MODULE 4

ENERGY PERFORMANCE ASSESSMENT

FOR EQUIPMENT AND UTILITY SYSTEMS
HVAC SYSTEM
▪ Example 5.2
▪ In an air-handling unit (AHU), the filter area is 1.5 m2 while air velocity is 2.2 m/s. The
inlet air has an enthalpy of 77 kJ/kg. At the outlet of AHU, air has an enthalpy of 59
kJ/kg. The density of air of 1.3 kg/m3. TR of the air-handling unit is assessed as follows:
▪ TR of AHU = (Enthalpy difference x density x area x velocity x 3600) / (4.187 x 3024)
▪ = (77-59) x 1.3 x 1.5 x 2.2 x 3600 / (4.187x 3024)
▪ = 21.96 TR
PROBLEM 5.5

If the compressor of centrifugal chiller consumes 205 kW, the steam consumption for VAM
is 1620 kg/Hr. Calculate the following:
1. Refrigeration load delivered (TR) for both systems?
2. Condenser Heat load (TR) for both systems?
3. Compare auxiliary power consumption for both systems, give reason?
▪ A pharmaceutical unit had installed a centralized refrigeration system of 120 TR Capacity several
years ago. The refrigeration system operates 24 hours a day, 200 days per annum and the average
electricity cost is BDT. 4.5/ kWh. The following are the key operational parameters.
• Compressor operating current and power factor : 153 amps. 0.9 pf
• Condenser pump operating current and power factor: 43 amps, 0.88 pf
• Chiller pump operating current and power factor : 25 amps, 0.9 pf
• CT fan operating current and power factor : 20 amps. 0.65 pf
• ΔT across the chiller (evaporator) : 3.5OC • Chilled water flow : 23 Lit / Sec
• Total head developed by chiller pump : 35 mtrs. • Condenser water flow : 41 Lit / Sec
• Total head developed by condenser pump : 30 mtrs.
PS: all the motors operate at 415 Volts and efficiency of 90%
Calculate:
• The power consumed by the compressor, condenser pump, chiller pump and CT fan.
• TR developed by the system
• Specific power consumption i.e. overall kW/TR and COP and Energy Efficiency ratio (EER)
▪ Combined efficiency (motor and pump) of condenser and chiller pumps
 CURRENT STATUS CALCULATION
Solution
▪ Present Condition:
▪ Compressor Motor Power : 99 kW
▪ Condenser Pump Motor Power : 27.2 kW
▪ Chiller Pump Motor Power : 16.2 kW
▪ CT Fan : 9.4 kW
▪ Total Power : 151.8 kW
▪ TR Devp : (23 * 3600 * 3.5 / 3024) = 95.83
▪ Sp. Power : 1.58 kW/ TR
▪ Compressor kW/TR : 99/95.83 : 1.03 kW/TR
▪ COP : 3.516/kW/TR = 3.41
▪ EER :=COP*3.413=11.63
▪ Condenser pump efficiency : 44.4%
▪ Chiller pump efficiency : 48.8%
 PROPOSAL FOR IMPROVEMENT
▪ The unit proposes to replace the existing condenser and chilled water pumps with efficient pumps
having a combined efficiency of 65%. Also the unit goes in for condenser cleaning by which the
power consumption of compressor has reduced by 10%.
▪ Calculate:
▪ • Specific power consumption i.e. overall kW/ TR and COP and Energy Efficiency ratio (EER)
▪ • The envisaged power consumption of the compressor, condenser and chiller pump
▪ • Hourly energy savings (compressor, condenser and chilled water pump)
▪ • Annual energy and equivalent monetary savings (compressor, condenser and chilled water
pump)
▪ Proposed condition: Compressor Motor Power : 89 kW
▪ Condenser Pump Motor Power : 18.6 KW
▪ Chiller Pump Motor Power : 12.2 kW
▪ CT Fan : 9.4 kW Total Power : 129.2 kW
▪ TR Devp : 95.83 TR i.e. (23 LPS * 3600 Sec * 3.5 / 3024)
▪ Sp. Power : (Total kW/Ton of cooling effect) 1.35 kW/ TR
▪ Compressor kW/TR : 89/95.83 = 0.93 kW/TR
▪ COP : 3.516 / 0.93 = 3.78 & EER : 12 / 0.93 = 12.90
PERFORMANCE CALCULATION OF VAPOR
ABSORPTION REFRIGERATION SYSTEM
▪ Coefficient of performance, COP at evaporator side

For Steam heated Vapour Absorption

Chilling Packages,

For Direct Fuel-Fired Vapour Absorption

Chilling Packages,
COEFFICIENT OF PERFORMANCE, COP AT CONDENSER (WATER
COOLED) SIDE: FOR STEAM HEATED SYSTEM

At condenser, Refrigeration effect heat = Heat rejected in the condenser — Thermal energy input
Qe = Qc — Qin
 COEFFICIENT OF PERFORMANCE, COP AT CONDENSER (WATER
COOLED) SIDE: FOR DIRECT FUEL FIRED PACKAGES

At condenser, Refrigeration effect heat = Heat rejected in the condenser — Heat input
 PROBLEM
▪ In an air conditioning system of a food processing industry, the cold air flow rate is 20,000 m3/hr
at a density of 1.2 kg/m3 .The inlet and outlet enthalpy of the air are 105 kJ/kg and 80 kJ/kg. The
COP of the existing vapor compression system is 3.75. The efficiency of the motor coupled with
the compressor is 90%.
▪ The management wants to install a Vapor Absorption System (VAR).The saturated steam for VAR
will be supplied either from a new waste heat boiler to be installed with the existing DG sets or
from the existing FO fuel fired boiler. The plant is operating for 8000 hr/annum. The investment of
VAR system is BDT. 20 lakhs. The investment for waste heat boiler is BDT. 6 lakhs. The power cost
is BDT. 6/kWh.
▪ As an energy auditor which one of the following options will you recommend to the management?
Option-1: Supply steam from the existing FO fuel fired boiler to VAR system and avoid the
investment of waste heat boiler
Option-2: Supply steam from the waste heat boiler, which needs an investment in addition to VAR
system
▪ The steam consumption per TR will be 5.5 kg/TR. The cost ofF0 is BDT. 32,000/ ton. The
evaporation ratio of the existing FO fired boiler is 14. Neglect losses in transmission of steam and
chilled water.
=8000 hrs*296.22BDT
COMPREESED AIR
SYSTEM
 EXAMPLE 1.4
▪ In a medium sized engineering industry a 340 m3/hr reciprocating
compressor is operated to meet compressed air requirement at 7 bar. The
compressor is in loaded condition for 80% of the time. The compressor draws
32 kW during load and 7 kW during unload cycle.
▪ After arresting the system leakages the loading time of the compressor came
down to 60%. Calculate the annual energy savings at 6000 hours of operation
per year.
Solution:
▪ Average power consumption with 80% loading = [0.8 x 32 + 0.2 x 7] = 27 kW
▪ Average power consumption with 60% loading after leakage reduction = [0.6
x 32 + 0.4 x 7] = 22 kW saving in electrical power = 5 KW
▪ Yearly savings = 5 x 6000 = 30,000 kWH
 BOILER EFFICIENCY CALCULATION
▪ Test Conditions and Precautions for Indirect Method Testing
o Standby losses. Efficiency test is to be carried out, when the boiler is
operating under a steady load. Therefore, the efficiency test does not
reveal standby losses, which occur between firing intervals.
o Blow down loss. The amount of energy wasted by blow down varies over a
wide range.
▪ Soot blower steam. The amount of steam used by soot blowers is variable
and depends on the type of fuel.
o Auxiliary equipment energy consumption. The efficiency test does not
account for the energy usage by auxiliary equipment, such as burners,
fans, and pumps.
▪ Flue gas sampling location
o Flue gas sampling point for temperature and gas analysis is ideally carried
out at zone of maximum gas flow which normally coincides with zone of
maximum temperature.
 BOILER EFFICIENCY BY INDIRECT METHOD
▪ Heat loss due to radiation and convection:
▪ Normally surface loss and other unaccounted losses is assumed based on the type and size of
the boiler as given below
o For industrial fire tube / packaged boiler = 1.5 to 2.5%
o For industrial water tube boiler = 2 to 3%
o For power station boiler = 0.4 to 1%

▪ It can be calculated if the surface area of boiler and its surface temperature are measured
and using the following formula:
EXAMPLE
PROBLEM
The following are the data collected for a boiler using coal as the fuel. Find out the
boiler efficiency by indirect method.
CALCULATION OF
LOSS DUE TO
FLUE GAS
EXCESS AIR
CALCULATION
THEORETICAL AIR REQUIREMENT
COGENERATION AND TURBINES (GAS, STEAM)
 GAS TURBINE AND HEAT RECOVERY STEAM
GENERATOR PERFORMANCE
▪ Overall plant heat rate, kCal/kWh = Overall plant fuel rate kg/kWh × GCV of fuel kCal kg
EXAMPLE PROBLEM
SOLUTION
PROBLEM
SOLUTION
PROBLEM
▪ A gas engine-based trigeneration plant operates in two modes:
▪ Power and heating mode (10 hours per day) :
▪ Pel= 650 kW of electricity and 325 kg/h of steam with enthalpy addition of 530 kcal/kg of steam
&EUF heat = 0.85
▪ Power and cooling mode (14 hours per day) :
▪ Pel = 650 kW of electricity and chilling load of 213 TR for absorption chillers &EU Fcool = 0.73
▪ Calorific value of natural gas = 8500 kcal/Sm3
▪ Average operating days/year = 330
▪ Alternator efficiency = 0.95
▪ The energy loss in the flue gas and that in the cooling water is same as engine power output and
other losses are negligible
a) What is the average plant energy utilization factor
b) Calculate the useful energy produced daily by the trigeneration plant in Gcal
c) Determine the daily plant natural gas requirements based on average energy utilization factor
▪ The plant proposes to install a 60 TR hot water driven Vapour absorption chiller with a COP of 0.5
using waste heat from jacket cooling water. Check if it is feasible with supporting calculations.
PUMPING SYSTEM
▪ A refrigeration plant is located at
+0.00 level and the process plant
condensers are located at +15 m
level. A cooler having a design
pressure drop of 1.9 kg/cm2 is
located at the 0.00 level (ground
level). Other relevant data can be
taken from the earlier section. The
schematic of the plant is shown in
Figure
▪ The step-by-step approach for determining system resistance curve is as follows:
▪ Step-1: Divide system resistance into static and dynamic head
o Determination of static head Static head (condenser floor height): 15 m
o Determination of dynamic head Dynamic Head = Total Head – Static Head Dynamic head
o = (54–15) = 39m

▪ Step-2 Check maximum resistance in the different circuits

Cooler circuit offers the maximum resistance followed by condenser circuit.

Step 3: Draw system resistance curve
▪ Condenser loop is chosen as it offers maximum resistance as well as
has static head.
▪ At 100% flow, Static head: 15 m and Dynamic head at full load: 39 m
▪ Compute system resistance at different flow rates namely 75%, 50%
and 25%.
▪ Step 4 - Plot the calculated system resistance against various flows (Figure
8.2).
▪ If the system curve is plotted in the pump curves provided by the vendor,
operating duty point can be determined and check whether pump is
operating at design efficiency and deviations if any can be assessed.
 HEAT EXCHANGERS
Heat duty of the exchanger can be calculated either on the hot side fluid or
cold side fluid.
▪ Heat Duty for Hot fluid, Qh = W xCph x (Ti–To)
▪ Heat Duty for Cold fluid, Qc = wx Cpc x (to–ti)
Overall heat transfer coefficient, U
▪ Heat exchanger performance is normally evaluated by determining the
overall heat transfer coefficient U. The overall heat transfer equation is given
by the following relation:
▪ Q = U x A x LMTD
o Where,
o Q is the rate of heat transfer in kCal/hr
o A is the outside surface area of heat exchanger in m2
o LMTD is the logarithmic mean temperature difference in 0C
o U is the overall heat transfer coefficient kCal/hr/m2/ oC)
▪ Logarithmic Mean
Temperature Difference
(LMTD)
▪ The LMTD is Logarithmic
Mean Temperature
Difference, used to
determine the temperature
driving force for heat
transfer in heat exchangers.
It is determined by the
relationship of the fluid
temperature differences at
the terminals of the heat
exchanger.
 CORRECTION FACTOR FOR LMTD TO ACCOUNT
FOR CROSS-FLOW
▪ The LMTD correction factor is a function of the temperature effectiveness and the number of
tube and shell passes and is correlated as a function of two dimensionless temperature ratios.
Let R and P be the two dimensionless parameters used to calculate LMTD correction factor
defined by the following equations:
▪ R = Ta – Tb / tb – ta and P =tb – ta / Ta – ta
o Where,
o Ta = inlet temperature of shell-side fluid
o Tb = outlet temperature of shell-side fluid
o ta = inlet temperature of tube-side fluid
o tb = outlet temperature of tube-side fluid
Where, N Number of shell-side passes
S, α Parameters used to calculate LMTD correction factor defined by equations above.
Corrected LMTD= F x LMTD
Overall Heat Transfer Coefficient U = Q / (A x Corrected LMTD)
PROBLEM
SOLUTION
Calculation of Thermal data:
Heat Transfer Area = 3.1416×Tube dia (m)× Total Tube Length (m)= 264.55 m2
1.Heat Duty:
o Hot fluid, Q = 719800 x 2.847 x (145 –102) / 3600 = 24477.4 kW
o Cold Fluid, Q = 881150 x 4.187 x (49 – 25.5) / 3600 = 24083.4 kW
2. Hot Fluid Pressure Drop
Pressure Drop = Pi – Po = 4.1 – 2.8 = 1.3 bar g.
3. Cold Fluid Pressure Drop
Pressure Drop = pi – po = 6.2 – 5.1 = 1.1 bar g.
4. Temperature range hot fluid
Temperature Range ΔTh = Thi – Tho = 145 – 102 = 43 o C.
5. Temperature Range Cold Fluid
1. Temperature Range ΔTc = Tco – Tci = 49 – 25.5 = 23.5 0C.
6. LMTD
LMTD, Counter Flow = (96 – 76.5) / ln (96 / 76.5) = 85.9 0C
7. LMTD Correction Factor, F, to account for Cross flow:
▪ Computing the Parameters below:
▪ R = (Ta-Tb) / (tb-ta) = (145-102) / (49-25.5) = 1.83
▪ P = (tb-ta) / (Ta-ta) = (49-25.5) / (145-25.5) = 0.20
▪ For R ≠ 1:
▪ α = {(1-RP) / (1-P)} (1/N) = {(1- 1.83x0.2) / (1-0.2)} (1/1) = 0.793
▪ S = (0.793-1) / (0.793-1.83) = 0.20
▪ F = 0.977
8. Corrected LMTD = F x LMTD = 0.977 x 85.9 = 83.9 DegC.
9. Overall Heat Transfer Co-efficient
U = Q/ (A x Corrected LMTD) = 24477.4/ (264.55 x 83.9) = 1.104 kW/m2. K
 OBSERVATIONS
▪ Heat Duty: Actual duty differences will be practically negligible as these
duty differences could be because of the specific heat capacity deviation
with the temperature. Also, there could be some heat loss due to radiation
from the hot shell side.
▪ Pressure drop: Also, the pressure drop in the shell side of the hot fluid is
reported normal (only slightly less than the design figure). This is attributed
with the increased average bulk temperature of the hot side due to
decreased performance of the exchanger.
▪ Temperature range: As seen from the data the deviation in the temperature
ranges could be due to the increased fouling in the tubes (cold stream),
since a higher pressure drop is noticed.
▪ Heat Transfer coefficient: The estimated value has decreased due to
increased fouling that has resulted in minimized active area of heat transfer.
▪ Physical properties: If available from the data or Lab analysis can be used for
verification with the design data sheet as a cross check towards design
considerations.
FINANCIAL MANAGEMENT: REVENUE COST
& PROFIT
▪ REVENUE – THIS IS THE TOTAL MONEY YOU EARN FROM SELLING YOUR
PRODUCT OR SERVICE. THINK OF IT AS ALL THE CASH COMING IN BEFORE YOU
PAY ANY EXPENSES.
▪ EXAMPLE: IF YOU SELL 100 BURGERS AT $5 EACH, YOUR REVENUE IS 100 × $5 =
$500.
▪ COST – THIS IS THE MONEY YOU SPEND TO MAKE OR PROVIDE YOUR
PRODUCT/SERVICE. IT INCLUDES MATERIALS, SALARIES, RENT, ETC.
▪ EXAMPLE: IF EACH BURGER COSTS YOU $3 TO MAKE (INGREDIENTS, LABOR,
ETC.), YOUR TOTAL COST FOR 100 BURGERS IS 100 × $3 = $300.
▪ PROFIT – THIS IS WHAT YOU GET TO KEEP AFTER PAYING ALL YOUR COSTS. IT’S
CALCULATED AS:
PROFIT = REVENUE - COST
▪ EXAMPLE: $500 (REVENUE) - $300 (COST) = $200 (PROFIT)
EBIT (Earnings Before Interest and Taxes)
▪ This is your profit before paying interest on loans and taxes.
▪ It shows how much your business is earning from its core operations.
▪ Formula:
EBIT = Revenue - Operating Costs (excluding interest & taxes)
Example:
Revenue: $1,000; Operating Costs (salaries, rent, materials, etc.): $600
EBIT = $1,000 - $600 = $400 (this is your earnings before paying interest and taxes).
EBITDA (Earnings Before Interest, Taxes, Depreciation, and Amortization)
This is similar to EBIT but also excludes depreciation and amortization (which are non-
cash expenses).
It gives a clearer picture of your business’s actual cash earnings.
Formula:
EBITDA = EBIT + Depreciation + Amortization
Example:
Suppose depreciation is $50 and amortization is $30.
EBITDA = $400 (EBIT) + $50 + $30 = $480.
 AMORTIZATION VS DEPRECIATION

Amortization
▪ Used for intangible assets like patents, copyrights, and trademarks
▪ Gradually writes off the cost of an intangible asset over its useful life
▪ Helps businesses reduce their taxable income and tax obligations
Depreciation
▪ Used for tangible assets like manufacturing equipment, inventory, and
business vehicles
▪ Gradually writes off the cost of a tangible asset over its useful life
▪ Helps businesses reduce their taxable income and tax obligations
▪ Fixed Costs
▪ One-time expenses like purchasing capital
equipment.
▪ Includes design, planning, installation, and
commissioning.
▪ Not affected by inflation or discount rates (unless
paid in installments).
▪ Variable Costs & Savings
▪ Recurring costs paid over time, such as:
▪ Raw materials, taxes, insurance, leases,
energy, maintenance, and labor.
▪ Variable savings refer to reductions in these costs
over the project's life.
▪ Increased costs = negative cash flow, while cost
reductions or savings = positive cash flow.
▪ Key Takeaway:
▪ Total project cost = Fixed Costs + Variable Costs
The passage highlights that both fixed and
variable costs contribute to overall operating
costs, and savings from a project impact cash flow
positively.
 MARGINAL ANALYSIS
▪ Marginal analysis helps decide how Problem:
much to invest in energy-saving The marginal cost and cost savings of insulating a
projects and compare projects with home with different thicknesses of insulation are as
different costs. follows:
▪ "Marginal" means the last unit added
(e.g., last LED bulb or solar panel).
▪ The rule: Keep adding units as long
as marginal savings > marginal
cost.
▪ Once the cost of adding one more unit
exceeds the savings it generates, stop
investing.
▪ This ensures optimal spending on
energy projects for maximum returns
FIND: THE PROPER AMOUNT OF INSULATION TO INSTALL
▪ Solution: The proper amount is where marginal cost is equal to marginal cost
savings. This occurs between the eighth and ninth inch of insulation.
Therefore, if the home owner installs the eighth inch, the cost savings are
greater than the costs. However, if the ninth inch is installed, the cost of that
inch is greater than the cost savings it generates. Since insulation cannot be
bought by the half-inch the least profitable inch to install in this case is the
eighth inch. Therefore, eighth inches of insulation should be installed.
▪ To achieve maximum accuracy marginal analysis should be applied to the
smallest possible units. For example it is better to analyze solar collector by
square foot that by 32 ft2 collector
▪ The concept of equating costs with savings can also be estimated using
break-even analysis concept.
BREAK-EVEN ANALYSIS
▪ Break-even Point Calculation
▪ Used to determine when a project recovers its costs.
BREAK-EVEN ANALYSIS
▪ If the electricity bought from a utility company costs an
average of BDT 4.5/kWh, and energy cost of generator is BDT
3.5/ kWh, the breakeven point for the generator will be,:
▪ when the average output is 50 kW is given by:
o 4.5 ×50 ×n=(900000+150000)+(3.5 ×50 ×n)
o n = 21000 hours

▪ If the average output is 70 kW, the break-even point is given by:

o 4.5 ×70 ×𝑛=(9,00,000+1,50,000)+(3.50 ×70 ×𝑛)
o n = 15000 hours

▪ Thus, increasing the average output of the generator

significantly reduces the break-even time for the project. This
is because the capital investment (i.e. the generator) is being
better utilized.
TIME VALUE OF MONEY
▪ The Time Value of Money (TVM) means money today is worth more than
the same amount in the future because of its potential to earn interest or
returns.
▪ Why?
▪ Inflation reduces purchasing power over time.
▪ Investment growth allows money today to earn interest or profits.
▪ Opportunity cost—you could use today's money for something valuable now.
▪ Example:
▪ If you have $100 today, you can invest it at 5% interest and have $105 next
year.
▪ But if you get $100 next year, it’s worth less than $100 today because you lost
the chance to earn interest.
INTEREST RATE

Key Difference:
•Simple Interest grows in a straight line.
•Compound Interest grows exponentially (faster over time).
FINANCIAL ANALYSIS TECHNIQUES
▪ Simple Payback Period (SPP) represents, as a first approximation; the time (number
of years) required to recover the initial investment (First Cost), considering only
the Net Annual Saving
RETURN ON INVESTMENT (ROI)
▪ ROI measures the annual return from a project as a percentage of
the initial capital cost. It takes into account the cash flows over the
project’s life and adjusts for the time value of money (using the
discount rate) to express the return in annual terms.
 BENEFIT-COST ANALYSIS (BCA)
▪ Benefit-Cost Analysis (BCA) is a method used to evaluate and compare the benefits and costs of
a project or investment to determine its economic feasibility. It helps decide whether the benefits
of a project justify the costs.
▪ How It Works:
▪ Identify Benefits: List all the positive outcomes or savings the project will bring, such as increased
revenue, cost savings, or environmental benefits.
▪ Identify Costs: List all the costs required to implement the project, including both initial capital
costs and ongoing operating costs.
▪ Compare Benefits to Costs: Calculate the Benefit-Cost Ratio (BCR) to determine if the project is
worth pursuing.

▪ If BCR > 1, the benefits outweigh the costs, making the project economically viable.
▪ Net Present Value (NPV)
▪ When making financial decisions, businesses and investors need to evaluate multiple future cash
flows, both incoming (revenues) and outgoing (costs). The Net Present Value (NPV) method
converts all these future cash flows to their present values using a discount rate. The total present
value of all cash flows (both inflows and outflows) is then summed up.

▪ If NPV > 0, the investment is considered profitable. If NPV < 0, it means the investment is not
worthwhile.
▪ Conclusion
• Compounding is used to find how much a present amount will grow in the future.
• Discounting is used to find how much a future amount is worth today.
• NPV converts all future cash flows to the present value to compare them properly and make
better financial decisions.
Initial investment is 1,000,000
THANK YOU