:
o Please check that this question paper contains 8 printed pages
o Please check that this question paper contains 64 questions
o The question paper is divided into three sections
o All questions in all three sections are compulsory
o All parts of a question should be answered at one place

Section – I: OBJECTIVE TYPE

1. The energy intensity of countries that rely on import of carbon-intensive goods when
compared with those producing it, would in all probability be

a) Higher b) Lower c) Almost equal d) No correlation

O
2. If a 2 KW immersion heater is used to heat 30litres of water at 30 C, what would be the
temperature of water after 15 minutes? Assume no losses in the system
O O O
a) 87.3 C b) 44.3 C c) 71.3 C d) none of the above
3. Which of the following statements regarding ECBC are correct?

i) ECBC defines the norms of energy requirements per cubic metre of area
ii) ECBC does not encourage retrofit of Energy conservation measures
iii) ECBC prescribes energy efficiency standards for design and construction of
commercial and industrial buildings
iv) One of the key objectives of ECBC is to minimize life cycle costs
(construction and operating energy costs)

a) i b) ii c) iiii d) iv
4. Verification and Check-verification under PAT will be carried out by
a) Designated consumers
b) Accredited energy auditors
c) Certified energy auditor
d) Empanelled accredited energy auditors

5. Which of the following enhances the energy efficiency in buildings?

a) Light pipes

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Bureau of Energy Efficiency
 Paper 1 –Set A Reg with Solutions

b) Triple glaze windows

c) Building integrated solar photovoltaic panels
d) All of the above
6. Energy is consumed by all sectors of the economy but at different proportions. Which
sector in India is the largest consumer?

a) Agriculture c). Commercial

b) Industrial d). Domestic
7. M & V audit under PAT is carried out

a) Immediately after the baseline audit

b) Every year following the baseline audit
c) At the end of each PAT cycle
d) Before the baseline audit
8. Which of the following sector is not covered under PAT?

a) Chlor-alkali c) Cement
b) Aluminum d) Commercial buildings

9. A solar _______ is connected and packaged in a solar _________, which in turn is

linked with others in sequence in a solar _________.

a) module, cell, array

b) array, module, sequence
c) module, array, sequence
d) cell, module, array
10. At Standard Atmospheric Pressure, specific enthalpy of saturated water, having
o
temperature of 50 C will be _________ kCal/kg.

a) 1 b) 50 c) 100 d) none of the above

11. In a drying process product moisture is reduced from 60% to 30%. Inlet weight of the
material is 200 kg. Calculate the weight of the outlet product.

a) 80 b) 120.5 c) 114.3 d) none of the above

12. Which among the following factor(s) is most appropriate for adopting EnMS?

a) To improve their energy efficiency

b) To reduce costs
c) To increase productivity
d) Systematically manage their energy use
13. Which energy source releases the most climate-altering carbon pollution per kg?

a). Oil b). Coal c). Rice husk d). Bagasse

14. What is the future value of Rs.1000/- after 3 years, if the interest rate is 10%

a) Rs. 1331 b) Rs.1610 c) Rs.3221 d) none of the above

15. Having a documented energy policy in industry

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Bureau of Energy Efficiency
 Paper 1 –Set A Reg with Solutions

a) Satisfies regulations
b) Reflects top management commitment
c) Indicates availability of energy audit skills
d) None of the Above
16. Red wood seconds is a measure of

a) Density b) Viscosity c) Specific gravity d) Flash point

17. Which amongst the following sources of electricity has the highest installed capacity in
India ?

a)Gas b) Nuclear c) Oil d) Renewables

18. Energy Intensity is the ratio of

a) Fuel Consumption / GDP

b) GDP/Fuel Consumption
c) GDP/ Energy Consumption
d) Energy Consumption / GDP
19. If Heat Rate of Power plant is 3000 kCal/kWh then efficiency of Power plant will be

a) 28.67% b) 35% c) 41% d) None of the above

20. In a solar thermal power station , molten salt which is a mixture of 60% sodium nitrate
and 40% potassium nitrate is used. It is preferred as it provides an efficient low cost
medium to store _______

a. Electrical energy
b. Thermal energy
c. Kinetic energy
d. Potential energy
21. For every 10°C rise in temperature, the rate of chemical reaction doubles. When the
temperature is increased from 30°C to 70°C, the rate of reaction increases __________
times.
a) 8 b) 64 c) 16 d) none of the above
22. The producer gas is basically

a. CO, H2 and CH4

b. Only CH4
c. CO and CH4
d. Only CO and H2
23. The essential elements of monitoring and targeting system is
a) Recording b) Reporting
c) Controlling d) All of the above
24. One energy saving certificate ( ESCerts) under PAT is equivalent to

a. one ton of carbon

b. one MWh of electricity
c. one ton of coal
d. one ton of Oil equivalent
25. In an industry the billed electricity consumption for a month is 5.8 lakh kWh. The fixed
electricity consumption of the plant is 30000kWh and with a variable electricity

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 Paper 1 –Set A Reg with Solutions

consumption of 11 kWh/ton. Calculate the production of the industry

a) 50000 tonnes b) 60000 tonnes c) 58000 tonnes d) None of the above

26. If the reactive power drawn by a particular load is zero it means the load is operating
at
a) Lagging power factor b) Unity power factor
c) Leading power factor d) none of the above

27. Capital cost are associated with

a) Design of Project
b) Installation and Commissioning of Project
c) Operation and Maintenance cost of project
d) both a and b
28. Any management would like to invest in projects with

a) Low IRR b) Low ROI

c) Low NPV of future returns d) none of the above
29. The kilowatt-hour is a unit of

a) power b) work c) time d) force.

30. Which among the following is a green house gas?

a) Sulphur Dioxide b) Carbon Monoxide

c) NO2 d) Methane
31. The internal rate of return is the discount rate for which the NPV is

a. Positive
b. Zero
c. Negative
d. Less than 1
32. Greenhouse effect is caused by natural affects and anthropogenic effects. If there is no
natural greenhouse effect, the Earth's average surface temperature would be around
__________°C.

a) 0 b) 32 c) 14 d) - 18
o
33. The quantity of heat required to raise the temperature of a given substance by 1 C is
known as:

a) sensible heat b) specific heat c) heat capacity d) latent heat

34. The Metric Tonne of Oil Equivalent (MTOE) value of 125 tonnes of coal having GCV of
4000 kcal/kg is

a) 40 b) 50 c) 100 d) 125
35. A mass balance for energy conservation does not consider which of the following
a. Steam
b. water
c. Lubricating oil
d. Raw material
36. A sling psychrometer is capable of measuring

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Bureau of Energy Efficiency
 Paper 1 –Set A Reg with Solutions

a) only dry bulb temperature b) only wet bulb temperature

c) both dry and wet bulb temperature d) absolute humidity
37. Which of these is not true of payback period
a. Simple to calculate
b. Considers cash flow beyond the payback period
c. Shorter the period the better
d. Does not take into account, time value of money
38. To judge the attractiveness of any investment, the energy auditor must consider
a) Initial capital cost b) Net operating cash inflows
c) salvage value d) all the above
39. In a cumulative sum chart if the graph is going up, it means
a. Energy consumption is going up
b. Energy consumption is going down
c. Specific energy consumption is coming down
d. No inference can be made
40. Doppler effect principle is used in the following instrument
a) lux meter b) ultrasonic flow meter
c) infrared thermometer d) flue gas analyzer

In a coal fired boiler, hourly consumption of coal is 1300 kg. The ash content in the coal
41. is 6%. Calculate the quantity of ash formed per day. Boiler operates 24 hrs/day.
a) 216 kg b) 300 kg c) 1872 kg d) none of the above
42. Liquid fuel density is measured by an instrument called
a) Tachometer b) hygrometer c) anemometer d) none of the above
43. A comparison of the trapping of heat by CO2 and CH4 is that
a) CH4 traps 21 times more heat in the atmosphere than does CO 2
b) CO2 traps 21 times more heat in the atmosphere than does CH4
c) the same amount of heat is trapped by both CO2 and CH4
d) none of the above
44. Diagrammatic representation of input and output energy streams of an equipment or
system is known as
a) mollier diagram b) sankey diagram
c) psychrometric chart d) balance diagram
45. ISO 50001:2011 provides a framework of requirements for organizations to:
a) Develop a policy for more efficient use of energy b) Measure the results
c) Fix targets and objectives to meet the policy d) all of the above
46. A three phase induction motor is drawing 16 Ampere at 440 Volts. If the operating power
factor of the motor is 0.90 and the motor efficiency is 92%, then the mechanical shaft
power output of the motor is

a) 12.04 kW b) 10.09 kW
c) 10.97 kW d) None of the above
47. Absolute pressure is
a. Gauge pressure
b. Gauge pressure + Atmospheric pressure
c. Atmospheric pressure
d. Gauge pressure - Atmospheric pressure
48. In a chemical process two reactants A (300 kg) and B (400 kg) are used. If conversion is
50% and A and B react in equal proportions, the mass of the product formed is.

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a) 300 kg b) 350 kg
c) 400 kg d) none of the above
49. What is the expected power output in watts from a wind turbine with 6m diameter rotor,
a coefficient of performance 0.45, generator efficiency 0.8,a gear box efficiency 0.90
and wind speed of 11m/sec

a. 4875 watts
b. 1100 watts
c. 7312 watts
d. 73.12 kW
50. The lowest theoretical temperature to which water can be cooled in a cooling tower is
a. Difference between DBT and WBT of the atmospheric air
b. Average DBT and WBT of the atmospheric air
c. DBT of the atmospheric air
d. WBT of the atmospheric air

…….……. End of Section – I ………..….

Section – II: SHORT DESCRIPTIVE QUESTIONS

S-1 Give relationship between Absolute and Gauge pressures. Give 4

different units used in pressure measurement.

Ans Absolute pressure is zero-referenced against a perfect vacuum, so it is equal to gauge

pressure plus atmospheric pressure.
Gauge pressure is zero-referenced against ambient air pressure, so it is equal to absolute
pressure minus atmospheric pressure. (Negative signs are usually omitted)
Absolute Pressure = Prevailing Atmospheric Pressure + Gauge Pressure

(NOTE: also please refer guide book-1 pg-70)

The four units of pressure measurement are:

i) Pascal
2
ii) kg / cm
iii) Atmospheric
iv) mm of mercury
v) Meters of water column
2
vi) Pounds / inch

S- 2 A plant is using 6 tonnes / day of coal to generate steam . The calorific of coal is 3300
kcal/kg. The cost of coal is Rs 4200/tonne . The plant substitutes coal with agro-residue ,
as a boiler fuel, which has a calorific value of 3100 kcal/kg and costs Rs 1800/tonne.
Calculate the annual cost savings at 300 days of operation, assuming the boiler efficiency
remains same at 72% for coal and agro residue as fuel.

Ans Useful energy to generate steam by 6 tonnes of coal per day

= 6000 x 3300 x 0.72 = 14256000 kcal/day

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To deliver 14256000kcal/day , daily amount of rice husk required

= 14256000 =6387 kg/day
3100 x 0.72
Daily saving = 6000 x 4200 - 6387 x 1800
1000 1000
= 25200-11497
= Rs 13703/-
Annual saving=13703 x 300
=Rs 41,10,900/-

S–3 Explain how an ESCO model works?

ESCOs are usually companies that provide a complete energy project service, from assessment
to design to construction or installation, along with engineering and project management services
and financing.
The ESCO will usually offer the following performance contract options.
 Fixed fee
 Shared Savings
 Guaranteed savings

S-4 The annual fuel cost of boiler operation in a plant is Rs.10 Lakhs. The boiler with 65%
efficiency is now replaced by a new one with 78% efficiency. What is the annual cost
savings?

Existing efficiency =65%

Proposed efficiency=78%

Annual fuel cost =Rs. 10 Lakhs

Annual cost savings = annual fuel cost *( 1-(EffO/EffN))

= 10 x ((1-(0.65/0.78))

=Rs. 1,66,667 per annum

S-5 A tank containing 600 kg of kerosene is to be heated from 10°C to 40°C in 20 minutes,
using 4 bar (g) steam. The kerosene has a specific heat capacity of 2.0 kJ/kg °C over that
temperature range. Latent heat of steam (hfg) at 4.0 bar g is 2108.1 kJ/kg. The tank is well
insulated and heat losses are negligible.
Determine the steam flow rate in kg/hr.
O O
Q = 600 kg x 2 kJ/kg C x (40-10) C/(1200)

= 30 kJ/sec

Therefore mass of steam = 30 kJ/sec x3600 / 2108 .1 kJ/kg

= 51.23 kg/h

S–6 Feed water is provided to a boiler at 70oC from the feed water tank. The
temperature of condensate water returning to the tank is 86oC, while the

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temperature of makeup water is 27oC. Determine the amount of condensate water

that can be recovered?

 Performing a mass & heat balance yields,

 (i.e : mCpdTCond + mCpdT MakeUp = mCpdT FeedWater)
(27)(x)(1)+ (86)(1-x)(1) = 70(1)(1)
 Therefore, x = 0.27 or 27% of make-up water.
 Hence, condensate recovered = 73%
S–7 In a textile manufacturing unit, wet cloth is dried in a stenter. The cloth entering the stenter
has a moisture of 52% while that leaving the stenter is 96% dry. If the production rate
(output) from the stenter is 200 Kg/hr, what is the quantity of steam required per hour, if
the steam enters the stenter with an enthalpy of 660 kcal/kg. The condensate leaving the
o
stenter is at 170 C Consider drying to take place at atmospheric pressure where the latent
heat of water is 540 Kcal/Kg.

Stenter output = 200 kgs/hr

Bone dry cloth in output = 200 X 0.96 = 192 kgs.

Moisture in output = 8 kgs.

Moisture in input = 52%
Bone dry cloth in input = 48%
Total weight of input cloth = 192/0.48 = 400 kg/hr
Quantity of water evaporated = 400 – 200 = 200 kg/hr
O
Assuming sensible heat in steam at 170 C = 170 kcal/kg
Quantity of steam required = (200 X 540)/(660 – 170)
= 220.4 kg/hr

S–8 The average monthly electricity consumption in an Aluminium producing unit is 12.35 lac
kWh. The other energy sources used in the manufacturing process are Furnace oil (GCV-
9660 kcal/Ltr) and HSD (GCV-9410 kcal/Ltr). If the annual fuel oil consumption is 5760 kL of
Furnace oil (sp. gr. 0.92) and 720 kL of HSD (sp. gr. 0.88), determine if the unit qualifies as a
Designated Consumer under the EC Act?
7
1 Mtoe = 1 x 10 kcal
Annual electrical energy consumption = 12.35 x 12 = 148.2 lac Kwh
5 7
Equivalent heat energy = (148.2 x 10 x 860)/(1 x 10 )
= 1274.52 Mtoe _(i)
Annual Furnace oil consumption = 5760 kL
7
Equivalent heat energy = (5760 x 1000 x 9660)/(1 x 10 )
= 5564.16 Mtoe _(ii)
Annual HSD consumption = 720 kL
7
Equivalent heat energy = (720 x 1000 x 9410)/(1 x 10 )
= 677.52 Mtoe _(iii)

Total annual energy consumption = 1274.52 + 5564.16 + 677.52

= 7516.2 Mtoe

To be a designated consumer, the minimum annual energy consumption (in aluminium sector)

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should be 7500 Mtoe. As the plant exceeds this threshold limit, it qualifies to be a designated
consumer.

…….……. End of Section – II ………..….

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Section – III: LONG DESCRIPTIVE QUESTIONS

L-1 The integrated paper plant has produced 119366 MT of paper during the year 2012-13. The
management has implemented various energy conservation measures as part of PAT scheme
and reduced the specific energy consumption from 53 GJ/ tonne of product to 50 GJ/tonne of
product. The actual production during the assessment year (2014-15) is 124141 MT. Calculate
the plant energy performance and state your inference.

Ans Reference year production = 119366 MT

Reference year specific energy consumption = 53 GJ/tonne of product
Assessment year production = 124141 MT
Assessment year specific energy consumption = 50 GJ/tonne of product

production factor = (124141 / 119366) = 1.04

= 53 x 119366 = 6326398 GJ

=50 x 124141 = 6207050 GJ

= 6326398 GJ x 1.04 =6579454 GJ

= ((6579454 - 6207050 ) / 6579454 )x 100

= 5.66%

Inference : plant energy performance is positive and hence the plant is achieving energy savings.

L – 2 a) A 20 kW, 415V, 38A, 4 pole, 50 Hz, 3 phase rated squirrel cage induction motor has a full
load
efficiency and power factor of 88% and 0.85 respectively. An energy auditor measures the

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following
operating data of the motor.
1) Supply voltage= 408V
2) Current drawn= 30A
3) PF=0.83
Find out the following at motor operating conditions.
1) Power input in kW
2) % motor loading

b) List five energy saving measures in your home

Ans a) 1) Power input = 1.732*408*30*0.83

= 17.60 kW
2) % motor loading = [17.60/(20/0.88)]*100
= [(17.60/22.73)]*100
= 77.43%

b)
 Replacement of inefficient electric lamps with efficient electric lamps
 Using star labeled household appliances like A/c’s, Refrigerator,Lamps,Fans
 Using Solar water heating systems for hot water requirements to minimize use of electric
geysers
 Using Solar PV systems for electricity generation
 Proper ventilation maximizing the use of natural light
 Switching off all equipment when not required
 Using pressure cooker for cooking food
 Maximizing the use of low fire burner (SIM) in the gas stove

o o o
Using A/Cs at setpoint of 21 C-23 C instead of 16 C
 Placing the fridge so that the rear ( condenser coils ) are located where there is proper air
flow.

L-3 The cash flows in two different energy conservation projects are given in the table below. Please help
the management of an infrastructure company to decide which project to invest in as the
management is interested in investing in only one project. The company is likely to consider any
project which gives a minimum return on investment of 18%. Please justify your choice.
(Amount in Rs.)
Project A Project B
Investment 17,50,000/- 12,00,000/-
Year Expenses Savings Expenses Savings
1 4,00,000 4,50,000
2 4,00,000 4,00,000
3 4,00,000 3,50,000
4 4,00,000 3,00,000
5 1,00,000 6,00,000 2,50,000
6 6,00,000 2,00,000
7 6,00,000 1,16,650

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8 3,80,300

Ans As the investments required in both the cases as well as their durations are different, the prudent
method to compare the two projects would be to calculate their NPV.

0 1
a) NPV of Project A @ 18% = (-1750000 / (1+0.18) ) + (4,00,000 / (1+0.18) ) +(4,00,000 /
2 3 4 5
(1+0.18) ) +(4,00,000 / (1+0.18) )+ (4,00,000 / (1+0.18) ) +((6,00,000-100000) / (1+0.18) )+
6 7 8
(6,00,000 / (1+0.18) ) +(6,00,000 / (1+0.18) )+ (3,80,300 / (1+0.18) ) = 57,367

0 1
b) NPV of Project B @ 18% = (-1200000 / (1+0.18) ) + (4,50,000 / (1+0.18) ) +(4,00,000 /
2 3 4 5
(1+0.18) ) +(3,50,000 / (1+0.18) )+ (3,00,000 / (1+0.18) ) +(2,50,000 / (1+0.18) )+ (2,00,000 /
6 7
(1+0.18) ) +(1,16,650 / (1+0.18) ) = 57370

Since both the projects are having the same NPV at 18%, both the projects are worth
considering. However, by increasing the rate 20% if one of the projects shows higher NPV, that
project would be the choice between the two.
0 1 2
c) NPV of Project A @ 20% = (-1750000 / (1+0.2) ) + (4,00,000 / (1+0.2) ) +(4,00,000 / (1+0.2) )
3 4 5
+(4,00,000 / (1+0.2) )+ (4,00,000 / (1+0.2) ) +((6,00,000-100000) / (1+0.2) )+ (6,00,000 /
6 7 8
(1+0.2) ) +(6,00,000 / (1+0.2) )+ (3,80,300 / (1+0.2) ) = (-) 56734

0 1 2
d) NPV of Project B @ 20% =(-1200000 / (1+0.2) ) + (4,50,000 / (1+0.2) ) +(4,00,000 / (1+0.2) )
3 4 5 6
+(3,50,000 / (1+0.2) )+ (3,00,000 / (1+0.2) ) +(2,50,000 / (1+0.2) )+ (2,00,000 / (1+0.2) )
7
+(1,16,650 / (1+0.2) ) = 3.86

As the NPV of project B at 20% is higher than that of Project A, Project B is recommended.

L – 4 The energy consumption pattern in a steel re rolling mill over 8 month period is provided in the table
below;
Month Production (Tons) Coal Consumption (Tons)
1 488 422
2 553 412
3 455 411
4 325 363
5 488 438
6 585 426
7 455 414
8 419 396

Estimate,

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i) Fixed energy consumption in the Mill.

ii) Expected coal consumption for a production of 600 Tons/month.

Ans To establish the relationship between Production & Coal consumption, it is necessary to derive the
best-fit line for which the following normal equation are used ( see page 218 of book 1)
Cn +m∑x =∑y
2
c∑x + m∑x =∑xy

2
n x y x xy
1 488 422 238144 205936
2 553 412 305809 227836
3 455 411 207025 187005
4 325 363 105625 117975
5 488 438 238144 213744
6 585 426 342225 2 9 10
7 455 414 20702 188370
8 419 396 175561 165924
Total 3768 3282 1819558 1556000

Therefore, the normal equations become;

8c + 3768m = 3282 ……….i

3768c + 1819558m = 1556000 ……… ii

c = (3282 -3768m) / 8

Substituting in Eq. ii,

m = 0.23 and

c = 316

The best-fit straight line equation is;

y = 0.23x + 316

i) The fixed energy consumption in the Mill = 316 Tons of coal/month

ii) The expected coal consumption for a production of 600 Tons,

= 0.23 X 600 + 316 = 454 Tons

L-5 Explain PAT Scheme and its potential impact?

Ans Perform, Achieve and Trade (PAT) Scheme is a market based mechanism to enhance cost
effectiveness of improvements in energy efficiency in energy-intensive large industries and
facilities, through certification of energy savings that could be traded.

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The key goal of PAT scheme is to mandate specific energy efficiency improvements for the most
energy intensive industries. The scheme builds on the large variation in energy intensities of
different units in almost every sector. The scheme envisages improvements in the energy intensity
of each unit covered by it. The energy intensity reduction target mandated for each unit is dependent
on its operating efficiency: the specific energy consumption reduction target is less for those who are
more efficient, and is higher for the less-efficient units.

Further, the scheme incentivizes units to exceed their specified SEC improvement targets. To
facilitate this, the scheme provides the option for industries who achieve superior savings to receive
energy savings certificates for this excess savings, and to trade the additional certified energy
savings certificates with other designated consumers(energy intensive industries notified as
Designated Consumers under the Energy Conservation Act and included under PAT Scheme)
who can utilize these certificates to comply with their specific energy consumption reduction targets.
Energy Savings Certificates (ESCerts) so issued will be tradable at Power Exchanges. The scheme
also allows units which gain ESCerts to bank them for the next cycle of PAT, following the cycle in
which they have been issued. The number of ESCerts which would be issued would depend on the
quantum of energy saved over and above the target energy savings in the assessment year (for
st
1 Cycle of PAT, assessment year is 2014-15).

After completion of baseline audits, targets varying from unit to unit ranging from about 3 to 7% have
been set and need to be accomplished by 2014-15 and after which new cycle with new targets will be
proposed. Failing to achieve the specific energy consumption targets in the time frame would attract
penalty for the non-compliance under Section 26 (1A) of the Energy Conservation Act, 2001
(amended in 2010). For ensuring the compliance with the set targets, system of verification and
check-verification will be carried out by empanelment criteria of accredited energy auditors.

In a particular drying operation, it is necessary to hold the moisture content of feed to

a calciner to 15% (w/w) to prevent lumping and sticking. This is accomplishing by
L- 6
mixing the feed having 30% moisture (w/w) with recycle stream of dried material
having 3% moisture (w/w). The dryer operation is shown in fig below. What fraction of
the dried product must be recycled?

Let
F indicate quantity of feed

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R indicate quantity of recycle

P indicate quantity of product
Based on solid content at Mixer
0.7F + 0.97R = 0.85 (F + R)
Hence R =1.25 F ………………..(1)
Based on solid content at Drier
0.85 (F + R) = 0.97 (P + R)
0.85 (F + 1.25F) = 0.97 P + (0.97 x 1.25 F)
1.91 F = 0.97 P + 1.21F
0.7 F = 0.97 P
Hence F = 1.386 P ………………(2)
Substituting (2) in (1) for obtaining Recycle quantity in terms of Product
R = (1.25 x 1.386 P) = 1.7325 P ……………..(3)
Product plus Recycle is
P+R = (P + 1.7325 P) = P(1 + 1.7325) = 2.7325 P ……...(4)
R (as a fraction of dried product) = {(1.7325 P) / (2.7325 P)} x (100)
= 63.4%

…….……. End of Section – III ………..….

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18th NATIONAL CERTIFICATION EXAMINATION

FOR
ENERGY MANAGERS & ENERGY AUDITORS – September, 2017

PAPER – 1: GENERAL ASPECTS OF ENERGY MANAGEMENT & ENERGY

AUDIT

Date: 23.09.2017 Timings: 0930-1230 HRS Duration: 3 HRS Max. Marks: 150

General instructions:
o Please check that this question paper contains 11 printed pages
o Please check that this question paper contains 64 questions
o The question paper is divided into three sections
o All questions in all three sections are compulsory
o All parts of a question should be answered at one place

Section – I: OBJECTIVE TYPE Marks: 50 x 1 = 50

(i) Answer all 50 questions

(ii) Each question carries one mark
(iii) Please hatch the appropriate oval in the OMR answer sheet with Black Pen, as per
instructions

1. Which of the following is not a greenhouse gas ?

a) CFCs b) SO2 c) PFC d) SF6

2. Bio-gas generated through anaerobic process mainly consists of

a) only methane b) methane and carbon dioxide

c) only ethane d) none of these
3. In a boiler, fuel substitution of coal with rice husk results in

a) energy conservation
b) energy efficiency
c) both energy conservation and energy efficiency
d) carbon neutrality
4. A building intended to be used for commercial purpose will be required to follow
Energy conservation building code under Energy Conservation Act, 2001 provided its

a) connected load is 120 kW and above

b) contract demand is 100 kVA and above
c) connected load is100 kW and above or contract demand is 120 kVA and above

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d) connected load is 500 kW and contract demand is 600 kVA

5. Which of the following is not a part of energy audit as per the Energy Conservation
Act, 2001?

a) monitoring and analysis of energy use

b) verification of energy use
c) submission of technical report with recommendations
d) ensuring implementation of recommended measures followed by
review
6. Which of the following criteria is a responsibility of Designated Consumer?

a) designate or appoint an accredited Energy Auditor

b) adhere to stipulated energy consumption norms and standards as notified
c) submit the status of energy consumption information every three years
d) conduct energy audit through a certified energy auditor periodically
7. Which of the following is an energy security measure?

a) fully exploiting domestic energy resources

b) diversifying energy supply source
c) substitution of imported fuels for domestic fuels to the extent possible
d) all of the above
8. Which of the following statements are true?

i) reactive current is necessary to build up the flux for the magnetic field of
inductive devices
ii) some portion of reactive current is converted into work
iii) the cosine of angle between kVA and kVAr vector is called power factor
iv) the cosine of angle between kW and kVA vector is called power factor

a) i & iv b) ii & iii c) i & iii d) iii & iv

9. Which of the following statements regarding evacuated tube collectors (ETC) are true?

i) ETC is used for high temperatures upto 150oC

ii) because of use of vacuum between two concentric glass tube, higher amount
of heat is retained in ETC
iii) heat loss due to conduction back to atmosphere from ETC is high
iv) performance of evacuated tube is highly dependent upon the ambient temperature

i & iii b) ii & iii c) i & iv d) i & ii

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10. Which among the following has the highest flue gas loss on combustion due
to Hydrogen in the fuel?

a) natural gas b) furnace oil c) coal d) light diesel oil

11. Assume CO equivalent emissions by the use of a 60 W incandescent lamp are of the
2
order of 60 g/hr. If it is replaced by a 5 W LED lamp then the equivalent CO 2
emissions will be
a) nil b) 5 g/hr c) 12 g/hr d) 300 g/hr
12. The benchmarking parameter for a vapour compression refrigeration system is

a) kW / kg of refrigerant used b) kcal / m3 of chilled water

c) BTU / Ton of Refrigeration d) kW / Ton of Refrigeration
13. The rate of energy transfer from a higher temperature to a lower temperature is
measured in

a) kcal b) Watt c) Watts per second d) none of the above.

14. Energy sources which are inexhaustible are known as

a) commercial energy b) primary energy

c) renewable energy d) secondary energy
15. 1 kg of wood contains 15% moisture and 5% hydrogen by weight. How much water
is evaporated during complete combustion of 1kg of wood?

a) 0.6 kg b) 200 g c) 0.15 kg d) none of the above

16. The internal rate of return is the discount rate for which the NPV is
a) positive b) zero c) negative d) less than 1
17. As per Energy Conservation Act, 2001, a BEE Certified Energy Manger is required to
be appointed/designated by the

a) state designated agencies b) all industrial consumers

c) designated consumers d) electrical distribution licensees
18. The process of capturing CO2 from point sources and storing them is called .

a) carbon sequestration b) carbon sink

c) carbon capture d) carbon adsorption
19. Which of the following has the highest specific heat?

a) lead b) mercury c) water d) alcohol

20. The retrofitting of a variable speed drive in a plant costs Rs 2 lakh. The annual savings

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is Rs 0.5 lakh. The maintenance cost is Rs. 5,000/year. The return on investment is

a) 25% b) 22.5% c) 24% d) 27.5%

21. is a statistical technique which determines and quantifies the
relationship between variables and enables standard equations to be established for
energy consumption.

a) linear regression analysis b) time-dependent energy analysis

c) moving annual total d) CUSUM

22. The power generation potential in mini hydro power plant for a water flow of 3 m3/sec
with a head of 14 meters and with a system efficiency of 55% is

a) 226.6 kW b) 76.4 kW c) 23.1 kW d) none of the above

23. Which of the following two statements are true regarding application of Kaizen for
energy conservation?

i) Kaizen events are structured for reduction of only energy wastes

ii) Kaizen events engage workers in such a way so that they get involved in
energy conservation efforts
iii) Implementation of kaizen events takes place after review and approval of
top management
iv) In a Kaizen event, it may happen that small change in one area may
result in significant savings in overall energy use

a) ii & iv b) i & iii c) iii & iv d) i & iv

24. The electrical power unit Giga Watt (GW) may be written as

a) 1,000,000 MW b) 1,000 MW c) 1,000 kW d) 1,000,000 W

25. The producer gas basically consists of

a) Only CH4 b) CO & CH4 c) CO, H2 & CH4 d) Only CO & H2

26. Which of the following statements is correct regarding ‘float’ for an activity?

a) Time between its earliest start time and earliest finish time
b) Time between its latest start time and latest finish time
c) Time between latest start time and earliest finish time
d) Time between earliest finish time and latest finish time
27. The Energy Conservation Act,2001 requires that all designated consumers should get
energy audits conducted periodically by

a) certified energy manager b) certified energy auditor

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c) accredited energy auditor d) state Designated Agencies

28.
The term missing in the following equation (kVA) 2 = (kVA cos phi) 2 + ( ? )2 is

a) cos phi b) sin phi c) kVA sin phi d) kVArh

29.
2000 kJ of heat is supplied to 500 kg of ice at 0oC. If the latent heat of fusion of ice is
335 kJ/kg then the amount of ice in kg melted will be

a) 1.49 b) 83.75 c) 5.97 d) None of the above

30. An electric heater draws 5 kW of power for continuous hot water generation in an
industry. How much quantity of water in litres per min can be heated from 30oC to
85oC ignoring losses?.

a) 1.3 b) 78.18 c) 275 d) none of the above

31. An electric heater consumes 1000 Joules of energy in 5 seconds. Its power rating is:

a) 200 W b) 1000 W c) 5000W d) none of the above

32. The quantity of heat required to raise the temperature of a given substance by 1 oC is
known as:

a) sensible heat b) specific heat c) heat capacity d) latent heat

33. Which of the following parameters is not considered for external Bench Marking?

a) scale of operation b) energy pricing

c) raw materials and product quality d) vintage of technology
34. A sling psychrometer is used to measure :

a) only dry bulb temperature b) only wet bulb temperature

c) both a & b d) relative humidity
The number of moles of water contained in 36 kg of water is ------------
35.
a) 2 b) 3 c) 4 d) 5
36. A process electric heater is taking an hour to reach the desired temperature while
operating at 440 V. It will take ------- hours to reach the same temperature if the supply
voltage is reduced to 220 V.

a) 2 b) 3 c) 4 d) 5
37. In a manufacturing plant, following data are gathered for a given month: Production -
1200 pieces; specific energy consumption - 1000 kWh/piece; variable energy
consumption - 950 kWh/piece. The fixed energy consumption of the plant for the month

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is -------

a) 6,000 kWh b)10,000 kWh c) 12,000 kWh d) 60,000 kWh

38. Which of the following GHGs has the longest atmospheric life time?

a) CO2 b) CFC
c) Sulfur Hexafluoride (SF6) d) perfluorocarbon (PFC)
39. The component of electric power which yields useful mechanical power output is known
as

a) apparent power b) active power

c) reactive power d) none of the above
40. An oil fired boiler is retrofitted to fire coconut shell chips. Boiler thermal efficiency
drops from 82% to 70%. What will be the percentage change in energy consumption to
generate the same output

a) 12% increase b) 14.6% increase

c) 17.1% decrease d) 17.1% increase
41. A three phase induction motor is drawing 16 Ampere at 440 Volts. If the operating
power factor of the motor is 0.90 and the motor efficiency is 92%, then the mechanical
shaft power output of the motor is

a) 12.04 kW b) 10.09 kW
c) 10.97 kW d) None of the above
42. The energy conversion efficiency of a solar cell does not depend on

a) solar energy insolation b) inverter

c) area of the solar cell d) maximum power output
43. To maximize the combustion efficiency, which of the following in the flue gas needs to
be done?

a) maximize O2 b) maximize CO2 c) minimize CO2 d) maximize CO

44. An indication of sensible heat content in air-water vapour mixture is

a) wet bulb temperature b) dew point temperature

c) density of air d) dry bulb temperature
45. Which of the following is false?

a) electricity is high-grade energy

b) high grade forms of energy are highly ordered and compact
c) low grade energy is better used for applications like melting of metals rather
than heating water for bath

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 Paper 1 –Set A with Solutions

d) the molecules of low grade energy are more randomly distributed than the molecules
of carbon in coal
46. Which of the following is not applicable to liquid fuels?

a) the viscosity of a liquid fuel is a measure of its internal resistance to flow.

b) the viscosity of all liquid fuels decreases with increase in its temperature
c) higher the viscosity of liquid fuels, higher will be its heating value
d) viscous fuels need heat tracing
47. The cost of replacement of inefficient chiller with an energy efficient chiller in a plant
was Rs. 10 lakh .The net annual cash flow is Rs 2.50 lakh .The return on investment is:

a) 18% b) 20% c) 15 % d) none of the above

48. Which one is not an energy consumption benchmark parameter?

a) kcal/kWh of electricity generated b) kg/ oC.

c) kW/ton of refrigeration d) kWh/kg of yarn
49. The contractor provides the financing and is paid an agreed fraction of actual savings
achieved. This payment is used to pay down the debt costs of equipment and/or services.
This is known as

a) traditional contract b) extended technical guarantee/service

c) performance Contract d) shared savings performance contract
50. In project financing ,sensitivity analysis is applied because

a) almost all the cash flow methods involve uncertainty

b) of the need to assess how sensitive the project to changes in input parameters
c) what if one or more factors are different from what is predicted
d) all the above situation

…….……. End of Section – I ………..….

Section – II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40

(i) Answer all Eight questions

(ii) Each question carries Five marks

S-1 Explain Time of Day (TOD) Tariff and how it is beneficial for the power system and
consumers?
Ans  In Time of the Day Tariff (TOD) structure incentives for power drawl during off-
peak hours and disincentives for power drawl during peak hours are built in. Many

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 Paper 1 –Set A with Solutions

electrical utilities like to have flat demand curve to achieve high plant efficiency.
 ToD tariff encourage user to draw more power during off-peak hours (say during
11pm to 5 am, night time) and less power during peak hours. Energy meter will
record peak, off-peak and normal period consumption, separately.
 TOD tariff gives opportunity for the user to reduce their billing, as off peak hour
tariff is quite low in comparison to peak hour tariff.
 This also helps the power system to minimize in line congestion, in turn higher
line losses and peak load incident and utilities power procurement charges by
reduced demand
…………………..5 marks
( each point consider 1.5 marks)

S- 2 In a chemical factory where dyes are made, wet cake at 30 OC consisting of 60% moisture
is put in a dryer to obtain an output having only 5% moisture, at atmospheric pressure. In
each batch about 120 kgs of material is dried.

a. The quantity of moisture removed per batch.

b. What is the total quantity ( sensible & latent) of heat required to evaporate the
moisture, if the latent heat of water is 540 kcal/kg at atmospheric conditions,
Ignore heat absorbed by the solids

c. Find the quantity of steam required for the drying process (per batch), if steam
at 4 kg/cm2 is used for generating hot air in the dryer and the dryer efficiency
is 80%. Latent heat of steam at 4 kg/cm2 is 520 kcal/kg.

Ans Given that

 Qty of material dried per batch - 120 Kgs
 Moisture at inlet - 60%
a. The quantity of moisture removed per batch.

 Water quantity in a wet batch - 120 x 0.6 = 72 Kgs.

 Quantity of bone dry material - 120 – 72 = 48 Kgs.
 Moisture at outlet - 5%
 Total weight of dry batch output - 48/0.95 = 50.5 Kgs.
 Equivalent water in a dry batch - 50.5 - 48 = 2.5 Kgs.

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 Paper 1 –Set A with Solutions

 Total water removed in drying - 72 – 2.5 = 69.5 Kgs./batch

…………………….1.5 marks
b. The total quantity of heat required to evaporate the moisture.

To evaporate the moisture at atmospheric pressure, the material has to be first heated
up to 100 OC.

The total heat required would be;

Sensible heat - 72 x 1 x (100 – 30) = 5040 Kcal/batch

Latent heat - 69.5 x 540 = 37530 Kcal/batch
Total heat required - 5040 + 37530 = 42570 Kcal/batch

…………………….2 marks

c. The quantity of steam required for the drying process

Dryer Efficiency - 80%

Heat input to dryer - 42570/0.8 = 53212.50 Kcal/batch

Latent heat in 4 Kg/cm2 steam - 520 Kcal/Kg
Steam quantity required - 53212.50 / 520 = 102.3 Kgs / batch
…………………….1.5 marks
S-3 Explain PAT scheme and why it is a market based mechanism?

Ans Perform, Achieve and Trade (PAT) Scheme is a market based mechanism to
enhance cost effectiveness of improvements in energy efficiency in energy-intensive
large industries and facilities, through certification of energy savings that could be
traded. The genesis of the PAT mechanism flows out of the provision of the
Energy Conservation Act, 2001 (amended in 2010).

The key goal of PAT scheme is to mandate specific energy efficiency improvements for
the most energy intensive industries in sectors as listed below.
Sector
1. Aluminium
2. Cement
3. Chlor-Alkali
4. Fertilizer
5. Iron and Steel
6. Pulp and Paper
7. Textile
8. Thermal Power Plant

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The energy intensity reduction target mandated for each unit is depended on its operating
efficiency and the specific energy consumption reduction target is less for those who are
more efficient and more for the less efficient units.

Further, the scheme incentivizes units to exceed their specified SEC improvement
targets. To facilitate this, the scheme provides the option for industries who achieve
superior savings to receive energy savings certificates for this excess savings, and to
trade the additional certified energy savings certificates with other designated
consumers who can utilize these certificates to comply with their specific energy
consumption reduction targets. Energy Savings Certificates (ESCerts) so issued will be
tradable at Power Exchanges. The scheme also allows units which gain ESCerts to bank
them for the next cycle of PAT, following the cycle in which they have been issued.
The number of ESCerts which would be issued would depend on the quantum of energy
saved over and above the target energy savings in the assessment year.

After completion of baseline audits, targets varying from unit to unit ranging from about 3
to 7% are set and need to be accomplished during the 3 year cycle; after which new cycle
with new targets will be proposed. Failing to achieve the specific energy consumption
targets in the time frame would attract penalty for the non-compliance under Section
26 (1A) of the Energy Conservation Act, 2001 (amended in 2010). For ensuring the
compliance with the set targets, system of verification and check-verification will be
carried out by empanelment criteria of accredited energy auditors.
…………………….5 marks

Refer Book 1: Pg no 40-41

S-4 Give a short description about Availability Based Tariff (ABT).
Ans Introduction of availability based tariff(ABT) and scheduled interchange charges for
power was introduced in 2003 for interstate sale of power , have reduced voltage and
frequency fluctuation
 It is a performance-based tariff system for the supply of electricity by
generators owned and controlled by the central government.
 It is also a new system of scheduling and dispatch, which requires both
generators and beneficiaries to commit to day - ahead schedule.
 It is a system of rewards and penalties seeking to enforce day ahead pre-
committed schedules, though variations are permitted if notified one and a half
hours in advance.
 The order emphasizes prompt payment of dues , non-payment of prescribed
charges will be liable for appropriate action.
…………………….5 marks

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 Paper 1 –Set A with Solutions

S –5 In a heat treatment shop, steel components are heat-treated in batches of 80 Tons. The
heat treatment cycle is as follows;

 Increase temperature from 30 OC to 850 OC in 3 hours.

 Maintain 850 OC for 1 hour (soaking time).

 Cool the material to 60 OC in 4 hours.

a) Calculate the efficiency of the furnace, if the specific heat of steel is 0.12 kcal/kg OC
and fuel oil consumption per batch is 1400 litres.

 GCV of fuel oil - 10200 kcal/kg,

 Cost of fuel oil - Rs. 46,000/kL,
 Sp. gr. of fuel oil - 0.92.

b) Due to high cost of oil, the plant management decides to convert to a lower operating
cost LPG fired furnace lined on the inside with ceramic fibre insulation and with an
operating efficiency of 80%, for same requirement. The investment towards installation of
the new furnace is Rs. 50 lakhs. Calculate the Return on Investment, if the plant operates
two batches per day and 250 days in a year.

 Cost of LPG - Rs. 75/kg,

 GCV of LPG - 12500 kcal/kg.

Ans Quantity of steel treated per batch - 80 Tons

a. Efficiency of Furnace:
Useful heat supplied to steel - 80000 x 0.12 x (850 – 30)
= 7872000 kcal/batch
…………………….1 mark
Total heat supplied by fuel - 1400 x 0.92 x 10200
= 13137600 kcal/batch
Efficiency of Furnace - 7872000/12067824 = 59.9%
…………………….1 mark
b. Return on Investment (RoI):
Cost of operating fuel oil furnace - 1400 x 46 = Rs. 64400/batch
Efficiency of new LPG furnace - 80%
Heat supplied in new LPG furnace - 7872000/0.8

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 Paper 1 –Set A with Solutions

= 9840000 kcal/batch
Equivalent LPG consumption - 9840000/12500
= 787.2 kg/batch
…………………….1 mark
Cost of operating LPG Furnace - 787.2 x 75
=Rs. 59040/batch
Cost saving per batch - 64400 – 59040 =Rs. 5360/-
Annual cost saving - 5360 x 2 x 250
=Rs. 26,80,000/-
…………………….1 mark
Investment for new furnace - Rs. 50 Lakhs
Return on Investment (RoI) - (26.8/50)*100 = 53.6%
…………………….1 mark
S–6 In a 100 TPD Sponge Iron plant, the sponge iron is fed to the Induction melting furnace,
producing molten steel at 88% yield. The Energy consumption details are as follows:

Coal Consumption : 130 TPD

GCV of coal : 4500 kcal/kg
Power Purchased from Grid : 82400 kWh / day
Specific Energy consumption for Kiln producing Sponge Iron: 120 kWh / ton sponge iron
82400 kWh/day
from Grid

Factory Boundary
Electricity for
Induction Melting
130 TPD Coal Furnace
4500 kcals/kg
100 TPD Sponge
Induction Melting
Iron Ore Iron Molten
Sponge Iron Kiln Furnace
steel
Grid Electricity Yield: 88%
120 kWh/t of
Sponge iron

Calculate the following

1. Specific Energy Consumption of Induction melting furnace in terms of kWh/ton of
molten steel

2. Specific Energy Consumption of the entire plant, in terms of kcal/kg of molten

steel (product).

3. Total Energy Consumption of Plant in Tons of Oil Equivalent (TOE )

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 Paper 1 –Set A with Solutions

Ans a) Specific Energy Consumption of Induction Melting Furnace

Molten Steel Production from the Induction melting furnace per day

= 100 x 88/100 = 88 TPD

Total Energy Consumption of the Plant = 82400 kWh

Electrical Energy Consumption in Sponge Iron Making = 120 x 100

= 12000 kWh per day

Electrical Energy Consumption in Induction Melting Furnace = 82400-12000

= 70400 kWh/day
…………………….1 mark
Specific Energy Consumption of Induction Melting Furnace= 70400 / 88
= 800 kWh/ton of molten steel
…………………….1 mark
b)Total Energy Consumption of the Plant:

(82400x860) + (130x1000x4500) = (70864000+585000000)

= 655864000 kcal/day
…………………….1 mark
Specific Energy Consumption in terms of kcal/kg of Molten metal
=655864000/88000 =7453 kcal/kg of molten metal
…………………….1 mark
c) Total Energy consumption of Plant in ToE

= 655864000/107 = 65.586 ToE

…………………….1 mark
S- 7 A manufacturing industry plans to improve its energy performance under PAT through
implementation of an energy conservation scheme. After implementation, calculate the
Plant Energy Performance (PEP) with 2015-16 as the reference year. What is your
inference?
Given that:
 The current year (2016-17 ) Annual Production – 28,750 T ,
 Current year (2016-17 ) Annual Energy Consumption– 23,834 MWh,
 Reference year (2015-16 ) production - 34,000 T,
 Reference year (2015-16 ) Energy consumption - 27,200 MWh.

Ans Production factor (PF) = 28750/34000 = 0.846

…………………….1 mark
Ref year equivalent energy (RYEE) = Ref Year Energy Use (RYEU) x PF
= 27,200 x 0.846= 23011MWh

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 Paper 1 –Set A with Solutions

…………………….1 mark
PEP = (RYEE – current year energy)/RYEE = (23011 – 23834)/23011
= (-) 0.0369 ie (-) 3.7 %
…………………….1.5 marks
Since the PEP is negative, it implies that the energy conservation measure did not yield
reduction in energy consumption, action to be taken to improve the plant performance.
…………………….1.5 marks
S-8 List down any five Designated Consumers notified under the Energy Conservation Act.
Ans
(1) Aluminium, (2) Cement, (3) Chloralkali, (4) Fertiliser, (5) Steel, (6) Pulp & Paper,
(7)Thermal Power Plants, (8) Textile, (9) Railways.
…………………….5 marks
( any 5 of the above and each one carries one mark)

…….……. End of Section – II ………..….

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60

(i) Answer all Six questions

(ii) Each question carries Ten marks

L - 1 Saturated steam at 1 atm is discharged from a turbine at 1200 kg/h. Superheated steam at 300 0C
and 1 atm is needed as a feed to a heat exchanger. To produce it, the turbine discharge stream is
mixed with superheated steam at 400 0C, 1 atm and specific volume of 3.11 m3/kg.
Calculate the amount of superheated steam at 300 0C produced and the volumetric flow rate of the
400 0C steam.

Ans Solution
1. Mass balance of water
1200 + m1 = m2 ………………………………………… (1)
…………………….1 mark
2. Energy balance
(1200 kg/h)(2676 kJ/kg) + m1(3278 kJ/kg)

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 Paper 1 –Set A with Solutions

= m2(3074 kJ/kg) ……………………………………. …. (2)

…………………….1 mark
Eqs. (1) and (2) are solved simultaneously
3211200 + 3278m1 = (1200 + m1)3074
m1 = 2341.2 kg/h
m2 = 1200 + 2341.2 = 3541.2 kg/h (superheated steam produced)

…………………….4 marks
o
3. Volumetric flow rate of 400 C steam
The specific volume of steam at 400 C and 1 atm is 3.11 m3/kg. The volumetric flow rate is
calculated as follows:
(2341.2 kg/h)(3.11 m3/kg)
= 7281.1 m3/h
…………………….4 marks
L –2 The energy consumption and production patterns in a chemical plant over a 9 month period is provided in
the table below;
Month 1 2 3 4 5 6 7 8 9
Production in Tonnes / month 493 297 381 479 585 440 234 239 239
Energy Consumption MWh /month 78.2 75.7 76.3 76.1 78.1 70.7 73.7 64.4 72.1

th
Estimate the cumulative energy savings at end of the 9 month and give your inference on the
result ? ( consider 9 month data for evaluation for predicted energy consumption)
Ans It is required to use the equations Y= mX + C and
nC + mΣX = ΣY
cΣX + mΣX2 = ΣXY

X=
Y =Energy
Production in
Month Consumption MWh X2 XY
Tonnes /
/month
month
1 493 78.2 243049 38574.12
2 297 75.7 88209 22479.51
3 381 76.3 145161 29076.88
4 479 76.1 229441 36436.09
5 585 78.1 342225 45671.42
6 440 70.7 193600 31110.53
7 234 73.7 54756 17240.63
8 239 64.4 57121 15402.96
9 239 72.1 57121 17228.98
3387 665.3 1410683 253221

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 Paper 1 –Set A with Solutions

Therefore, the normal equations become;

9c + 3387m = 665.3 ……….i

3387C + 1410683m = 253221.1 ……… ii

…………………….2 marks

c = (665.3-3387m)/9

Substituting in Eq. ii,

m = 0.021 and

c = 66.1

The best-fit straight line equation is;

y = 0.021x + 66.1
…………………….3 marks

Production in E cal Y =
Tonnes / 0.021x + Difference
Month month x Eactual 66.1 CUSUM
1 493 78.2 76.45 1.75 1.75
2 297 75.7 72.34 3.36 5.11
3 381 76.3 74.10 2.20 7.31
4 479 76.1 76.16 -0.06 7.25
5 585 78.1 78.39 -0.28 6.97
6 440 70.7 75.34 -4.64 2.33
7 234 73.7 71.01 2.69 5.01
8 239 64.4 71.12 -6.72 -1.71
9 239 72.1 71.12 0.98 -0.73
…………………….4 marks
Since the CUSUM value at the end of 9th month is negative, the plant has not achieved any net energy
savings and action has to be taken to determine reason for no performance of the encon option.
…………………….1 mark
L - 3 Explain the following
a) Dry Bulb Temperature and Wet bulb Temperature

b) Maximum Demand and Power Factor

c) Gross Calorific Value & Net Calorific Value

d) 5S & Return of Investment (ROI)

e) CUSUM

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 Paper 1 –Set A with Solutions

Ans a) Dry Bulb Temperature and Wet bulb Temperature

• Dry bulb Temperature is an indication of the sensible heat content of air-water vapour
mixtures
• Wet bulb Temperature is a measure of total heat content or enthalpy. It is the temperature
approached by the dry bulb and the dew point as saturation occurs.
…………………….2 marks
b) Maximum Demand and Power Factor

• Maximum demand is maximum KVA or KW over one billing cycle

• Power Factor Cos = kW/ KVA or kW = kVA cos 
…………………….2 marks
c) Gross Calorific Value & Net calorific Value:

• Gross calorific value assumes all vapour produced during the combustion process is fully
condensed.

• Net calorific value assumes the water leaves with the combustion products without being
fully condensed.

• The difference being the latent heat of condensation of the water vapour produced during
the combustion process.

…………………….2 marks

d) 5S:

Housekeeping. Separate needed items from unneeded items. Keep only what is
immediately necessary item on the shop floor.

Workplace Organization. Organize the workplace so that needed items can be easily and
quickly accessed. A place for everything and everything in its place.

Cleanup. Sweeping, washing, and cleaning everything around working area immediately.

Cleanliness. Keep everything clean in a constant state of readiness.

Discipline. Everyone understands, obeys, and practices the rules when in the plant.

…………………….1 mark( any one of the above is sufficient)

d) Return on Investment:

ROI expresses the annual return from project as % of capital cost.

This is a broad indicator of the annual return expected from initial capital investment,

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 Paper 1 –Set A with Solutions

expressed as a percentage.

…………………….1 mark

e) Cumulative Sum (CUSUM) Technique:

• Difference between expected or standard consumption with actual consumption data

points over baseline period of time.

• Follows a fixed trend unless something (energy saving measure, deterioration in

performance..) happens

• Helps calculation of savings/losses till date after changes

…………………….2 marks

L – 4 Answer the following

Chose the correct
S. No Statement answer OR
Fill-in-the-blanks
1 Fyrite measures CO2, O2 and SO2 True/False

2 Ultrasonic Flow Meter uses the principle of____& ____ Fill in the blanks
Non Contact Infrared Thermometer cannot measure
3 True/False
temperature of objects placed in hazardous places
To measure the RPM of a Flywheel, ______ type of RPM
4 meter is used and for a visible shaft-end _______ type of Fill in the blanks
RPM meter is used.
In a switch yard, _____ instrument is used to identify the
5 Fill in the blanks
loose joints and terminations
Every Designated Consumer shall have its first energy
audit conducted by ________ Energy Auditor within
6 Fill in the blanks
______ months of notification issued by the Central
Government

280 kcal/ hr is equivalent to _____Watts and 3.5 bar is

7 Fill in the blanks
equivalent to __________kPa

8 One metric ton of oil equivalent is to ________MW Fill in the blanks

1 kg of Coal, consisting of 30% of Carbon produces

9 Fill in the blanks
______ kg of CO2

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 Paper 1 –Set A with Solutions

In a gasification system the reduction zone is above the

10 True/False
combustion zone

Ans Chose the correct

Sr
Statement answer OR Solution
No
Fill-in-the-blanks
1 Fyrite measures CO2, O2 and True/False False
SO2
2 Ultrasonic Flow Meter uses the Fill in the blanks Transit Time; Doppler
principle of____& ____ Effect

3 Non Contact Infrared True/False False

Thermometer cannot measure
temperature of objects placed in
hazardous places
4 To measure the RPM of a Fill in the blanks Stroboscope; Tachometer
Flywheel, ______ type of RPM
meter is used and for a visible
shaft-end _______ type of
RPM meter is used.
5 In a switch yard, _____ Fill in the blanks Thermal imager or IR gun
instrument is used to identify
the loose joints and
terminations
6 Every Designated Consumer Fill in the blanks Accredited ; 18 months
shall have its first energy audit
conducted by ________ Energy
Auditor within ______ months
of notification issued by the
Central Government
7 Fill in the blanks 325.6 Watts;
280 kcal/ hr is equivalent to
(280x4.187x1000/3600)
_____Watts and 3.5 bar is
350 kPa (3.5 x100)
equivalent to __________kPa
8 One metric ton of oil equivalent Fill in the blanks 11.62 MW
is to ________MW (1x1000x10000/(860x1000)
9 1 kg of Coal, consisting of 30% Fill in the blanks 1.1
of Carbon produces ______ kg [(44/12)x(0.3]
of CO2
10 In a gasification system the True/False False
reduction zone is above the
combustion zone

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 Paper 1 –Set A with Solutions

…………………….10 marks(each one carries one mark)

L - 5 A company has to choose between two projects whose cash flows are as indicated
below;

Project 1:

i. Investment – Rs. 15 Lakhs

ii. Annual cost savings – Rs. 4 lakhs.

iii. Bi-annual maintenance cost – Rs. 50,000/-

iv. Reconditioning and overhaul during 5th year: 6 lakhs

v. Life of the project – 8 years

vi. Salvage value – Rs. 5 lakhs

Project 2:

vii. Investment – Rs. 14 Lakhs

viii. Annual cost savings – Rs. 3.5 lakhs.

ix. Annual Maintenance cost – Rs. 20,000/-

x. Reconditioning and overhaul during 4th year: 5 lakhs

xi. Life of the project – 8 years

xii. Salvage Value- 2 lakhs

Which project should the company choose? The annual discount rate is 12%.

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 Paper 1 –Set A with Solutions

Ans
Year Project 1 Project 2
Outgo Saving NPV Outgo Saving NPV
0 15.0 0 =-15.0 14.0 0 = -14
1 0 4.0 = (4 / (1+.12)1 0.2 3.5 = (3.3 / (1+.12)1
= 3.571 = 2.95
2 0.5 4.0 = (3.5 / (1+.12)2 0.2 3.5 = (3.3 / (1+.12)2
= 2.79 = 2.63
3 0 4.0 = (4 / (1+.12)3 0.2 3.5 = (3.3 / (1+.12)3
= 2.84 = 2.35
4 0.5 4.0 = (3.5 / (1+.12)4 5 3.5 = (-1.5 / (1+.12)4
= 2.22 = -0.95
5 6 4.0 = (-2 / (1+.12)5 0.2 3.5 = (3.3 / (1+.12)5
= -1.13 = 1.87
6 0.5 4.0 = (3.5 / (1+.12)6 0.2 3.5 = (3.3 / (1+.12)6
= 1.77 = 1.67
7 0 4.0 = (4 / (1+.12)7 0.2 3.5 = (3.3 / (1+.12)7
= 1.81 = 1.49
8 0.5 9 (4+5) = (8.5 / (1+.12)8 0.2 5.5 = (5.3 / (1+.12)8
= 3.43 (3.5+2) = 2.14
NPV = + 2.301 @12% = + 0.15

NPV Project 1 is higher than Project 2. Hence project 1 is preferred.

…………………….10 marks

A project activity has several components as indicated below;

L- 6 S. Activity Preceded by Duration (in

No. Weeks)
1 A - 8
2 B A 6
3 C A 12
4 D B 4
5 E D 5
6 F B 12
7 G E& F 9
8 H C 8
9 I F&H 5
10 J I&G 6

d. Prepare a PERT chart, estimate the duration of the project and identify the
critical path.

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 Paper 1 –Set A with Solutions

e. What are the Earliest Start, Latest Start and Total Float of activity ‘H’?

f. What would be the project duration if activity ‘H’ got delayed by 3 weeks?

Ans PERT Diagram based on Activity on Arrow

D 5 G

B 5 9 J, 6

A 6 F,12 I ,5

8 c
12 H 8

OR

PERT Diagram based on Activity on Node

…………………….6 marks

a. Critical Path: A-B-F-G-J

…………………….1 mark

b. Estimated Project Duration: 41 weeks

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 Paper 1 –Set A with Solutions

…………………….1 mark

c. For activity H, Early Start is 20, Latest Start is 22 and Total Float is 2 weeks.

…………………….1 mark

d. Project duration will be 42 weeks- a delay of 1 week.

…………………….1 mark

…….……. End of Section – III ………..….

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 Paper 1 SET A

19th NATIONAL CERTIFICATION EXAMINATION

FOR
ENERGY MANAGERS & ENERGY AUDITORS – SEPTEMBER, 2018

PAPER – 1: GENERAL ASPECTS OF ENERGY MANAGEMENT & ENERGY AUDIT

Section – I: OBJECTIVE TYPE Marks: 50 x 1 = 50

(i) Answer all 50 questions

(ii) Each question carries one mark
(iii) Please hatch the appropriate oval in the OMR answer sheet HB pencil only, as per instructions

1. Which among the following is not a renewable source of energy?

a) Bagasse b) Rice husk
c) Nuclear d) Wind

2. What is shale Oil?

a) Sedimentary rock containing solid bituminous materials
b) Heavy black viscous oil combination of clay, sand, water and bitumen
c) A form of naturally compressed peat
d) combustible brownish-black sedimentary rock

3. Which of the following has the lowest energy content in terms of MJ/kg
a) LPG b) Diesel
c) Bagasse d) Furnace oil

4. _________ and _______ consume major share of Natural Gas consumption in India.
a) Domestic sector and Transport sector
b) Transport sector and Fertilizer Industry
c) Power Generation and Fertilizer Industries
d) Domestic Sector and Fertilizer Industries

5. The sector consuming major share of energy in India

a) Agriculture Sector b) Transport Sector
c) Industrial Sector d) Domestic Sector

6. Which of the following designated consumer has the lowest energy intensity?
a) Aluminium b) Iron and Steel
c) Cement d) Chlor alkali

7. Which of the following is not a Demand Side Management measure?

a) Implementing Time of the Day (ToD) Electricity Tariff
b) Maximizing fossil fuel based energy utilization
c) Replacement of inefficient electrical appliances
d) Use of ice bank system

8. Which of the following does not meet the Designated Consumer criteria?
a) Pulp and Paper Industries with minimum annual energy consumption of 30,000
TOE.
b) Cement Industries with minimum annual energy consumption of 30,000 TOE.
c) Chlor- Alkali Industries with minimum annual energy consumption of 7500
TOE.

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Bureau of Energy Efficiency
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d) Textile Industries with minimum annual energy consumption of 3000 TOE.

9. The kW or HP of a motor given on the name plate indicates

a) The shaft output of the motor at part load
b) The shaft output of the motor at full load
c) The input power to the motor at the best efficiency point
d) The input power to the motor at any load

10. Which of the following has the highest Specific Heat?

a) Steel b) Aluminium
c) Copper d) Water

11. Heat transfer in an air cooled condenser occurs predominantly by

a) conduction b) convection
c) radiation d) none of the above

12. Definition of Energy Audit as per EC Act does not include:

a) Creation of an Energy Management System (EnMS)
b) evaluation of Techno-economics
c) Verification, monitoring and analysis of energy use
d) Action plan required for energy saving

13. The ISO standard for Energy Management System is

a) ISO 14001
b) ISO 50001
c) ISO 9001
d) ISO 18001

14. To arrive at the relative humidity at a point we need to know___________of air

a) dry bulb temperature b) wet bulb temperature
c) enthalpy d) both a & b

15. As per Energy Conservation Act, 2001 appointment of BEE Certified Energy Manger is
mandatory for
a) all State designated agencies b) all large Industrial consumers
c) all designated consumers d) all commercial buildings

16. A waste heat recovery system requires Rs. 50 lakhs investment and Rs. 2 lakhs per year
to operate and maintain. If the annual savings is Rs. 22 lakhs, the payback period will
be
a) 2.28 years b) 2.5 years
c) 3 years d) 10 years

17. What is the heat content of the 200 liters of water at 5000C in terms of the basic unit
of energy in Kilo Joules
a) 30000 b) 23880
c) 10000 d) 41870
Note: 1 Mark is awarded to all candidate who have attempted this question.

18. Which of the following GHGs has the longest atmospheric life time
a) CH4 b) SF6
c) CFC d) PFC

19. Which of the following is used for non-contact measurement of temperature

a) Thermocouples b) Infrared Thermometer
c) Leaf type contact probe d) All of the above

20. The force field analysis in energy action planning considers

a) Positive forces only b) negative forces only
c) Both negative and positive forces d) no forces

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21. Which of the following equation is used to calculate the future value of the cash flow?
a) NPV (1 – i)n b) NPV / (1 – i)n
n
c) NPV (1 + i) d) NPV/ (1 + i)n

22. For investment decision, ROI must always be _____ prevailing interest rate.
a) Lower than b) Higher than
c) Equal to d) No relation

23. Large scattering on production versus energy consumption trend line indicates
a) Poor process control b) Inefficient equipment
c) Inefficient process d) None of the above

24. Frequency of energy audit for designated consumers is______

a) once in a year b) once in two years
c) once in three years d) Once in five years

25. The rotor axis is aligned with the wind direction in a wind mill by ________ control
a) yaw b) pitch
c) disc break d) all of the above

26. Producer gas basically comprises of

a) CO, H2 and CH4 b) Only CH4
c) CO and CH4 d) Only CO and H2

27. The lowest theoretical temperature to which water can be cooled in a cooling tower is
a) Difference between DBT and WBT of the atmospheric air
b) Average DBT and WBT of the atmospheric air
c) DBT of the atmospheric air
d) WBT of the atmospheric air

28. In a solar thermal power station Molten salt is preferred as it provides an efficient low
cost medium to store ______ energy
a) Electrical b) Thermal
c) Kinetic d) Potential

29. From Voltage, Amps and Power factor given in the name plate of a motor, one can
calculate ________.
a) Rated output power b) Shaft power
c) Rated input power d) Both (b) & (c)

30. RPM of an electric motor is measured using ___.

a) Ultrasonic meter b) Stroboscope
c) Lux meter d) Rotameter

31. If asset depreciation is considered, then net operating cash inflow would be
a) lower b) higher
c) no effect d) none of the above

32. Which of the following comes under Capital cost in a project?

a) Design cost b) Installation cost
c) Commissioning cost d) All of the above

33. Energy consumption per GDP is termed as ___.

a) Energy factor b) Energy intensity
c) Energy Efficiency index d) All of the above

34. A three phase induction motor is drawing 10 Ampere at 440 Volts. If the operating

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 Paper 1 SET A

power factor of the motor is 0.9 and the efficiency of the motor is 95%, then the
mechanical shaft power of the motor is
a) 3.76 KW b) 4.18 KW
c) 6.51 KW d) 7.21 KW

35. For an activity in a project, Latest start time is 8 weeks and Latest finish time is 12
weeks. If the earliest finish time is 9 weeks, Slack time for the activity is ____.
a) 3 weeks b) 4 weeks
c) 1 week d) none of the above

36. The amount of CO2 produced in complete combustion of 18 Kg of Carbon is ______.

a) 50 b) 44
c) 66 d) 792

37. Which mode of heat transfer does not require medium?

a) Natural convection b) Forced convection
c) Radiation d) Conduction

38. If the fixed energy consumption of a company is 2000 kWh per month and the line slope
of the energy (y) versus production (x) chart is 0.3, then the energy consumed in kWh
per month for a production level of 60,000 tons/month is _______.
a) 16,000 KWh b) 18,000 KWh
c) 22,000 KWh d) none of the above

39. Which technique takes care of time value of money in evaluation?

a) payback period b) IRR
c) NPV d) Both (b) & (c)

40. The heat rate of a power plant is expressed as

a) kWh/kg of steam b) kCal/kWh
c) kg of steam / kg of fuel d) kWh / kVA

41. Which equipment does not come under mandatory labelling program?
a) Room Air conditioners b) Frost free refrigerator
c) Induction motors d) Distribution transformer

42. Furling speed of wind turbine indicates ____

a) Cut out speed b) Cut in speed
c) Rated speed d) None of the above

43. One Silicon cell in a PV module typically produces

a) 0.5 V b) 1 V
c) 2 V d) 12 V

44. The input to a fuel cell is.

a) Electricity b) Hydrogen
c) Oxygen d) All of the above

45. The production factor is defined as the ratio of

a) current year production to the reference year production
b) current year production to the reference month production
c) reference month production to the current month production
d) reference year production to the current year production

46. To reduce the distribution losses within a plant, the capacitors should be located
a) Closest to the load b) Farthest from the load
c) In the substation d) Before the billing meter

47. Absolute pressure is measured as

a) Gauge pressure – Atmospheric pressure

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b) Gauge pressure + Atmospheric pressure

c) Gauge pressure / Atmospheric pressure
d) none of the above

48. The dryness (x) fraction of superheated steam is taken as

a) x= 0 b) x= 0.9
c) x= 0.87 d) x= 1

49. When the evaporation of water from a wet substance is zero, the relative humidity of
the air is likely to be
a) 0% b) 100%
c) 50% d) unpredictable

50. Which of the following type of collector is used for low temperature systems?
a) Flat plate collector b)Line focusing parabolic collector
c) Parabolic trough collector d) None of the above

…….……. End of Section – I ………..….

Section – II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40

(i) Answer all Eight questions

(ii) Each question carries Five marks

S-1 List five equipment and appliances covered under Standards and Labelling program.

Ans Refer BEE Guide Book 1- Page No 37

S- 2 State true or false (each carries 1 mark)

a) When it is raining, there is a substantial difference between the dry and wet bulb
temperatures.
b) The specific gravity of light diesel oil is given in kg/m3
c) The major constituent of LNG is propane
d) Evaporative cooling of space requires use of refrigerant R134a
e) HSD needs preheating to increase viscosity

Ans a) False b) False

c) False d) False
e) False

S – 3 For installing a recuperator in a furnace, the plant has assessed the following time estimates
Optimistic Time : 2.5 weeks
Most Likely Time : 3 weeks
Pessimistic Time : 3.5 weeks

Find out the “Expected Time”, “Standard Deviation” and “Variance” to complete the activity
(2 +1.5+1.5 Marks)

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 Paper 1 SET A

Ans Expected time = (Optimistic Time + 4 X Most Likely Time + Pessimistic Time) / 6
= (2.5 + 4 x 3 + 3.5)/ 6
=3
Standard Deviation = (3.5-2.5)/6 = 1/6 = 0.167
Variance = {(PT-OT/6)} 2
= 1 / 36 = 0.0278

S-4 A thermal power plant uses 0.72 kg of coal to generate one KWh of electricity. If the coal
contains 38% carbon by weight, calculate the amount of CO2 emission/KWh under complete
combustion.

Ans Amount of carbon present in coal = 0.72 * 38/100

= 0.2736 kg

As per chemical reaction, C + O2 = CO2

1 kg of carbon generates 44/12 kg of carbon dioxide (CO2) under complete combustion
Amount of CO2 generation while generating one KWh of electricity
= 0.2736 * 44/12
= 1.0032 Kg/KWh

S-5 A solar photovoltaic power plant is installed with 350 Watts panel of size 1.5 m x 1.5 m in
roof top area of a building having dimension of 9 m x 10m. If solar insolation is 1,000 W/m2,
calculate the panel conversion efficiency?

Ans Area of solar cell = 1.5 x 1.5

= 2.25 m2
Efficiency = (350 /(2.25 x 1000)) x 100
= 15.6 %

S – 6 A paint drier requires 75.4 m3/min of air at 93°C, which is heated in a steam-coil unit. How
many kg of steam at 4 bar does this unit require per hour? The density of air is 1.2 kg/m3
and specific heat of air is 0.24 kcal/kg°C. The ambient temperature is 32°C.

Pressure bar Temperature °C Enthalpy kcal/kg

Water Evaporation Steam

4 143 143 510 653

Ans Solution:
Air flow rate = 75.4 m3/min * 60 = 4524 m3/hr
Air flow rate = 4524 * 1.2
= 5428.8 kg/hr

Sensible heat of air = m * Cp * ∆T

= 5428.8 * 0.24 * (93-32)
= 79477.6 kcal/hr

Latent heat of Steam = 510 Kcal/kg

Steam required = 79477.6 / 510
Steam required = 156 kg/hr

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 Paper 1 SET A

S – 7 An ESCO company is required to invest in a waste heat recovery project, which is expected to
yield an annual saving of Rs.10,00,000 and the life of the equipment is 7 years. If the ESCO
expects 30% IRR on this project, calculate the investment required to be made.

Ans The PV of the Annual Savings of Rs.1,000,000 per year:

Investment 1000000 1000000 1000000 1000000 1000000 1000000 1000000
0=− + + + + + + +
(1+ 0.3)0 (1+ 0.3)1 (1+ 0.3)2 (1+ 0.3)3 (1+ 0.3)4 (1+ 0.3)5 (1+ 0.3)6 (1+ 0.3)7

or
Investment = Rs.1,000,000/year (P/AIN Factor)
= Rs.1,000,000/year (2.8021) = Rs. 2,802,100

Thus, we can pay Rs.2,802,100 for the Waste Heat Exchanger and still have a positive NPV.

S – 8 In a textile plant monthly energy consumption is 7,00,000 kWh of electricity , 40 kL of

furnace oil ( specific gravity=0.92) for thermic fluid heater, 360 tonne of coal for steam boiler
and 10 kL of HSD ( specific gravity= 0.885) for material handling equipment. Compute the
energy consumption in terms of Metric Tonne of Oil Equivalent (MTOE) for the plant.

Given Data: (1 kWh = 860 kcal, GCV of coal= 3450 kCal/kg, GCV of furnace oil= 10,000
kcal/kg, GCV of HSD= 10,500 kcal/kg, GCV of rice husk= 3100 kcal/kg, 1 kg oil equivalent
= 10,000 kcal)

Ans Aggregate Energy Use=

(40000 x0.92x 10000) + (360000 x 3450) + (7, 00,000 x 860) + (10,000x 0.885 x 10,500).

MTOE = (36.8 x 107) + (124.2 x 107) + (60.2 x 107)+ (9.2925 x 107)

107
= 230.5 Metric Tonnes of Oil Equivalent per month

Energy consumption of the textile plant = 230.5 x 12 = 2766 MTOE

…….……. End of Section – II ………..….

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 Paper 1 SET A

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60

(i) Answer all Six questions

(ii) Each question carries Ten marks

L-1 Describe the stages of Gasification of Biomass process with a pictorial diagram and
reaction equations?

Ans Refer BEE Guide Book 1- Page No 275-276

L–2 a. Explain briefly three types of Performance Contracting? (6 Marks)

b. What are the drawbacks of ESCO? (4 Marks)

Ans Refer BEE Guide Book 1- Page No.178

L - 3 a) Write down the steps for computing energy savings using CUSUM over a period.
(4 Marks),
b) Develop a table using a CUSUM technique to calculate energy savings for 8 months
period for a production level of 2000 MT per month. Refer to field data given in the
table below. (6 marks)
Month Actual SEC kWh/MT Predicted SEC kWh/MT

May 1225 1250

June 1227 1250

July 1240 1250

Aug 1245 1250

Sep 1238 1250

Oct 1257 1250

Nov 1248 1250

Dec 1264 1250

Ans a) Steps for CUSUM analysis:

Refer BEE Guide Book 1 Page No. 229

b) Estimate the savings accumulated from use of the heat recovery system.

Month Actual SEC Predicted SEC Difference CUSUM

kWh/MT kWh/MT (Actual SEC - Savings
Predicted SEC) kWh/MT
kWh/MT

May 1225 1250 -25 -25

June 1227 1250 -23 -48

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 Paper 1 SET A

July 1240 1250 -10 -58

Aug 1245 1250 -5 -63

Sep 1238 1250 -12 -75

Oct 1257 1250 +7 -68

Nov 1248 1250 -2 -70

Dec 1264 1250 +14 -56

Positive savings i.e. savings in energy consumption over a period of eight months are
56 x 2000 = 112,000 kWh

L–4 In a Chlor-Alkali plant, an evaporator was designed to concentrate 500 kg of liquor

containing solids of 7% w/w (weight by weight) to 45% solids w/w in the output. Presently
the output from evaporator has 30% solids w/w. The energy manager suggested
overhauling the evaporator to achieve the design rate of solids w/w in the output. Calculate
the percentage improvement in water removal in the evaporator after overhauling of the
evaporator.

Ans Amount of feed (input) to the evaporator = 500 Kg

Concentration of solids in feed = 7 wt%
Amount of solids in feed (input) = 500 * 7 /100
= 35 Kg

Present scenario :
Concentration of solids in product (output) = 30 wt% = 0.3
Mass balance across the evaporator :
Amount of product (output) from the evaporator = 35 / 0.3
= 116.7 Kg

Water vapour removed from the evaporator is = 500 – 116.7

= 383.3 Kg

Design scenario :
Concentration of solids in product (output) = 45 wt% = 0.45
Mass balance across the evaporator :
Amount of product (output) from the evaporator = 35 / 0.45
= 77.8 Kg

Water vapour removed from the evaporator is = 500 – 77.8

= 422.2 Kg

Incremental water removal achieved is = 422.2 – 383.3

= 38.9 Kg

% increase in water removal = 38.9 / 383.3 * 100

% improvement in water removal after overhaul = 10.14 %

L -5 A process plant is planning to implement a waste heat recovery project. The various
activities from procurement to commissioning are given in the table below along with their
duration and dependency.

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 Paper 1 SET A

Activity Predecessor Time in Weeks

A. - 3
B. - 5
C. A 4
D. A 6
E. C 5
F. C 3
G. B&D 2
H. D&E 1
I. F,G,H 2
a) Construct a PERT/CPM network diagram for the above project. (5 Marks)
b) Compute the earliest start, earliest finish, latest start, latest finish and slack for all the
activities (3 Marks)
c) Compute the project duration. (1 Mark)
d) Identify the critical activities and the critical path(s). (1 Mark)

Ans a) PERT/CPM network diagram for a project

b) Early start (ES), Early Finish (EF), Latest start (LS), Latest finish (LF) and slack for all
the activities.

Slack
Activity Duration ES EF LS LF (LS-ES) or
(LF-EF)
A 3 0 3 0 3 0

B 5 0 5 6 11 6

C 4 3 7 3 7 0

D 6 3 9 5 11 2

E 5 7 12 7 12 0

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 Paper 1 SET A

F 3 7 10 10 13 3
13
G 2 9 11 11 2

H 1 12 13 12 13 0

I 2 13 15 13 15 0
X1 and X2 are dummy activities

c) Critical Path : A- C- E- H- I

d) Total time on critical path (project duration) : 15 weeks

L- 6 A medium size chemical plant receives electricity from grid and also generates electricity
from coal based Captive Power Plant (CPP). Coal is also used for process requirements. The
fine coal from CPP is sold to neighboring plant. The annual energy details are given below:

Electricity purchased from grid 5 MU

Electricity exported to grid 11 MU
Power generation from CPP 36 MU
Power Supplied from CPP to Process 25 MU
plant
Fine coal sold to neighboring unit 1000 ton
Coal used for process plant 5000 ton
GCV of coal 4500 kcal/kg
Heat rate of CPP 3500 kcal/kWh
Annual Operating Hours 7200
Calculate
a. Energy usage in TOE (Tons of oil equivalent) (5 Marks)
b. Coal used in CPP (3 Marks)
c. Calculate the CPP operating power in MW. (2 Marks)

Ans Energy usage in TOE (Tons of oil equivalent)

➢ Grid electricity Imported = (5 x 106 kWh)x (860 kcal/kWh) = (+) 43 x 108 kcals/year
➢ 6
Power generated from CPP = (36 x 10 kWh)x (3500 kcal/kWh) = (+) 1260 x 108 kcals/year
➢ 3
Coal imported for process = (5000 x 10 kg)x (4500 kcal/kg) = (+) 225 x 108 kcals/year
➢ 6
Power exported to grid = (11 x 10 kWh)x (3500 kcal/kWh) = (-) 385 x 108 kcals/year
➢ Coal fines exported to neighbour = (1000 x 10 kg)x (4500 kcal/kg) = (-) 45 x 108 kcals/year
3

➢ Net annual energy consumption = ( 43+1260+225)-(385+45) = (+) 1098 x 108 kcals/year

a. Energy usage in TOE = (1098 x 108 kcals/year) / (107) = 10980 MTOE

(1 MTOE = 107 kcals)
b. Coal used in CPP = ( (36 x 106 kWh)x (3500 kcal/kWh) ) / (4500 kcal/kg)
= 28 x 106 kgs Coal/ Year
= (28 x 106 )/103 = 28000 Tons Coal/ Year
c. Calculate the CPP operating MW = (36 x 106 kWh/year)/ (7200 hrs/year)
= 5000 kW
= 5 MW
…….……. End of Section – III ………..….

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 Paper 1 Code : GREEN

20th NATIONAL CERTIFICATION EXAMINATION

FOR
ENERGY MANAGERS & ENERGY AUDITORS – September, 2019

PAPER – 1: General Aspects of Energy Management & Energy Audit

Section - I: OBJECTIVE TYPE Marks: 50 x 1 = 50

1. Which of the following statement is true regarding the EC act?

a) Designated consumers have to appoint Energy managers with prescribed qualifications.
b) State Designated Agencies have to appoint Energy auditor with prescribed qualifications.
c) Designated consumer has to get an energy audit conducted by a certified energy Manager.
d) Designated consumer has to get an energy audit conducted by the State Designated Agency
2. Which of the following statements regarding evacuated tube collectors (ETC) are true?
i) ETC can reach high temperatures upto 150oC
ii) because of vacuum between two concentric glass tubes, higher amount of heat is retained
in ETC
iii) heat loss due to conduction back to atmosphere from ETC is high
iv) performance of evacuated tube is highly dependent upon the ambient temperature

a) i & iii b) ii & iii c) i & iv d) i & ii

3. How much power you would expect to generate from a river-based mini hydropower with flow of
40 litres/second, head of 12 metres and system efficiency of 55%.

a) 872 kW b) 2.59 KW c) 264 kW d) none of the above

4. Which among the following has the highest flue gas loss on combustion due to Hydrogen in the
fuel?
a) Natural gas b) furnace oil c) coal d) light diesel oil
5. Energy in one Tonne of Oil Equivalent (toe) corresponds to
a) 4.187 GJ b) 1.162 MWh c) 10,000 kcal d) none of the above
6. Assume CO2 equivalent emissions by the use of a 40 W fluorescent lamp are of the order of 60
g/hr. If it is replaced by a 20 W LED lamp then the equivalent CO2 emissions will be
a) nil, as LED does not emit CO2 b) 30 g/hr
c) 20 g/hr d) 1200 g/hr

7. Under the Energy Conservation Act, the designated consumer is required to get the mandatory
energy audit conducted by
a) certified energy manager b) certified energy auditor
c) accredited energy auditor d) BEE
8. Stroboscope is an instrument for measuring
a) steam flow b) composition of flue gas c) speed d) pressure
9. The benchmarking parameter for a vapour compression refrigeration system is
a) kW / kg of refrigerant used b) kcal / m3 of chilled water
c) BTU / TR d) kW / TR
10. If 1 kWh of electrical energy is used to heat 10 kg of ice at 0o C, what will be the temperature of
water after melting? (Latent heat of fusion of ice is 80 kcal/kg)
a) 0oC b) 6oC c) 86oC d) none of the above
11. If feed of 15 tonnes per hour at 6% concentration is fed to an evaporator, the product obtained at
1
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30% concentration is equal to ____ tonnes per hour.

a) 3 b) 9 c) 0.9 d) 4.5
12. The discount rate is used as an input in determining _________.
a) NPV b) IRR c) payback period d) all of the above
13. The rate of energy transfer from a higher temperature to a lower temperature is measured in

a) kcal b) Watt c) Watts per second d) none of the above.

14. The cost of an economizer is Rs. 2 lakhs. The simple payback period (SPP) in years considering
annual savings of Rs 1,10,000 and annual maintenance cost of Rs 10,000 is ___________.
a) 1.8 b) 2.5 c) 2 d) 0.5
15. 1 kg of wood contains 15% moisture and 5% hydrogen by weight. How much water is evaporated
during complete combustion of 1kg of wood?
a) 0.6 kg b) 200 g c) 0.15 kg d) none of the above

16. In an industry the average electricity consumption is 10 lakh kWh for a given period. The average
production is 90,000 tons with a specific electricity of 10 kWh/ton for the same period. The fixed
electricity consumption for the plant is
a) 1,00,000 kWh b) 9,90,000 kWh c) 10,000 kWh d) none of the above
17. The cost of retrofitting a humidification system with an energy efficient one costs Rs. 20 lakhs.
The net annual cash flow is Rs. 5 lakhs. The return on investment is ________.
18% b) 25% c) 15% d) 33.33%
18. The theoretical amount of electricity required to heat 500 litres of brine solution with a specific
gravity of 1.2 and specific heat of 1 kcal/kg K from 30oC to 70 oC through resistance heating
is_________
a) 27.9 kWh b) 23.3 kWh c) 20 kWh d) none of the above

19. In project financing, sensitivity analysis is applied because

a) almost all the cash flows involve uncertainly
b) it evaluates how sensitive the project is to change in the input parameters
c) it assesses the impact of ‘what if one or more factors are different from what is predicted’
d) it is applicable to all the above situations
20. A process requires 120 kg of fuel with a calorific value of 4800 kcal/kg for heating with a system
efficiency of 82 %. The loss would be____________.
576000 kcal b) 472320 kcal c) 103680 kcal d) 480000 kcal
21. Which of the following is true?
a) The internal rate of return is the discount rate for which the NPV is Zero
b) NPV is the internal rate of return for which the discount rate is Zero
c) The discount rate is the internal rate of return for which NPV is positive
d) NPV is the discount rate for which internal rate of return is positive

22. Having energy policy _____________

a) satisfies regulations b) shows top management commitment
c) indicates energy audit skills d) Ensures ISO 50001 certification
23. Which of the following is not true of fuels cells?
a) they consume electricity b) they are fuelled by hydrogen
c) they have an electrolyte c) produce water and heat

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24. Which of the following has the highest Reserve to Production (R/P) ratio in India?
a) Lignite
b) Petroleum
c) Coal
d) Natural gas
25. SI unit for energy is_____________
a) Watt
b) Kilogram
c) Newton
d) Joule
26. Which of the following has the lowest energy content in terms of MJ/kg?
a) LPG
b) Diesel
c) Furnace Oil
d) Coal
27. Which of the following industries has the highest Specific Electrical Energy Consumption?
a) Aluminum
b) Sugar
c) Paper & Pulp
d) Cement
28. Select the wrong statement.
a) Energy Efficiency and Energy Conservation are distinct and interrelated
b) Unscheduled power interruption is an Energy conservation measure
c) Productivity improvements leads to energy conservation
d) Energy Efficiency is an integral part of energy conservation
29. _____ in Centre and______ _ in States are mandated to implement the provisions of The Energy
Conservation Act, 2001
a) BEE and NPC
b) BEE and DISCOM
c) BEE and SERC
d) BEE and SDA
30. Energy Conservation Building Code (ECBC) sets;
a) Minimum Energy Efficiency Standards for design and Construction of Buildings
b) Green Building Rating System
c) Municipal DSM Regulations
d) Incentives for energy efficient buildings
31. Which of the following is one of the schemes of BEE under Energy Conservation Act ?
a) Standards and Labelling
b) Availability based Tariff
c) Standard of Performance of DISCOMs
d) Renewable Energy Certificates
32. Which one of the following is not a Designated Consumer category under PAT ?
a) Paper and Pulp Industries
b) Cement Plants
c) Chlor Alkali Plants
d) Sugar Plants
33. Steam contains 10% moisture by mass, its dryness fraction x is _________.

0.1 b. 1 c. 0.9 d. None of the above

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34. Which of the following has highest Global Warming Potential?

a) SF6
b) CO2
c) CH4
d) N2O
35. Which of the following is not true?
a) Primary energy is converted to secondary energy in industries
b) Secondary energy is converted to primary energy in industries
c) Coal is primary energy
d) Electricity is secondary energy
36. Which primary energy is used as a feedstock in fertilizer industry?
a) Steam b) Natural gas c) Electricity d) All of the above
37. Bio-gas generated through anaerobic process mainly consists of
a) only methane b) Methane and carbon dioxide c) only ethane d) only carbon dioxide
38. Which of the following statements are true?
i) Rice husk is a source of secondary energy
ii) nuclear energy is non-renewable energy
iii) electricity is basically a convenient form of primary energy
iv) steam is a convenient form of secondary energy
a) (ii) & (iii) b) (i) & (iii) c) (ii) & (iv) d) (ii) & (i)
39. Trillion cubic meters is a unit normally used for
a) Crude oil b) Lignite c) Bituminous coal d) Natural Gas
40. Which of the following will have maximum value when expressed as MTOE (Metric Tonne of Oil
Equivalent)?
a) 1000 tonnes of furnace oil b) 10,000 kWh of electrical energy
b) 1000 tonnes of bituminous coal d) 1000 tonnes of lignite
41. Which of the following is not true of natural gas?
a) Requires more excess air compared to oil b) Consists mainly of methane
c) Becomes liquefied when cooled to -161oC d) All of the above
42. In a boiler, substitution of coal with rice husk will definitely lead to__________.
a) energy conservation b) energy efficiency
c) both energy conservation and energy efficiency d) GHG reduction
43. For determining the Energy intensity at the national level, which of the following are not
required?
(i) Gross domestic product
(ii) Total final consumption,
(iii) R/P ratio in years
(iv) Prevailing prices of various fuels

a) (i) & (iv) b) (i) & (ii) c) (iii) & (iv) d) (i) & (iv)
44. A building intended to be used for commercial purpose will be required to follow Energy
conservation building code under Energy Conservation Act, 2001 provided its

a) connected load is 120 kW and above

b) contract demand is 100 kVA and above
c) connected load is 100 kW and above or contract demand is 120 kVA and above
d) connected load is 500 kW and contract demand is 600 kVA
45. Which of the following is true of DSM?

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a) results in energy and/or demand reduction

b) enables end-users to better manage their load curve
c) can improve the profitability of power supply company
d) All of the above
46. An induction motor with 11 kW rating and a rated power factor of 0.9 in its name plate means
a) it will draw 12.22 kW at full load
b) it will draw 11 kW at full load
c) it will draw 9.9 kW at full load
d) it will deliver 11 kW at full load
47. The unit used for determining a designated consumer is ___________.
a) million tonnes of oil equivalent per year
b) metric tonnes of oil equivalent per month
c) metric tonnes of oil equivalent per year
d) million tonnes of oil equivalent per month
48. Which of the following statements are true regarding simple payback period?
a) considers impact of cash flow even after payback period
b) takes into account the time value of money
c) considers cash flow throughout the project life cycle
d) determines how quickly invested money is recovered
49. Global warming will not result in___________

a) melting of the ice caps b) increasing sea levels

c) severe damage to ozone layer in stratosphere d) unpredictable climate patterns
50. The process of capturing CO2 from point sources and storing them is called __________.

a) carbon sequestration b) carbon sink c) carbon Capture d) carbon absorption

……. End of Section – I …….

Section – II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40

Each question carries Five marks

An industry intends to invest Rs. 5,00,000 in a new energy saving project.

S1
The cash flows expected are:
Year 1 : Rs.2,00,000
Year 2 : Rs.3,00,000
Year 3 : Rs.2,00,000

The expected return is 10%. Evaluate the Net Present Value and comment on the feasibility of the
project?
Solution:

NPV = {-500,000+(200,000/1.10)+[300,000/(1.1)2]+[200,000/(1.1)3]}
= (-500,000+181818+247934+150263)
= 80015
NPV is positive Rs. 80,015; therefore, the proposed investment for the new energy saving project is
viable and attractive.
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Write short note on any one of the following.

S2
a) ISO 50001 (Book 1 - Page 157 & 158)
b) Energy Security (Book 1 - Page 20 to 22)

A continuous centrifuge separates 36,000 kg of whole milk containing 4% fat in 6-hour period into
S3
skim milk with 0.40% fat and cream with 40 % fat. Find out the flow rates of whole milk, cream and
skim milk using mass balance.

Ans:

Mass in

Total mass flow of whole milk = 36000/6

= 6000 kg per hour.
Fat per hour = 6000 x 0.04
= 240 kg/hr.
Therefore, Water plus solids other than fat = (6000-240) kg per hr.
= 5760 kg per hr.
Mass out :
Let the mass of cream be X kg then its total fat content is 0.40X.

The mass of skim milk is (6000 - X) and its total fat content is 0.0040 (6000 - X)
Material balance on fat:
Fat in = Fat out 6000 x 0.04
= 0.0040(6000 - X) + 0.40X; solving this,
X = 545 kg/hr

So that the flow of cream is 545 kg / hr and skim milk (6000- 545) = 5455 kg/hr.
A water pumping station fills a tank at a fixed rate. The head and flow rate are constant and hence
S-4
the power drawn by the pump is always same. The pump delivers 80 litres per second. The power
consumption was measured as 84 kW.

Calculate the energy consumption for pumping 2880 kL of water to the reservoir.
Ans
Time taken to pump water in hours = (2880 × 103 lit)
(80 lit/s x 3600 sec/hr)

= 10 hours

Power required to pump water = 84 kW

Energy consumption = 84 x 10 hrs = 840 kWh

S-5 A conveyor delivers coal with a width of 0.9 m and coal bed height of 0.15 m at a speed of 0.8 m/s.
Determine the coal delivery in tons per hour considering the coal density as 1.1 ton/m3.

Ans Volume of coal delivered per hour = Area x Length travelled per second
= 0.9 m x 0.15 m x 0.8 m/s

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= 0.108 m3/s = 0.108 x 3600 = 388.8 m3/hr

Coal delivery rate = 388.8 m3/hr x 1.1 t/m3
= 427.7 tonnes/hr

S-6 In a textile industry, 25,000 kg/hr water is currently being heated from 28 oC to 80 oC by indirect
heating of steam in dyeing machines.
It is proposed to recover heat from the hot effluent and generate hot water at 45 oC which would be
further raised to 80 oC by steam.
Estimate the reduction in steam in kg/hr considering the latent heat of steam as 520 kcal/kg in both
the cases.

Ans Ans:
Without heat recovery
Heating required (Q1) = mCpT
= 25000 x 1 x (80-28)
= 13,00,000 kcal/hr
Steam required = 13,00,000 / 520
= 2500 kg/hr
After heat recovery
Heating required (Q2) = 25000 x 1 x (80 – 45)
= 8,75,000 kcal/hr
Steam required = 8,75,000 / 520
= 1682.7 kg/hr
Reduction in steam required = 2500 - 1682.7 = 817.3 kg/hr
Briefly explain the difference between flat plate collector and evacuated tube collector.
S7
(Book 1, Page 264-265)

a) What is solar constant and solar insolation? (3 Marks)

S8
b) Which of them determines the amount of electrical energy that can be produced per unit area of
solar panel on any given day? (2 Marks)
(Book 1, Page 263 – 264)

------- End of Section - II ---------

Section – III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60

(i) Answer all Six questions

(ii) Each question carries Ten marks

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L1 a) A sample of fuel being used in a boiler is found to contain 40% carbon and 23% ash. The
refuse obtained after combustion is analyzed and found to contain 7% carbon & the rest is
ash. Compute the percentage of the original carbon in fuel which remains as unburnt in the
refuse. (5 Marks)
b) During an ESP performance evaluation study, the inlet gas stream to ESP is 2,89,920
Nm3/hr and the dust loading is 5,500 mg/ Nm3. The outlet gas stream from ESP is
3,01,100 Nm3/hr and the dust loading is 110 mg/Nm3. How much fly ash is collected in
the system in kg/hr? (5 Marks)

a)
Let the quantity of Refuse sample = 100 kg
Amount of unburnt Carbon in Refuse = 7 kg
Amount of Ash in the Refuse = 93 kg
Total ash in the fuel that has come into the Refuse = 23% of fuel

93 kg of Ash corresponds to 23% ash in the fuel

Therefore, quantity of total raw fuel = 93 / 0.23
= 404.35 kg
Quantity of original Carbon in the fuel = 0.40 x 404.35
= 161.74 kg
Quantity of unburnt fuel in Refuse = 7 kg

%age of the original carbon unburnt in the refuse = (7 / 161.74) x 100

= 4.32%
b)
Based on Mass balance,
Inlet gas stream dust = Outlet Gas stream dust + Fly ash

i) Inlet gas stream flow = 289920 NM3/hr

Dust Concentration = 5500 mg/NM3
Inlet dust quantity = 289920 x 5500
-----------
1000000
= 1594.56 kg/hr

ii) Outlet dust quantity = 301100 (NM3/hr) x 110 (mg/NM3) x 1

----------
1000000
= 33.12 kg/hr

iii) Fly Ash = Inlet gas stream dust – Outlet gas stream dust
= 1594.56 – 33.12 = 1561.44 kg/hr
L2 Write short note on any two of the following. (Each 5 Marks)

a) Benchmarking, (Book 1, Page 98 – 100)

b) DSM, (Book 1, Page 38)
c) TPM, (Book 1, Page 154 – 155)

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L3 Based on the following network diagram, identify the total number of paths with duration, critical path,
and float for each path.
1 Mark each path x 5 = 5 Mark
1 Mark for identifying critical path = 1 Mark
1 Mark for float of each path x 4 = 4 Mark

Solution:

The above network diagram has five paths; the paths and their duration are as follows:

Start -> A -> B -> C-> End, duration: 46 days.

Start ->D -> E ->F -> End, duration: 33 days.
Start -> D -> B -> C -> End, duration: 41 days.
Start -> G ->H ->I -> End, duration: 28 days.
Start -> G -> E ->F -> End, duration: 31 days.

Since the duration of the first path given below is the longest, it is the critical path.

Start -> A -> B -> C-> End, duration: 46 days.

The float on the critical path is zero.

The float for the second path “Start ->D -> E ->F -> End” =
Duration of the critical path – duration of the path “Start ->D -> E ->F -> End” = 46-33 = 13
Hence, the float for the second path is 13 days.

Using the same process, we can calculate the float for other paths as well.
Float for the third path = 46 – 41 = 5 days.
Float for the fourth path = 46 – 28 = 18 days.
Float for the fifth path = 46 – 31 = 15 days.
L4 In the washing process of an automobile plant, electricity is being used to heat 5000 litres/hr of water
by 8 0C. The industry is planning to convert from Electrical heating to LPG heating.
Other Parameters:
Annual operating hours = 6000 hours
Efficiency of indirect heating with LPG = 85%
Efficiency of electrical heating = 95%
Calorific value of LPG = 12,000 kcal/kg
Landed cost of LPG = Rs.60/kg
Cost of electricity = Rs.8/kWh

a) If electrical heating is replaced with LPG heating, with an investment is Rs.15 lakhs, compute the
simple payback period. ( 6 Marks)
b) Also, calculate the CO2 emissions in both the cases considering the emission factors for LPG as 3

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tons of CO2/Ton of LPG and Electricity as 0.81 tons of CO2/MWh. (4 Marks)

Ans a).
Water flow rate = 5000 Litres/hr
Temperature rise = 8 oC
Useful Heat Required = (5000 x 1 X 8)
= 40,000 kcal/hr
Equivalent LPG consumption = 40000/(12000 x 0.85)
= 3.92 kg/hr
Hourly Cost of Operating with LPG = 3.92 x 60
= Rs.235 / hr
Equivalent electricity consumption = 40000/ (860 * 0.95)
= 48.96 kW
Hourly Cost of operating with electricity = 48.96 x 8
= Rs.391.68/ hr
Difference in hourly operating cost = Rs. (391.68 – 235)
= Rs.156.68/ hr
Annual monetary savings = Rs.156.68/ hr x 6000 hrs/yr
= Rs.9,40,080/ yr
Investment = Rs.15,00,000
Simple payback period = Rs. 15,00,000/ Rs.9,40,080/ yr
= 1.6 yr
a)
Annual CO2 Emission with electrical heating = 48.96 kW x 6000 x (0.81 kg CO2/kWh)
= 237946 kg CO2/ yr
= 237.95 tonnes CO2/ yr
Annual CO2 emission with LPG heating = 3.92 kg LPG/hr x 6000 hr/yr x
(3 kg CO2/kg LPG)
= 70560 kg CO2/yr
= 70.6 tonnes CO2/yr
Thus, by converting from electricity to LPG use, there is a huge advantage, not only in operating
cost but also in reduced CO2 emissions.
L5 A company has got the following two energy saving project investment options:
Option A:
Investment envisaged is Rs. 40 lakhs with an annual return of Rs. 12 lakhs; life of the project is 5
years. Calculate IRR.
Option B:
A project having IRR of 12%
Which option should the company select?
Option A:

Investment = Rs. 40 lakh

Annual Return = Rs. 12 lakh
Life of project = 5 years

0 = (-) 40 + (12) [1/ (1 + i)1 + 1/ (1 + i)2 + 1/ (1 + i)3 + 1/ (1 + i)4 + 1/ (1 + i)5 ]

IRR = 15.24 %

Based on IRR, the Option A has higher IRR value and the company may opt for Option A.

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L6 Match the following:

1. Biomass a. Radiation
2. CNG b. Distribution Loss Reduction
3. HVDS c. Oxidation
4. Cement d. Sankey Diagram
5. Combustion e. ISO 50001
6. Energy Balance f. Designated consumer
7. kWh/ton of product g. Transport
8. Objectives, targets & action h. Carbon neutral
plans
9. Performance Contracting i. Benchmarking
10. Surface Heat Loss j. ESCO

(Each 1 Mark)
Solution:

1. Biomass : Carbon neutral

2. CNG : Transport
3. HVDS : Distribution Loss Reduction
4. Cement : Designated consumer
5. Combustion : Oxidation
6. Energy Balance : Sankey Diagram
7. kWh/ton of product : Benchmarking
8. Objectives, targets & action plans : ISO 50001
9. Performance Contracting : ESCO
10. Surface Heat Loss : Radiation

-------- End of Section - III ---------

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