PAPER SOLUTION

SUMMER 2023
Subject code : 3110006
Subject Name : Basic Mechanical Engineering
Q-1 : (a) Classify boiler according to (i) relative position of hot gases and water (ii) axis of the
shell (iii) method of water circulation. Give example of each.
Ans : (i) 1) fire tube boiler : hot gases pass through fire tubes which are surrounded by water.
2) water tube : water flow inside the tubes and the hot flue gases flow outside the tubes.
(ii) 1) vertical boiler : the axis of the shell is vertical
2) horizontal boilers the axis of the shell is horizontal.
3) inclined boiler the axis of the boiler is inclined.
(iii) 1) forced circulation boiler : water is circulated by pumps which is driven by water.
2) natural circulation boiler : water is circulated by natural convection currents which
are set up due to the temperature difference produced by the application of heat.

(b) Make a list of (i) natural fuels and (ii) artificial fuels. write to effective points to reduce
global warming.
Ans : (i) Natural Fuels
1) Wood : biomass obtained from trees and plants.
2) Coal : formed from the remains of ancient plants
3) Natural gas : mainly composed of methane extraced from the Earth.
4) Crude oil : derived from the remains of marine organisms.

(ii) Artificial fuels :

1) Synthetic fuels : produce through chemical process like synthetic gasoline.
2) Hydrogen : often produce through electrolysis or reforming process.
• Effective points to reduce global warming :
1) Transmitter to renewable energy : shift from fossil fuels to renewable sources like
solar , wind and hydro power to significantly reduce carbon emissions.
2) Energy efficiency : Implement and promote energy efficient particles and technologies
across industries transportation and household to decrease overall energy
consumption.
(c) (i) Determine the amount of heat required to raise the temperature of a steel workpiece
from 40°C to 160°C. The mass of workplace is 20 kg and specific heat is 460 J/kg K.
Ans:- Given data :
m = 20 kg
C = 460 J/kg K
T1 = 40°C

Click :Youtube Channel | Free Material | Download App

 T2 = 160°C
Q=?

Q = mc ∆T
Q = mc ( T2 — T1 )
Q = 20 × 460 × (160—40)
Q = 20 × 460 × 120
Q = 1104000 J
Q = 1104 kJ
(ii) A temperature cycle has four processes the heat and work interaction with
surrounding is given in the table determine the world done during process 3 - 4
what will be the change in international energy after completion of one cycle.

Process Heat transfer (kJ) Work done (kJ)

1-2 +925 +70
2-3 -110 -50
3-4 -770 ?
4-5 +220 +170
Ans:- The work done during process 3 – 4 is ;

∮𝑄 = ∮𝑊

∴ 𝑄1 + 𝑄2 + 𝑄3 + 𝑄4 = 𝑊1 + 𝑊2 + 𝑊3 + 𝑊4
∴ 925 − 110 − 770 + 220 = 70 − 50 + 𝑊3 + 170
∴ 𝑊3 = 75 𝑘𝐽

→ The change in internal energy is :

𝑄 = 𝑊 + ∆𝑈
∴ ∆𝑈 = 𝑄 − 𝑊 ∴ [𝑄 = 𝑊]
∴ ∆𝑈 = 0

Q-2 : (a) State the function of the following in boilers:

Ans:
1) Feed check valve : the controls the supply of water to the boiler and prevents the back flow
of water from the boiler when the feed pump pressure is less than boiler
pressure or pump is stopped.
2) Blow of cock : It performs to functions ;
1. To discharge mud sediment and heavy deposits periodically which settle
down at the bottom of the boiler while the boiler is in operation.
2. To empty the boiler for cleaning and repair.

3)Fusible Plug : It is used to protect the boiler against damage due to overheating caused by
low water level in the boiler.

Click :Youtube Channel | Free Material | Download App

(b) Identify the following properties of materials.

Ans :
1) This this property is desirable 4 parts subjected to impact loads vibrations .
— Toughness.
2) By this property gold , silver can be flattered or bent without cracking when hammered.
— malleability
3) This property is useful for material subjected to high temperature like boiler and turbines.
— strength
4) this property is necessary for material to be used in making springs.
— resilience

(c) Air has a volume of 0.15 𝒎𝟑 , pressure is 1.5 bar and temperature 107°C. It is compressed
at constant pressure and their its volume becomes 0.11 𝒎𝟑 determine (i) temperature at
the end of compression (ii) work done during compression,(iii) change in internal
energy. 𝑪𝒑 = 𝟏. 𝟎𝟎𝟓𝒌𝑱/𝒌𝒈 𝑲, 𝑪𝒗 = 𝟎. 𝟕𝟏𝟖 𝒌𝑱/𝒌𝒈 𝑲
Ans :
Given data:
𝑉1 = 0.15 𝑚 3
𝑇2 = 107°𝐶 = 380 𝐾
𝑃1 = 𝑃2 = 𝑃 = 1.5 × 105
𝑉2 = 0.11 𝑚 3
𝑇2 =? , 𝑤 = ? , ∆𝑈 =?
𝐶𝑃 = 1.005 𝑘𝐽/𝑘𝑔 𝐾
𝐶𝑣 = 0.718 𝑘𝐽/𝑘𝑔 𝐾

(i)
𝑉1 𝑇1
=
𝑉2 𝑇2
𝑉2
∴ 𝑇2 = 𝑇1 .
𝑉1
380 × 0.11
𝑇2 =
0.15
𝑇2 = 278.66 𝐾
(ii)
𝑊 = 𝑃(𝑉2 − 𝑉1 )
= 1.5 × 105 (0.11 − 0.15)
= −6000 𝐽
(iii)
𝑃1 𝑉1 = 𝑚𝑅𝑇1
𝑃1 𝑉1
𝑚=
𝑅𝑇1

Click :Youtube Channel | Free Material | Download App

 1.5 × 105 × 0.15
𝑚=
0.287 × 380 × 107
𝑚 = 0.2 𝑘𝑔

∆𝑈 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 )
= 0.2 × 0.718 × (278.66 − 380)
= −14.55 𝑘𝐽

(c) 0.35𝒎𝟑 of gas at 10 bar, 157°C expends the adiabatically to 4 bar. It is then compressed
isothermally to the its original volume. Find five temperature and pressure of gas.
Cp =0.996 kJ/kg K , C v= 0.703 kJ/kg K.
Ans :
→ Given data :
𝑉1 = 0.35 𝑚 3 = 𝑉3
𝑃1 = 10 𝑏𝑎𝑟
𝑃2 = 4 𝑏𝑎𝑟
𝑇2 = 430 𝐾 = 157°𝐶
𝐶𝑃 = 0.996 𝑘𝐽/𝑘𝑔 𝐾
𝐶𝑣 = 0.703 𝑘𝐽/𝑘𝑔 𝐾

→
𝐶𝑝
𝑟=
𝐶𝑣
0.996
=
0.703
= 1.4
→ For adiabatically process 𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝛾 𝛾
∴ 𝑃1 𝑉1 = 𝑃2 𝑉2
𝑉2 𝛾 𝑃1
( ) =
𝑉1 𝑃2
1
𝑉2 𝑃1 𝛾
=( )
𝑉1 𝑃2
1
𝑃 𝛾
∴ 𝑉2 = ( 1) × 𝑉1
𝑃2
1
10 1.4
𝑉2 = ( ) × 0.35
4
𝑉1 = 0.875 𝑚 3
𝑇 𝑉 𝛾−1
→ ( 1) = ( 2)
𝑇2 𝑉1

Click :Youtube Channel | Free Material | Download App

 430 0.673 1.4−1
=( )
𝑇2 0.35
430
𝑇2 =
1.3
𝑇3 = 330.77 𝐾
→ Here for isothermal process 𝑇2 = 𝑇3
∴ 𝑇2 = 330.77 𝐾
→ For isothermal process 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃2 𝑉2 = 𝑃3 𝑉3
𝑃2 𝑉2
𝑃3 =
𝑉3
(4 × 0.673)
𝑃3 =
0.35
𝑃3 = 7.69 𝑏𝑎𝑟

Q-3 : (a) State the need of couplings along with the purposes they serve in machineries.
Ans : couple in is a device used to connect two shafts together at the ends for transmitting power. shafts
are usually available up to 7 m in length due to difficulty in fabrication and transportation. Long shafts
is obtained by joining to or more shafts by coupling.
Shafts coupling are used in machinery for following purpose :
1. To connect shafts of units that are manufactured separately such as motor and generator
and to provide for disconnection for repairs and alteration.
2. To joint shafts having slide misalignment.
3. To reduce the transmission of shock loud from one shaft to another.
4. To provide protection against overloads.
5. To alter the vibration characteristic of rotating units.

(b) Draw a schematic diagram of vapour compression refrigeration system State the
function of each main component.
Ans :

Click :Youtube Channel | Free Material | Download App

 Function of main parts :
1. Compressor : To remove the vapour from the evaporator and increase its pressure and
temperature upto it can be condensed in the condenser. pressure of refrigerant coming from
compressor should be such that the saturation temperature of vapour higher than the
temperature of cooling medium in condenser. so that high pressure in the condenser.
2. The function of condenser is to facilitate a heat transfer surface through which heat transfer
take place from the hot refrigerant vapour to the condensing medium. In domestic
refrigerator condensing medium is atmospheric air.
3. Expansion valve or device : the function of expansion valve is to meter the proper amount
of liquid refrigerant and reduce pressure of liquid refrigerant entering the evaporator. Hence
liquid will vaporize in the evaporator and the desired low temperature and absorb heat from
the space.
4. Evaporator : an evaporator provides a heat transfer surface through which low temperature
liquid refrigerant can absorb heat from space and vaporized.

(c) steel is initially dry saturated at 9 bar in each the following page determine.
Determine,
1) its dryness fraction if it loses 50 kJ/kg heat at constant pressure,
2) the degree of superheat if it receives 150 kJ/kg heat at constant pressure.

Ans : (i) For dryness fraction ,

→ The enthalpy of dry saturated steam
ℎ𝑔 = ℎ𝑓 + ℎ𝑓𝑔
= 742.6 + 2029 .5
ℎ𝑔 = 2772 .1 𝑘𝐽/𝑘𝑔
→ After so kJ/kg heat lose , ℎ = 2772.1 𝑘𝐽/𝑘𝑔
→ ℎ = ℎ𝑓 + 𝑥ℎ𝑓𝑔
2722.1 = 742.6 + 𝑥(2029 .5)

(ii) for degree of superheat,

→ degree of superheat = (𝑇𝑠𝑢𝑝 − 𝑇𝑠𝑢𝑏 ) = ∆𝑇
→ ℎ𝑔 = 2772.1 𝑘𝐽/𝑘𝑔
→ After receives 150 kJ/kg heat
ℎ = 2922 .1 𝑘𝐽/𝑘𝑔
→ ℎ = ℎ𝑔 + 𝐶∆𝑇
2922.1 = 2772 .1 + 2.1 . ∆𝑇
∆𝑇 = 71.42 °𝐶

Q-3 : (a) Draw a half-sectional view of the sleeve (muff) coupling.

Ans :

Click :Youtube Channel | Free Material | Download App

 (b) Draw a labeled diagram of split air-conditioner. Why does the split air-conditioner
consume more power than a window air-conditioner of same capacity?
Ans :

→ The split air-conditioner consume more power than a window air-conditioner of same
capacity because power consumption is higher as two separate motors are required. One
each for operating blower in indoor unit and in outdoor unit. also power construction
increase due to longer length of copper tubes.

(c) Draw a neat self-explanatory diagram of separating calorimeter. During the test on a
separating calorimeter, 2.2 kg moisture was collected and 16 kg of steam left the
calorimeter. Calculate dryness fraction of steam.
Ans :
→ 𝑚 𝑤 = 2.2 𝑘𝑔
→ 𝑚 𝑠 = 16 𝑘𝑔
𝑚𝑠 16
→ Dryness fraction of steam is,𝑥 = = = 0.879
𝑚𝑠 +𝑚𝑤 16+2.2

Click :Youtube Channel | Free Material | Download App

Q-4 : (a) Classify clutches.

Ans : Clutch

(b) Draw a schematic diagram of reciprocating pump. Show the variation of discharge of
water with crank angle for single-acting reciprocating pump.
Ans :

Click :Youtube Channel | Free Material | Download App

 → The crank moves from 0° to 180° as shown in figure this creates vacuum in the cylinder on
the left side of piston causing the suction valve to open. The liquid enters the cylinder and
fills it.
→ The crank moves from 180° to 360° . This cause is increase of pressure in the left side of
cylinder. The delivery valve open and liquid is forced to delivery pipe.

(c) A Carnot cycle works with isentropic compression ratio of 6 and isothermal expansion
ratio of 2. The volume of air at the beginning of isothermal expansion is 0.2 m³. If T max
and Pmax is limited to 600 K and 20 bar respectively. determine:
(i) minimum pressure during the cycle,
(ii) thermal efficiency of cycle. Show the T-s and p-v diagram.
Ans :

→ Given data :-
𝑉1 = 0.2 𝑚 3
𝑇𝑚𝑎𝑥 = 𝑇1 = 𝑇2 = 600𝐾
𝑃𝑚𝑎𝑥 = 20 𝑏𝑎𝑟

𝑉4 𝑉3
→ Isentropic compression ratio = = 6=
𝑉1 𝑉2

𝑉2 𝑉3
→ Isothermal expansion ratio = =2=
𝑉1 𝑉4

→ 𝑃𝑚𝑖𝑛 = ?
𝜂 =?

Click :Youtube Channel | Free Material | Download App

 (i) For minimum pressure ,
𝑉4
=6
𝑉1
𝑉4 = 6 × 0.2
𝑉4 = 1.2 𝑚 3

𝑉2
=2
𝑉1
𝑉2 = 2 × 0.2
𝑉2 = 0.4 𝑚 3

𝑉3
=6
𝑉2
𝑉3 = 6 × 0.4
𝑉3 = 2.4 𝑚 3

→ For process 1-2

𝑃1 𝑉1 = 𝑃2 𝑉2
20 × 0.2 = 𝑃2 × 0.4
𝑃2 = 10 𝑏𝑎𝑟

→ For process 2-3

𝛾 𝛾
𝑃2 𝑉2 = 𝑃3 𝑉3
𝑉2 𝛾
𝑃3 = 𝑃2 × ( )
𝑉3
0.4 1.4
𝑃3 = 10 × ( )
2.4
𝑃3 = 0.81 𝑏𝑎𝑟

→ Here minimum pressure 𝑃𝑚𝑖𝑛 = 𝑃3

𝑃𝑚𝑖𝑛 = 0.81 𝑏𝑎𝑟

(ii) for thermal efficiency,

→ For process 2-3 ,

𝛾−1 𝛾−1
𝑇2 𝑉2 = 𝑇3 𝑉3
0.4 1.4−1
𝑇3 = 600 × ( )
2.4
𝑇3 = 293.01 𝐾

𝑇𝑚𝑖𝑛
𝜂 = 1−
𝑇𝑚𝑎𝑥
293.01
𝜂 = 1−
600
𝜂 = 0511
𝜂 = 51.1%

Q-4 : (a) State the function of brakes and classify them.

Click :Youtube Channel | Free Material | Download App

Ans : Function : Brakes is a mechanical device which produce frictional resistance against moving
machine member in order to slow down or stop down motion of machine

→ Classification of brakes:

(b) Draw a labeled figure of centrifugal pump indicating all main components.
Ans:

(c) The compression ratio of an ideal air standard diesel cycle is 15. Heat supplied
at constant pressure is 1470 kJ/kg of air: Show the cycle on p-v diagram and
determine the cycle efficiency if inlet conditions are 300 K and 1 bar.
Ans :
→ 𝑟 = 15

Click :Youtube Channel | Free Material | Download App

 𝑄 = 1470 𝑘𝐽/𝑘𝑔
𝑇1 = 300𝐾
𝑝 = 1 𝑏𝑎𝑟
𝛾 = 1.4
𝐶𝑝 = 1.005 𝑘𝐽/𝑘𝑔
→ 𝑇2 = 𝑇1 𝑟 𝛾−1
𝑇2 = 300 × 15𝛾−1
𝑇2 = 886.25 𝐾

→ 𝑄𝑠 = 𝐶𝑝 (𝑇3 − 𝑇2 )
𝑄𝑠
= 𝑇3 − 𝑇2
𝐶𝑝
𝑄𝑠
𝑇3 = + 𝑇2
𝐶𝑝
1470
𝑇3 = + 886.25
1.005
𝑇3 = 2348.93 𝐾

𝑇3
→ 𝜌=
𝑇2
2348.93
𝜌=
886.25
𝜌 = 2.65

→ The cycle efficiency is

1 𝜌𝛾 − 1
𝜂 = 1 − 𝛾−1 [ ]
𝑟 𝛾(𝜌 − 1)
1 (2.65) 1.4 − 1
𝜂 =1− [ ]
(1.5) 1.4−1 1.4(2.65 − 1)
1 2.9133
𝜂 =1− [ ]
2.9541 2.31
𝜂 = 1 − 0.4269
𝜂 = 0.5730

→ P-V Diagram :-

𝑃2 = 𝑃3 2 3

𝑃4 4

𝑃1 1

𝑉2 𝑉3 𝑉1 = 𝑉4

Click :Youtube Channel | Free Material | Download App

Q-5 : (a) Enlist various belt drives. Name any three belt materials.
Ans : List of belt drives
• Open belt drive
• cross belt drive
• Quarter turn belt drive
• Belt drive with idler pulley
• Fast and loose pulley drive fast
• Stepped or cone pulley drive
• Compound belt drive.
Materials used for belts :
1. Leather
2. Rubber
3. Canvas
4. Balata

(b) Outline the technical meaning of following terms used in air compressor.
Ans :
1. Single acting compressor : single acting compressor compresses air during only one
of the distance movement unlike a double acting compressor
that compresses air during both strokes.
2. Single stage compressor : single stage compressor compresses air in one step within a
single cylinder suitable for moderate pressure application.
3. Pressure ratio : It is the ratio of absolute discharge pressure to the absolute inlet
Pressure.
4. Swept volume : swept volume in an air compressor is the total air volume displaced by
the piston in one full stroke cycle.
5. Volumetric efficiency : it is the ratio of actual volume of air taken in the cylinder to the
swept volume of the compressor.

(c) Indicated power of a 6-cylinder 4-stroke engine is 150 kW at an average piston 07 speed
of 300 m/min. Stroke to bore ratio is 1.25. If mean effective pressure is 650 kN/m²,
determine crankshaft speed.
What is a flywheel and what is its function in IC engine?
Ans :
𝐿
→ = 1.25
𝑑
𝐿 = 1.25𝑑
𝐼. 𝑃 = 150𝐾𝑊
𝐾𝑁
𝑃𝑚 = 650 2
𝑚
𝑉𝜌 = 300 𝑚/𝑚𝑖𝑛
No of cylinder = 6
→ 𝑉𝑝 = 2𝐿𝑁
𝑉𝜌 300
𝑁= =
2𝐿 2 × 1.25𝑑
120
𝑁=
𝑑
𝑃 𝐿𝐴𝑁
→ 𝐼. 𝑃 = 𝑚 × 𝑁𝑜 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
60000×2

Click :Youtube Channel | Free Material | Download App

 𝑃𝑚 𝐿𝐴𝑁 × 6
=
60000 × 2

𝜋𝑑 2
𝑃𝑚 × 𝐿 × ( )×𝑁 ×6
4
=
60000 × 2

𝑃𝑚 × 𝐿 × 𝜋 × 𝑑 2 × 𝑁
=
20000 × 4

𝑃𝑚 × 1.25 × 𝜋 × 𝑑 2 × 120
𝐼. 𝑃 =
4 × 20000

650 × 1.25 × 3.14 × 𝑑 2 × 120 × 103

150 =
80000

150 = 3826 .8 𝑑 2

𝑑 2 = 0.039

𝑑 2 = 0.19𝑚

→ Crankshaft speed is,

120
𝑁=
𝑑

120
𝑁=
0.19

𝑁 = 631 𝑟𝑝𝑚

→ Flywheel is a heavy wheel mounted on the crankshaft of the engine. It minimizes cyclic
variation in speed by storing the energy during power stroke and same is released
during other stroke.

Q-5 : (a) Match the following:

Ans :

A B-Ans
Spiral gears Non -Parallel non-intersecting shafts
Bevel gears Intersecting shafts
Worm gears Shaft axes at right angle &non-intersecting
Helical gears Parallel shafts

(b) Define with regard to compressors: mean effective pressure, indicated power, 04
brake power, mechanical efficiency
Ans :
1. Mean effective pressure (P m) :-
For compressor mean effective pressure can be calculated by workdone per cycle.

Click :Youtube Channel | Free Material | Download App

 𝑤𝑜𝑟𝑘𝑑𝑜𝑛𝑒/𝑐𝑦𝑐𝑙𝑒
Pm =
𝐿𝐴

Where , LA = swept volume m³/cycle

2. Indicated power :-
The power required to compress the air is called the indicated power or cycle Power or
air power.

𝑤𝑜𝑟𝑘𝑑𝑜𝑛𝑒/𝑚𝑖𝑛
I.p.. = 𝐾𝑊
60×1000

3. Brake power (B.P.)

The power required to drive the compressor is called the brake power or shaft power of
compressor.
2𝜋𝑁𝑇
B.P. =
60000

4. Mechanical efficiency (𝜂𝑚 ) :

The ratio of indicated power to the break power of compressor is called mechanical
efficiency.
𝐼.𝑃.
𝜂𝑚 =
𝐵.𝑃.

(C) A 4-cylinder 4-stroke petrol engine has a bore 60 mm and a stroke of 90 mm. The rated
speed is 2800 rpm and torque is 55 N-m Fuel consumption is 6.75 liter/hr. Specific
gravity of petrol is 0.76 and calorific value is 44200 kJ/kg. Calculate brake power, brake
mean effective pressure, brake thermal efficiency and brake specific fuel consumption.
Ans :
→ 𝑑 = 60𝑚𝑚 = 60 × 10 −3 𝑚
𝑙 = 90𝑚𝑚 = 90 × 10 −3 𝑚
𝑁 = 2800𝑟𝑝𝑚
𝑇 = 55 𝑁𝑚
𝑙𝑖𝑡𝑒𝑟
𝑓𝑢𝑒𝑙 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 = 6.75
ℎ𝑟
𝜌𝑓 = 0.76
𝐶. 𝑉 = 44200 𝐾𝐽/𝐾𝑔

→ 𝑚𝑓 = 𝑓𝑢𝑒𝑙 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 × 𝜌𝑓

= 6.75 × 0.76

= 5.13 𝐾𝑔/ℎ

→ The brake power is,

2𝜋𝑁𝑇
𝐵. 𝑃 =
60000

2 × 3.14 × 2800 × 55
=
60000

= 16.11 𝐾𝑤

Click :Youtube Channel | Free Material | Download App

→ The Brake mean effective pressure is,

𝑃𝑚𝑏 𝐿𝐴𝑁
𝐵. 𝑃 = × 𝑁𝑜 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
60000

𝑃𝑚𝑏 𝐿𝐴𝑁
𝐵. 𝑃 = ×4
60000 × 2

𝐵. 𝑃 × 60000 × 2
𝑃𝑚𝑏 =
(𝐿𝐴𝑁 × 4)

𝐵. 𝑃 × 60000 × 2
𝑃𝑚𝑏 = 𝜋
𝐿 × 𝑑2 × 𝑁 × 4
4
𝐵. 𝑃 × 60000 × 2 × 4
𝑃𝑚𝑏 =
𝐿 × 𝜋 × 𝑑2 × 𝑁 × 4

16.11 × 60000 × 2
𝑃𝑚𝑏 =
90 × 10 −3 × 3.14 × (60) 2 × 10−6 × 2800

𝑃𝑚𝑏 = 6.786 × 105 𝑃𝑎

𝑃𝑚𝑏 = 678.6 𝐾𝑝𝑎

→ Breake thermal efficiency is

𝐵. 𝑃
𝜂𝑏𝑡 =
𝑚𝑓 × 𝐶. 𝑉

16 .11
=
5.13
( ) × 44200
3600

16.11 × 3600
=
5.11 × 44200

= 0.2567

= 25.67%

→ Breake specific fuel consumption is ,

𝑚𝑓
𝐵𝑆𝐹𝐶 =
𝐵. 𝑃

5.13
=
16.11

𝐾𝑔
0.3171
𝐾𝑤

*****

Click :Youtube Channel | Free Material | Download App