Gas Turbines for Aircraft Propulsion Chapter 9.9-10

Gas Turbines for Aircraft Propulsion

Gas turbines

The turbojet engine consists of three main sections: the diffuser, the gas generator, and the nozzle. The diffuser placed before the compressor decelerates the incoming air relative to the engine. A pressure rise known as the ram effect is associated with this deceleration.

Gas turbines

The gas generator section consists of a compressor, combustor, and turbine, with the same functions as the corresponding components of a stationary gas turbine power plant.The gases leave the turbine at a pressure significantly greater than atmospheric and expand through the nozzle to a high velocity before being discharged to the surroundings.

Gas turbines

The working fluid is air modeled as an ideal gas. The diffuser, compressor, turbine, and nozzle processes are isentropic, and the combustor operates at constant pressure.

In an actual engine, there would be increases in specific entropy across the diffuser, compressor, turbine, and nozzle.

Gas turbines

Process a–1 shows the pressure rise that occurs in the diffuser as the air decelerates isentropically through this component.Process 1–2 is an isentropic compression.Process 2–3 is a constant-pressure heat addition.Process 3–4 is an isentropic expansion through the turbine during which work is developed.Process 4–5 is an isentropic expansion through the nozzle in which the air accelerates and the pressure decreases.

Gas turbines

In a typical thermodynamic analysis of a turbojet on an air-standard basis, the following quantities might be known: the velocity at the diffuser inlet, the compressor pressure ratio, and the turbine inlet temperature (at 3). The objective of the analysis would be to determine the velocity at the nozzle exit. Once the nozzle exit velocity is determined, the thrust is determined by applying Newton's second law of motion in a form suitable for a control volume

Example 9.12 Air enters a turbojet engine at 11.8 lbf/in.2, 430°R, and an inlet velocity of 620 miles/h (909.3 ft/s). The pressure ratio across the compressor is 8. The turbine inlet temperature is 2150°R and the pressure at the nozzle exit is 11.8 lbf/in.2

The work developed by the turbine equals the compressor work input. The diffuser, compressor, turbine, and nozzle processes are isentropic, and there is no pressure drop for flow through the combustor.

For operation at steady state, determine the velocity at the nozzle exit and the pressure at each principal state. Neglect kinetic energy at the exit of all components except the nozzle and neglect potential energy throughout.

Air enters a turbojet engine at 11.8 lbf/in.2, 430°R, and an inlet velocity of 620 miles/h (909.3 ft/s). The pressure ratio across the compressor is 8. The turbine inlet temperature is 2150°R and the pressure at the nozzle exit is 11.8 lbf/in.2

Each component is analyzed as a control volume at steady state. The control volumes are shown on the sketch by dashed lines.The diffuser, compressor, turbine, and nozzle processes are isentropic.There is no pressure drop for flow through the combustor.The turbine work output equals the work required to drive the compressor.

Except at the inlet and exit of the engine, kinetic energy effects can be ignored. Potential energy effects are negligible throughout.The working fluid is air modeled as an ideal gas. Analysis: To determine the velocity at the exit to the nozzle, the mass and energy rate balances for a control volume enclosing this component reduce at steady state to give

Except at the inlet and exit of the engine, kinetic energy effects can be ignored. Potential energy effects are negligible throughout.The working fluid is air modeled as an ideal gas. Analysis: To determine the velocity at the exit to the nozzle, the mass and energy rate balances for a control volume enclosing the nozzle are applied.

Energy rate balances for a control volume enclosing the nozzle

The inlet kinetic energy is dropped by the assumption that kinetic energy effects can be ignored except at the inlet and exit of the engine.

Solving for V5

With the operating parameters specified, the determination h4 and h5 values is accomplished by analyzing each component in turn, beginning with the diffuser.

The pressure at each principal state can be evaluated as a part of the analyses required to find the enthalpies h4 and h5.

Energy rate balance for a control volume enclosing the diffuser gives

Table A-22E

T=420 R, h = 100.32 btu/lbT=440 R, h = 105.11 btu/lb

ha (at 430 R) =102.7 Btu/lb

Table A-22Eh = 114.69 btu/lb, pr = 0.9182 h = 119.48 btu/lb, pr = 1.0590Interpolation at h = 119.2 btu/lb, pr1 = 1.051

Table A-22ET=420 R, pr = 0.5760 T=440 R, pr = 0.6776 Interpolation pa (at 430 R) = 0.6268

The flow through the diffuser is isentropic, so pressure p1 is

p1 = 19.79 lbf/in.2

Using the given compressor pressure ratio, the pressure at state 2 is

p2 = 8(19.79 lbf/in.2) = 158.3 lbf/in.2

The flow through the compressor is also isentropic. Thus

Table A-22Epr = 7.761, h = 211.35 Btu/lb pr = 8.411, h = 216.26 Btu/lbInterpolation h2 (at pr = 8.408) = 216.2 Btu/lb

At state 3 the temperature is given as T3 = 2150°R.

Table A-22ET= 2150°R, h = 546.54 Btu/lb

From Table A-22E, h3 = 546.54 Btu/lb. By assumption of no pressure drop for flow through the combustor, p3 = p2. The work developed by the turbine is just sufficient to drive the compressor. That is

Solving for h4

h4 = 449.5 Btu/lb

Table A-22Eh = 436.12 btu/lb, pr = 101.98h = 449.71 btu/lb, pr = 114.0Interpolation at h4 = 449.5 btu/lb, pr4 = 113.8

The expansion through the turbine is isentropic, so

With p3 = p2 = 158.3 lbf/in.2 and pr3 data from

Table A-22Eat h3 = 546.54 btu/lb, pr3 = 233.5

The expansion through the nozzle is isentropic to p5 = 11.8 lbf/in.2 thus

From Table A-22E, h5 = 265.8 Btu/lb, which is the remaining specific enthalpy value required to determine the velocity at the nozzle exit.

Using the values for h4 = 449.5 Btu/lb and h5 = 265.8 Btu/lb

determined above, the velocity at the nozzle exit is  

Combined Gas Turbine–Vapor Power Cycle

A combined power cycle couples two power cycles such that the energy discharged by heat transfer from one cycle is used partly or wholly as the input for the other cycle.

The stream exiting the turbine of a gas turbine is at a high temperature. This high-temperature gas stream can be used by the combined cycle shown, involving a gas turbine cycle and a vapor power cycle.

The two power cycles are coupled so that the heat transfer to the vapor cycle is provided by the gas turbine cycle, which may be called the topping cycle.

The combined cycle has the gas turbine's high average temperature of heat addition and the vapor cycle's low average temperature of heat rejection, and thus a thermal efficiency greater than either cycle would have individually.

For many applications combined cycles are economical, and they are increasingly being used worldwide for electric power generation.

The thermal efficiency of the combined cycle is

is the net power developed by the gas turbine

is the net power developed by the vapor cycle

is the total rate of heat transfer to the combined cycle, including additional heat transfer, if any, to superheat the vapor entering the vapor turbine.

For steady-state operation, negligible heat transfer with the surroundings, and no significant changes in kinetic and potential energy, energy balance around the heat exchanger gives

Combined cycle performance can be analyzed using mass and energy balances. To complete the analysis, however, the second law is required to assess the impact of irreversibilities and the true magnitudes of losses.

Combined cycle performance can be analyzed using mass and energy balances. To complete the analysis, however, the second law is required to assess the impact of irreversibilities and the true magnitudes of losses.

Among the irreversibilities, the most significant is the exergy destroyed by combustion. About 30% of the exergy entering the combustor with the fuel is destroyed by combustion irreversibility.

Definition of exergy:Exergy is the maximum theoretical work obtainable from an overall system consisting of a system and the environment as the system comes into equilibrium with the environment (passes to the dead state).

The exergy of a system, E, at a specified state is given by the expression                                                

The exergy of a system, E, at a specified state is given by the expression                                                

U, KE, PE, V, and S denote, respectively, internal energy, kinetic energy, potential energy, volume, and entropy of the system at the specified state.

U0, V0, and S0 denote internal energy, volume, and entropy, respectively, of the system when at the dead state. In this chapter kinetic and potential energy are evaluated relative to the environment.

Thus, when the system is at the dead state, it is at rest relative the environment and the values of its kinetic and potential energies are zero.

Death state: the system comes into equilibrium with the environment—that is, as the system passes to the dead state.

Neglecting changes in kinetic and potential energies, the net rate of exergy increase for an open system is given by

3fE 2fE = mg[(h3 h2) To(s3 s2)]

2 3

Q in

Com bustor

mg = mass flow rate through the systemTo = surrounding temperature

Example 9.13: A combined gas turbine–vapor power plant has a net power output of 45 MW. Air enters the compressor of the gas turbine at 100 kPa, 300K, and is compressed to 1200 kPa.

The isentropic efficiency of the compressor is 84%. The condition at the inlet to the turbine is 1200 kPa, 1400 K. Air expands through the turbine, which has an isentropic efficiency of 88%, to a pressure of 100 kPa.

The air then passes through the interconnecting heat exchanger and is finally discharged at 400 K. Steam enters the turbine of the vapor power cycle at 8 MPa, 400°C, and expands to the condenser pressure of 8 kPa. Water enters the pump as saturated liquid at 8 kPa. The turbine and pump of the vapor cycle have isentropic efficiencies of 90 and 80%, respectively.

A combined gas turbine–vapor power plant has a net power output of 45 MW. Air enters the compressor of the gas turbine at 100 kPa, 300K, and is compressed to 1200 kPa (at 2).

The isentropic efficiency of the compressor is 84%. The condition at the inlet to the turbine is 1200 kPa, 1400 K. Air expands through the turbine, which has an isentropic efficiency of 88%, to a pressure of 100 kPa (at 4).

The air then passes through the interconnecting heat exchanger and is finally discharged at 400 K.

Steam enters the turbine of the vapor power cycle at 8 MPa, 400°C, and expands to the condenser pressure of 8 kPa (at 8).

Water enters the pump as saturated liquid at 8 kPa. The turbine and pump of the vapor cycle have isentropic efficiencies of 90 and 80%, respectively.

Let T0 = 300 K, p0 = 100 kPa.

Gas Turbine Vapor Cycle

State h (kJ/kg) s° (kJ/kg · K) State h (kJ/kg) s (kJ/kg · K)

1 300.19 1.7020 6 183.96 0.5975

2 669.79 2.5088 7 3138.30 6.3634

3 1515.42 3.3620 8 2104.74 6.7282

4 858.02 2.7620 9 173.88 0.5926

5 400.98 1.9919

From the temperature of relative pressure, the enthalpy and entropyat 9 states can be determined and listed in the following table:

State 1: T1 = 300 K h1 = 300.19 kJ/kg, pr1 = 1.3860State 2: p2 = 1200 kPa

For adiabatic compression from p1 to p2:

Therefore pr2: = pr1

1 2

1 2r r

p p

p p

2

1

p

p= 1.3860

1, 200

100= 16.6320

From Table A-22pr = 16.28, h = 607.02 kJ/kgpr = 17.30, h = 617.53 kJ/kg

From Table A-22pr = 16.28, h = 607.02 kJ/kgpr = 17.30, h = 617.53 kJ/kg

h2,isen = 607.02 + (16.632 16.28)*(617.53 607.02)/(17.30 16.28)

h2,isen = 610.647 kJ/kg

h2 = h1 + (h2,isen h1)/0.84 = 300.19 + (610.647 300.19)/0.84

h2 = 669.7817 kJ/kg

State 3: T3 = 1400 K h3 = 1515.42 kJ/kg, pr3 = 450.5State 4: p = 100 kPaAdiabatic expansion from 1,200 kPa to 100 kpa

34

4 3

rr pp

p p pr4 = p4

3

3

rp

p = 100

450.5

1,200 = 37.5417

From Table A-22pr = 37.35, h = 767.29 kJ/kgpr = 39.27, h = 778.18 kJ/kg

h4,isen = 767.29 + (37.5417 37.35)*(778.18 767.29)/(39.27 37.35)h4,isen = 768.3773 kJ/kg

h4 = h3 + 0.88*(h4,isen h3) = 1515.42 + 0.88*(768.3773 1515.42)h4 = 858.0224 kJ/kg

State 7: T7 = 400oC, p7 = 8 MPaFrom steam table or CATT2 program (Computer Aided Thermodynamic Tables 2)

h7 = 3138.3 kJ/kg, s7 = 6.3633 kJ/kgK

Isentropic expansion to state 8, p8 = 8 kPa h8,isen = 1989.9 kJ/kg

h8 = h7 + 0.90*(h8,isen h7) = 3138.3 + 0.90*(1989.9 3138.3)h8 = 2104.7 kJ/kg

State 9: Saturated liquid at 8 kPa: h9 = 173.86 kJ/kg, s9 = 0.59254 kJ/kgK

State 9: Saturated liquid at 8 kPa: h9 = 173.86 kJ/kg, s9 = 0.59254 kJ/kgK

Isentropic compression to state 6: 8 MPa Þh6,isen = 181.9 kJ/kg

h6 = h9 + (h6,isen h9)/0.8 = 173.86 + (181.9 173.86)/0.8

h6 = 183.91 kJ/kg

Heat exchanger

5 4

6 7

m g

m v

Apply energy balance to the heat exchanger:

mg(h4 h5) = mv(h7 h6)

4 5

7 6

v

g

m h h

m h h

=

858.02 400.98

3138.3 183.96

= 0.1547

1

2 3

4

Com pressor Turb ine

W p W t

The net power developed by the gas turbine is given by

gW = mg[(h3 h4) (h2 h1)]

7

8Turbine

W t

Condenser

9

6

W p

Vaporcycle

The net power developed by the steam turbine is given by

vW = mv[(h7 h8) (h6 h9)]

The net power developed by the gas turbine-vapor plant is given by

netW = gW + vW

netW = mg[(h3 h4) (h2 h1)] + mv[(h7 h8) (h6 h9)]

netW = mg{[(h3 h4) (h2 h1)] + (mv/mg)[(h7 h8) (h6 h9)]}

45 MW = mg{[(h3 h4) (h2 h1)] + (0.1547)[(h7 h8) (h6 h9)]} h1 = 300.19 kJ/kg, h2 = 669.79 kJ/kg h3 = 1515.42 kJ/kg, h4 = 858.02 kJ/kg h6 = 183.96 kJ/kg, h7 = 3138.30 kJ/kg h8 = 2104.74 kJ/kg, h9 = 173.88 kJ/kg Solve for mg to obtain mg = 100.87 kg/s

mg = 100.87 kg/s mv = (0.1547) mg = 15.6 kg/s

The net power developed by the gas turbine is given by

gW = mg[(h3 h4) (h2 h1)] = (100.87) [(h3 h4) (h2 h1)]

h1 = 300.19 kJ/kg, h2 = 669.79 kJ/kg h3 = 1515.42 kJ/kg, h4 = 858.02 kJ/kg

gW = 29.03 MW

The net power developed by the steam turbine is given by

vW = mv[(h7 h8) (h6 h9)] = (15.6) [(h7 h8) (h6 h9)]

h6 = 183.96 kJ/kg, h7 = 3138.30 kJ/kg h8 = 2104.74 kJ/kg, h9 = 173.88 kJ/kg

vW = 15.97 MW

The net rate of exergy increase of the air passing through the combustor is

3fE 2fE = mg[(h3 h2) To(s3 s2)]

3fE 2fE = mg[(h3 h2) To(s3o s2

o Rln(p3/p2)]

mg = 100.87 kg/s, h2 = 669.79 kJ/kg, p2 = p3

h3 = 1515.42 kJ/kg, To = 300 K, s2o = 2.5088 kJ/kgK, s3

o = 3.3620 kJ/kgK

3fE 2fE = 59.48 MW

2 3

Q in

Com bustor

The net rate of exergy increase because of the water passing through

the condenser is

8fE 9fE = mv[(h8 h9) To(s8 s9)]

8

Condenser

9

Q out

mv = 15.6 kg/s, To = 300 K h8 = 2104.74 kJ/kg, h9 = 173.88 kJ/kg s8 = 6.7282 kJ/kgK, s9 = 0.5926 kJ/kgK

8fE 9fE = 1.41 MW