Gas Power Cycles
9.1 ❐ Introduction
The gas as the working fluid does not undergo any phase change. The engines working on the gas cycles may be either cyclic or non-cyclic. The working fluid is assumed to obey all laws of a perfect gas.
In most of the gas power cycles, the working fluid consists mainly of air. Therefore, it is convenient to analyse the gas power cycles by devising an idealized cycle known as Air Standard Cycle, in which the working fluid is pure air.
The following assumptions are made in the air-standard cycles:
- The air as the working fluid follows the perfect gas law: pV = mRT.
- The working fluid is homogeneous throughout and no chemical reaction takes place.
- The specific heats of air do not vary with temperature.
- The mass of air in the cycle remains fixed.
- The combustion process is replaced by an equivalent heat addition process.
- The exhaust process is replaced by an equivalent heat rejection process.
- All processes are internally reversible.
The thermal efficiency, ηth, of air standard cycle is given by
The thermal efficiency is also called air-standard efficiency (ηa)
Relative efficiency, (ηr)= Actual thermal efficiency/Air standard efficiency.
9.2 ❐ Piston-cylinder Arrangement
The piston-cylinder arrangement is shown in Fig. 9.1. The various definitions are:
Bore (d): is the cylinder or piston diameter.
Stroke (L): is the distance moved by the piston in one direction. It is equal to twice the crank radius.
Top Dead centre (T.D.C.): is the extreme position of the piston near to the head of cylinder. It is also called inner dead centre (I.D.C) position
Bottom Dead centre (B.D.C.): is the extreme position of the piston opposite to the head of
cylinder. It is also called outer dead centre (O.D.C) position
Clearance Volume (Vc): is the minimum volume of clearance between the cylinder head and the piston at T.D.C. position.
Swept Volume (Vs): is the maximum volume swept by the piston in moving from T.D.C. to B.D.C. position or vice versa.
Cylinder Volume (V): is the sum of clearance volume and swept volume
Compression ratio (r): is the ratio of cylinder volume to the clearance volume. Thus
Clearance ratio (c): is the ratio of clearance volume to stroke volume. Thus
Mean effective pressure (pm): is the average pressure which, if acted on the piston during the entire power or outward stroke, would produce the same work output as the network output for the actual cyclic process. Thus
Work done per cycle,
Work done per minute = pm L An
n = number of strokes per minute.
= N, for two stroke engines.
= N/2, for four stroke engines.
N = rpm
where pm is in kPa.
9.3 ❐ Carnot Cycle
The Carnot cycle consists of two reversible isotherms and two reversible adiabatics, as shown in Fig. 9.2. For one kg of gas (T1 > T2), we have
Figure 9.2 Carnot cycle: (a) p-v diagram, (b) T-s diagram
9.4 ❐ Stirling Cycle
This cycle works on the principle of regeneration by using a regenerator within the engine itself which would store the rejected heat energy during heat rejection process and supply the same during heat addition process. The Stirling cycle consists of two reversible isotherms and two reversible isochors. The p − v and T − s diagrams are shown in Fig. 9.3. We note that heat addition process 2−3 and heat rejection process 4−1 cancel each other, since the energy taken from the regenerator is returned in the latter. An alternative interpretation is to consider the regenerator as a part of the system so that the heat exchange with the surroundings involves only the source at temperature T1 and the sink at temperature T2. The heat supplied per cycle is, therefore, during the process from 3−4 and heat rejection is during the process 1−2. For one kg of ideal gas (T1 > T2), we have
But v4 = v3, v1 = v2 and hence
Generally there is loss of heat due to radiation and poor conductivity of gas. If ηr is the efficiency of the regenerator, then
Heat supplied = RT1 ln r + (1 − ηr) cv (T1 − T2)
Heat rejected = RT2 ln r + (1 − ηr) cv (T1 − T2)
9.5 ❐ Ericsson Cycle
This cycle also works on the principle of regeneration. It consists of two reversible isothermals and two reversible isobars. The p-v and T-s diagram are shown in Fig. 9.4. The heat rejected during the heat rejection process 4 – 1 at constant pressure is stored in the regenerator and the same is supplied during heat addition process 2 – 3. Assuming 100% efficiency of the regenerator, we have
Figure 9.4 Ericsson cycle: (a) p-v diagram, (b) T-s diagram
9.6 ❐ Atkinson Cycle
This cycle has two reversible adiabatics (isentropics), one isobar and one isochore. The p-v and T-s diagrams are shown in Fig. 9.5. For one kg of the working fluid, we have
Heat supplied, qs = cv (T3 – T2)
Heat rejected, qr = cp (T4 – T1)
Net work done, wnet = qs − qr = cv (T3 −T2) − cp (T4 −T1)
Let compression ratio,
From compression process 1 − 2, we have
For constant volume heat addition process 2 − 3, we have
Figure 9.5 Atkinson cycle: (a) p-v diagram, (b) T-s diagram
Substituting the values of T2, T3 and T4, we get
9.7 ❐ Otto Cycle (constant volume cycle)
This cycle is composed of four internally reversible processes, two adiabatic and two constant volume processes. The p-v and T-s diagrams are shown in Fig. 9.6. The various processes are:
Process 1−2: Isentropic compression.
Process 2−3: Constant volume heat addition.
Process 3−4: Isentropic expansion.
Process 4−1: Constant volume heat rejection.
This cycle is used for spark ignition (petrol) engines.
Consider 1 kg of air flowing through the cycle. Since the air in the cylinder acts as a closed system, from first law of thermodynamics for isentropic compression and expansion, we have
For constant volume heat supplied and rejection processes, since w = 0,
Heat supplied qs = q2−3 = cv(T3 – T2)
Heat rejected qr = q4−1 = cv (T4 – T1)
Work done per cycle, wnet = qs – qr = cv(T3 – T2) – cv(T4 – T1)
Air standard (or thermal) efficiency =
where pressure ratio
The air standard efficiency of Otto cycle depends on compression ratio only and increases as compression ratio increases (Fig. 9.7). In actual engine working on Otto cycle, the compression ratio varies from 5 to 8. This engine is used for spark ignition engines working on petrol.
Mean Effective Pressure (m.e.p.). It may be defined as the ratio of work done to the displacement volume of piston.
Work done, w = cv[(T3 – T2) – (T4 – T1)]
Displacement volume, vs = v1 – v2
9.8 ❐ Diesel Cycle
This cycle is used for compression ignition internal combustion engines working on diesel oil. The p-v and T-s diagrams are shown in Fig. 9.8. It consists of four internally reversible processes two adiabatic, one constant pressure and one constant volume. The various processes are:
Process 1–2: Isentropic compression of air.
Process 3–4: Isentropic expansion of air.
Process 4−1: Heat rejection at constant volume.
Considering 1 kg of air.
Heat supplied, qs = q2–3 = cp (T3 – T2)
Heat rejected, qr = q4–1 = cv (T4 – T1)
Work done per cycle, w = qs – qr = cp (T3 – T2) – cv (T4 – T1)
Let = compression ratio
Let = cut off ratio
Figure 9.8 Diesel cycle: (a) p-v diagram, (b) T-s diagram
The thermal efficiency of diesel cycle increases as compression ratio increases but decreases as cut-off ratio increases (Fig. 9.9). The thermal efficiency of diesel cycle is less than that of Otto cycle. The compression ratio for diesel cycle varies from 14 to l8.
Mean effective pressure
Figure 9.9 Diesel cycle thermal efficiency v’s compression ratio
9.9 ❐ Dual Cycle
In the dual cycle, part of the heat is supplied at constant volume and the rest at constant pressure. This is also called mixed or limited pressure cycle. The p-v and T-s diagrams are shown in Fig. 9.10. The various processes are:
Process 1−2: Isentropic compression of air
Process 2−3: Constant volume heat addition
Process 3−4: Constant pressure heat addition
Process 4−5: Isentropic expansion of air
Process 5−1: Constant volume heat rejection
Considering 1 kg of working fluid,
Heat supplied, qs = cv (T3 − T2) + cp (T4 − T3)
Heat rejected, qr = cv (T5 − T1)
Work done per cycle, w = qs – qr = cv (T3 – T2) + cp (T4 – T3) – cv (T5 – T1)
Let, = compression ratio
Then from process 1−2, we have
From process 2 – 3,
Figure 9.10 Dual cycle: (a) p-v diagram, (b) T-s diagram
From process 4 – 5,
The variation of thermal efficiency of dual cycle is shown in Fig. 9.11.
Mean effective pressure,
Figure 9.11 Variation of thermal efficiency of dual cycle with compression ratio
9.10 ❐ Brayton Cycle
The air standard Brayton or Joule cycle is a constant pressure cycle used in gas turbine power plants. The p-v and T-S diagrams are shown in Fig 9.12. It consists of the following processes:
1−2: Isentropic compression in the compressor
2−3: Constant pressure heat addition.
3−4: Isentropic expansion of air
4−4: Constant pressure heat rejection
Consider 1 kg of working fluid
From first law of thermodynamics for steady flow, rejecting ΔKE and ΔPE, we have
Heat added, qs = h3 – h2 = cp (T3 – T2)
Heat rejected qr = h4 – h1 = cp (T4 – T1)
Net work done by turbine, wnet = qs – qr = cp [(T3 − T2) − (T4 − T1)]
Also work done by turbine, wt = h3 − h4 = cp (T3 − T4)
Work consumed by compressor, wc= h2 − h1 = cp (T2 − T1)
Figure 9.12 Brayton (or Joule) cycle: (a) p-v diagram, (b) T-s diagram
From isentropic compression process 1 – 2, we have
and from isentropic expansion process 3 – 4,
The variation of thermal efficiency v’s pressure ratio is shown in Fig. 9.13. The thermal efficiency increases with increasing values of pressure ratio. This cycle is used in gas turbines.
wt = work done by turbine
wc = work supplied to compressor
Pressure ratio for maximum work,
9.11 ❐ Comparison between Otto, Diesel, and Dual Cycles
The comparison parameters selected are:
- Equal compression ratio and heat input.
- Constant maximum pressure and heat input.
- Constant maximum pressure and work output.
- Constant maximum pressure and temperature.
- Equal compression ratio and heat rejection.
- Equal compression ratio and heat input. The p-v and T-s diagrams for the three cycles are shown in Fig.9.14. The cycles have been represented as follows:
That cycle will be more efficient which rejects the least amount of heat after expansion, because
Thus the order of efficiencies is:
- Constant maximum pressure and heat input. The p-v and T-s diagrams for Otto and Diesel cycles are shown in Fig. 9.15 for constant maximum pressure and heat input
Area a – 2 − 3 – b = Area a – 2′ – 3′ –b′ for same heat input.
Thus cycle will be more efficient that rejects the least amount of heat.
(Area a – 1 − 4 –b) < (Area a – 1 – 4′ –b′)
Thus the Diesel cycle is more efficient than the Otto cycle. A similar argument will show that the dual cycle performance falls between the other two. Thus
- Constant maximum pressure and work output. The p-v and T-s diagrams are shown in Fig. 9.15 for maximum pressure and output.
Area 1 – 2 – 3 – 4 = Area 1 – 2′ – 3′ – 4′ for same work output
The cycle will be more efficient which rejects least amount of heat.
(Area a – 1 – 4 – b) < (Area a – 1′ – 4′ – b′)
Hence the Diesel cycle is more efficient than the Otto cycle. Thus
- Constant maximum pressure and temperature. The p-v and T-s diagrams are shown in Fig. 9.16 for same maximum pressure and temporal are.
All the cycles reject equal amount of heat.
Thus cycle will be more efficient which has higher heat addition.
Hence, the Diesel cycle is more efficient than the Otto cycle. Thus
- Equal compression ratio and heat rejection. The p-v and T-s diagrams are shown in Fig. 9.17, for equal compression ratio and heat rejection
For same qr, if qs is more, η is more.
Figure 9.17 Comparison for equal compression ratio and heat rejection: (a) p-v diagram, (b) T-s diagram
A Carnot cycle has lowest temperature and pressure of 20°C and 1 bar. The pressures are: 4 bar after isothermal compression; 12 bar after isentropic compression and 6 bar after isothermal heat addition. Calculate (a) the highest temperature in the cycle, (b) thermal efficiency of the cycle, (c) mean effective pressure, and power developed with 150 cycles per minute.
Given: p4 = 1 bar, T2 = 293 K, p2 = 12 bar, p1 = 4 bar, p3 = 6 bar (Refer to Fig. 9.2)
Stroke volume, vs= v4 – v2 = 0.8409 – 0.0958 = 0.7451 m3/kg
- Highest temperature in the cycle,
- Heat supplied,
= 79.67 kJ/kg
Heat rejected,= − 58.07 kJ/kg
Net work done, wnet = qs – qr = 79.67 – 58.07 = 21.60 kJ/kg
- Mean effective pressure =
- Power developed = wnet × rpm/60 = 21.6 × 150/60 = 54 kW.
Compression ratio = 3, lower temperature = 30°C, speed = 50 rpm, higher temperature = 550°C, regenerative efficiency =90%, initial pressure = 1 bar, heat added = 2300 kJ/min.
Calculate (a) work done per kg, (b) thermal efficiency, (c) mean effective pressure, and (d) indicated power.
Refer to Fig. 9.18.
- Work done = R(T1 – T2) ln r
= 0.287 (823 – 300) ln 3
= 163.956 kJ/kg
Heat added = RT1 ln r + (1 – ηreg) cv(T1 – T2)
= 0.287 × 823 × ln 3 + (1 – 0.9) × 0.718 × (823 – 303)
= 296.829 kJ/kg
- Thermal efficiency
- Air flow rate
- Working cycles/min =
Mass of air per working cycle,
Mean effective pressure,
- Indicated power
An engine working on Atkinson cycle takes in air at 1 bar and 25°C. The air is compressed isentropically by a compression ratio of 6:1. The heat is added at constant volume increasing the final pressure to 20 bar. Now, the air is expanded isentropically to 1 bar. The heat is rejected at constant pressure Calculate (a) pressure and temperature at various points, (b) work done per kg of air, and (c) cycle efficiency
Refer to Fig. 9.19.
- p1 = 1 bar,T1 = 25+ 273 = 298 K
T2 = T1rγ – 1 = 298 ×60.4 = 610.2 K
p2 = p1rγ = 1 × 6 1.4 = 12.286 bar
p3 = 20 bar
Heat added, qs = cv(T3 – T2) = 0.718 (993.3 – 610.2) = 279.4 kJ/kg
Heat rejected, qr = cp(T4 – T1) = 1.005 (422.7 – 298) = 124.7 kJ/kg
- Work done, w = qs − qr = 279.4 – 124.7 = 154.7 kJ/kg
- Thermal efficiency =
Assume γ = 1.4.
Refer to Fig. 9.6.
- Compression, ratio,
- Air standard efficiency,
In an Otto cycle, air at 15°C and 1 bar is compressed adiabatically until the pressure is 15 bar. Heat is added at constant volume until the pressure rises to 40 bar. Calculate (a)the air-standard efficiency, (b) the compression ratio, and (c) the mean effective pressure for the cycle. Assume cv = 0.718 kJ/kg. K, γ = 1.4 and R = 8.314 kJ/kmol.K.
The Otto cycle is shown in Fig. 9.20.
For isentropic process 1 – 2,
Air standard efficiency,
For constant volume process 2 – 3:
Heat supplied qs = cv(T3 – T2) = 0.718 (1669.7 – 629.3) = 747.03 kJ/kg
Work done, w = ηa qs = 0.539 × 747.03 = 402.65 kJ/kg
Mean effective pressure,
Specific volume, v1 – v2 = 0.7064 m3/kg
A gas engine working on Otto cycle has a cylinder of diameter 220 mm and stroke 300 mm. The clearance volume is 1600 cc. Find the air standard efficiency.
Assume cp = 1.004 kJ/kg. K and cv = 0.718 kJ/kg.K for air.
Air standard efficiency,
- From Fig. 9.6, we have
p2 = 1 × 9.5182 = 9.5182 barp3 = 20 bar
A petrol engine with compression ratio of 5 develops 20kW indicated power and consumes 8 litres of fuel per hour. The specific gravity of fuel is 0.78 and its calorific value is 44 MJ/kg. Calculate the indicated thermal efficiency and relative efficiency. Take = γ 1.4.
Air standard efficiency,
Fuel consumption = 8 × 0.78 × 1 = 6.24 kg/h
Indicated thermal efficiency,
A diesel engine has a compression ratio of 20 and cut-off takes place at 5% of the stroke. Find the air-standard efficiency. Assume γ = 1.4.
Refer to Fig. 9.8.
A diesel cycle operates at a pressure of 1 bar at the beginning of compression and the volume is compressed to th of the initial volume. Heat is supplied until the volume is twice that of the clearance volume. Calculate the mean effective pressure of the cycle. Assume γ = 1.4.
With reference to Fig. 9.8,
Swept volume, vs = v1 − v2 = (r −1) v2 = (15−1) v2 =14 v2
The mean effective pressure of an ideal diesel cycle is 10 bar. If the initial pressure is 1 bar and the compression ratio is 14, determine the cut off ratio and the air standard efficiency. Assume γ = 1.4.
See Fig. 9.8.
Work done per cycle
We take ρ = 2.64
In an engine working on the diesel cycle, the air-fuel ratio by weight is 50:1. The temperature of air at the beginning of combustion is 40°C and the compression ratio is 19. What is the ideal efficiency of the engine. Calorific value of fuel is 42 MJ/kg. Assume cv = 0.717 kJ/kgK, cp = 1.004 kJ/kgK.
Refer to Fig. 9.8.
Process 1 – 2:
Process 2 – 3:
An air standard diesel cycle has a compression ratio of 16. The pressure at the beginning of compression stroke is 1 bar and the temperature is 20°C. The maximum temperature is 1430°C. Determine the thermal efficiency and the mean effective pressure for this cycle. Take γ = 1.4.
Refer to Fig. 9.8.
Process 2 – 3:
An engine works on dual combustion cycle, the compression ratio being 11. The pressure at the commencement of combustion is 1 bar and the temperature is 90°C. The maximum pressure in the cycle is 50 bar and the constant pressure heat addition continues for of the stroke. Calculate the work done per kg of air and the ideal thermal efficiency. Assume, cv = 0.718 kJ/kgK, cp = 1.005 kJ/kgK.
Refer to Fig. 9.21.