POWER PLANT ENGINEERING 3

42 PSP eBook | Power Plant Engineering 3

h2 = hg at 45C
h2 = 204.87 kJ/kg.

h3 = hf at 45C
h3 = 79.71 kJ/kg.

h4 = hf at 45C
h4 = 79.71 kJ/kg.

(i) COPref = Q = (h1 − h4 )
W (h2 − h1 )

= 174.62 − 79.71 = 94.91
204.87 −174.62 30.25

= 3.14

Example 2.3

A refrigeration machine operates between a condenser temperature of 45C and a refrigerant
temperature of -15C using R12 refrigerant. Determine the actual COPref for the refrigerant.

Solution:
The modification continues the evaporation process until point 1 becomes dry saturation.

PSP eBook | Power Plant Engineering 3 43

h1 = hg at -15C
h1 = 180.97 kJ/kg

h3 = h4 = hf at 45C
h3 = h4 = 79.71 kJ/kg

From the diagram, s1=s2 and s2 is in the superheated zone.

S1 = Sg at -15C
S1 = 0.7051 kJ/kgK
S2 = Sg at 45C is at superheated vapor between 0 to 15K.

From the interpolation:

h2 = 204.87 +  0.7051 − 0.6811  (216.74 − 204.87)
 0.7175 − 0.6811

h2 = 212.70 kJ/kg.

(i) COPref = Q = (h1 − h4 )
W (h2 − h1 )

= 180.97 − 79.71 = 101.26
212.70 −180.97 31.73

= 3.191

Example 2.4

A refrigeration machine operates between a condenser temperature of 45C and a refrigerant
temperature of -15C using R12 refrigerant. Determine the actual COPref for the refrigerant:

Solution:
Modifications whereby the liquid at condition 3 is allowed to undergo sub-cooling to 5C
before entering the expansion valve.

44 PSP eBook | Power Plant Engineering 3

h1 = hg at -15C

h1 = 180.97 kJ/kg

h3 = h4 = hf at 40C
h3 = h4 = 74.59 kJ/kg

From the diagram, s1=s2 and s2 is in the superheated zone.

S1 = Sg at -15C
S1 = 0.7051 kJ/kgK
S2 = Sg at 45C is at superheated vapor between 0 to 15K.

From the interpolation:

h2 = 204.87 +  0.7051 − 0.6811  (216.74 − 204.87)
 0.7175 − 0.6811

h2 = 212.70 kJ/kg.

(i) COPref = Q = (h1 − h4 )
W (h2 − h1 )

= 180.97 − 74.59 = 106.38
212.70 −180.97 31.73

= 3.35

PSP eBook | Power Plant Engineering 3 45

Tutorial Refrigerant System

Question 1
An ammonia vapor – compression refrigerator operates between an evaporator pressure of
2.077 bars and a condenser pressure of 12.37 bars. The vapor is dry saturated at entry to the
compressor. There is no under cooling in the condenser and isentropic compression may be
assumed. Calculate:

(i) COPref
(ii) Refrigerating effect per unit mass

Question 2
A heat pump uses ammonia as the refrigerant operates between saturation temperature of
6 and 38C. The refrigerant is compressed isentropically from dry saturation and there is 6K
of under cooling in the condenser. Calculate:

(i) COPref
(ii) The mass flow of refrigerant per kilowatt power input
The heat available per kilowatt power input.

Question 3
An ammonia refrigerator operates between evaporating and condensing temperature of -16
and 50C. The vapor is dry saturated at the compressor inlet, and there is no under cooling of
the condensate. Calculate:

(i) The refrigerating effects
(ii) The mass flow rate per kilowatt of refrigeration capacity.
(iii) The power input per kilowatt of refrigerating capacity
(iv) COPref

46 PSP eBook | Power Plant Engineering 3

Question 4
In a refrigerating plant using R12 the vapor leaves the evaporator dry saturated at 1.826 bars
and is compressed to 7.449 bars. The temperature of the vapor leaving the compressor is
45C. The liquid leaves the condenser at 25C and is throttled to the evaporator pressure.
Calculate:

(i) The refrigerating effects
(ii) The specific work input
(iii) The COPref

Question 5
A vapor compression refrigerator using R 12 works between the temperature limits of -20C
and 25C. The refrigerant leaves the compressor in dry saturated condition. If the liquid
refrigeration is undercooled to 20C before entering the throttle valve, determine:

(i) Work required to drive the compressor
(ii) Refrigerating effect
(iii) COPref

Question 6
In a vapor compression refrigerating cycle using Freon 12 as refrigerant, the temperature in
the evaporator is -10C and the pressure in the condenser is 13.66 bars. The vapor is dry
saturated at the inlet to the compressor and the liquid which is under cooled by 10C enters
the expansion valve of the cycle.

(i) Sketch the cycle with the aid of temperature – entropy (T-s)
(ii) The rate of flow of refrigerant in kg/s if 15kW of heat to be rejected in the cycle.
(iii) The power input to the compressor
(iv) The coefficient of performance

3CHAPTER PSP eBook | Power Plant Engineering 3 47

GAS TURBINE

Objectives:
At the end of this chapter, students should be able to:

• Define Gas Turbine system and its main components
• Sketch the block diagram and temperature-entropy (T-s) diagram.
• Perform calculation related to Gas Turbine.

48 PSP eBook | Power Plant Engineering 3

3.0 Gas Turbine

Gas turbine power plant can be divided into three categories:
1) Open cycle
2) Closed cycle
3) Combined cycle

Figure 3.1 Open Cycle

Figure 3.2 Closed Cycle

PSP eBook | Power Plant Engineering 3 49

Figure 3.3 Combined Cycle

3.1 Main Components of Gas Turbine
3.1.1 Compressor

• The air compressor used in gas turbines is of rotary type mainly axial flow turbines.
• It draws air from the atmosphere and compressed to the required pressure.
• This compressed air is then transferred to the combustion chamber.

3.1.2 Combustion Chamber

• The compressed air from the air compressor is drawn to combustion chamber.
• The fuel is injected to the air and then ignited in the combustion chamber.
• It increases the pressure and temperature of the air instantaneously.

3.1.3 Turbine

• The high pressure and temperature air is expanded in the turbine.
• Turbine is also of rotary type.
• During the expansion, the heat energy in the gas is converted into mechanical energy.
• This mechanical energy is again converted into electrical energy by using generator.

50 PSP eBook | Power Plant Engineering 3

3.2 Working of Open Cycle Gas Turbine

• The most basic gas turbine unit is the one operating on the open cycle in which a rotary
compressor and a turbine are mounted on a common shaft as shown above.

• Air is drawn from the atmosphere into the compressor and compressed to pressure
of 300 to 400 kN/m2.

• The compressed air is then entered into the combustion chamber where the energy is
supplied by spraying fuel into the air and ignited by hot gasses.

• The hot gasses expand through the turbines to produce the mechanical power.
• Then the burned gasses are exhausted to the atmosphere.
• Then the fresh air is drawn into the compressor for the next cycle.

3.3 Brayton Cycle or Joule Cycle

• The Brayton cycle is the theoretical cycle for gas turbines.
• It consists of two reversible adiabatic processes and two constant pressure processes.
• This cycle is also called constant pressure cycle.
• Process 1 - 2s: represents the ideal isentropic process between the same pressure P1

and P2.
• Process 3 - 4s: represents the ideal isentropic expansion process between the pressure

P2 and P1.
• Line 1 - 2: represents irreversible adiabatic compression.
• Line 2 - 3: represents constant pressure heat supply in the combustion chamber.
• Line 3 - 4: represents irreversible adiabatic expansion.

PSP eBook | Power Plant Engineering 3 51

Figure 3.4 T-s diagram for Brayton Cycle

Applying the flow equation to each part of the cycle, we have the following for unit mass:

• For compressor ( )= Cpair T2' − T1 (1)
Work input (2)
( )= Cpgas T3 − T2'
• For combustion chamber
Heat supplied

• For turbine ( )= Cpgas T3 − T4' (3)
Work output

• Network output ( ) ( )= Cpgas T3 − T4' − Cpair T2' − T1

• Thermal efficiency = Network output / Heat supplied

( ) ( )C pgas T3 − T4' − C pudara T2' − T1
( )= C pgas T3 − T2'

(4)

52 PSP eBook | Power Plant Engineering 3

• Compressor isentropic efficiency, ( )= C pair T2 − T1
( )C pair T2' − T1

• Turbine isentropic efficiency, ( )= C pgas T3 − T4'
( )C pgas T3 − T4

3.4 Open Cycle Gas Turbine

Figure 3.5 Open Cycle for Brayton Cycle

3.4.1 T-s Diagram for Basic Turbine Gas

Figure 3.6 T-s Diagram for Brayton Cycle

PSP eBook | Power Plant Engineering 3 53

3.4.2 Power Turbine

• In the example below, the turbine is arranged to drive the compressor and to develop
net work.

• It is sometimes more convenient to have two separate turbines, one of which drives
the compressor while the other provides the power output.

• The first, or high pressure (HP) turbine, is known as the compressor turbine.
• The second, or low pressure (LP) turbine, is called the power turbine.

3.5 Two Stage Turbine Gas

Figure 3.7 Two Stage Turbine Gas

Applying the flow equation to each part of the cycle, we have the following for unit mass:

• For compressor ( )= Cpair T2' − T1
Work input

• For combustion chamber = ( )Cpgas T3' − T2'
Heat supplied

• For turbine ( )Cpair T3 − T4'
Work output
=

54 PSP eBook | Power Plant Engineering 3

Net Work output = ( )Cpair T4' − T5'

• Thermal efficiency = Network output / Heat supplied

= ( )C pgas T4' − T5'

• Compressor isentropic efficiency, = ( )C pgas T3 − T2'

• HPT isentropic efficiency, = ( )C pair T2 − T1
( )C pair T2' − T1
• LPT isentropic efficiency =
( )C pgas T3 − T4'
( )C pgas T3 − T4

( )C pgas T4' − T5'
( )C pgas T4' − T5

3.6 Reheat

• The expansion process is very frequently performed in two separate turbine stages.
• The HP turbine driving the compressor and the LP turbine providing the useful power

output.
• The work output of the LP turbine can be increased by raising the temperature at inlet to

this stage.
• This can be done by placing a second combustion chamber between the two turbines

stages to heat the gases leaving the HP turbine.

Figure 3.8 Turbine Gas with Reheater

PSP eBook | Power Plant Engineering 3 55

Applying the flow equation to each part of the cycle, we have the following for unit mass:

• For compressor ( )= Cpair T2' − T1
Work input

• For combustion chamber ( ) ( )= Cpgas T3 − T2' + Cpgas T5 − T4'
Heat supplied
( )= Cpgas T3 − T4'
• For turbine ( )= Cpgas T5 − T6'
Work output

Network output

• Thermal efficiency = Network output / Heat supplied

Compressor isentropic efficiency, ( )C pgas T5 − T6'
• HPT isentropic efficiency,
• LPT isentropic efficiency ( ) ( )= C pgas T3 − T2' + C pgas T5 − T4'

( )= C pair T2 − T1
( )C pair T2' − T1

(( ))=
C pgas T3 − T4'
C pgas T3 − T4

( )C pgas T5 − T6'
( )= C pgas T5 − T6

56 PSP eBook | Power Plant Engineering 3

3.7 Intercooler (Turbine Gas)

• Work Input Figure 3.9 Turbine Gas with Intercooler
• Heat Supplied
= ( ′ − ) + ( ′ − )

( )= C pgas T5 − T4'

Example 3.1

A gas turbine plant operates with a pressure ratio of 9:1. The plant consists of two units of
compressor (with complete and ideal intercooling), a combustion chamber and a turbine. Air
enters the compressor at 30°C with a flow rate of 20kg/s. the maximum temperature in the
cycle is 600°C. Sketch the layout and T-s diagrams of the plant.
Calculate the:
a) Plant efficiency.
b) Work ratio.
c) Net output power in Kw
Neglect the mass of fuel.

Take: γ = 1.40 and cp = 1.005kJ/kg.K for air
γ = 1.33 and cp = 1.11 kJ/kg.K for gases

PSP eBook | Power Plant Engineering 3 57

Solution

a. Diagram sketch
i. Component arrangement diagram.

S

b. Calculate:

Given:

P4 = P5 = 9
P1 P6 1

Intercooler process:

P2 = P4 = 93
P1 P3 1=1

T1 = T3 = 30°C = 303 K

T2 = T4

T5 = 600°C = 873 K

m = 20 kg/s

γa = 1.40 and cpa = 1.005kJ/kg.K

γg = 1.33 and cpg = 1.11 kJ/kg.K

58 PSP eBook | Power Plant Engineering 3

i. Cycle efficiency.

ηTH = Net Work
Heat supplied

η =TH W − WTURBINE COMPRESSOR

Q SUPPLIED

T2 PP21  −1 
T1
= 



PP21 γ −1γ

T2 = T1  



 T2 = 303  3 1.4 −11.4 K

T2 = 414.72 K

T5 PP65  −1 
T6
= 



T6 = T5 γ

P5 γ −1
P6
 
 
 

 T6 = 873 K
9 0.331.33

T6 873 K
= 1.725

T6 = 506.11 K

For 1 kg of air,

Total work done by compressor, Wp:

 WP = 2  Cpa  (T2 − T1)
 WP = 2 1.005  (414.72 − 303) kJkg

 WP = 224.58 kJkg

Total work done by turbine, WT:

 WT = Cpg  (T5 − T6 )
 WT = 1.11 (873 − 506.11) kJkg

 WT = 407.25 kJkg

PSP eBook | Power Plant Engineering 3 59

Heat supplied, QSUPPLIED:

( ) Q SUPPLIED = Cpg  T5 − T4
 Q SUPPLIED = 1.11 (873 − 414.72) kJkg

 Q SEUPPLIED = 508.69 kJkg

η =TH W − WTURBINECOMPRESSOR

Q SUPPLIED

ηTH = 407.25 − 224.58
508.69

ηTH = 0.359 @ 35.9%

ii. Work ratio.

Work ratio = Net Work
Gross Work

Nisbah kerja = W − WTURBINECOMPRESSOR

WTURBINE

Nisbah kerja = 407.25 − 224.58
407.25

Nisbah kerja = 0.4485

iii. Net Work in unit kW.

 WNET = WTURBINE − WCOMPRESSOR
 WNET = 407.25 − 224.58 kJkg
 WNET = 182.67 kJkg
 WNET = 182.67  20 kJkg  kg s = kJs = kW
 WNET = 3653.4 kW

60 PSP eBook | Power Plant Engineering 3

Tutorial Turbine Gas

Question 1
A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7. The
minimum and maximum temperatures in the cycle are 300 and 1150K. Assume an isentropic
efficiency of 75% for the compressor and 82% for the turbine. Determine:

(i) The air temperature at the turbine exit.
(ii) The network output.
(iii) Heat supplied
(iv) The thermal efficiency
(v) Work ratio

Question 2
A gas turbine power plant is operating on an ideal Brayton cycle has a pressure ratio of 8. The
gas temperature is 300K at the compressor inlet and 1300K at the turbine inlet. Assuming a
compressor efficiency of 80% and a turbine efficiency of 85%. Determine:

(i) The gas temperature at the exits of the compressor and the turbine.
(ii) The back work ratio.
(iii) The thermal efficiency.

Question 3
A gas turbine unit takes in air at 17C and 1.01 bars and the pressure ratio is 8/1. The
compressor is driven by the HP turbine and the LP turbine drives a separate power shaft. The
isentropic efficiencies of the compressor, and the HP and LP turbines are 0.8, 0.85, and 0.83
respectively. The maximum temperature is 627C. Calculate:

(i) The pressure and temperature of the gases entering the power turbine.
(ii) The net power developed by the unit per kg/s mass flow rate.
(iii) The work ratio
(iv) Heat supplied
(v) Cycle efficiency

PSP eBook | Power Plant Engineering 3 61

Question 4
A gas turbine has an overall pressure ratio of 5 and a maximum cycle temperature of 550C.
The turbine drives the compressor and an electric generator, the mechanical efficiency of the
drive being 97%. The ambient temperature is 20C and air enters the compressor at a rate of
15 kg/s; the isentropic efficiencies of the compressor and turbine are 80% and 83%.
Neglecting changes in kinetic energy, the mass flow rate of fuel, and all pressure losses.
Calculate:

(i) The power output
(ii) The cycle efficiency
(iii) The work ratio
For problem above Cp and  may be taken as 1.005 kJ/kg K and 1.4 for air, and as 1.15 kJ/kg
K and 1.333 for combustion and expansion processes.

Question 5
In a marine gas turbine unit, a HP stage turbine drives the compressor, and a LP stage turbine
drives the propeller through suitable gearing. The overall pressure ratio is 4/1, the mass flow
rate is 60 kg/s, the maximum temperature is 650C, and the air intake conditions are 1.01 bar
and 25C. The isentropic efficiencies of the compressor, HP turbine, and LP turbine are 0.8,
0.83, and 0.85 respectively, and the mechanical efficiency of both shafts is 98%. Neglecting
kinetic energy changes, and the pressure loss in combustion, Calculate:

(i) The pressure between turbine stages
(ii) The cycle efficiency
(iii) The shaft power.
For problem above Cp and  may be taken as 1.005 kJ/kg K and 1.4 for air, and as 1.15 kJ/kg K
and 1.333 for combustion and expansion processes.

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Question 6

In a gas turbine plant working on the Brayton cycle, the air at the inlet is at 15C, 0.1MPa. The

pressure ratio is 5:1 and the maximum temperature of gasses supplied to high pressure

turbines is 650C. The gases expand in high pressure turbine are then reheated to the initial

temperature before expanding in low pressure turbine. The efficiency of compressor and two

turbines are 80% and 85% respectively. If the mass flow rate of air is 5 kg/sec. Determine:

(i) maximum power that can be obtained from this plant

(ii) heat supplied

(iii) overall efficiency

(iv) work ratio

Assume: air = 1.4

gas = 1.333
Cpair =
1.005 kJ/kg K

Cpgas = 1.15 kJ/kg K

4CHAPTER PSP eBook | Power Plant Engineering 3 63

INTERNAL COMBUSTION ENGINE

Objectives:
At the end of this chapter, students should be able to:

• Define Internal Combustion Engine
• State the difference between Four Stroke Engine and Two Stroke Engine.
• Perform calculation related to Internal Combustion Engine.

64 PSP eBook | Power Plant Engineering 3

4.0 Introduction

Internal-combustion engine, any of a group of devices in which the reactants
of combustion (oxidizer and fuel) and the products of combustion serve as the working fluids
of the engine. Such an engine gains its energy from heat released during the combustion of
the no reacted working fluids, the oxidizer-fuel mixture.

4.1 The Process of Four Stroke Engine Cycle

4.1.1 Suction/Intake Stroke

In this stroke, the piston moves from TDC to BDC [(Top Dead Centre – the farthest position of
piston to the crankshaft) to (Bottom Dead Centre – the nearest position of piston to the
crankshaft)].
The piston moves downward sucking the air-fuel mixture from the intake valve.

Key points:
Intake valve – open
Exhaust valve – closed
Crankshaft rotation – 180°

4.1.2 Compression Stroke

Here, the piston moves from BDC to TDC compressing the air-fuel mixture. The momentum
of flywheel helps the piston to move upwards.

Key points:
Intake valve – closed
Exhaust valve – closed
Crankshaft rotation – 180° (total = 360°)

PSP eBook | Power Plant Engineering 3 65

Figure 4.1 Four Stroke Cycle Engine

4.1.3 Power Stroke

The second rotation of crankshaft has begun as it completes one full rotation during
compression stroke. Power stroke begins with the expansion of air-fuel mixture ignited with
the help of spark plug. Here, the piston moves from TDC to BDC. This stroke produces
mechanical work to rotate the crankshaft.
Key points:
Intake valve – closed
Exhaust valve – closed
Crankshaft rotation – 180° (total = 540°)

66 PSP eBook | Power Plant Engineering 3

4.1.4 Exhaust Stroke

Again, the momentum of flywheel moves the piston up from BDC to TDC thereby driving the
exhaust gases outside through the exhaust valve.
Key points:
Intake valve – closed
Exhaust valve – open
Crankshaft rotation – 180° (total = 720°)

Figure 4.2 Proses of Four Stroke Cycle

PSP eBook | Power Plant Engineering 3 67

4.2 Working of 2 Stroke Engine

4.2.1 Down Stroke

First the piston is moved downside from TDC to BDC to let the fresh air enter the combustion
chamber. The fresh air-fuel mixture gets into the combustion chamber through crankcase.
Crankshaft rotation – 180°.

4.2.2 Up Stroke

Here happens all the magic. The piston is pushed up from BDC to TDC. The fuel-air mixture
gets compressed & spark plug ignites the mixture. As the mixture gets expanded, the piston
moves down. During up stroke, the inlet port is opened. While this inlet port is opened, the
mixture gets sucked inside the crankcase. When the mixture is pushed up into the combustion
chamber during the previous up stroke, a partial vacuum is created as no mixture is left behind
in the crankcase. This mixture is ready to go into the combustion chamber during down stroke
but remains in the crankcase until the piston goes up till TDC. Crankshaft rotation – 360°.

Two strokes get completed along with one power cycle.

From the 2nd down stroke onwards the exhaust gases get expelled out from one side while a
fresh mixture enters the combustion chamber simultaneously due to partial vacuum created
in the combustion chamber after removal of exhaust gases. This is the beauty of the engine.
Both things happen at the same time which makes it a 2-stroke engine.

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Figure 4.3 Two Stroke Engine Cycle

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4.3 Difference Between 4 Stroke and 2 Stroke Engine

4 STROKE ENGINE 2 STROKE ENGINE

4 stroke engine completes 2 rotations of 2 stroke engine completes 1 rotation of

crankshaft after completing one cycle. crankshaft after completing one cycle.

Power is produced once every 4 strokes of Power is produced once during 2 strokes of

the piston. the piston.

Engine design is a bit complicated due to 2 stroke engine has ports which makes its
valve mechanism which is operated through design simpler.
gear & chain mechanism.

No need of adding oil or lubricant to fuel. Addition of oil is required.

Top side of the piston is flat. A bump or protuberance may be needed on
top side of piston.

Mixture remains only in the combustion Air-fuel mixture enters through inlet port &
chamber. travels to combustion chamber passing
through crankcase.

4 stroke engines are heavier. 2 stroke engines are lighter comparatively.

4 stroke engines make less noise. 2 stroke engines are louder comparatively.

4.3.1 Advantages of 4 Stroke Engine

i. More torque: In general, 4 stroke engines always make extra torque than 2 stroke
engines at low RPM. Although 2 stroked ones give higher torque at higher RPM, but it
has a lot to do with fuel efficiency.

ii. More fuel efficiency: 4 stroke engines have greater fuel efficiency than 2 stroke ones
because fuel is consumed once every 4 strokes

iii. Less pollution: As power is generated once every 4 strokes & also as no oil or lubricant
is added to the fuel; 4 stroke engine produces less pollution.

70 PSP eBook | Power Plant Engineering 3

iv. More durability: We all know that more the engine runs, quicker it wears out. 2 stroke
engines are designed for high RPM. If an engine can go for 10000 rpm’s before it wears
out; a 4-stroke engine with 100 rpm will run for 100 minutes than the other 2 stroke
engine which has a higher rpm of 500 & will run for only 20 minutes.

v. No extra addition of oil: Only the moving parts need lubrication intermediately. No
extra oil or lubricant is added to fuel.

4.3.2 Disadvantages of 4 Stroke Engine

i. Complicated design: A 4 stroke engine has complex valve mechanisms operated &
controlled by gears & chain. Also, there are many parts to worry about which makes
it harder to troubleshoot.

ii. Less powerful: As power gets delivered once every 2 rotations of crankshaft (4
strokes), hence 4 stroke is less powerful.

iii. Expensive: A four stroke engine has much more parts than 2 stroke engines. So, they
often require repairs which leads to greater expense.

4.3.3 Advantages of 2 Stroke Engine

i. Simple design & construction: It doesn’t have valves. It simply has inlet & outlet ports
which makes it simpler.

ii. More powerful: In 2 stroke engines, every alternate stroke is power stroke unlike 4
stroked one in which power gets delivered once every 4 strokes. This gives a significant
power boost. Also, the acceleration will be higher & power delivery will be uniform
due to same reason.

PSP eBook | Power Plant Engineering 3 71

iii. Position doesn’t matter: 2 stroke engines can work in any position as lubrication is
done through the means of fuel (as the fuel passes by through whole cylinder &
crankcase).

4.3.4 Disadvantages of 2 Stroke Engine

i. Less fuel efficiency: For every alternate power stroke, fuel gets consumed every
alternate stroke. This makes the engine less fuel efficient although it results in uniform
power delivery.

ii. Oil addition could be expensive: Two-stroke engines require a mix of oil in with the
air-fuel mixture to lubricate the crankshaft, connecting rod and cylinder walls. These
oils may empty your pockets.

iii. More pollution: 2 stroke engine produces a lot of pollution. The combustion of oil
added in the mixture creates a lot of smoke which leads to air pollution.

iv. Wastage of fuel: Sometimes the fresh charge which is going to undergo combustion
gets out along with the exhaust gases. This leads to wastage of fuel & also power
delivery of the engine gets effected.

v. Improper combustion: The exhaust gases often get trapped inside the combustion
chamber. This makes the fresh charge impure. Therefore, maximum power doesn’t
get delivered because of improper incomplete combustion.

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4.4 Internal Combustion Engine Formula

4.4.1 Indicated Mean Effective Pressure (IMEP)

• The indicated mean effective pressure of an engine is obtained from the indicator diagram
drawn with the help of an engine indicator.

IMEP =  a   s
l

Pi = Indicated Mean Effective Pressure (kN/m2) or bar
a= Area of the diagram (m2)
l= Length of the diagram (m)
s= Scale of the pressure / Scale of indicator spring (kN/m2)

4.4.2 Indicated Power (I.P)

• The indicated power is the power developed by the engine cylinder.
• It is based on the information obtained from the indicator diagram of the engine.

I.P = IMEP  A  L  N N (four stroke cycle engine)
2  60

I.P = Indicated Power
Pi = Indicated Mean Effective Pressure
A= Area of piston
N= Speed of the engine
n= Number of working strokes per minute
L= Length of stroke

I.P = IMEP  A  L  N N (two stroke cycle engine)
60

PSP eBook | Power Plant Engineering 3 73

4.4.3 Brake Power (BP)

• The brake power is the power available at the crank shaft.
• The brake power of an I.C engine is usually measured by means of brake mechanism.

BP = 2  N  T

BP = T

T = Fxr

BP = Brake Power (kW or W)
(Nm)
T = Torque (N)
(m)
F = Brake Load (r.p.s)
(rad/sec)
r = Length of arm

N = Speed of the engine

= Angle velocity

4.4.4 Brake Mean Effective Pressure (BMEP)

• Brake mean effective pressure, the average (mean) pressure which, if imposed on the
pistons uniformly from the top to the bottom of each power stroke, would produce the
measured brake power output.

BMEP = BP  2
A  L  N N (bar or N/m2)

4.4.5 Friction Power (FP)

• The power, which is lost in overcoming the engine friction, is known as frictional power.
FP = IP – BP

74 PSP eBook | Power Plant Engineering 3

4.4.6 Efficiencies of Internal Combustion Engine

• The efficiency of an engine is defined as the ratio of work done to the energy supplied to
an engine.

4.4.7 Mechanical Efficiency (M)

• It is the ratio of brake power (B.P) to the indicated power (I.P)
M = Brake Power / Indicated Power

BP
= IP

4.4.8 Indicated Thermal Efficiency

• It is the ratio of the heat equivalent to one kW hour to the heat in the fuel per I.P hour.

IT = IP
mf C

IP = Indicated Power (kW or W)
mf = Mass of fuel consumed (kg/s)
C= Calorific value of fuel (kJ/kg)

4.4.9 Brake Thermal Efficiency

• It is the ratio of the heat equivalent to one kW hour to the heat in fuel per BP hour.

• It is also known as overall thermal efficiency of the engine.

BT = BP (kW or W)
BP =
mf C
Brake Power

mf = Mass of fuel consumed (kg/s)
C= Calorific value of fuel (kJ/kg)

PSP eBook | Power Plant Engineering 3 75

4.4.10 Specific Fuel Consumption

• The ratio mf / BP is known as specific fuel consumption (SFC)

SFC = m f (kg/kWh)
BP

4.4.11 Volumetric Efficiency

• It is the ratio of actual volume of charge admitted during the suction stroke at cylinder
to the swept volume of the piston.

V = Volume of charge admitted at cylinder / Swept volume of the piston

V = va
vs

4.4.12 Heat Balance Sheet

• The complete record of heat supplied, and heat rejected during a certain time (say one
minute) by an I.C engine is entered in a tabulated form known as heat balance sheet.

• The following value are required to complete the heat balance sheet of an I:C engine:

4.4.12.a Heat supplied by the fuel

Let mf = mass of fuel supplied (kg/min)
C
= Lower calorific value of the fuel (kJ/kg)
Heat in fuel supplied
= mf x C (kJ/min)

76 PSP eBook | Power Plant Engineering 3

4.4.12.b Heat absorbed in Brake Power

BP = 2  N  T = T

4.4.12.c Heat rejected to the cooling water

• The mass of cooling water, circulating through the cylinder jackets, as well as its inlet

and outlet temperature are measured to determine the heat rejected to the cooling

water.

Let mw = Mass of cooling water supplied (kg/min)

Cw = Specific heat of water (4.18 kJ/kgK)

T1 = Inlet temperature (K)

T2 = Outlet temperature (K)

Then heat rejected to cooling water (kJ/min)

= mW  cW  (T1 − T2 )

4.4.12.d Heat carried away by exhaust gases

• The mass of exhaust gases may be obtained by adding together the mass of fuel supplied
and the mass of air supplied.

• The mass of air supplied may be measured by an orifice or it may be calculated from the
analysis of the exhaust gases.

• The temperature of the exhaust gases is also measured.

Let mg = mass of exhaust gasses produced (kg/min)
Cg = specific heat of exhaust gases (kJ/kgK)
t = rise in temperature (K)

Heat carried away by exhaust gases
= mg x cg x t

PSP eBook | Power Plant Engineering 3 77

4.4.12.e Unaccounted heat

• There is always some loss of heat due to friction, leakage, radiation etc, which cannot be
determined experimentally.

• To complete the heat balance sheet, this loss is obtained by the difference of heat
supplied by fuel and heat absorbed in BP, cooling water and exhaust gases.

Heat balance sheet is prepared as given below:

Particulars Heat In Heat Out
No kJ %

Total heat supplied
1 Heat absorbed in BP
2 Heat rejected to the cooling water
3 Heat carried away by exhaust gases
4 Unaccounted heat

Example 4.1

A 4-stroke 4-cylinder petrol engine has an engine stroke spacing of 92 mm rotating at a speed
of 2500 ppm. At that speed the applied load is 100 N at an arm distance of 340 mm while the
indicated minimum pressure value is 6.75 bar and the indicated power value is 13.68 kW. The
fuel consumption rate of gasoline is 6.15, the fuel used has a comparative gravity value of
0.735 and a low caloric value of 44200 and. Specify the values of the following values:
a. Engine torque
b. Brake power
c. Frictional power
d. Cylinder diameter
e. Piston speed
f. Mechanical efficiency

78 PSP eBook | Power Plant Engineering 3

g. Minimum brake pressure
h. Brake thermal efficiency, ηBT
i. Indicated thermal efficiency

Solution:

a. Torque, T.
T = Fr
T = 100  340 x10−3 Nm
T = 34 Nm

b. Brake power, BP.

BP = T  ω

BP = 34  2π 2500 W
60

BP = 8901 W

BP = 8.901kW

c. Frictional power, FP
FP = IP — BP
FP = 13.68 — 8.901 kW
FP = 4.78 kW

d. Cylinder diameter

IP = PmVsNn
2

IP = PmxAxLxNxn
2

( )13.68x103 =  2500 
6.75x105 x A x 0.092 x  60  x 4

2

( )13.68x103 =  2500 
6.75x105 x A x 0.092 x  60  x4

2

13.68x103 = 5175000A

A = 13.68x103 m2
5175000

A = 2.6435x10 -3 m2

PSP eBook | Power Plant Engineering 3 79

A = πd2
4

d= 4A m
π

d = 2.6435 x10−3 m

d = 0.05141 m @ 51.41 mm

e. Piston speed

VPISTON = 2 x π x N x jarak lejang

VPISTON =2xπx 2500 x 0.092 ms
60

VPISTON = 7.67 m
s

f. Mechanical efficiency

ηMEK = BP
IP

ηMEK = 8.9
13.68

ηMEK = 0.65 @ 65%

g. Minimum brake pressure (Pb).

BP = Tω = Pb x Vs x N x n
2

T = Pb x A x Lx N x n
2

Pb = 2 x T x A x L 1 N x n
x

Pb = 2 x 8.9 x π(0.05141)2 1 2500 kN
0.092 m2
x x x4
4 60

Pb = 2x8.9x31.42 kN
m2

Pb = 559.24 kN @ 5.5924 bar
m2

80 PSP eBook | Power Plant Engineering 3

h. Brake thermal efficiency, ηBP

ηBT = B.P
Q BEKALAN

Q BEKALAN = mf  Nilai Kalorifik Bahanapi

mf = 6.15 10−3  0.735 103 kg
3600 s

mf = 1.2556 10−3 kg
s

Q BEKALAN = 1.2556 10−3  44200 kW

Q BEKALAN = 55.5 kW

ηBT = 8.9
55.5

ηBT = 0.1604 @ 16.04 %

i. Indicated thermal efficiency, ηIP

η IP = I.P
Q BEKALAN

ηBT = 13.68
55.5

ηBT = 0.2327 @ 23.27 %

Example 4.2

A quality governed four-stroke, single-cylinder engine has a bore of 146 mm and stroke of 280
mm. At 475 rev/min and full load on the friction brake is 433 N, and the torque arm is 0.45 m.
The indicator diagram gives a net area of 578 mm2 and the length of 70 mm with a spring
stiffness of 0.815 bar per mm. Calculate:

i. Indicated Power (IP)
ii. Brake Power (BP)
iii. Mechanical efficiency

PSP eBook | Power Plant Engineering 3 81

Solution:

Given:
4 stroke engine, 1 selinder.
d bore = 146 mm = 0.146 m
L stroke = 280 mm = 0.280 m
N = 475 N
 = 433 N
r = 0.45
net area, a = 578 mm2
length, ℓ = 70 mm
spring stiffness, s = 0.815 bar per mm

i. Indicated power, I.P. = Pi ALNn
2

Luasgambarajah tertunjuk  pegas spring  mm2  bar 
= Panjang tapak gambarajah tertunjuk  mm = bar
Pi 
 mm 

 

Pi = 578  0.815 bar
70

Pi = 6.7296 bar @ 6.7296 102 kN
m2

Vs = A L = π (d)2 L m3
4

Vs = π (0.146)2  0.28 m3
4

Vs = 4.688 10-3 m3

I.P. = Pi ALNn
2

6.7296  4.688  10 -3  475  1  kN  m3 kN.m 
60  m2 s kW
I.P. = = = 
2 s


I.P. = 12.49kW

82 PSP eBook | Power Plant Engineering 3

ii. Brake Power, B.P. = 2πNT = ωT

T =Fr
T = 433 0.45 N.m
T = 194.85 N.m

B.P. = 2π  475  194.85 N.m = Watt
60  s

B.P. = 9690 Watt @ 9.69 kW

iii. Mechanical efficiency, Mek.,

ηmech = bp
ip

ηmech = 9.69
12.49

ηmech = 0.776 @ 77.6%

Example 4.3

In an engine test conducted, it was found that the data collected are as follows: -
Test time taken: 30 minutes
Speed: 1750 ppm
Brake Torque: 330 Nm
Fuel Consumption: 9.35 kg
Caloric value of fuel: 42300
Cooling water consumption rate: 483 kg
Temperature between cooling water: 17ºC and 77ºC
Air consumption rate: 182 kg
Exhaust temperature: 486ºC
Atmospheric temperature: 17ºC

Calculate:
i. The Brake Power (BP)
ii. The specific fuel consumption of the inner brake and the indicated thermal efficiency
if the mechanical efficiency is 83%.

PSP eBook | Power Plant Engineering 3 83

iii. Assuming that the specific charge heat for the exhaust gas is 1.25, construct an
energy balance table in kJ/min

Given a specific heat capacity for water = 4.18

Solution:

i. Brake Power, B.P. = T = 2NT

B.P = 2π  1750 x330 W
60

B.P = 60 475.66 W @ 60.475 kW

ii. Specific Fuel Consumption,

. . . = [ ⁄ ℎ = ⁄ . ℎ ]


. . . = 9.35 × 2 ⁄ . ℎ
60.475

. . . = 0.309 ⁄ . ℎ

iii. Indicated power I.P. = PmLANn or ηmek = B.P
2 I.P

ηmek = B.P
I.P

I.P. = B.P.
ηmek

I.P. = 60.475 kW
0.83

I.P. = 72.8 kW

Indicated thermal efficiency, IP,


=

= ×

= 9.35 × 2 42300 [ ⁄ × ⁄ = ]
3600 ×

84 PSP eBook | Power Plant Engineering 3

= 219.725
72.8

= 219.725
= 0.3313 @ 33.13%

Heat Balance

Heat Supply from fuel : Q= m x Calorific value = 219.735 kJ

Distribution :

 Brake Power = B.P. = Tω = 2π  N T = 60.475 kJ

 Heat rejected to the cooling water: Q = mCW ( C  TCW OUT )− TCWIN

Q CW = 483 x4.18x(77 − 17)kJ
30  60

Q CW = 483 x4.18x60 kJ
60

Q CW = 67.298 kJ

( ) Heat carried away by exhaust gases: Q EKZOS = mEKZOS  CGAS  TEKZOS − TAIR IN

Jumlah jisim gas ekzos = mA + mF

Q EKZOS = 182 + 9.35 x1.25x(486 − 17) kJ
30  60

Q EKZOS = 112 179 kJ
1800

Q EKZOS = 62.32 kJ

 Surrounding heat:
Q surrounding = Qsupply—( ++ )
Qsurounding = 219.735—( 60.475+67.298+62.32 ) kJ
Qsurrounding = 29.642 kJ

Heat Balance Sheet PSP eBook | Power Plant Engineering 3 85
Q supply = 219.735 kJ
100 %
Distribution
1 B.P. = 60.475 kJ 27.52 %
2 QCW = 67.298 kJ 30.63 %
3 QEKZOS = 62.32 kJ 28.36 %
4 Qserounding = 29.642 kJ 13.49 %
100 %
Total = 219.735 kJ

Tutorial Internal Combustion Engine

Question 1
A single cylinder, two stroke petrol engine develops 4 kW indicated power. Find the average
speed of the piston, if the mean effective pressure is 6.5 bar and piston diameter is 100 mm.

Question 2

In a laboratory experiment, the following observations were noted during test of a four

stroke Diesel engine:

Area of indicator diagram = 420 mm2

Length of indicator diagram = 62 mm

Spring number = 1.1 bar/mm

Diameter of piston = 100 mm

Length of stroke = 150 mm

Engine speed = 450 r.p.m

Determine:

(i) Indicated mean effective pressure

(ii) Indicated power

86 PSP eBook | Power Plant Engineering 3

Question 3
A four-cylinder petrol engine has a bore of 57 mm and a stroke of 90mm. Its rated speed is
2800 rev/min and it is tested at this speed against a brake which has a torque arm of 0.356m.
The net brake load is 155 N and the fuel consumption is 6.74 liter/hour. The specific gravity
of the petrol used is 0.735 and it has a lower calorific value, Qnet,v of 44 200 kJ/kg. Calculate
for this speed:

(i) The engine torque
(ii) The bmep
(iii) The brake thermal efficiency
(iv) The specific fuel consumption

Question 4

A test is performed on a 4-stroke single cylinder engine with the engine speeding at 1200 rpm.

The following data were obtained.

Engine cylinder bore = 0.15m

Brake Power = 31.42 kW

Specific fuel consumption = 0.7877 kg/kW hour

The calorific value of fuel = 42300 kJ/kg

Indicated Power = 39.28 kJ/s

Calculate:

(i) The area of cylinder

(ii) Torque

(iii) Fuel flow rate

(iv) Mechanical efficiency

(v) Indicated thermal efficiency

PSP eBook | Power Plant Engineering 3 87

Question 5:

A four-stroke six-cylinder CI engine is tested against a water brake dynamometer which B.P =

WN/17x103 in kW, where W is the brake load in newton and N is the speed of the engine in

the r.p.m. The air consumption was measured by means of a sharp-edged orifice. During the

test following observations were taken:

Bore = 10 cm

Stroke = 14 cm

Speed = 2500 rpm

Brake load = 480 N

Barometer reading = 76 cm Hg

Orifice diameter = 3.3 cm

Co-efficient of discharge of orifice= 0.62

Pressure drop across orifice = 14 cm of Hg

Room temperature = 25oC

Fuel Consumption = 0.32 kg/min

Calculate following:
(i) The volumetric efficiency
(ii) The brake means effective pressure (b.m.e.p)
(iii) The engine torque
(iv) The brake specific fuel consumption (b.s.f.c)

Question 6
A six cylinder spark ignition engine works on 4 stroke cycle. The bore of each cylinder is 70mm
and the stroke 100mm. The clearance volume per cylinder is 67c.c and calorific value of fuel
is 45000kJ/kg. At a speed of 3300 r.p.m the fuel consumption is 18.5kg/h and the torque
developed is 135Nm. Calculate:

(i) The brake power
(ii) The brake means effective pressure
(iii) The brake thermal efficiency
(iv) Relative efficiency on brake power basis

88 PSP eBook | Power Plant Engineering 3

Question 7:
The following data were obtained from a test on a single cylinder, 4 strokes, diesel engine:

Cylinder bore : 15cm
25cm
Stroke : 450mm2
50mm
Area of indicator diagram : 1.2mm for a pressure of 9.80665 N/cm2
400rpm
Length of indicator diagram : 225Nm
3kg/h
Indicator spring rating : 44200kJ/kg
4kg/min
Engine speed : 42oC
: 4.1868 kJ/kgK
Brake torque :

Fuel consumption :

Calorific value of fuel :

Cooling water flow rate :

Cooling water temperature rise :

Specific heat of cooling water

Calculate:
(i) The mechanical efficiency
(ii) The brake thermal efficiency
(iii) The specific fuel consumption
(iv) Draw heat balance in kW

PSP eBook | Power Plant Engineering 3 89

Question 8

A test on a single cylinder,4 stroke diesel engine, having a 180mm bore and a 360mm stroke

gave the following result:

Speed : 290r.p.m

Brake torque : 392Nm

Indicated mean effective pressure : 7.2bar

Oil consumption : 3.5kg/h

Cooling water flow : 270kg/h

Cooling water temperature rise : 36oC

Air fuel ratio by weight : 25

Exhaust gas temperature : 415oC

Barometric pressure : 1.013bar

Room temperature : 21oC

Calorific value of fuel : 45200kJ/kg

Calculate:
(i) The indicated thermal efficiency
(ii) The volumetric efficiency
(iii) Heat balance in kJ/min
Take R=0.287kJ/kgK and Cp for dry exhaust gases = 1.005kJ/kgK

CHAPTER 590 PSP eBook | Power Plant Engineering 3

AIR STANDARD CYCLE

AIR STANDARD CYCLE

Objectives:
At the end of this chapter, students should be able to:

• Define Air Standard Cycle
• State the difference between Otto Cycle and Diesel Cycle.
• Perform calculation related to Air Standard Cycle.

PSP eBook | Power Plant Engineering 3 91

5.1 Constant Volume or Otto Cycle

Figure 5.1 Otto Cycle P-v and T-s Diagram

• The Otto cycle is the ideal air standard cycle for the petrol engine, the gas engine, and the
high-speed oil engine.

• Figure above (a) and (b) shows the theoretical P-v diagram and T-s diagram of this cycle
respectively.

• The point 1 represents that cylinder is full of air volume V1, Pressure P1 and absolute
temperature T1.

Line 1-2 represents the adiabatic compression of air due to which P1, V1, and T1 change to P2,
V2 and T2 respectively.

Line 2 -3 shows the supply of heat to the air at constant volume so that P2 and T2 change to
P3 and T3 (V3 being the same as V2).

Line 3 -4 represents the adiabatic expansion of the air. During expansion P3, V3 and T3 change
to a final value of P4, V4 or V1 and T4 respectively.

Line 4 -1 shows the rejection of heat by air at constant volume till original state (point 1)
reaches.