contents · 7.4 the otto cycle the otto cycle is also known as constant volume (isochoric) cycle...

Contents

7.1 Terminology Used in Gas Power Cycles .......................................................................................... 7.2

7.2 Air Standard Analysis ........................................................................................................................ 7.3

7.3 The Carnot Gas Power Cycle ............................................................................................................ 7.4

7.4 The Otto Cycle ................................................................................................................................... 7.6

7.5 The Diesel Cycle .............................................................................................................................. 7.10

7.6 The Dual Cycle ................................................................................................................................. 7.13

7.7 Comparison of Otto, Diesel and Dual Cycles................................................................................. 7.17

7.8 The Brayton Cycle ........................................................................................................................... 7.19

7.9 Solved Examples ............................................................................................................................. 7.24

7.10 References ....................................................................................................................................... 7.30

7.2 Prof. Bhavin J. Vegada, Department of Mechanical Engineering

Engineering Thermodynamics (3131905) | Unit-7 Gas Power Cycles

7.1 Terminology Used in Gas Power Cycles

Terminologies related to a reciprocating internal combustion

engine are shown in Fig.7.1.

a) Cycle: A cycle is defined as a repeated series of operations

occurring in a certain order.

b) Air Standard Cycle: The thermodynamic cycle with air as

the working fluid is called an air standard cycle.

c) Bore (𝑫): Diameter of the cylinder is known as the bore.

d) Stroke Length (𝑳): The distance between TDC & BDC is

called stroke length.

e) Clearance Volume (𝑽𝒄): The minimum volume formed in

the cylinder when the piston is at TDC is called the

clearance volume.

f) Swept Volume (𝑽𝒔): The volume displaced by the piston as

it moves from BDC to TDC or TDC to BDC is called the swept

volume.

𝑉𝑠 =𝜋

4𝐷2𝐿

g) Compression Ratio (𝒓): It is defined as a ratio of total cylinder volume to the clearance volume.

𝑟 = 𝑇𝑜𝑡𝑎𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑣𝑜𝑙𝑢𝑚𝑒

𝐶𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑣𝑜𝑙𝑢𝑚𝑒=

𝑉𝐶 + 𝑉𝑆

𝑉𝐶

h) Thermal Efficiency: It is the ratio of net work output to the heat input by the fuel.

ηth = Work output

Heat input =

Q1 − Q2

Q1

Where, Q1 = Heat addition and Q2 = Heat rejection

[Assuming no friction & heat losses, so W = Q1 − Q2 ]

i. Indicated thermal efficiency = Indicated Power/ Heat supplied by burning of fuel

𝜂𝑖𝑡ℎ =𝐼. 𝑃.

𝑚𝑓 × 𝐶. 𝑉.

Where, mf = mass of fuel supplied in Kg/sec and C.V. = calorific value of fuel in J/kg

ii. Brake thermal efficiency = Brake Power/ Heat supplied by burning of fuel

𝜂𝑏𝑡ℎ =𝐵. 𝑃.

𝑚𝑓 × 𝐶. 𝑉.

i) Mechanical Efficiency: It is defined as the ratio of brake power to indicated power. Mechanical

efficiency is an indicator of losses due to friction.

𝜂𝑚𝑒𝑐ℎ =𝐵. 𝑃.

𝐼. 𝑃.=

𝜂𝑏𝑡ℎ

𝜂𝑖𝑡ℎ

j) Air Standard Efficiency: The efficiency of an engine using air as the working medium is known as an

“Air standard efficiency” or “Ideal efficiency”.

Fig.7.1 – Engine Nomenclature

Prof. Bhavin J. Vegada, Department of Mechanical Engineering

Engineering Thermodynamics (3131905) | Unit-7 Gas Power Cycles 7.3

k) Relative Efficiency: The actual efficiency of a cycle is always less than the air standard efficiency of that cycle under ideal conditions. This is taken into account by introducing a new term “Relative efficiency”.

ηrelative = Actual thermal efficiency

Air standard efficiency

l) Mean Effective Pressure (𝒑𝒎):

The pressure variation versus volume inside the

cylinder of a reciprocating engine is plotted with the

help of an engine indicator. The resulting contour is

closed one and is referred to as indicator diagram as

shown in Fig.7.2.

The area enclosed by the contour is a measure of the

work output per cycle from the engine.

Mean effective pressure is defined as the average

pressure acting on the piston which will produce the

same output as is done by the varying pressure

during a cycle. Therefore,

Area of indicator loop = Area of rectangle abcd

The height of the rectangle than represents the mean effective pressure.

∴ pm = work done per cycle

swept volume=

Area of indicator loop

length of loop

Unit: bar or N/m2

Mean effective pressure is used as a parameter to compare the performance of reciprocating

engines of equal size.

An engine that has a large value of m.e.p. will deliver more net work and will thus perform better.

7.2 Air Standard Analysis

The actual gas power cycles are complex. To simplify the analysis, we utilize the following approximations,

commonly known as the air-standard assumptions:

1. The gas in the engine cylinder is a perfect gas i.e. it obeys the gas laws and has constant specific heat.

2. The compression and expansion processes are adiabatic and they take place without internal friction

i.e. these processes are Isentropic.

3. No chemical reaction takes place in the cylinder. Heat is supplied or rejected by bringing a hot body or

a cold body in contact with the cylinder at appropriate points during the process.

4. The engine operates in a closed cycle. The cylinder is filled with a constant amount of working medium

and the same fluid is used repeatedly.

The approach and concept of ideal air cycle help to…….

a. Indicate the ultimate performance i.e. to determine the maximum ideal efficiency of a specific

thermodynamic cycle.

b. Study qualitatively the influence of different variables on the performance of an actual engine.

c. Evaluate one engine relative to another.

Fig.7.2 – Engine Indicator Diagram



7.3 The Carnot Gas Power Cycle

A Carnot cycle is a hypothetical cycle consisting of four different processes: two reversible isothermal

processes and two reversible adiabatic (isentropic) processes.

According to Carnot theorem, “No cycle can be more efficient than a reversible cycle operating between the

same temperature limits.”

Assumptions made in the working of the Carnot cycle are:

Working fluid is a perfect gas.

Piston cylinder arrangement is weightless and does not produce friction during motion.

The walls of the cylinder and piston are considered perfectly insulated.

Compression and expansion are reversible.

The transfer of heat does not change the temperature of sources or sink.

Fig.7.3 – Schematic, p-V and T-s diagram of Carnot gas power cycle

The Carnot cycle has the highest possible efficiency and it consists of four simple operations as below:

Isothermal expansion (1 – 2):

The source of heat (H) is applied to the end of the cylinder and isothermal reversible expansion occurs at

a temperature 𝑇1 = 𝑇2 = 𝑇𝐻. During this process 𝑞𝑠 amount of heat is supplied to the system.

Adiabatic expansion (2 – 3):

Non conducting (insulating) cover (C) is applied to the end of the cylinder and the cylinder becomes

perfectly insulated. The adiabatic cover is brought in contact with the cylinder head. Hence no heat transfer

takes place. The fluid expands adiabatically and reversibly. The temperature falls from 𝑇𝐻 to 𝑇𝐿.



Isothermal compression (3 – 4):

The adiabatic cover is removed and sink (S) is applied to the end of the cylinder. The heat, 𝑞𝑟 is transferred

reversibly and isothermally at a temperature 𝑇𝐿 from the system to the sink (S).

Adiabatic compression (4 – 1):

The adiabatic cover is brought in contact with the cylinder head. This completes the cycle and system is

returned to its original state at 1. During the process, the temperature of the system is raised from 𝑇𝐿 to

𝑇𝐻.

7.3.1 The efficiency of Carnot Gas Cycle

Consider 1 kg of working substance (air) is enclosed in the cylinder. The T-s diagram of the Carnot cycle

is shown in Fig.7.3 (b) and the area under each reversible process curve on the T-s diagram represents the

heat transfer for that process.

Heat supplied during the isothermal process (1 – 2):

𝑞𝑠 = 𝑇𝐻(𝑠2 − 𝑠1) Eq. (7.1)

Heat rejected during isothermal compression (3 – 4)

𝑞𝑟 = 𝑇𝐿(𝑠3 − 𝑠4) Eq. (7.2)

Since process (2 – 3) and (4 – 1) are isentropic, the heat transfer from or to the system is zero

and 𝑠2 = 𝑠3 and 𝑠4 = 𝑠1.

Net Work Done,

𝑤𝑛𝑒𝑡 = 𝑞𝑠 − 𝑞𝑟 Eq. (7.3)

Thermal Efficiency,

𝜂 =𝑁𝑒𝑡 𝑤𝑜𝑟𝑘 𝑜𝑢𝑡𝑝𝑢𝑡

𝐻𝑒𝑎𝑡 𝑖𝑛𝑝𝑢𝑡=

𝑤𝑛𝑒𝑡

𝑞𝑠

∴ 𝜂 =𝑞𝑠 − 𝑞𝑟

𝑞𝑠= 1 −

𝑞𝑟

𝑞𝑠

∴ 𝜂 = 1 −𝑇𝐿(𝑠3 − 𝑠4)

𝑇𝐻(𝑠2 − 𝑠1)

∴ 𝜂 = 1 −𝑇𝐿(𝑠2 − 𝑠1)

𝑇𝐻(𝑠2 − 𝑠1)

∴ 𝜼 = 𝟏 −𝑻𝑳

𝑻𝑯 Eq. (7.4)

Where,

𝑇𝐻 = Maximum temperature of the cycle (K)

𝑇𝐿 = Minimum temperature of the cycle (K)

In Eq. (7.4), if temperature 𝑇𝐿 decreases, efficiency increases and it becomes 100% if the temperature 𝑇𝐿

becomes absolute zero; which is impossible to attain.

Note that the thermal efficiency of a Carnot cycle is independent of the type of the working fluid used (an

ideal gas, steam, etc.). It depends only on the temperature of the source and sink.

7.3.2 Limitations of Carnot Gas Cycle

The Carnot cycle is hypothetical.



The thermal efficiency of the Carnot cycle depends upon the absolute temperature of heat source

𝑇𝐻 and heat sink 𝑇𝐿 only, and independent of the working substance.

Practically it is not possible to neglect friction between piston and cylinder. It can be minimized

but cannot be eliminated.

It is impossible to construct cylinder walls which are a perfect insulator. Some amount of heat will

always be transferred. Hence perfect adiabatic process cannot be achieved.

The isothermal and adiabatic processes take place during the same stroke. Therefore the piston

has to move very slowly for the isothermal process and it has to move very fast during the

remaining stoke for the adiabatic process which is practically not possible.

The output obtained per cycle is very small. This work may not be able to overcome the friction of

the reciprocating parts.

7.4 The Otto Cycle

The Otto cycle is also known as Constant Volume (Isochoric) cycle was successfully applied by a German

scientist Nikolaus A. Otto to produce a successful 4 – stroke cycle engine in 1876.

Fig.7.4 – p-V and T-s diagrams of the Otto cycle

This thermodynamic cycle is operated with isochoric (constant volume) heat addition and consists of two

reversible adiabatic (Isentropic) processes and two constant volume processes. Fig.7.4 shows the Otto

cycle plotted on p–V and T–s diagram.

Isentropic Compression Process (1 – 2):

At point 1 cylinder is full of air with a volume 𝑉1, pressure 𝑝1 and temperature 𝑇1.

Piston moves from BDC to TDC and an ideal gas (air) is compressed isentropically to state point 2 through

compression ratio,

𝑟 = 𝑉1

𝑉2

Constant Volume Heat Addition Process (2 – 3):

Heat is added at constant volume from an external heat source. The pressure rises at constant volume

and the ratio 𝑟𝑝 𝑜𝑟 𝛼 = 𝑃3

𝑃2 is called pressure ratio.



Isentropic Expansion Process (3 – 4):

The increased high pressure exerts a greater amount of force on the piston and pushes it towards the

BDC. Expansion of working fluid takes place isentropically and work is done by the system. The volume

ratio V4

V3 is called isentropic expansion ratio.

Constant Volume Heat Rejection Process (4 – 1):

Heat is rejected to the external sink at constant volume. This process is so controlled that ultimately the

working fluid comes to its initial state 1 and the cycle is repeated.

Many petrol and gas engines work on a cycle which is a slight modification of the Otto cycle.

This cycle is called a constant volume cycle because the heat is supplied to air at constant volume.

7.4.1 Thermal Efficiency of an Otto Cycle

Consider a unit mass of air undergoing a cyclic change.

Heat supplied during the process 2 – 3,

𝑞𝑠 = 𝐶𝑉(𝑇3 − 𝑇2)

Heat rejected during process 4 – 1,

𝑞𝑟 = 𝐶𝑉(𝑇4 − 𝑇1)

Net Work Done,

∴ 𝑤𝑛𝑒𝑡 = 𝑞𝑠 − 𝑞𝑟

∴ 𝑤𝑛𝑒𝑡 = 𝐶𝑉 (𝑇3 − 𝑇2) − 𝐶𝑉(𝑇4 − 𝑇1)

Thermal Efficiency,

𝜂 = 𝑁𝑒𝑡 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒

𝐻𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑=

𝑤𝑛𝑒𝑡

𝑞𝑠

= 𝐶𝑉 (𝑇3 − 𝑇2) − 𝐶𝑉(𝑇4 − 𝑇1)

𝐶𝑉 (𝑇3 − 𝑇2)

∴ 𝜼 = 𝟏 − (𝑻𝟒 − 𝑻𝟏)

(𝑻𝟑 − 𝑻𝟐) Eq. (7.5)

For isentropic compression process (1 – 2),

𝑇2

𝑇1 = (

𝑉1

𝑉2)𝛾−1

= 𝑟𝛾−1

∴ 𝑻𝟐 = 𝑻𝟏𝒓𝜸−𝟏 Eq. (7.6)

For constant volume heat addition (2 – 3),

𝑝2

𝑇2=

𝑝3

𝑇3

∴ 𝑇3 = 𝑇2

𝑝3

𝑝2

∴ 𝑻𝟑 = 𝑻𝟏𝒓𝜸−𝟏𝜶 Eq. (7.7)

For isentropic expansion process (3 – 4),



𝑇4

𝑇3 = (

𝑉3

𝑉4)𝛾−1

∴ 𝑇4 = 𝑇3 (𝑉3

𝑉4)𝛾−1

∴ 𝑇4 = 𝑇3 (𝑉2

𝑉1)𝛾−1

(∵ 𝑉3 = 𝑉2 , 𝑉4 = 𝑉1)

∴ 𝑇4 = 𝑇1𝑟𝛾−1𝛼

1

𝑟𝛾−1

∴ 𝑻𝟒 = 𝑻𝟏𝜶 Eq. (7.8)

Now put the value of 𝑇2, 𝑇3 𝑎𝑛𝑑 𝑇4 in Eq. (7.5), we get,

𝜂𝑜𝑡𝑡𝑜 = 1 − (𝑇1𝛼 − 𝑇1)

𝑇1𝑟𝛾−1𝛼 − 𝑇1 𝑟𝛾−1

∴ 𝜂𝑜𝑡𝑡𝑜 = 1 − 𝑇1(𝛼 − 1)

𝑇1𝑟𝛾−1(𝛼 − 1)

∴ 𝜼𝒐𝒕𝒕𝒐 = 𝟏 − 𝟏

𝒓𝜸−𝟏 Eq. (7.9)

Eq. (7.9) is known as the air standard efficiency of the Otto cycle.

7.4.2 Effect of Compression Ratio on Performance

It is clear from the Eq. (7.9) that, the efficiency increases with the

increase in the value of 𝑟 (as γ is constant). We can have

maximum efficiency by increasing 𝑟 to a considerable extent, but

due to practical difficulties, its value is limited to 8.

In actual engines working on Otto cycle, the compression ratio

varies from 5 to 8 depending upon the quality of fuel.

At compression ratios higher than this, the temperature after

combustion becomes high and that may lead to spontaneous and

uncontrolled combustion of fuel in the cylinder.

The phenomenon of uncontrolled combustion in the petrol engine

is called detonation and it leads to poor engine efficiency and in

structural damage of engine parts.

Fig.7.5 shows the variation of air standard efficiency of the Otto cycle with compression ratio.

7.4.3 Mean Effective Pressure of an Otto Cycle

Net work done per unit mass of air,

𝑤𝑛𝑒𝑡 = 𝑞𝑠 − 𝑞𝑟 = 𝐶𝑉 (𝑇3 − 𝑇2) − 𝐶𝑉(𝑇4 − 𝑇1) Eq. (7.10)

Swept volume,

𝑣𝑠 = 𝑣1 − 𝑣2 = 𝑣1 (1 − 𝑣2

𝑣1) =

𝑅𝑇1

𝑝1(1 −

1

𝑟)

∴ 𝑣𝑠 = 𝑅𝑇1

𝑝1𝑟(𝑟 − 1) Eq. (7.11)

Fig.7.5 – Variation of efficiency with 𝑟



Mean effective pressure,

𝑝𝑚 = 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒

𝑠𝑤𝑒𝑝𝑡 𝑣𝑜𝑙𝑢𝑚𝑒

∴ 𝑝𝑚 = 𝐶𝑉 (𝑇3 − 𝑇2) − 𝐶𝑉(𝑇4 − 𝑇1)

𝑅 𝑇1

𝑝1𝑟 (𝑟 − 1)

∴ 𝑝𝑚 = 𝐶𝑉

𝑅

𝑝1𝑟

(𝑟 − 1)[(𝑇3 − 𝑇2) − (𝑇4 − 𝑇1)

𝑇1] Eq. (7.12)

Now from Eq. (7.6), Eq. (7.7) and Eq. (7.8),

𝑇2 = 𝑇1 𝑟𝛾−1

𝑇3 = 𝑇1 𝛼 𝑟𝛾−1

𝑇4 = 𝑇1 ∙ 𝛼

Substituting all these temperature values in Eq. (7.12), We get,

𝑝𝑚 = 𝐶𝑉

𝑅

𝑝1 𝑟

(𝑟 − 1)[(𝑇1 𝛼 𝑟𝛾−1 − 𝑇1 𝑟

𝛾−1) − (𝑇1 𝛼 − 𝑇1 )

𝑇1]


𝑅

𝑝1 𝑟

(𝑟 − 1)[𝑇1 𝑟

𝛾−1(𝛼 − 1) − 𝑇1 (𝛼 − 1)

𝑇1]


𝑅

𝑝1 𝑟

(𝑟 − 1)[(𝑟𝛾−1 − 1)(𝛼 − 1)]

∴ 𝒎𝒆𝒑 = 𝒑𝟏 𝒓

(𝒓 − 𝟏)(𝜸 − 𝟏)[(𝒓𝜸−𝟏 − 𝟏)(𝜶 − 𝟏)] Eq. (7.13)

7.4.4 Condition for Maximum Work of an Otto Cycle

For a unit mass of air,

𝑤𝑛𝑒𝑡 = 𝑞𝑠 − 𝑞𝑟

∴ 𝑤𝑛𝑒𝑡 = 𝐶𝑉 (𝑇3 − 𝑇2) − 𝐶𝑉 (𝑇4 − 𝑇1)

∴𝑤𝑛𝑒𝑡

𝐶𝑉 = 𝑇3 − 𝑇2 − 𝑇4 + 𝑇1 Eq. (7.14)

We know that,

𝑇2 = 𝑇1𝑟𝛾−1

𝑇4 = 𝑇3 1

𝑟𝛾−1= 𝑇3

𝑇1

𝑇2 (∵

𝑇2

𝑇1= 𝑟𝛾−1)

So

𝑤𝑛𝑒𝑡

𝐶𝑉 = 𝑇3 − 𝑇2 −

𝑇3𝑇1

𝑇2+ 𝑇1 Eq. (7.15)

The intermediate temperature 𝑇2 for maximum work output can be obtained by differentiating the above

equation with respect to 𝑇2 & setting the derivatives equal to zero.

∴ 1

𝐶𝑉

𝑑𝑤𝑛𝑒𝑡

𝑑𝑇2= −1 +

𝑇1𝑇3

𝑇22 = 0 (𝑓𝑜𝑟 𝑚𝑎𝑥 𝑤𝑜𝑟𝑘)

[

𝐶𝑃

𝐶𝑉= 𝛾,

𝐶𝑃 − 𝐶𝑉 = 𝑅,

𝐶𝑉 (𝐶𝑃

𝐶𝑉− 1) = 𝑅,

∴𝐶𝑉

𝑅 =

1

𝛾 − 1 ]



∴ 𝑇22 = 𝑇1𝑇3

∴ 𝑻𝟐 = √𝑻𝟏𝑻𝟑 Eq. (7.16)

Similarly for temperature 𝑻𝟒,

𝑤𝑛𝑒𝑡

𝐶𝑉 = 𝑇3 −

𝑇1 ∙ 𝑇3

𝑇4− 𝑇4 + 𝑇1

∴ 1

𝐶𝑉

𝑑𝑤𝑛𝑒𝑡

𝑑𝑇4=

𝑇1𝑇3

𝑇42 − 1 = 0 (𝑓𝑜𝑟 𝑚𝑎𝑥 𝑤𝑜𝑟𝑘)

∴ 𝑻𝟒 = √𝑻𝟏𝑻𝟑 Eq. (7.17)

Thus for maximum work,

𝑻𝟐 = 𝑻𝟒 = √𝑻𝟏𝑻𝟑 Eq. (7.18)

i.e. the intermediate temperature 𝑇2 & 𝑇4 must be equal for maximum work.

Maximum work,

𝑤𝑚𝑎𝑥 = 𝐶𝑉(𝑇3 − 𝑇2 − 𝑇4 + 𝑇1)

∴ 𝑤𝑚𝑎𝑥 = 𝐶𝑉(𝑇3 − √𝑇1𝑇3 − √𝑇1𝑇3 + 𝑇1)

∴ 𝒘𝒎𝒂𝒙 = 𝑪𝑽(𝑻𝟑 + 𝑻𝟏 − 𝟐√𝑻𝟏𝑻𝟑) Eq. (7.19)

7.5 The Diesel Cycle

The Diesel cycle was discovered by a German engineer Dr. Rudolph Diesel in 1890. It is also known as

constant pressure (Isobaric) cycle because heat is supplied to the air at constant pressure.

Fig.7.6 – p-V and T-s diagram of the Diesel cycle

Isentropic Compression Process (1 – 2):

At point 1 cylinder is full of air with volume 𝑉1, pressure 𝑝1 and temperature 𝑇1.

Piston moves from BDC to TDC and an ideal gas (air) is compressed isentropically to state point 2 through

compression ratio,

𝑟 = 𝑉1

𝑉2



Constant Pressure Heat Addition Process (2 – 3):

Heat is added at constant pressure from an external heat source. The heat supply is stopped at point 3

which is called the cut – off point and the volume ratio 𝜌 = V3

V2 is called cut off ratio.


Expansion of working fluid takes place isentropically and work done by the system. The volume ratio V4

V3

is called isentropic expansion ratio.


In this process, heat is rejected at constant volume to the heat sink.

7.5.1 Thermal Efficiency of the Diesel Cycle

Consider a unit mass of air undergoing a cyclic change.

Heat supplied during process 2 – 3,

𝑞𝑠 = 𝐶𝑃(𝑇3 − 𝑇2)


𝑞𝑟 = 𝐶𝑉(𝑇4 − 𝑇1)

Net Work Done,


∴ 𝑤𝑛𝑒𝑡 = 𝐶𝑃(𝑇3 − 𝑇2) − 𝐶𝑉(𝑇4 − 𝑇1)

Thermal Efficiency,



𝑤𝑛𝑒𝑡

𝑞𝑠

∴ 𝜂 = 𝐶𝑃(𝑇3 − 𝑇2) − 𝐶𝑉(𝑇4 − 𝑇1)

𝐶𝑃(𝑇3 − 𝑇2)

∴ 𝜂 = 1 − 𝐶𝑉

𝐶𝑃

(𝑇4 − 𝑇1)

(𝑇3 − 𝑇2)

∴ 𝜼 = 𝟏 − 𝟏

𝜸

(𝑻𝟒 − 𝑻𝟏)

(𝑻𝟑 − 𝑻𝟐) Eq. (7.20)


𝑇2

𝑇1 = (

𝑉1

𝑉2)𝛾−1

= 𝑟𝛾−1

∴ 𝑻𝟐 = 𝑻𝟏 𝒓𝜸−𝟏 Eq. (7.21)

For constant pressure heat addition process (2 – 3)

𝑉2

𝑇2=

𝑉3

𝑇3

∴ 𝑇3 = 𝑇2

𝑉3

𝑉2



∴ 𝑇3 = 𝑇2 𝜌 ( ∵ 𝜌 = V3

V2)

∴ 𝑻𝟑 = 𝑻𝟏 𝒓𝜸−𝟏𝝆 Eq. (7.22)


𝑇4

𝑇3 = (

𝑉3

𝑉4)𝛾−1

∴ 𝑇4 = 𝑇3 (𝑉3

𝑉4×

𝑉2

𝑉2)𝛾−1

∴ 𝑇4 = 𝑇3 (𝜌

𝑟)𝛾−1

( ∵ 𝑟 =𝑉1

𝑉2 𝑎𝑛𝑑 𝜌 =

𝑉3

𝑉2)

∴ 𝑇4 = 𝑇1 𝑟𝛾−1𝜌 ×

𝜌𝛾−1

𝑟𝛾−1

∴ 𝑻𝟒 = 𝑻𝟏 𝝆𝜸 Eq. (7.23)

Now put the value of 𝑇2, 𝑇3 𝑎𝑛𝑑 𝑇4 in Eq. (7.20), we get,

𝜂 = 1 − 1

𝛾

(𝑇4 − 𝑇1)

(𝑇3 − 𝑇2)

∴ 𝜂 = 1 − 1

𝛾

(𝑇1 𝜌𝛾 − 𝑇1)

(𝑇1 𝑟𝛾−1𝜌 − 𝑇1 𝑟𝛾−1)

∴ 𝜂 = 1 − 1

𝛾[

𝑇1(𝜌𝛾 − 1)

𝑇1𝑟𝛾−1(𝜌 − 1)]

∴ 𝜼 = 𝟏 − 𝟏

𝒓𝜸−𝟏 [

(𝝆𝜸 − 𝟏)

𝜸(𝝆 − 𝟏)] Eq. (7.24)

7.5.2 Effect of Cut-off Ratio on Performance

Eq. (7.24) shows, the efficiency of the Diesel cycle

depends upon the compression ratio (𝑟)and cutoff ratio

(𝜌) and hence upon the quantity of heat supplied.

Fig.7.7 shows the air standard efficiency of the Diesel

cycle for various cut off ratio.

Further,

𝐾 = 𝜌𝛾 − 1

𝛾(𝜌 − 1)

reveals that with an increase in the cut-off ratio (𝜌) the

value of factor K increases.

That implies that for a diesel engine at constant compression ratio, the efficiency would increase with a

decrease in 𝜌 and in the limit ρ → 1, the efficiency would become,

1 − 1

𝑟𝛾−1

Fig.7.7 – Variation in efficiency with cut-off ratio



Since the factor K = ργ− 1

γ(ρ− 1) is always greater than unity, the Diesel cycle is always less efficient than a

corresponding Otto cycle having the same compression ratio.

However, Diesel engine operates on a much higher compression ratio (14 to 18) compared to those for

S.I. Engines operating on Otto cycle.

High compression ratios for Diesel engines are not only for high efficiency but also to prevent diesel knock;

a phenomenon which leads to uncontrolled and rapid combustion in diesel engines.

7.5.3 Mean Effective Pressure of the Diesel cycle


𝑤𝑛𝑒𝑡 = 𝑞𝑠 − 𝑞𝑟 = 𝐶𝑝 (𝑇3 − 𝑇2) − 𝐶𝑉(𝑇4 − 𝑇1) Eq. (7.25)

Swept volume,

𝑣𝑠 = 𝑣1 − 𝑣2 = 𝑣1 (1 − 𝑣2

𝑣1) =

𝑅𝑇1

𝑝1(1 −

1

𝑟)


𝑝1𝑟(𝑟 − 1) Eq. (7.26)


𝑝𝑚 = 𝑁𝑒𝑡 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒

𝑠𝑤𝑒𝑝𝑡 𝑣𝑜𝑙𝑢𝑚𝑒=

𝑤𝑛𝑒𝑡

𝑣𝑠

∴ 𝑝𝑚 = 𝐶𝑃 (𝑇3 − 𝑇2) − 𝐶𝑉(𝑇4 − 𝑇1)

𝑅 𝑇1

𝑝1𝑟 (𝑟 − 1)


𝑅

𝑝1𝑟

(𝑟 − 1)[𝛾(𝑇3 − 𝑇2) − (𝑇4 − 𝑇1)

𝑇1] Eq. (7.27)

Now from Eq. (7.21), Eq. (7.22) and Eq. (7.23),

𝑇2 = 𝑇1 𝑟𝛾−1

𝑇3 = 𝑇1 𝑟𝛾−1𝜌

𝑇4 = 𝑇1 𝜌𝛾

By putting the values of 𝑇2, 𝑇3 and 𝑇4 in Eq. (7.27), we get,

∴ 𝑚𝑒𝑝 = 𝐶𝑉

𝑅

𝑝1𝑟

(𝑟 − 1)[𝛾(𝑇1 𝑟

𝛾−1𝜌 − 𝑇1 𝑟𝛾−1) − (𝑇1 𝜌

𝛾 − 𝑇1)

𝑇1]

∴ 𝒎𝒆𝒑 = 𝒑𝟏𝒓

(𝜸 − 𝟏)(𝒓 − 𝟏)[𝜸𝒓𝜸−𝟏( 𝝆 − 𝟏) − (𝝆𝜸 − 𝟏)] Eq. (7.28)

(∵𝐶𝑉

𝑅 =

1

𝛾 − 1)

7.6 The Dual Cycle

The p-V diagram of actual I.C. engine is more closely described by the air standard dual cycle, in which the

heat addition occurs partly at constant volume and partly at constant pressure. This cycle is also known

as the Semi-Diesel Cycle or Limited Pressure Cycle. The p-V and T-s diagram of the dual cycle is shown in

Fig.7.8.



Adiabatic Compression Process (1 – 2):

At point 1 cylinder is full of air with a volume 𝑉1, pressure 𝑝1 and temperature 𝑇1. Piston moves from BDC

to TDC and an ideal gas (air) is compressed isentropically to state point 2 through compression ratio,

𝑟 = 𝑉1

𝑉2

Fig.7.8 – p-V and T-s diagram of air standard dual cycle

Constant Volume Heat Addition Process (2 – 3):

Heat is added at constant volume from an external heat source. The pressure rises at constant volume

and the ratio 𝑟𝑝 𝑜𝑟 𝛼 = 𝑃3

𝑃2 is called pressure ratio.

Constant Pressure Heat Addition Process (3 – 4):

Heat is added at constant pressure from an external heat source. The heat supply is stopped at point 4

which is called the cut-off point and the volume ratio 𝜌 = V4

V3 is called cut off ratio.


Expansion of working fluid takes place isentropically and work done by the system. The volume ratio V5

V4

is called isentropic expansion ratio.


In this process heat is rejected at constant volume to the heat sink.

7.6.1 Thermal Efficiency of the Dual Cycle

Consider unit mass of air undergoing the cyclic change.

Heat supplied during process (2 – 3) and (3 – 4),

𝑞𝑠 = 𝑞2−3 + 𝑞3−4

∴ 𝑞𝑠 = 𝐶𝑉(𝑇3 − 𝑇2) + 𝐶𝑃(𝑇4 − 𝑇3)

Heat rejected during process (5 – 1),

𝑞𝑟 = 𝐶𝑉(𝑇5 − 𝑇1)

Net work done,




∴ 𝑤𝑛𝑒𝑡 = 𝐶𝑉(𝑇3 − 𝑇2) + 𝐶𝑃(𝑇4 − 𝑇3) − 𝐶𝑉(𝑇5 − 𝑇1)

Thermal efficiency,



𝑤𝑛𝑒𝑡

𝑞𝑠

∴ 𝜂 = 𝐶𝑉(𝑇3 − 𝑇2) + 𝐶𝑃(𝑇4 − 𝑇3) − 𝐶𝑉(𝑇5 − 𝑇1)

𝐶𝑉(𝑇3 − 𝑇2) + 𝐶𝑃(𝑇4 − 𝑇3)

∴ 𝜂 = 1 − (𝑇5 − 𝑇1)

(𝑇3 − 𝑇2) + 𝛾(𝑇4 − 𝑇3) Eq. (7.29)


𝑇2

𝑇1 = (

𝑉1

𝑉2)𝛾−1

= 𝑟𝛾−1

∴ 𝑻𝟐 = 𝑻𝟏 𝒓𝜸−𝟏 Eq. (7.30)

For constant volume heat addition process (2 – 3),

𝑝2

𝑇2=

𝑝3

𝑇3

∴ 𝑇3 = 𝑇2

𝑝3

𝑝2

∴ 𝑻𝟑 = 𝑻𝟏𝒓𝜸−𝟏𝜶 Eq. (7.31)

For constant pressure heat addition process (3 – 4),

𝑉3

𝑇3=

𝑉4

𝑇4

∴ 𝑇4 = 𝑇3

𝑉4

𝑉3

∴ 𝑇4 = 𝑇3 𝜌 ( ∵ 𝜌 = V4

V3)

∴ 𝑻𝟒 = 𝑻𝟏 𝒓𝜸−𝟏𝜶𝝆 Eq. (7.32)

For adiabatic expansion process (4 – 5),

𝑇5

𝑇4 = (

𝑉4

𝑉5)𝛾−1

∴ 𝑇5 = 𝑇4 (𝑉4

𝑉5×

𝑉2

𝑉2)𝛾−1

∴ 𝑇5 = 𝑇4 (𝑉4

𝑉1×

𝑉2

𝑉3)𝛾−1

(∵ 𝑉5 = 𝑉1 𝑎𝑛𝑑 𝑉2 = 𝑉3)

∴ 𝑇5 = 𝑇4 (𝜌

𝑟)𝛾−1

( ∵ 𝑟 =𝑉1

𝑉2 𝑎𝑛𝑑 𝜌 =

𝑉3

𝑉2)

∴ 𝑇5 = 𝑇1 𝑟𝛾−1𝛼𝜌 ×

𝜌𝛾−1

𝑟𝛾−1

∴ 𝑻𝟓 = 𝑻𝟏𝜶𝝆𝜸 Eq. (7.33)



Now put the value of 𝑇2, 𝑇3, 𝑇4 𝑎𝑛𝑑 𝑇5 in Eq. (7.29), we get,

∴ 𝜂 = 1 − (𝑇1 𝛼 𝜌𝛾 − 𝑇1)

(𝑇1 𝑟𝛾−1𝛼 − 𝑇1 𝑟𝛾−1) + 𝛾(𝑇1 𝑟𝛾−1𝛼𝜌 − 𝑇1 𝑟𝛾−1𝛼)

∴ 𝜂 = 1 − (𝜌𝛾𝛼 − 1)

[𝑟𝛾−1 {(𝛼 − 1) + 𝛾𝛼(𝜌 − 1)}]

∴ 𝜼 = 𝟏 − 𝟏

𝒓𝜸−𝟏 [

(𝜶𝝆𝜸 − 𝟏)

(𝜶 − 𝟏) + 𝜸𝜶(𝝆 − 𝟏)] Eq. (7.34)

It can be seen from Eq. (7.34) that the thermal efficiency of a Dual cycle can be increased by

supplying a greater portion of heat at constant volume (high value of 𝛼) and smaller portion at

constant pressure (low value of 𝜌).

In the actual high-speed Diesel engines operating on this cycle, it is achieved by early fuel injection

and an early cut-off.

It is to be noted that Otto and Diesel cycles are special cases of the Dual cycle.

If 𝝆 = 𝟏 (𝑽𝟑 = 𝑽𝟒)

Hence, there is no addition of heat at constant pressure. Consequently, the entire heat is supplied

at constant volume and the cycle becomes the Otto cycle. By substituting 𝜌 = 1 in Eq. (7.34), we

get,

𝜂 = 1 −1

𝑟(𝛾−1)= 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝑂𝑡𝑡𝑜 𝑐𝑦𝑐𝑙𝑒

Similarly if 𝜶 = 𝟏, the heat addition is only at constant pressure and cycle becomes Diesel cycle.

By substituting 𝛼 = 1 in Eq. (7.34), we get,

𝜂 = 1 − 1

𝑟𝛾−1 [

(𝜌𝛾 − 1)

𝛾(𝜌 − 1)] = 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝐷𝑖𝑒𝑠𝑒𝑙 𝑐𝑦𝑐𝑙𝑒

7.6.2 Mean Effective Pressure of the Dual Cycle


𝑤𝑛𝑒𝑡 = 𝐶𝑉(𝑇3 − 𝑇2) + 𝐶𝑝 (𝑇4 − 𝑇3) − 𝐶𝑉(𝑇5 − 𝑇1) Eq. (7.35)

Swept volume,

𝑣𝑠 = 𝑣1 − 𝑣2 = 𝑣1 (1 − 𝑣2

𝑣1) =

𝑅𝑇1

𝑝1(1 −

1

𝑟)


𝑝1𝑟(𝑟 − 1) Eq. (7.36)


𝑝𝑚 = 𝑁𝑒𝑡 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒

𝑆𝑤𝑒𝑝𝑡 𝑣𝑜𝑙𝑢𝑚𝑒=

𝑤𝑛𝑒𝑡

𝑣𝑠

∴ 𝑝𝑚

= 𝐶𝑉(𝑇3 − 𝑇2) + 𝐶𝑝 (𝑇4 − 𝑇3) − 𝐶𝑉(𝑇5 − 𝑇1)

𝑅 𝑇1

𝑝1𝑟 (𝑟 − 1)



∴ 𝑝𝑚

= 𝐶𝑉

𝑅

𝑝1𝑟

(𝑟 − 1)[(𝑇3 − 𝑇2) + 𝛾(𝑇4 − 𝑇3) − (𝑇5 − 𝑇1)

𝑇1]

Now from Eq. (7.30), Eq. (7.31), Eq. (7.32) and Eq. (7.33),

𝑇2 = 𝑇1 𝑟𝛾−1

𝑇3 = 𝑇1 𝑟𝛾−1𝛼

𝑇4 = 𝑇1 𝑟𝛾−1𝛼𝜌

𝑇5 = 𝑇1 𝛼 𝜌𝛾


𝑅

𝑝1𝑟

(𝑟 − 1)[𝛾(𝑇1 𝑟

𝛾−1𝛼 − 𝑇1 𝑟𝛾−1) + 𝛾(𝑇1 𝑟

𝛾−1𝛼𝜌 − 𝑇1 𝑟𝛾−1𝛼) − (𝑇1 𝛼𝜌𝛾 − 𝑇1)

𝑇1]

∴ 𝒎𝒆𝒑 = 𝒑𝟏𝒓

(𝜸 − 𝟏)(𝒓 − 𝟏)[( 𝜶 – 𝟏)𝒓𝜸−𝟏 + 𝜸𝜶𝒓𝜸−𝟏(𝝆 − 𝟏) − (𝜶𝝆𝜸 – 𝟏)] Eq. (7.37)

7.7 Comparison of Otto, Diesel and Dual Cycles

Following are the important variable factors which are used as a basis for comparison of the cycles:

Compression ratio

Maximum pressure

Heat supplied

Heat rejected and

Net work.

7.7.1 For the Same Compression Ratio and the Same Heat Input

Fig.7.9 – p-V and T-s diagram for same compression ration and same heat input

We know that,

𝜂 = 1 −𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑

𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑= 1 −

𝑞𝑟

𝑞𝑠 Eq. (7.38)

The quantity of heat rejected from each cycle is represented by the appropriate area under the line 4 – 1

on the T–s diagram.

From Eq. (7.38) and Fig.7.9; it is clear that the cycle which has the least heat rejected will have the highest

efficiency.

∴ 𝜼𝑶𝒕𝒕𝒐 > 𝜼𝑫𝒖𝒂𝒍 > 𝜼𝑫𝒊𝒆𝒔𝒆𝒍



7.7.2 Same Maximum Pressure and Temperature

When pressure is the limiting factor in engine design, it becomes necessary to compare the air standard

cycles on the basis of the same maximum pressure & temperature.

Fig.7.10 – p-V and T-s diagram for same maximum pressure and temperature

Here the Otto cycle must be limited to low compression ratio to fulfill the condition that point 3 (same

maximum pressure & temperature) is to be a common state for all the three cycles.

We know that,



𝑞2

𝑞1 Eq. (7.39)

From Eq. (7.39) and Fig.7.10; it is clear that the heat rejected is same for all the three cycles. Hence with

the same heat rejected, the cycle with greater heat addition is more efficient.

∴ 𝜼𝑫𝒊𝒆𝒔𝒆𝒍 > 𝜼𝑫𝒖𝒂𝒍 > 𝜼𝑶𝒕𝒕𝒐

7.7.3 For Constant Maximum Pressure and Heat Input

Fig.7.11 shows the Otto and Diesel cycles on p-V and T-s diagrams for constant maximum pressure and

heat input respectively.

Fig.7.11 – p-V and T-s diagram for constant maximum pressure and heat input

For the constant maximum pressure, points 3 and 3’ must lie on the constant pressure line.



Also for the same heat input, the areas under the line 2 − 3 and 2′ − 3′ on the T-s plot must be equal.

Now,



𝑞2

𝑞1 Eq. (7.40)

Hence from Eq. (7.40) and Fig.7.11; it is clear that, for the same amount of heat supplied, the cycle with

less heat rejected has a higher value of thermal efficiency.

∴ 𝜼𝑫𝒊𝒆𝒔𝒆𝒍 > 𝜼𝑫𝒖𝒂𝒍 > 𝜼𝑶𝒕𝒕𝒐

7.8 The Brayton Cycle

The Brayton cycle, also known as Joule cycle is a constant pressure cycle for a perfect gas. It is a

theoretical cycle on which constant pressure gas turbine works.

The schematic, p-V and T-s diagram of closed Brayton cycle is shown in Fig.7.12.

Fig.7.12 – Schematic, p-V, and T-s diagram of closed Brayton cycle

The various operations are as follows:

Isentropic Compression (1 – 2):

The air is compressed isentropically from the lower pressure 𝑝1 to the higher pressure 𝑝2, the temperature

rising from 𝑇1 to 𝑇2 due to compression. No heat flow takes place.



Constant Pressure Heat Addition (2 – 3):

The compressed air is passed through a heat exchanger, where heat is externally supplied to it at constant

pressure. Heat flows into the system which increases the volume from 𝑉2 to 𝑉3 and temperature from 𝑇2

to 𝑇3. The pressure remains constant at 𝑝2.

Isentropic Expansion (3 – 4):

Isentropic expansion of high pressure & high-temperature air takes place in the turbine during which the

work is done by the system. The air is expanded isentropically from 𝑝2 to 𝑝1, the temperature falling from

𝑇3 to 𝑇4. No heat flow takes place.

Constant Pressure Heat Rejection (4 – 1):

The air at state point 4 is passed through a heat exchanger and heat is rejected at constant pressure. The

volume decreases from 𝑉4 to 𝑉1 and the temperature from 𝑇4 to 𝑇1. The pressure remains constant at 𝑝1.

7.8.1 Thermal Efficiency of Closed Brayton Cycle

Consider unit mass of air,

Heat supplied during process 2 – 3,

𝑞𝑠 = 𝐶𝑃(𝑇3 − 𝑇2)


𝑞𝑟 = 𝐶𝑃(𝑇4 − 𝑇1)

Work done,


∴ 𝑤𝑛𝑒𝑡 = 𝐶𝑃(𝑇3 − 𝑇2) − CP(T4 − T1)

Thermal efficiency,

𝜂 = 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒


𝑤𝑛𝑒𝑡

𝑞𝑠

∴ 𝜂 = 𝐶𝑃(𝑇3 − 𝑇2) − CP(T4 − T1)

𝐶𝑃(𝑇3 − 𝑇2)

∴ 𝜼 = 𝟏 − (𝑻𝟒 − 𝑻𝟏)

(𝑻𝟑 − 𝑻𝟐) Eq. (7.41)

Take pressure ratio,

𝑟𝑝 =𝑃2

𝑃1=

𝑃3

𝑃4 Eq. (7.42)


𝑇2

𝑇1= (

𝑃2

𝑃1)

𝛾−1𝛾

= (𝑟𝑝)𝛾−1𝛾 Eq. (7.43)


𝑇3

𝑇4= (

𝑃3

𝑃4)

𝛾−1𝛾

= (𝑟𝑝)𝛾−1𝛾 Eq. (7.44)

Thus from Eq. (7.43) and Eq. (7.44),



𝑇2

𝑇1=

𝑇3

𝑇4

∴𝑇4

𝑇1=

𝑇3

𝑇2 Eq. (7.45)

Now from Eq. (7.41),

𝜂 = 1 − (𝑇4 − 𝑇1)

(𝑇3 − 𝑇2)

∴ 𝜂 = 1 − 𝑇1 (

𝑇4

𝑇1− 1)

𝑇2 (𝑇3

𝑇2− 1)

= 1 − 𝑇1 (

𝑇3

𝑇2− 1)

𝑇2 (𝑇3

𝑇2− 1)

(∵ 𝐸𝑞. (7.45))

∴ 𝜂 = 1 − 𝑇1

𝑇2

∴ 𝜂 = 1 − (𝑃1

𝑃2)

𝛾−1𝛾

(∵ 𝐸𝑞. (7.43))

∴ 𝜼 = 𝟏 − (𝟏

𝒓𝒑)

𝜸−𝟏𝜸

Eq. (7.46)

Eq. (7.46) shows that, the thermal efficiency of an ideal

Brayton cycle depends on the pressure ratio (𝒓𝒑) across

the compressor and the specific heat ratio (𝜸) of the

working fluid. The thermal efficiency increases with both

of these parameters.

A plot of thermal efficiency versus the pressure ratio is

shown in Fig.7.13 for 𝛾 = 1.4, which is the standard value

for air at room temperature.

The highest temperature in the cycle is at the end of the

combustion process (State 3), which is limited by the

metallurgical limitations of the turbine blades. This also

limits the pressure ratio.

For a fixed turbine inlet temperature 𝑇3 , the net work

output increases with the pressure ratio up to a certain

limit and then starts to decrease.

Therefore, there should be a compromise between the pressure ratio (thus thermal efficiency) and the net

work output. With less work output, a larger mass flow rate (thus a larger system) is needed to maintain

the same power output, which may not be economical. Hence it is desirable to operate the engine at the

optimum compressor pressure ratio that yields the maximum work per unit mass

7.8.2 The Actual Brayton Cycle

The open cycle gas turbine works on the actual Brayton cycle. In Open cycle gas turbine, the products of

combustion coming out from the turbine are exhausted to the atmosphere as they cannot be used

anymore. The working fluids (air and fuel) must be replaced continuously as they are exhausted into the

atmosphere.

Fig.7.13 – Effect of pressure ratio on efficiency



In practice, it is not possible to achieve either isentropic compression or isentropic expansion because of

internal friction, turbulence and leakage.

If the pressure drop is neglected in the combustion chamber, the actual Brayton cycle on the T-s diagram

is shown by process 1-2’-3-4’ in Fig.7.15.

In the actual cycle, the temperatures at the end of compression and at the end of expansion are higher

than in an ideal case for the same pressure ratio.

Efficiency of Compressor,

𝜂𝑐 =𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑟𝑖𝑠𝑒

𝐴𝑐𝑡𝑢𝑎𝑙 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑟𝑖𝑠𝑒

Efficiency of Turbine,

𝜂𝑡 =𝐴𝑐𝑡𝑢𝑎𝑙 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦

𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦

∴ 𝜼𝒕 =𝒉𝟑 − 𝒉𝟒

′

𝒉𝟑 − 𝒉𝟒=

𝑪𝒑(𝑻𝟑 − 𝑻𝟒′ )

𝑪𝒑(𝑻𝟑 − 𝑻𝟒)=

𝑻𝟑 − 𝑻𝟒′

𝑻𝟑 − 𝑻𝟒 Eq. (7.48)

7.8.3 Pressure Ratio for Maximum Net Work

Take pressure ratio,

𝑟𝑝 =𝑃2

𝑃1=

𝑃3

𝑃4 Eq. (7.49)


𝑇2

𝑇1= (

𝑃2

𝑃1)

𝛾−1𝛾

= (𝑟𝑝)𝛾−1𝛾 Eq. (7.50)


𝑇3

𝑇4= (

𝑃3

𝑃4)

𝛾−1𝛾

= (𝑟𝑝)𝛾−1𝛾 Eq. (7.51)

Thus from Eq. (7.50) and Eq. (7.51),

Fig.7.14 – Schematic of Actual Brayton Cycle

Fig.7.15 – T-s diagram of Actual Brayton Cycle

∴ 𝜼𝒕 =𝒉𝟐 − 𝒉𝟏

𝒉𝟐′ − 𝒉𝟏

=𝑪𝒑(𝑻𝟐 − 𝑻𝟏)

𝑪𝒑(𝑻𝟐′ − 𝑻𝟏)

=𝑻𝟐 − 𝑻𝟏

𝑻𝟐′ − 𝑻𝟏

Eq. (7.47)



∴𝑻𝟐

𝑻𝟏=

𝑻𝟑

𝑻𝟒= (𝒓𝒑)

𝜸−𝟏𝜸 = 𝒙 Eq. (7.52)

Actual compression work,

𝑤𝑐𝑎𝑐𝑡= ℎ2

′ − ℎ1 =ℎ2 − ℎ1

𝜂𝑐=

𝐶𝑝(𝑇2 − 𝑇1)

𝜂𝑐

Actual turbine work,

𝑤𝑡𝑎𝑐𝑡= ℎ3 − ℎ4

′ = 𝜂𝑡(ℎ3 − ℎ4) = 𝐶𝑝(𝑇3 − 𝑇4)𝜂𝑡

Actual net work,

𝑤𝑛𝑒𝑡𝑎𝑐𝑡= 𝑤𝑡𝑎𝑐𝑡

− 𝑤𝑐𝑎𝑐𝑡

∴ 𝑤𝑛𝑒𝑡𝑎𝑐𝑡= 𝐶𝑝(𝑇3 − 𝑇4)𝜂𝑡 −

𝐶𝑝(𝑇2 − 𝑇1)

𝜂𝑐

∴ 𝑤𝑛𝑒𝑡𝑎𝑐𝑡= 𝐶𝑝𝜂𝑡𝑇3 (1 −

𝑇4

𝑇3) −

𝐶𝑝𝑇1

𝜂𝑐(𝑇2

𝑇1− 1)

∴ 𝑤𝑛𝑒𝑡𝑎𝑐𝑡= 𝐶𝑝𝜂𝑡𝑇3 (1 −

1

𝑥) −

𝐶𝑝𝑇1

𝜂𝑐

(𝑥 − 1) (∵ 𝐸𝑞. (7.52))

For maximum work, differentiate above equation w.r.t. 𝑥 while keeping 𝑇1 & 𝑇3 constants,

∴𝑑𝑤𝑛𝑒𝑡𝑎𝑐𝑡

𝑑𝑥= 0

∴ 𝐶𝑝𝜂𝑡𝑇3 (1

𝑥2) −

𝐶𝑝𝑇1

𝜂𝑐= 0

∴ 𝑥 = √𝜂𝑡𝜂𝑐

𝑇3

𝑇1

∴ (𝑟𝑝)𝛾−1𝛾 = (𝜂𝑡𝜂𝑐

𝑇3

𝑇1)

12⁄

∴ 𝑟𝑝𝑜𝑝𝑡= (𝜂𝑡𝜂𝑐

𝑇3

𝑇1)

𝛾2(𝛾−1)⁄

∴ 𝒓𝒑𝒐𝒑𝒕= (𝜼𝒕𝜼𝒄

𝑻𝒎𝒂𝒙

𝑻𝒎𝒊𝒏)

𝜸𝟐(𝜸−𝟏)⁄

Eq. (7.53)

For Ideal cycle,

𝒓𝒑𝒐𝒑𝒕= (

𝑻𝒎𝒂𝒙

𝑻𝒎𝒊𝒏)

𝜸𝟐(𝜸−𝟏)⁄

For maximum work, the temperature 𝑇1 at compressor entry should be as low as possible and the

temperature 𝑇3 at the entry to the turbine should be as high as possible.

The compressor inlet temperature is normally at atmospheric temperature (say 288K at sea level), while

the turbine inlet temperature is decided by metallurgical considerations (the maximum value of about

1000K that the metal can withstand).



The performance of an actual gas turbine plant depends upon both the pressure ratio and the temperature

ratio.

Performance of the Brayton cycle can be improved by using multi-stage compression with inter-cooling,

multi-stage expansion with reheating and regeneration.

7.9 Solved Examples

Ex. 7.1 [GTU; Jun-2010; 3 Marks]

An engine uses 6.5 Kg of oil per hour of calorific value of 30,000 kJ/Kg. If the Brake power

of engine is 22 kW and mechanical efficiency is 85% calculate (a) indicate thermal efficiency

(b) Brake thermal efficiency (c) Specific fuel consumption in Kg/B.P/hr.

Solution: Given Data:

�̇�𝑓 = 6.5 𝑘𝑔 ℎ𝑟⁄

𝐶. 𝑉.= 30000 𝑘𝐽 𝑘𝑔⁄

𝐵. 𝑃.= 22𝑘𝑊

𝜂𝑚 = 85%

To be Calculated:

a) 𝜂𝑖𝑡ℎ =?

b) 𝜂𝑏𝑡ℎ =?

c) 𝐵𝑆𝐹𝐶 =?

Indicated Power,

𝜂𝑚 =𝐵. 𝑃.

𝐼. 𝑃.

∴ 𝐼. 𝑃. =22

0.85= 25.8824 𝑘𝑊

Indicated Thermal Efficiency:

𝜂𝑖𝑡ℎ =𝐼. 𝑃.

�̇�𝑓 × 𝐶. 𝑉.

∴ 𝜂𝑖𝑡ℎ =25.8824

(6.5

3600) × 30000

∴ 𝜼𝒊𝒕𝒉 = 𝟎. 𝟒𝟕𝟕𝟖 = 𝟒𝟕. 𝟕𝟖%

Break Thermal Efficiency:

𝜂𝑏𝑡ℎ =𝐵. 𝑃.

�̇�𝑓 × 𝐶. 𝑉.

∴ 𝜂𝑏𝑡ℎ =22

(6.5

3600) × 30000

∴ 𝜼𝒃𝒕𝒉 = 𝟎. 𝟒𝟎𝟔𝟐 = 𝟒𝟎. 𝟔𝟐%

Break Specific Fuel Consumption:

𝐵𝑆𝐹𝐶 =�̇�𝑓(𝑘𝑔 ℎ𝑟⁄ )

𝐵. 𝑃. (𝑘𝑊)

∴ 𝐵𝑆𝐹𝐶 =6.5

22

∴ 𝑩𝑺𝑭𝑪 = 𝟎. 𝟐𝟗𝟓𝟓 𝒌𝒈 𝒌𝑾𝒉⁄



Ex. 7.2 [Ex 13.3; P. K. Nag]

In a Carnot cycle, the maximum pressure and temperature are limited to 18 bar and 410°C.

The ratio of isentropic compression is 6 and isothermal expansion is 1.5. Assuming the

volume of the air at the beginning of isothermal expansion as 0.18 m3, determine :

(a) The temperature and pressures at main points in the cycle.

(b) Change in entropy during isothermal expansion.

(c) Mean thermal efficiency of the cycle.

(d) Mean effective pressure of the cycle.

(e) The theoretical power if there are 210 working cycles per minute.

Solution:

Given Data:

𝑝1 = 18 𝑏𝑎𝑟

𝑇𝐻 = 𝑇1 = 𝑇2 = 410°𝐶 𝑉4

𝑉1= 6

𝑉2

𝑉1= 1.5

𝑉1 = 0.18 𝑚3

No. of cycles = 210 /min

Assume, 𝛾 = 1.4 for air

To be Calculated:

a) 𝑇𝐿 , 𝑝2, 𝑝3, 𝑝4 =?

b) ∆𝑆 =?

c) 𝜂𝑡ℎ =?

d) 𝑝𝑚 =?

e) 𝑃 =?

Temperatures & Pressures at the main point of the cycle:

For Process 4-1 (Isentropic Compression),

𝑇𝐿

𝑇𝐻 = (

𝑉1

𝑉4)𝛾−1

∴ 𝑇𝐿 = 683 × (1

6)1.4−1

∴ 𝑻𝑳 = 𝑻𝟑 = 𝑻𝟒 = 𝟑𝟑𝟑. 𝟓𝟒𝟗𝟒 𝑲

Also,

𝑝1𝑉1𝛾

= 𝑝4𝑉4𝛾

∴ 𝑝4 = 𝑝1 × (𝑉1

𝑉4)𝛾

∴ 𝑝4 = 18 × (1

6)1.4

∴ 𝒑𝟒 = 𝟏. 𝟒𝟔𝟓 𝒃𝒂𝒓

For Process 1-2 (Isothermal Expansion),

𝑝1𝑉1 = 𝑝2𝑉2

∴ 𝑝2 = 𝑝1 × (𝑉1

𝑉2)

∴ 𝑝2 = 18 × (1

1.5)

∴ 𝒑𝟐 = 𝟏𝟐 𝒃𝒂𝒓

For Process 2-3 (Isentropic Expansion),



𝑝2𝑉2𝛾

= 𝑝3𝑉3𝛾

∴ 𝑝3 = 𝑝2 × (𝑉2

𝑉3)𝛾

∴ 𝑝3 = 𝑝2 × (𝑉1

𝑉4)𝛾

(∵𝑉4

𝑉1=

𝑉3

𝑉2)

∴ 𝑝3 = 12 × (1

6)1.4

∴ 𝒑𝟑 = 𝟎. 𝟗𝟕𝟔𝟕 𝒃𝒂𝒓

Change in Entropy:

From T-S diagram,

𝑄𝑠 = 𝑇𝐻 × (𝑆2 − 𝑆1)

∴ ∆𝑆 = 𝑆2 − 𝑆1 =𝑄𝑠

𝑇𝐻=

𝑝1𝑉1 ln𝑉2

𝑉1

𝑇𝐻

∴ ∆𝑆 =18 × 105 × 0.18 × ln 1.5

683

∴ ∆𝑺 = 𝟏𝟗𝟐. 𝟑𝟒𝟑𝟔 𝑱/𝑲

Thermal Efficiency of the Cycle:

Heat Supplied,

𝑄𝑠 = 𝑝1𝑉1 ln𝑉2

𝑉1

∴ 𝑄𝑠 = 𝑇𝐻 × (𝑆2 − 𝑆1)

∴ 𝑄𝑠 = 683 × 192.3436

∴ 𝑄𝑠 = 131370.6788 𝐽 = 131.370 𝑘𝐽

Heat Rejected,

𝑄𝑟 = 𝑝4𝑉4 ln𝑉3

𝑉4

∴ 𝑄𝑟 = 𝑇𝐿 × (𝑆3 − 𝑆4) = 𝑇𝐿 × (𝑆2 − 𝑆1)

∴ 𝑄𝑟 = 333.5494 × 192.3436

∴ 𝑄𝑟 = 64156.0923 𝐽 = 64.1561 𝑘𝐽

Efficiency,

𝜂 =𝑄𝑠 − 𝑄𝑟

𝑄𝑠=

131.370 − 64.1561

131.370

∴ 𝜼 = 𝟎. 𝟓𝟏𝟏𝟔 = 𝟓𝟏. 𝟏𝟔%

Mean Effective Pressure of the Cycle:

𝑉4

𝑉1= 6 &

𝑉2

𝑉1= 1.5

Also,

𝑉4

𝑉1=

𝑉3

𝑉2



∴𝑉3

𝑉2×

𝑉2

𝑉1= 6 × 1.5

∴𝑉3

𝑉1= 9

Swept Volume,

𝑉𝑠 = 𝑉3 − 𝑉1 = 9𝑉1 − 𝑉1 = 8𝑉1

∴ 𝑉𝑠 = 8 × 0.18

𝑉𝑠 = 1.44 𝑚3

Mean Effective Pressure,

𝑝𝑚 =𝑁𝑒𝑡 𝑊𝑜𝑟𝑘

𝑆𝑤𝑒𝑝𝑡 𝑉𝑜𝑙𝑢𝑚𝑒=

𝑄𝑠 − 𝑄𝑟

𝑉𝑠

∴ 𝑝𝑚 =131370.6788 − 64156.0923

1.44

∴ 𝒑𝒎 = 𝟒𝟔𝟔𝟕𝟔. 𝟕𝟗𝟔𝟏 𝑷𝒂

Power of the Engine:

𝑃 =𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒

𝐶𝑦𝑐𝑙𝑒×

𝑁𝑜. 𝑜𝑓 𝐶𝑦𝑐𝑙𝑒

𝑆𝑒𝑐

∴ 𝑃 = (131370.6788 − 64156.0923) ×210

60

∴ 𝑷 = 𝟐𝟑𝟓𝟐𝟓𝟏. 𝟎𝟓𝟐𝟖 𝑾 = 𝟐𝟑𝟓. 𝟐𝟓𝟏 𝒌𝑾

Ex. 7.3 [GTU; Jan-2015; 7 Marks]

In an I C Engine working with the Otto cycle, the cylinder diameter is 250mm and a stroke is

375mm. If the clearance volume is 0.00263m3, and the initial pressure and temperature are

1bar and 50°C, calculate (a) The air standard efficiency and (b) Mean effective pressure of

the cycle. The maximum cycle pressure is limited to 25bar.

Solution:

Given Data:

𝐷 = 0.250 𝑚

𝐿 = 0.375 𝑚

𝑉𝑐 = 0.00263 𝑚3

𝑝1 = 1 𝑏𝑎𝑟

𝑇1 = 323 𝐾

𝑝3 = 25 𝑏𝑎𝑟

To be Calculated:

a) 𝜂 =?

b) 𝑝𝑚 =?

Swept Volume,

𝑉𝑠 =𝜋

4𝐷2𝐿

∴ 𝑉𝑠 =𝜋

40.2502 × 0.375



∴ 𝑉𝑠 = 0.0184 𝑚3

Total Volume,

𝑉1 = 𝑉𝑠 + 𝑉𝑐

∴ 𝑉1 = 0.0184 + 0.00263

∴ 𝑉1 = 0.02103 𝑚3

Compression Ratio,

𝑟 =𝑉1

𝑉2=

0.02103

0.00263

∴ 𝑟 = 7.9961

Air Standard Efficiency:

𝜂 = 1 −1

𝑟𝛾−1

∴ 𝜂 = 1 −1

(7.9961)1.4−1

∴ 𝜼 = 𝟎. 𝟓𝟔𝟒𝟔 = 𝟓𝟔. 𝟒𝟔%


𝑇2

𝑇1 = (𝑟)𝛾−1

∴ 𝑇2 = 323 × (7.9961)1.4−1

∴ 𝑇2 = 741.9144 𝐾

And,

𝑝1𝑉1𝛾

= 𝑝2𝑉2𝛾

∴ 𝑝2 = 𝑝1 × (𝑉1

𝑉2)𝛾

∴ 𝑝2 = 1 × (7.9961)1.4

∴ 𝑝2 = 18.3666 𝑏𝑎𝑟

For Process 2-3 (Constant Volume Heat Addition)

𝑇3

𝑇2=

𝑃3

𝑃2 (∵ 𝑉2 = 𝑉3)

∴ 𝑇3 = 741.9144 ×25

18.3666

∴ 𝑇3 = 1009.869 𝐾

Mass of Air,

𝑝1𝑉1 = 𝑚𝑅𝑇1

∴ 𝑚 =1 × 105 × 0.02103

0.287 × 103 × 323

∴ 𝑚 = 0.02268 𝑘𝑔

Heat Supplied,

𝑄𝑠 = 𝑚𝐶𝑣(𝑇3 − 𝑇2)

∴ 𝑄𝑠 = 0.02268 × 0.718 × 103 × (1009.869 − 741.9144)

∴ 𝑄𝑠 = 4363.437 𝐽 = 4.3634 𝑘𝐽



Net Work,

𝜂 =𝑊𝑛𝑒𝑡

𝑄𝑠

∴ 𝑊𝑛𝑒𝑡 = 0.5646 × 4363.437

∴ 𝑊𝑛𝑒𝑡 = 2463.5965 𝐽

Mean Effective Pressure:

𝑝𝑚 =𝑊𝑛𝑒𝑡

𝑉𝑠

∴ 𝑝𝑚 =2463.5965

0.0184

∴ 𝑝𝑚 = 133891.1141 𝑃𝑎

∴ 𝒑𝒎 = 𝟏. 𝟑𝟑𝟖𝟗 𝒃𝒂𝒓

Ex. 7.4 [GTU; Nov-2011; 7 Marks]

In an air standard diesel cycle the compression ratio is 16. At the beginning of isentropic

compression the temperature is 15 °C and pressure is 0.1 MPa. Heat is added until the

temperature at the end of constant pressure process is 1480°C. Calculate: (a) cut off ratio,

(b) cycle efficiency and (c) M. E. P.

Solution:

Given Data:

𝑟 = 16

𝑝1 = 0.1 𝑀𝑃𝑎 = 1𝑏𝑎𝑟

𝑇1 = 288 𝐾

𝑇3 = 1753 𝐾

To be Calculated:

a) 𝜌 =?

b) 𝜂 =?

c) 𝑝𝑚 =?

Cut off Ratio:


𝑇2

𝑇1 = (𝑟)𝛾−1

∴ 𝑇2 = 288 × (16)1.4−1

∴ 𝑇2 = 873.0527 𝐾

For Process 2-3 (Constant Pressure Heat Addition)

𝑉2

𝑇2=

𝑉3

𝑇3 (∵ 𝑝2 = 𝑝3)

∴𝑉3

𝑉2=

1753

873.0527

∴ 𝝆 =𝑽𝟑

𝑽𝟐= 𝟐. 𝟎𝟎𝟕

Cycle Efficiency:

𝜂 = 1 − 1

𝑟𝛾−1 [

(𝜌𝛾 − 1)

𝛾(𝜌 − 1)]



∴ 𝜂 = 1 − 1

160.4 [

(2.0071.4 − 1)

1.4(2.007 − 1)]

∴ 𝜼 = 𝟎. 𝟔𝟏𝟑𝟒 = 𝟔𝟏. 𝟑𝟒%

Mean Effective Pressure:

Heat Supplied (per unit mass),

𝑞𝑠 = 𝐶𝑃(𝑇3 − 𝑇2)

∴ 𝑞𝑠 = 1.005(1753 − 873.0527)

∴ 𝑞𝑠 = 884.347 𝑘𝐽

𝑘𝑔

Net Work,

𝜂 =𝑤𝑛𝑒𝑡

𝑞𝑠

∴ 𝑤𝑛𝑒𝑡 = 0.6134 × 884.347

∴ 𝑤𝑛𝑒𝑡 = 542.4584 𝑘𝐽

𝑘𝑔

Swept Volume,

𝑉𝑠 = 𝑉1 − 𝑉2 = 𝑉1 (1 − 𝑉2

𝑉1)

∴ 𝑣𝑠 =𝑅𝑇1

𝑝1(1 −

1

𝑟)

∴ 𝑣𝑠 =287 × 288

1 × 105(1 −

1

16)

∴ 𝑣𝑠 = 0.7749 𝑚3

𝑘𝑔

Mean Effective Pressure,

𝑝𝑚 =𝑊𝑛𝑒𝑡

𝑉𝑠

∴ 𝑝𝑚 =542.4584 × 103

0.7749

∴ 𝑝𝑚 = 700036.6499 𝑃𝑎

∴ 𝒑𝒎 = 𝟕. 𝟎𝟎𝟎𝟑 𝒃𝒂𝒓

7.10 References

1) Mahesh M. Rathore, “Thermal Engineering”, 4th Edition, 2010, Tata McGraw-Hill Publication.

2) Yunus Cengel & Michael A. Boles, “Thermodynamics – An Engineering Approach”, 9th Edition, 2019,

McGraw-Hill Education (India) Pvt. Ltd.

3) Michel J. Moran, Howard N. Shapiro, Daisie D. Boettner & Margaret B. Bailey, “Fundamentals of

Engineering Thermodynamics” 9th Edition, 2018, Wiley India (P) Ltd.

4) R. K. Rajput, “Engineering Thermodynamics”, 3rd Edition, 2007, Laxmi Publications (P) Ltd.

contents · 7.4 the otto cycle the otto cycle is also known as constant volume (isochoric) cycle...

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