1. Gas Power Cycles

2. Multi-Cylinder Spark Ignition Engine

3. Power Cycles Ideal Cycles, Internal Combustion
Otto cycle, spark ignition
Diesel cycle, compression ignition
Brayton cycles
Combined cycle

4. Ideal Cycles Assumptions
Air is the working fluid, circulated in a closed loop, is an ideal gas
exhaust and air intake are substituted with heat transfer from the system to the surroundings
combustion is replaced by heat transfer from an external source to the system
all processes are internally reversible.
gas specific heat is constant
Cycle does not involve any friction
Pipes connecting components have no heat loss
Neglecting changes in kinetic and potential energy

5. Carnot Cycle

6. Carnot Cycle

7. Otto Cycle
Nicolaus August Otto the inventor of the four-stroke cycle was born on 14th June 1831 in Germany. In 1862 he began first experiments with four-strokes engines. The first four-stroke engines is shown. He died on 26th January 1891 in Cologne

8. Engine Terms Bottom-dead center (BDC) –
piston position where volume is maximum
Top-dead center (TDC) –
piston position where volume is minimum
Clearance volume –
minimum cylinder volume (VTDC = V2)
Compression ratio (r)
Displacement volume
ME 152 8

9. Engine Terms
Mean effective pressure (MEP)

10. Processes of Otto Cycle
Four internally reversible processes
1-2 Isentropic compression
2-3 Constant-volume heat addition
3-4 Isentropic expansion
4-1 Constant-volume heat rejection

11. Otto Cycle

13. Ideal Otto Cycle (cont.)
1st law for this cycle:
energy conversion efficiency is:

14. Ideal Otto Cycle (cont.)
for an isentropic process:
in case of an ideal gas:
Compression ratio

15. Ideal Otto Cycle (cont.)

22. Diesel Cycle
Rudolf Diesel (1858 – 1913) was born in Paris in After graduation he was employed as a refrigerator engineer. However, his true love was in engine design. In 1893, he published a paper describing an engine with combustion within a cylinder, the internal combustion engine. In 1894, he filed for a patent for his new invention, the diesel engine. He operated his first successful engine in 1897.

23. Spark or Compression Ignition
Spark (Otto), air-fuel mixture compressed (constant-volume heat addition)
Compression (Diesel), air compressed, then fuel added (constant-pressure heat addition)

24. Early CI Engine Cycle vs Diesel Cycle
FUEL
Fuel injected
at TC A I R
Fuel/Air
Mixture
Combustion
Products
Actual
Cycle
Intake
Stroke
Compression
Stroke
Power
Stroke
Exhaust
Stroke
Qin
Qout
Air
Diesel
Cycle TC BC
Compression
Process
Const pressure
heat addition
Process
Expansion
Process
Const volume
heat rejection
Process

25. Diesel Cycle 1-2......isentropic (Adiabatic) compression.
Addition of heat at constant pressure.
isentropic (Adiabatic) expansion.
Rejection of heat at constant volume.

26. Diesel Cycle Analysis Thermal efficiency Heat addition
(process 2-3, P = const)
Heat rejection (process 4-1, v = const)

28. Cold-Air Standard Thermal Efficiency
Otto Cycle
Diesel Cycle

29. Diesel Cycle

35. Brayton Cycle (Joule Cycle)
Usually used in gas turbines
Basis of jet engines

36. Operation of an open cycle gas turbine engine
2.8 Brayton cycle : The Ideal Cycle for Gas Turbine Engines
Operation of an open cycle gas turbine engine

37. Brayton Cycle Four internally reversible processes
1-2 Isentropic Compression (compressor)
2-3 Constant-pressure heat addition
3-4 Isentropic expansion (turbine)
4-1 Constant-pressure heat rejection

38. Brayton Cycle
Gas turbine cycle
Open vs closed system model

39. Brayton Cycle Analyze as steady-flow process So
With cold-air-standard assumptions

40. Brayton Cycle
Since processes 1-2 and 3-4 are isentropic, P2 = P3 and P4 = P1
where

41. Brayton Cycle

46. Actual Gas-Turbine Cycles
For actual gas turbines, compressor and turbine are not isentropic

47. Regeneration

48. Regeneration Regenerator Effectiveness
Use heat exchanger called regenerator
Counter flow
Regenerator Effectiveness

49. The Brayton Cycle: Reheat
A high pressure and a low pressure turbine are used in a Reheat Brayton Cycle .
A reheater is a heat exchanger that increases the power output without increasing the maximum operating temperature but it does not increase the efficiency of the cycle
The capital cost to build a reheater alone cannot be justified because the thermal efficiency does not increase.
      

50. Brayton with Intercooling, Reheat, & Regeneration

52. T1 = T3
T6 = T8
P2 = P4
P1 = P3

54. Decrease the total compression work
Improve the back work ratio
If the number of compression and expansion stages is increased, the ideal turbine cycle with intercooling, reheating and regeneration approaches the Ericsson cycle

55. The volumetric flow rate entering the cycle.
Example (9.74): Air enters the compressor of a gas turbine at 100 kPa, K. The air is compressed in two stages to 900 kPa, with intercooling to 300 K between the stages at a pressure of 300 kPa. The turbine inlet temperature is 1480 K and the expansion occurs in two stages, with reheat to 1420 K between the stages at a pressure of 300 kPa. The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
P (kPa)
100
900 Pr
h (kJ/kg)
The volumetric flow rate entering the cycle.
The thermal efficiency of the cycle.
The back work ratio.

56. The thermal efficiency of the cycle. The back work ratio. a b
Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine
The thermal efficiency of the cycle.
The back work ratio.
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
p (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
Find state a, Process 1 – a is
Isentropic compression
From the table haS = kJ/kg
Using the isentropic compressor efficiency:

57. The thermal efficiency of the cycle. The back work ratio. a b
Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine
The thermal efficiency of the cycle.
The back work ratio.
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
P (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
432.42
Find state a, Process b – 2 is
Isentropic compression
From the table h2S = kJ/kg
Using the isentropic compressor efficiency:

58. The thermal efficiency of the cycle. The back work ratio. a b a b
Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine
The thermal efficiency of the cycle.
The back work ratio.
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
p (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
432.42
1275.4
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
p (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
432.42
Find state a, Process 3 – c is
Isentropic expansion
From the table hcS = kJ/kg
Using the isentropic turbine efficiency:

59. Find state a, Process d – 4 is Isentropic expansion
Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
p (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
432.42
1275.4
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
p (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
432.42
1275.4
Find state a, Process d – 4 is
Isentropic expansion
From the table h4S = kJ/kg
Using the isentropic turbine efficiency:

60. The thermal efficiency of the cycle. The back work ratio.
Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
p (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
432.42
1275.4
The thermal efficiency of the cycle.
The back work ratio.
Determine the mass flow rate

61. The volumetric flow rate. The thermal efficiency of the cycle.
Example (9.74): Air enters the compressor of a gas turbine at 100 kPa, K. The air is compressed in two stages to 900 kPa, with intercooling to 300 K between the stages at a pressure of 300 kPa. The turbine inlet temperature is 1480 K and the expansion occurs in two stages, with reheat to 1420 K between the stages at a pressure of 300 kPa. The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
p (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
432.42
1275.4
The volumetric flow rate.
The thermal efficiency of the cycle.
The back work ratio.

62. The volumetric flow rate. The thermal efficiency of the cycle.
Example (9.74): Air enters the compressor of a gas turbine at 100 kPa, K. The air is compressed in two stages to 900 kPa, with intercooling to 300 K between the stages at a pressure of 300 kPa. The turbine inlet temperature is 1480 K and the expansion occurs in two stages, with reheat to 1420 K between the stages at a pressure of 300 kPa. The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
p (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
432.42
1275.4
The volumetric flow rate.
The thermal efficiency of the cycle.
The back work ratio.

63. The volumetric flow rate. The thermal efficiency of the cycle.
Example (9.74): Air enters the compressor of a gas turbine at 100 kPa, K. The air is compressed in two stages to 900 kPa, with intercooling to 300 K between the stages at a pressure of 300 kPa. The turbine inlet temperature is 1480 K and the expansion occurs in two stages, with reheat to 1420 K between the stages at a pressure of 300 kPa. The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine
State 1 a b 2 3 c d 4
T (K)
300
1480
1420
p (kPa)
100
900 pr
1.3860
568.8
478.0
h (kJ/kg)
300.19
432.42
1275.4
The volumetric flow rate.
The thermal efficiency of the cycle.
The back work ratio.