1020-Diploma in Mechanical Engineering - 4020340- III Semester - Thermal Engineering I -Unit 2 :Thermodynamic Air Cycle - Introduction & Carnot Cycle PPT Prepared By… U. ARAVIND, B.E., M.I.S.T.E., LECTURER / MECHANICAL ENGINEERING LAKSHMI AMMAL POLYTECHNIC COLLEGE K.R. NAGAR, KOVILPATTI - 628 503 THOOTHUKUDI DISTRICT, TAMIL NADU Mobile No: 9943244068 Email Id: aravindulaganathanmech@gmail.com aravindmech@lapc.in

1. 6 April 2022 U.Aravind, Lect/Mech, LAPC 1
Lakshmi Ammal Polytechnic College
Department of Mechanical Engineering
Subject Code : 4020340
Year & Semester : II / III
Scheme : N
By…
U.Aravind,
Lecturer/Mechanical Engineering,
LAPC, Kovilpatti.

2. 6 April 2022 U.Aravind, Lect/Mech, LAPC 2
Lakshmi Ammal Polytechnic College
Department of Mechanical Engineering
Subject Code : 4020340
Subject Name : Thermal Engineering I
Year & Semester : II / III
Scheme : N
Unit : 2
Thermodynamics Air Cycles

3. Thermodynamic Air Cycles:
• The thermodynamic cycles consists of
a series of thermodynamic operations,
which takes place in a certain order
and the initial conditions are restored
at the end of the process.
• These process can be plotted on p-V
diagram and T-s diagram.
6 April 2022 U.Aravind, Lect/Mech, LAPC 3

4. Thermodynamic Air Cycles:
• The area enclosed by the complete curve
in the p-V diagram gives Workdone
during the complete cycle.
• The area enclosed by the complete curve
in the T-s diagram gives Heat Transfer
during the complete cycle.
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5. Air Cycle & Air Standard Efficiency:
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6. Efficiency of Cycles:
Thermal Efficiency:
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8. Efficiency of Cycles:
Relative Efficiency:
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11. Thermodynamic Cycles
• Carnot Cycle(Constant Thermal)
• Otto Cycle(Constant Volume)
• Joule/Brayton Cycle(Constant
Pressure)
• Diesel Cycle
• Dual Combustion Cycle
• Rankine Cycle
• Stirling Cycle
• Ericssion Cycle
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12. I. Carnot Cycle:
• It is an imaginary cycle
• It is introduced by NL Sadi Carnot
(1824)
• This process consists of two
Isothermal & two Isentropic process
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13. I. Carnot Cycle:
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14. I. Carnot Cycle:
Process 1-2 : Isothermal Heat Addition
• Air is heated slowly at constant temperature
Process 2-3 : Adiabatic Expansion
• Air is expanded adiabatically and it does
work on piston
• Temperature reduced from T2 to T3
Process 3-4 : Isothermal Heat Rejection
• Air is cooled slowly at constant temperature
Process 4-1 : Adiabatic Compression
• Air is compressed adiabatically
• Temperature raised from T4 to T1
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15. 6 April 2022 U.Aravind, Lect/Mech, LAPC 15
• Air Standard Efficiency of Carnot
Cycle
𝛈𝐂 =(1 -
𝑻𝟏
𝑻𝟐
) x 100
𝛈𝐂 =(1 -
𝟏
𝒓γ−𝟏) x 100
• Efficiency
𝛈 =
𝑾
𝐐𝐒
x 100
𝑻𝟏 - Lower temperature
𝑻𝟐 - Higher temperature

16. Problems on Carnot
Cycle
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17. 1. A carnot engine works between 650 K and 310 K produces 150 KJ
of work. Find a) Thermal Efficiency b) Heat added during the
process.
Given Data:
• LowerTemperature, T1 = 310 K
• Higher Temperature, T2 = 650 K
• Work, W = 150 KJ
To Find:
• Thermal Efficiency, ηC
• Heat added, QS
Solution:
I. To Find Thermal Efficiency, ηC
• Air Standard Efficiency of Carnot Cycle
• 𝛈𝐂 = (1 -
𝑻𝟏
𝑻𝟐
) x 100
= (1 -
𝟑𝟏𝟎
𝟔𝟓𝟎
) x 100 17
6 April 2022 U.Aravind, Lect/Mech, LAPC
𝛈𝐂= 52.307 %

18. II. To Find Heat added, 𝑄𝑆
Air Standard Efficiency
•𝛈 =
𝑾
𝐐𝐒
x 100
•𝐐𝐒 =
𝑾
𝛈
x 100
=
𝟏𝟓𝟎
𝟓𝟐.𝟑𝟎𝟕
x 100
18
6 April 2022 U.Aravind, Lect/Mech, LAPC
𝐐𝐒= 286.78 KJ

19. 2. The temperature limits for a carnot cycle using air as working
fluid are 500⁰C and 10⁰C. Calculate the efficiency of the cycle and
the ratio of adiabatic expansion. Assume γ = 1.4 for air.
Given Data:
• LowerTemperature, T1 = 10°C = 10 + 273 = 283 K
• Higher Temperature, T2= 500°C = 500 + 273 = 773 K
• γ = 1.4
To Find:
• Efficiency, ηC
• Ratio of Adiabatic expansion, r
Solution:
I. To Find Thermal Efficiency, ηC
• Air Standard Efficiency of Carnot Cycle
• 𝛈𝐂 = (1 -
𝑻𝟏
𝑻𝟐
) x 100
= (1 -
𝟐𝟖𝟑
𝟕𝟕𝟑
) x 100 19
6 April 2022 U.Aravind, Lect/Mech, LAPC
𝛈𝐂= 63.389 %

20. II. To Find Ratio of adiabatic expansion
Air Standard Efficiency
•𝛈𝐂 =(1 -
𝟏
𝒓γ−𝟏) x 100
• 1 -
𝟏
𝒓γ−𝟏 =
𝛈𝐂
100
•
𝟏
𝒓γ−𝟏 = 1 -
𝛈𝐂
100
•
𝟏
𝒓𝟏.𝟒−𝟏 = 1 -
𝟔𝟑.𝟑𝟖𝟗
100
•
𝟏
𝒓𝟎.𝟒 = 0.36616
•
𝟏
0.36616
= 𝒓𝟎.𝟒
•𝒓𝟎.𝟒
= 2.731
•r = 2.731
𝟏
𝟎.𝟒 20
U.Aravind, Lect/Mech, LAPC
𝒓 = 12.33

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