Chapter 9 Gas Power Cycles – Thermal Engineering

Chapter 9

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:

  1. The air as the working fluid follows the perfect gas law: pV = mRT.
  2. The working fluid is homogeneous throughout and no chemical reaction takes place.
  3. The specific heats of air do not vary with temperature.
  4. The mass of air in the cycle remains fixed.
  5. The combustion process is replaced by an equivalent heat addition process.
  6. The exhaust process is replaced by an equivalent heat rejection process.
  7. 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.

Figure 9.1 Piston-cylinder arrangement

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

L = T.D.C. − B.D.C.

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

V = Vc +Vs

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

Where

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

Heat supplied,

Heat rejected,

Net work,

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

Heat supplied,

Heat rejected,

Thermal efficiency,

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 (T1T2)

Heat rejected = RT2 ln r + (1 − ηr) cv (T1T2)

Figure 9.3 Stirling cycle: (a) p-v diagram, (b) T-s diagram

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

Heat supplied,

Heat rejected,

Where

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 (T3T2)

Heat rejected, qr = cp (T4T1)

Net work done, wnet = qsqr = cv (T3T2) − cp (T4 T1)

Thermal efficiency,

where

Let compression ratio,

Explosion 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

From expansion process 3−4, we have

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

qw = Δu

For constant volume heat supplied and rejection processes, since w = 0,

q = Δu = cv ΔT

Figure 9.6 Otto cycle: (a) p-v diagram, (b) T-s diagram

Heat supplied       qs = q2−3 = cv(T3 – T2)

Heat rejected       qr = q4−1 = cv (T4T1)

Work done per cycle, wnet = qsqr = cv(T3T2) – cv(T4T1)

Air standard (or thermal) efficiency =

where

where pressure ratio

Figure 9.7 Otto cycle thermal efficiency v’s compression 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[(T3T2) – (T4T1)]

Displacement volume, vs = v1v2

M.E.P.,

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 2–3: Heat addition at constant pressure.

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 (T3T2)

Heat rejected, qr = q4–1 = cv (T4T1)

Work done per cycle, w = qsqr = cp (T3T2) – cv (T4T1)

Thermal efficiency,

Let = compression ratio

Now

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

Now

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 12: Isentropic compression of air

Process 23: Constant volume heat addition

Process 34: Constant pressure heat addition

Process 45: Isentropic expansion of air

Process 51: Constant volume heat rejection

Considering 1 kg of working fluid,

Heat supplied, qs = cv (T3T2) + cp (T4T3)

Heat rejected, qr = cv (T5T1)

Work done per cycle, w = qsqr = cv (T3T2) + cp (T4T3) – cv (T5T1)

Let, = compression ratio

Then from process 1−2, we have

T2 = T1rγ−1

From process 2 – 3,

Figure 9.10 Dual cycle: (a) p-v diagram, (b) T-s diagram

From process 3 – 4,

From process 4 – 5,

The variation of thermal efficiency of dual cycle is shown in Fig. 9.11.

Mean effective pressure,

Now

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

δq δw = dh

Heat added, qs = h3h2 = cp (T3T2)

Heat rejected qr = h4h1 = cp (T4T1)

Net work done by turbine, wnet = qsqr = cp [(T3T2) − (T4T1)]

Also work done by turbine, wt = h3h4 = cp (T3T4)

Work consumed by compressor, wc= h2h1 = cp (T2T1)

wnet = cp [(T3T4) − (T2T1)] = cp [(T3T2) − (T4 T1)]

Thermal efficiency,

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,

 

W = mcp [(T3T2) – (T4T1)]

 

= mcp [(T3T4) – (T2T1)]

 

Figure 9.13 Efficiency v’s pressure ratio in simple Brayton cycle

9.11 ❐ Comparison between Otto, Diesel, and Dual Cycles

The comparison parameters selected are:

  1. Equal compression ratio and heat input.
  2. Constant maximum pressure and heat input.
  3. Constant maximum pressure and work output.
  4. Constant maximum pressure and temperature.
  5. Equal compression ratio and heat rejection.

Figure 9.14 Comparison for equal compression ratio and heat input:
(a) p-v diagram, (b) T-s diagram

  1. 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:

  2. 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

    Figure 9.15 Comparison for equal maximum pressure and heat input:
    (a) p-v diagram, (b) T-s diagram

  3. 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

  4. 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.

     

    Now

    Figure 9.16 Comparison for equal maximum pressure and temperature: (a) p-v diagram, (b) T-s diagram

    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

  5. 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

Example 9.1

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.

Solution

Given: p4 = 1 bar, T2 = 293 K, p2 = 12 bar, p1 = 4 bar, p3 = 6 bar (Refer to Fig. 9.2)

Stroke volume, vs= v4v2 = 0.8409 – 0.0958 = 0.7451 m3/kg

  1. Highest temperature in the cycle,
  2. Heat supplied,
    = 79.67 kJ/kg

    Heat rejected,

    = − 58.07 kJ/kg

    Net work done, wnet = qsqr = 79.67 – 58.07 = 21.60 kJ/kg

    Thermal efficiency,

  3. Mean effective pressure =
  4. Power developed = wnet × rpm/60 = 21.6 × 150/60 = 54 kW.

Example 9.2

The following particulars refer to an engine working on Stirling cycle.

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.

Solution

Refer to Fig. 9.18.

  1. Work done = R(T1T2) ln r

    = 0.287 (823 – 300) ln 3

    = 163.956 kJ/kg

    Heat added = RT1 ln r + (1 – ηreg) cv(T1T2)

    = 0.287 × 823 × ln 3 + (1 – 0.9) × 0.718 × (823 – 303)

    = 296.829 kJ/kg

  2. Thermal efficiency
  3. Air flow rate
  4. Working cycles/min =

    Mass of air per working cycle,

    Figure 9.18

    Stroke volume Vs = V1V2 = 0.2694 − 0.0898 = 0.1796m3

    Mean effective pressure,

  5. Indicated power

Example 9.3

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

Solution

Refer to Fig. 9.19.

  1. 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(T3T2) = 0.718 (993.3 – 610.2) = 279.4 kJ/kg

    Heat rejected, qr = cp(T4T1) = 1.005 (422.7 – 298) = 124.7 kJ/kg

  2. Work done, w = qsqr = 279.4 – 124.7 = 154.7 kJ/kg
  3. Thermal efficiency =

    Figure 9.19

Example 9.4

In an engine working on ideal Otto cycle, the temperature at the beginning and end of compression are 45° C and 370°C. Find the compression ratio and air standard efficiency of the engine.

Assume γ = 1.4.

Solution

Refer to Fig. 9.6.

  1. Compression, ratio,
  2. Air standard efficiency,

Example 9.5

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.

Solution

The Otto cycle is shown in Fig. 9.20.

For isentropic process 1 – 2,

p1v1γ = pv2γ

Compression ratio,

Air standard efficiency,

Figure 9.20

For constant volume process 2 – 3:

Heat supplied qs = cv(T3T2) = 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, v1v2 = 0.7064 m3/kg

Example 9.6

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.

Solution

Swept volume,

Compression ratio,

Adiabatic index,

Air standard efficiency,

Example 9.7

The pressure limit in an Otto cycle are 1 bar and 20 bar. The compression ratio is 5. Calculate (a) thermal efficiency and (b) mean effective pressure. Assume γ = 1.4 for air.

Solution

  1. From Fig. 9.6, we have
    p2 = 1 × 9.5182 = 9.5182 bar
    p3 = 20 bar

Pressure Ratio,

Example 9.8

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.

Solution

Air standard efficiency,

Fuel consumption = 8 × 0.78 × 1 = 6.24 kg/h

Indicated thermal efficiency,

Relative efficiency

Example 9.9

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.

Solution

Refer to Fig. 9.8.

Cut-off ratio,

Example 9.10

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.

Solution

With reference to Fig. 9.8,

Swept volume, vs = v1v2 = (r −1) v2 = (15−1) v2 =14 v2

or

Process 1–2:

p2 = p3 = 49.31 bar

= 0.071 [49.31 + 122.55 – 79.27] = 6.685 bar

Example 9.11

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.

Solution

See Fig. 9.8.

Work output

Work done per cycle

or   f (ρ) = ρ1.4 − 9.024ρ + 6.738 = 0

We take     ρ = 2.64

Example 9.12

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.

Solution

Refer to Fig. 9.8.

Process 1 – 2:

T2 = 929.66 K

Process 2 – 3:

Example 9.13

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.

Solution

Refer to Fig. 9.8.

Process 2 – 3:

Example 9.14

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.

Solution

 

T1 = 90 + 273= 363K

 

Refer to Fig. 9.21.