This document defines important terms related to thermodynamic air cycles such as cylinder bore, stroke length, clearance volume, and compression ratio. It then summarizes several common thermodynamic cycles including Carnot, Otto, and Diesel cycles. For the Carnot cycle, it shows the calculations for work done and efficiency. For the Otto cycle, it outlines the four stages and provides equations to calculate efficiency based on compression ratio and temperature changes between stages.

1. Thermodynamic air Cycles
Important Terms
1. Cylinder bore: The inner diameter of the cylinder.
2. Stroke length: The distance between TDC and BDC.
3. Clearance volume: The volume occupied by the working fluid,
when piston reaches TDC.
4. Swept volume: The volume swept by the piston, when it
moves between TDC to BDC.
5. Total cylinder volume: The volume occupied by the working
fluid, when the piston is at the BDC.
6. Compression ratio: The ratio of total cylinder volume to the
clearance volume.
7. Mean effective pressure: It is the ratio of work done to the
stroke volume.

2. Efficiency of a cycle η =
Work done
Heat supplied
=
Heat supplied
Heat rejectedHeat supplied -
Types of Thermodynamic Cycles:
1. Carnot cycle 2.Stirling cycle
3. Ericsson cycle 4. Joule cycle
5. Otto cycle 6. Diesel cycle
7. Dual combustion cycle

3. First stage(Isothermal expansion):
Heat supplied = work by the air during isothermal expansion
Carnot Cycle
Q1-2 = p1v1loge(v2/v1) = mRT1loge(v2/v1) = mRT1loger, where r = Expansion ratio = v2/v1
Second stage(Isentropic expansion):
Decrease in internal energy = Work done by the air during adiabatic expansion
P2v2 – p3v3
γ-1=
mRT2 – mRT3
γ-1=
mR(T1 – T3)
γ-1=
IC
HB
CB
=

4. Fourth stage(Isentropic compression):
Increase in internal energy = Work done on the air during adiabatic compression
P1v1 – p4v4
γ-1=
mRT1 – mRT4
γ-1=
mR(T1 – T3)
γ-1=
Now Work done, W = Heat supplied – Heat rejected
= mRT1loger - mRT3loger = m R loger (T1 - T3)
And efficiency η =
Work done
Heat supplied
=
m R loger (T1 - T3)
mRT1loger
=
T1 - T3
T1
= 1-
T3
T1
Q3- 4 = p3v3loge(v3/v4) = mRT3loge(v3/v4) = mRT3loger
Third stage(Isothermal compression):
Heat rejected = Work done on the air during isothermal compression

5. A Carnot engine operates between two reservoirs at temperature T3 and T3 . The work
Output of the engine is 0.6 times the heat rejected. The difference in temperatures
Between the source and the sink is 200 K. Calculate the thermal efficiency, source
Temperature and the sink temperature.
We know thermal efficiency η=
Work done
Heat supplied
Work done
Work done + Heat rejected
=
Given W = .6×Heat rejected Q3- 4= 0.6
T1 - T2 = 200 K
Q3- 40.6
Q3- 40.6 + Q3- 4
= = .375 = 37.5%
==
T1 - T3
T1
And .375
200
T1
T3Now,T1 = 533.3K and = 333.3K

6. Otto Cycle
First stage(Isentropic expansion): The air is expanded reversibly and adiabatically from initial
Temperature T3 to a Temperature T4.
Second stage(Constant volume cooling): The air is cooled at constant volume from temperature
T4 to a Temperature T1. Heat rejected Q4- 1 = m cv(T4 – T1)
Third stage(Isentropic compression): The air is compressed reversibly and adiabatically from
Temperature T1 to a Temperature T2.
Fourth stage(Constant volume heating): The air is now heated at constant volume from
Temperature T2 to a Temperature T3. Heat absorbed Q2- 3 = m cv(T3 – T2)

7. We know that work done = Heat absorbed – Heat rejected
= m cv(T3 – T2) – m cv(T4 – T1)
Ideal or air standard Efficiency of a cycle η =
Work done
Heat absored
m cv(T3 – T2) – m cv(T4 – T1)
=
m cv(T3 – T2)
= 1 -
T4 – T1
T3 – T2
= 1 -
T4
T1
– 1T1
T3
T2
– 1T2
-----(i)
For isentropic expansion
v4/v3 = r = Expansion ratio
=
T3
T4 v3
v4
γ-1
1
r
γ-1
= -----------------------(ii)

8. For isentropic compression
v4/v3 = r = Expansion ratio = compression ratio = v1/v2
T2
T1 v2
v1
γ-1
1
r
γ-1
= -----------------------(iii)=
=
T4
T3
T1
T2
1
r
γ-1
= or =
T4
T1
T3
T2
1
r
γ-1
=
Substituting this value in equation ---------------(i) we get
η = 1-
T1
T2
= 1-
1
r
γ-1

9. 1. In an Otto cycle , the temperature at the beginning and end of the isentropic compression
are 316 K and 596 K respectively. Determine the air standard efficiency and the compression
ratio. Take γ = 1.4
Solution. Given T1 = 316 K, T2 = 596 K, γ = 1.4
Let r = Compression ratio = v1/v2
We know that for isentropic compression
T2
T1 v2
v1
γ-1
1
r
γ-1
==
r = 4.885
316
596
=
1
r
1.4 -1
η = 1-
1
r
γ-1 η = 1 -
1
4.885
1.4-1
Air standard efficiency
or
= .47 = 47%η