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Gas Power Cycles

CHAPTER 9

9-1 Basic Considerations in the Analysis of Power Cycles

Air Standard Cycles The Spark Ignition Engine

Modeling is a powerful engineering tool that provides great insight and simplicity at the expense of some loss in accuracy.

General classes of engines

• • • Reciprocating internal combustion engines.

Reciprocating compressors Reciprocating steam engines

Single vs. double action engines

Single action reciprocating engine

Double action reciprocating engine.

Indicator diagrams

An analog instrument that measures pressure vs. Displacement in a reciprocating engine. The graph is in terms of P and V.

Area is proportional to the work done per cycle

FIGURE 9-2 The analysis of many complex processes can be reduced to a manageable level by utilizing some idealizations.

FIGURE 9-6 P-v and T-s diagrams of a Carnot cycle.

9-2 The Caront Cycle and Its Value in Engineering

FIGURE 9-7 A steady-flow Carnot engine.

FIGURE 9-8 T-s diagram for Example 9 –1.

9-3 Air Standard Assumptions

• • • • The working fluid is air, which continuously circulates in a closed loop and always behaves as an ideal gas.

All the processes that make up the cycle are internally reversible.

The combustion process is replaced by a heat addition process from an external source.

The exhaust process is replaced by a heat-rejection process that restores the working fluid to its initial state.

9-4 An Overview of Reciprocating Engines

Air Standard Cycles The Spark Ignition Engine

FIGURE 9-10 Nomenclature for reciprocating engines.

FIGURE 9-11 Displacement and clearance volumes of a reciprocating engine.

FIGURE 9-12 The net work output of a cycle is equivalent to the product of the mean effective pressure and the displacement volume.

Indicator Diagrams

Work per cycle is represented in terms of a mean effective pressure and the displacement.

MEP = Mean effective pressure V

X = Displacement

Indicator Diagrams

To compute the work per cycle, use the MEP and the displacement.

Work = MEP x A x Displacement.

MEP = k i Y, where k i is a constant Y = the average ordinate on the indicator diagram.

X = displacement A = area of cylinder heat.

Work



k i YA

c a Ideal indicator diagram Spark ignition engine - The Otto cycle d e b

Processes: a-b Intake b-c Compression c-d Combustion (spark ignited) d-e Power Stroke e-f Gas Exhaust b-a Exhaust Stroke

Displacement Clearance volume

FIGURE 9-13 Actual and ideal cycles in spark-ignition engines and their P-v diagrams.

FIGURE 9-14 Schematic of a two stroke reciprocating engine.

The spark ignition engine The Otto cycle

The Otto cycle The air standard Otto cycle approximates the automotive internal combustion engine and aircraft engines.

Assume a quasistatic process, isentropic compression, and isentropic exhaust.

4 3 1

s = Constant

5 2 6

V Displacement

s = Constant

The Otto cycle

3 5 1

P-V diagram for the actual variable mass system.

2 6

4 3 5 2 6

V = Constant

This is the T-s diagram for the fixed mass system.

The air-standard Otto cycle

• • • • • • Real Process 1-2 Intake of fuel/air mixture at P 1 2-3 Compression to P 3 3-4 Combustion, V = cons. and increasing P 4-5 Expansion 5-6 Exhaust at constant V and falling pressure 6-1Exhaust at P 1 Air Cycle Approximation • • • • (Constant mass, closed internally reversible, cycle) 2-3 Isentropic compression 3-4 Heat addition at constant volume 4-5 Isentropic expansion 5-6 Heat rejection at constant volume

The reversible Otto cycle

• • • • Internally reversible processes Constant specific heats –

k = C p /C v = Constant

Polytropic compression and expansion, PV k = Const.

Treat air as an ideal gas.

3 Q H Thermodynamic analysis for the reversible Otto cycle 4

Q Q H L

 

C v

(

C v

(

5 

)



)

2 6 Q C 5

V = Constant

 

Q H



Q C Q H



1



4 

6 

3 (1)



3  4 

3  4    

5  6 

path



Q rev T

 3   4

C v dt T



3  4 

C v

ln   

3    ,

Thermodynamic analysis for the air-standard Otto Cycle



5  6 

C v

ln   

T T

5 6    

3 

T T

6 5 (2)

3 

T T

6 5

2 

3 

5    

v v

4 5   

 1

2    

3   

 1

Thermodynamic Analysis for the air-standard Otto Cycle

3 

T T

6 5    1 

Thermal efficiency

4  

3  

5 

6  1   

v v

6 3   1 

k r v



3    1 

r v

1 

Thermal efficiency

r v



3    1 

r v

1 

Key assumptions: (1) Internally reversible processes (2) Constant specific heats Important consequence: (1) Efficiency is independent of working fluid (2) Efficiency independent of temperatures

Efficiency of Air Standard Otto Cycle k = 1.3 = C p /C v

T Q H

4 3 5

W NET

2,6

Q C s

r v

Air Standard Otto Cycle

0.8

0.7

0.6

0.5

0.4

0.3

0.2

k = 1.4

k = 1.3

0.1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Pressure Ratio, Pv Se ri es 1 Se ri es 2

FIGURE 9-16 Thermal efficiency of the ideal Otto cycle as a function of compression ratio (k = 1.4).

FIGURE 9-18 The thermal efficiency of the Otto cycle increases with the specific heat ratio k of the working fluid.

The Otto cycle with real gases

• • Variable specific heats.

Efficiency based on internal energies obtained from the gas tables which take into account variable specific heats .

  1 

4  

3    (

r v

2 ,

Q H m TOT

)

3 2,6 Q H Q C 4 5

V = Constant

9-6 Diesel Cycle:

Air Standard Cycles The Compression Ignition Engine

The compression ignition engine - the Diesel cycle

Air Standard Compression Ignition Engine - The Diesel cycle

• • An IC power cycle useful in many forms of automotive transportation, railroad engines, and ship power plants.

Key assumptions are, – Constant specific heats, the ideal gas, and internally reversible processes.

c a d Ideal Indicator Diagram Compression Ignition Engine e b

Processes: a-b Intake b-c Compression c-d Combustion d-e Power Stroke e-f Gas Exhaust b-a Exhaust Stroke

Displacement Clearance volume

e a Ideal Indicator Diagram Compression Ignition Engine b

Displacement Clearance volume

c d

Processes: a-b Combustion

(P = Const.)

b-c Expansion

(s = Const)

c-d Exhaust

(V = Const)

d-e Exhaust

(P = Const)

e-d Intake

(P = Const)

d-a Compression

(s = Const)

FIGURE 9-21 T-s and P-v diagrams for the ideal Diesel cycle.

Advantages of the Diesel Cycle

•

Eliminates pre-ignition of the fuel-air mixture when compression ratio is high.

•

All heat transfer to a constant mass system.

The air-standard compression ignition engine

T W in

a d

p = Constant Q C

Q H

V = Constant W out s

d a

p = Constant Q H Q C

Thermal efficiency of the air standard Diesel cycle

V = Constant s



Q H

 1  

Q C C Q H p



T b Q C



C v

(

T c



T a

 

T d

)   1 

C v

(

T C p

(

T b c

 

T d T a

) )  1  (

T c k

(

T b

 

T d T a

) )

Key parameters for

the Diesel cycle

a b e c d

V Displacement Clearance volume

r v



v d v a r e



v b c r c



v v a b

a e b Thermal efficiency

  1    1

r v k

 1

   

k r c k



r c

  1 1    

V Displacement Clearance volume

a e b Displacement Clearance volume Cut Off Ratio:

r c



v b v a

c d

V Air-standard Diesel cycle

Thermal Efficiency

  1    1

r v k

 1    

k r c k



r c

  1 1    

Compression Ratio:

r v



v d v a

Expansion Ratio:

r e



v v b c

Important features of the Diesel cycle

• • At r c – = 1, the Diesel and Otto cycles have the same efficiency.

Physical implication for the Diesel cycle: No change in volume when heat is supplied.

– A high value of k compensates for this.

For r c > 1, the Diesel cycle is less efficient than the Otto cycle.

FIGURE 9-22 Thermal efficiency of the ideal Diesel cycle as a function of compression and cutoff ratios (k = 1.4).



Efficiency Comparisons (Approximate)

r c > 1

r

Effect of variable specific heats

•

Efficiency is based on internal energies and enthalpies obtained from the gas tables which take into account variable specific heats.

  1 

u a h a



u b



h b

Key terms and concepts

Air-standard Diesel Cycle Expansion ratio Pressure ratio Cut of ratio Compression ignition

Comparison of the Otto and Diesel cycles

Comparison of the Otto and Diesel cycles Comparison No. 1: (a) Same inlet state (P,V) (b) Same compression ratio, r v (c) Same Q H Key factor: Constant volume heat addition of the Otto cycle vs. constant pressure heat addition of the Diesel cycle.

c b Otto cycle with a specified inlet condition at “a” with a given compression ratio r v = v a /v b

Displacement

d a

c b Otto and Diesel cycles with same compression inlet conditions at “a” and the same compression ratio, r v .

c* d* d

Displacement

c b c* Diesel Cycle

  1    1

r v k

 1    

k r c k



r c

  1 1    

d* d a

Otto Cycle

  1 

r v

1 

First Law analysis of the heat addition process

 

 

Otto: Heat addition with V = 0 in process b

 

W = 0, and P and T increase c.

Diesel: Heat addition with P = Constant in process b



c*.

dW > 0, and P and T lower that in the Otto cycle.

V = Const.

T T c T c* P = Const.

b a c

Otto Cycle

Diesel Cycle

d* d

T-s diagrams for equal heat addition

T T c T c* Q H

b a c d c* d*

The areas under the process paths b



c and b



c* are equal under the assumption of equal heat addition, Q H .

Q H,Otto = Q H,Diesel

T T c T c*

b a Efficiency comparisons c d

c* d*

When Q H and r v are the same for both cycles,



OTTO

 

DIESEL

Q C,Otto < Q C,Diesel

Comparison of the Otto and Diesel cycles Comparison No. 2: (a) Same inlet state (P,V) (b) Same maximum P (c) Same Q H Comparison No. 2 is more practical when the “knocking” effect is considered. In the Otto cycle, about 11 atm are needed to achieve combustion with engine knock.

T Diesel Cycle

c* b* b a c

Otto Cycle

d d*

V = Const.



P = Const.

DIESEL

 

OTTO

Gas Power Systems - 3

The Dual Cycle

The Dual cycle

• • The dual cycle is designed to capture capture some of the advantages of both the Otto and Diesel cycles.

It it is a better approximation to the actual operation of the compression ignition engine.

p Q H,V

Q H,P

The Dual cycle

s = Constant

a d e

Q

C,V V

FIGURE 9-23 P-v diagram of an ideal dual cycle.

Q H,V Q H,P

c a d e

Q C,V

  1   1 

C V

(

T b

(

T b

 

C v T a

(

T d

)  

C T P

) (

T C T

(

T d a

)  

T e

)

(

T C



T b



T b

) )

Q H,V

b a

Q H,P

c d

r v



V V a e r c



V V b c

  1  1

r v k

 1   

r p

r P



P P a b r p r c k

 1 

kr p

 1 (

r c

 1 )   

Gas Power Systems - 4

The Gas Turbine

Overview

• • • Gas Turbines – The Carnot cycle as an “air standard” cycle The Ericcson Cycle The Brayton Cycle – The standard cycle – – The reheat cycle Inter-cooling

The modern gas turbine cycle

• • • Long-haul automotive power.

Large aircraft flight envelopes Commercial aircraft • Altitude (30,000 to 40,000 ft) • Long range and low specific fuel consumption (sfc) • Moderate speed and thrust

The modern gas turbine cycle

• • Military aircraft • High altitude ( > 40,000 ft) • Moderate range and high specific fuel consumption (sfc) • High speed and large thrust Relevant cycles – The Ericsson and Brayton cycles

General features of the aircraft gas turbine engine Combustor Fuel Intake Air Compressor Turbine Exhaust Section Compressor Drive Shaft, W comp

The air-standard turbine cycle

• • • • • Open system modeled as a closed system - fixed with fixed mass flow.

Air is the working fluid.

Ideal gas assumptions are applied.

Approximate the combustor as the high temperature source.

Internally reversible processes.

Ideal gas power cycles Air Standard Carnot Cycle Air Standard Ericcson Cycle Air Standard Brayton Cycle

Air-standard Carnot cycle

Air-standard Carnot cycle

Employ the same assumptions as for the air-standard turbine cycle.

p 1

Q H

p 2



1 

2 ,  1 

T C T H P

3 

4  1 

p 4

Q C

p 3

  1   

P P

4 1    1 



 1   

1    1 

  1   

3    1 



 1   

2   ( 1 

)

Limitations of the air-standard Carnot cycle

• • Heat addition at constant temperature is difficult and costly.

–

Work is required because fluid expands

Heat addition is limited because a large change in volume would imply a low mean pressure in the heat addition process.

–

Frictional effects might become too great if the mean pressure is too low.

FIGURE 9-26 T-s and P-v diagrams of Carnot, Stirling, and Ericsson cycles.

FIGURE 9-27 The execution of the Stirling cycle.

Air-standard Ericsson cycle

FIGURE 9-28 A steady-flow Ericsson engine.

The air-standard Ericsson cycle

• • Constant pressure heat addition and rejection Constant temperature compression and expansion

Q b-c T = Constant

Q a-b Q c-d

Q d-a v

Q a-b Q b-c T = Const.

Q c-d

a The Air Standard Ericcson Cycle

Q d-a v

T P = Const.

Q b-c

Q c-d

Q d-a

Q a-b

Q a-b Q b-c T = Const.

Q c-d

Thermal Efficiency Ericcson Cycle

W a-b

Q d-a

 

W NET Q Added

W c-d v



W c



d Q b



W a





Q c



The Brayton cycle

The Brayton Cycle

• •

Modern gas turbines operate on an open Brayton cycle.

– Ambient air is drawn at the inlet.

– Exhaust gases are released to the ambient environment.

The air standard Brayton cycle is a closed cycle.

– – All processes are internally reversible.

Air is the working fluid and assumed an ideal gas.

General features of the aircraft gas turbine engine Intake Air Fuel Combustor Exhaust Gases & Work Output, W turb Compressor Turbine Compressor Drive Shaft, W comp

The closed, air standard cycle for the gas turbine.

The open, actual cycle for the gas turbine.

Q H W COMP Q L W TURB

The gas turbine processes

• • • • Isentropic compression to T H Constant pressure heat addition at T H Isentropic expansion to T C Constant pressure heat rejection at T C

1 2

Q H

W COMP Q L p

W TURB Q H

Air Standard Brayton Cycle

S = Constant

Q C

4 Vv

Q H

Q C

4 3

s = Constant

Air Standard Brayton Cycle

T Q H

p 2 = p 3

2 1

s 1 = s 2

Q C s 3 = s 4 p 1 = p 4 s

T W IN

Q H

4 3

p 2 = p 3 W OUT

Thermal efficiency of the ideal Brayton cycle

p 1 = p 4 Q C

 

W OUT Q H



W in

 (

3 

4 )

3   

2 

1 

T W IN

2 1

Q H

p 2 = p 3 W OUT

Thermal Efficiency Ideal Brayton Cycle



Q C

p 1 = p 4 For an ideal gas, h-h 0 = C p (T - T 0 ). The compression and expansion process are polytropic with constant k. s

 1    

1    ( 1 

)

Cycle efficiency

• • • Ideal gas assumptions apply.

All processes internally reversible – Compressor and turbine efficiency are each 100%.

Assume fuel added in the combustor is a small percent (mass or moles) of total flow, and thus air properties provide a good estimate of cycle performance.

Gas Power Systems - 5 Real Turbine Performance

Overview

•

Internally irreversible Brayton cycles

– Case study •

Real turbine performance

Internally irreversible Brayton cycles

Real cycle analysis - Brayton cycle

• • • • Internal irreversibility arises in each process of the open cycle.

Compression (compressor efficiency) Heat addition Expansion (turbine efficiency)

Internal irreversibility will lower work output and decrease thermal efficiency. Greater entropy production will also result.

T,h

2 

1 2 

p 2 = p 3

3  4 

p 1 = p 4

s 3 = s 4

 4

3 4

Case Study

Case study

Given: Turbine and compressor efficiencies of 80%, and the operating data as given below. No internal irreversibility in the combustor. Find: For the actual cycle, the thermal efficiency, the ratio of W comp /W turb and entropy production.

c b’ Q H c b b’ d d’ W IN W OUT a a d’

Q C

a b b’ c d d ’ State T (R) 530 881 967 1700 1059 1192 15 P (p sia) 90 90 90 15 15 h (Btu / lbm ) 126.7

211.6

232.7

422.6

255.7

289.3

S 0 (Btu / lbm R) 0.5963

0.7191

0.7421

0.8876

0.7647

0.7946

Note: Values of h and s o are from the gas tables, and all initial data are supplied here but this is not necessarily the case in practice.

Process Calculations

Process a-b a-b’ b’-c c-d  h (Btu/lbm) 84.9

106 189.9

-166.9

W (Btu/lbm) -84.9

-106 0 166.9

Q (Btu/lbm) 0 0 189.9

0  s (Btu/lbmR) 0 0.0230

0.1455

0 c-d’ d’-a -133.3

-162.6

133.3

Note:

NET 

s i





TURB

 

s i o



 



h s

0 ,  27.3

ln  

P P j i

 

COMP

 

h s



0 -162.6

0 0.0299

-.1983

The pressure ratios for the isentropic and actual processes are the same, and therefore



s =



s o .

Thermal Efficiency and Work Ratio



CYCLE



W Q b net

' 

 133 .

3  106 189  27 .

3  0 .

144 189

W comp W turb

 106 133 .

3  0 .

795

Notice that here we have a high ratio of the work of compression to that of expansion, which is typical in gas turbine systems. The key variable is the turbine inlet temperature, T c , and this is to be made as large a possible. T c is limited by the material and turbine blade cooling capability.

Entropy Production

Process

Q (Btu/lbm)



s (Btu/lbm-R) Q/T (Btu/lbm-R)



(Btu/lbm-R)

a-b a-b’ b’-c c-d c-d’ d’-a NET 0 0 189.9

0 0 -162.6

27.3

0 0.0230

0.1455

0 0.0299

-0.1983

0 0 0 0.1455

0 0 -0.1983

-0.0528

0 0.0230

0 0 0.0299

0 0.0529

The entropy production rates are based on the application of the entropy balance for an open system to each process in the cycle. The heat addition and rejection processes are at the high and low temperatures respectively. Thus, these processes produce an entropy production from the external heat transfer process that is not included in the above calculations.

The Carnot Efficiency for the Cycle

Q H , T H



Carnot

 1 

T T H C

 0 .

68   0 .

144  1  530 1700

Q C , T C (General cycle representation)

Here, the difference in the efficiency calculations indicates a high level of external entropy production.

0.25



0.1

Efficiency Curves 

TURB

 

COMP

 1 0.9

2 4 Pressure Ratio, p 2 /p 1

0.8

0.7

T,h

2 

1 s’ 2 S 1 = S 2 p 2 = p 3 p 1 = p 4 S 3 = S 4

3  4 

s’

Brayton cycle efficiency is greatly dependent on the efficiencies of the turbine and compressor. The curves at the left are approximate.

FIGURE 9-29 An open-cycle gas-turbine engine.

FIGURE 9-30 A closed-cycle gas-turbine engine.

FIGURE 9-31 T-s and P-v diagrams for the ideal Brayton cycle.

FIGURE 9-32 Thermal efficiency of the ideal Brayton cycle as a function of the pressure ratio.

Gas Power Systems - 6 Cycle Improvements

Cycle Improvements

• • • Regeneration – Reduces heat input requirements and lowers heat rejected.

Inter-cooling – Lowers mean temperature of the compression process Reheat – Raises mean temperature of the heat addition process

For fixed values of T min and T max the net work of , the Brayton cycle first increases with the pressure ratio, then reaches a maximum at r

p (T

max

min

)

k/[2(k –

1)], and finally decreases.

FIGURE 9-36 The deviation of an actual gas turbine cycle from the ideal Brayton cycle as a result of irreversibilities.

The Brayton cycle with regeneration

FIGURE 9-38 A gas-turbine engine with regenerator.

The standard Brayton cycle shown as an open cycle.

Regeneration is accomplished by preheating the combustor air with the exhaust gas from the turbine.

1 2 5 2 1

W IN Q H

3 4

W OUT W COMP Q H

3 4

W TURB

T,h

Q H 3 x 4 2 y

Internal heat transfer, States 2 - x.

1 Q C

The maximum benefit of regeneration is obtained when the exhaust temperature T y is brought to temperature T 2 by the regenerative heat exchanger.

T,h

3 Q H

The ideal regenerator

x 2 y 4 The ideal regenerator will heat the combustion air up to T 4 1 Q C



CYCLE

 1  (

3  

h h

4 1 )

T,h

Q H 3

The actual regenerator

2 1





4 Actual internal heat transfer takes place between States 2 - x’.

y The effectiveness of the regenerator is the actual heat transfer divided by the maximum possible heat transfer.

Q C



C p

(

T x



C p

(



2 ' ) 

2 )  (

(

T x

 4 

2 

2 ) )

T,h

3 Q H 2



x y



Thermal efficiency

Regeneration reduces net heat input at the high temperature.



cycle

Q C

 (

3 

4 )

3   (

h h x

 2 

1 )

FIGURE 9-39 T-s diagram of a Brayton cycle with regeneration.

FIGURE 9-40 Thermal efficiency of the ideal Brayton cycle with and without regeneration.

The Brayton cycle with inter-cooling

The standard Brayton cycle with on stage of inter cooling

Q H

3 1 5 6

W COMP

W TURB

7 7’ 5 5’ 4 4’ 6 1

One stage of inter-cooling with turbine and compressor inefficiencies.

FIGURE 9-42 Comparison of work inputs to a single-stage compressor (1AC) and a two stage compressor with intercooling (1ABD).

FIGURE 9-43 A gas-turbine engine with two stage compression with intercooling, two stage expansion with reheating, and regeneration.

FIGURE 9-44 T-s diagram of an ideal gas-turbine cycle with intercooling, reheating, and regeneration.

FIGURE 9-45 As the number of compression and expansion stages increases, the gas-turbine cycle with intercooling, reheating, and regeneration approaches the Ericsson cycle.

The Brayton cycle with reheat

Q H

T,h



2  3 

4  5

3 5 4 6 2

Q c



6  1

s The re-heating of the fluid at an intermediate pressure raises the enthalpy input to the cycle without raising the temperature of the external thermal reservoir.

Example

This example compares several modifications to the basic Brayton cycle : multi-stage compression and expansion with reheat, inter-cooling and regeneration.

Description: In an air-standard gas turbine engine, air at 60 o F and 1 atm enters the compressor with a pressure ratio of 5:1. The turbine inlet temperature is 1500 o F, and the exhaust pressure is 1 atm.

Determine the ratio of turbine work to compressor work and the thermal efficiency under the following operating conditions.

Operating conditions: (a) The engine operates on an ideal Brayton cycle.

(b) The adiabatic efficiency of the turbine and compressor are 0.83 and 0.93, respectively.

(c ) The conditions of (b) hold and a regenerator with an effectiveness of 0.65 is introduced.

Operating conditions: (d) The conditions of (b) and (c ) hold, and an inter-cooler that cools the air to 60 o F at a constant pressure of 35 psia is added.

(e) The inter-cooler is removed and a re-heater the heats the fluid to 1500 o F at a constant pressure of 35 psia is added.

(f) The conditions of (e) hold and the inter cooler of (d) is put back.

Assumptions: (1) All cycles operate on the air-standard basis.

(2) Ideal gas (3) Constant specific heats; k = constant, C p = 0.24 BTU/lbm-R, k = 1.40 = constant.

(4) Internally reversible processes except where isentropic efficiency (adiabatic efficiency) is less than one.

(5) 1 atm = 14.7 psia.

Solution: (a) The engine operates as an ideal Brayton cycle.

2 1 3 4

T T

4 3  

3  4  



3 4    

k k

1  

C p



3  4 

1  2 

C p



1  2

State D ata State 1 2 3 4 P (psia) 14.7

73.5

14.7

T (R) 520 825 1960 1237

Process quantities Process Quantities 1-2 2-3 3-4 4-1 Total



h BTU/Lbm 73.1

272.4

-173 -172 0.5 (~0)



Q BTU/lbm 0 272.4

0 -172 100.32

Process quantities

3  4  

3  4 

C p



1  2  

1  2  73 .

1 

BTU

  173

lbm BTU W turbine W compressor



3  4

1  2  173 73 .

1  2 .

37 

cycle

 

W net



2  3 173 

3  4 73 ( 0 .

24 )( 1960 .



1 2   3

1  2 825 )   

2  3  

1  2 

2  3 0 .

367

lbm

(b) The adiabatic efficiency of the compressor and turbine are 0.83 and 0.92 respectively.

2 1



2’



3 4

0 .

92 ,

4’

4 

3 

3  4  

3    

k k

C p



3  4 

1  2



comp



C p



1  2  0 .

State D ata State 1 2 3 P (psia) 14.7

73.5



comp T

2   

14.7

W s

, 1  2 

1  2 

2  

comp T

2   

1 

887



turb



3  4

W s

, 3  4 

R T

4  

1295

T (R) 520 825 1960 1237

State D ata State 1 2’ 3 4’ P (psia) 14.7

73.5

3  4 

1  2   

3  4 

C p



3  

1  2  88 .

14.7



4 

BTU

  159 .

lbm BTU W turbine W compressor



3  4

1  2  159 .

5 88 .

8  1 .

81 

cycle

 

W net



W Q

2   3

2   3 159 .

5  88 .

8 ( 0 .

24 )( 1960 3   4  

1  887 )  2   0 .

28 

3  4   

1  2  

2   3

lbm

T (R) 520 887 1960 1295

(c) The conditions of (b) hold, and a regenerator with an effectiveness of 0.65 is introduced.

2 2’ x 3 y 4’

T T

4 3  

3  4    

    

k k

1  



3  4

C p



1  2 

T x

   (

T x

 (

4 '

2    

2  '  ' ) )

4    0 .

2   

1052 

1  2



  

comp

0 .

92  0 .

State D ata State 1 x’ 3 4’ P (psia) 14.7

73.5

3  4 

1  2   

3  4 

C p



3  

1  2  88 .

14.7



4 

BTU

  159 .

lbm BTU W turbine W compressor



3  4

1  2  159 .

5 88 .

8  1 .

81 

cycle

 

W net Q x

  3  159 .

3  4  

1 

Q x

  3  88 .

8 2  ( 0 .

24 )( 1960  1052 )   

3  4   

1  2  

h x

  3 0 .

lbm

T (R) 520 1052 1960 1295

(d) The conditions of (b) and (c ) hold, and an inter-cooler that cools the air to 60 o F at 35 psia is added.

T P = 35 psia 520 R

6 7’ x 1 5’ y 4’

T 520 R

6 7’ x 1 5’ 3

P = 35 psia

4’ y

Compute T 5’ ,T 7’ ,

temperature between state 7’ and x due to regeneration (7”).

State D ata State 7 7’ 7” 3 4’ 1 5 5’ 6 x y

5   

P P

1  667 5    

R k

1  

P (psia) 14.7

35 35 14.7

73.5

14.7

73.5

14.7

T (R) 520 667 697 520 643 668 616 1960 520 1295 888

5  

1   1 

5 

1 

6 7’ x 1 5’ 3

P = 35 psia

4’

3  4   159

1  5  

6  7  .

BTU

 78 /

lbm BTU

lbm W

turbine W compressor

 2 .

04  

C p

(

W net T

3 

7  )  0 .

(e) Eliminate inter-cooling but add one stage of reheat to 1500 o F.

  0 .

(f) Combine parts (d) and (e).

  0 .

W turbine W Compressor

 2 .

Summary of results:



(a) Ideal Cycle (b) Actual Basic Cycle (c) Actual Basic Cylce and Regenerator (



= 0.65) (d) Actual Cycle w ith Intercooling at 35 psia to 60 o F and part (c) (e) Actual Basic Cycle w ith Reheat at 35 psia and part (c) (f) Parts (c ) ,(d), and (e) combined 0.37

0.28

0.37

0.38

0.40

Work Ratio W t /W c 2.37

1.81

1.18

2.04

2.02

2.28

W t 173 160 160 160 178 178

FIGURE 9-48 Basic components of a turbojet engine and the T-s diagram for the ideal turbojet cycle.

[ Source: The Aircraft Gas Turbine Engine and Its Operation. © United Aircraft Corporation (now United Technologies Corp.), 1951, 1974.]

FIGURE 9-51 Energy supplied to an aircraft (from the burning of a fuel) manifests itself in various forms.

FIGURE 9-52 A turbofan engine.

[Source: The Aircraft Gas Turbine and Its Operation. © United Aircraft Corporation (now United Technologies Corp.), 1951, 1974.]

A modern jet engine used to power Boeing 777 aircraft. This is a Pratt & Whitney PW4084 turbofan capable of producing 84,000 pounds of thrust. It is 4.87 m (192 in.) long, has a 2.84 m (112 in.) diameter fan, and it weighs 6800 kg (15,000 lbm).

Photo Courtesy of Pratt&Whitney Corp.

FIGURE 9-54 A turboprop engine.

[Source: The Aircraft Gas Turbine Engine and Its Operation.

FIGURE 9-55 A ramjet engine.

[Source: The Aircraft Gas Turbine Engine and Its Operation.

FIGURE 9-57 Under average driving conditions, the owner of a 30-mpg vehicle will spend $300 less each year on gasoline than the owner of a 20-mpg vehicle (assuming $1.50/gal and 12,000 miles/yr).

Aerodynamic drag increases and thus fuel economy decreases rapidly at speeds above 55 mph.

(Source: EPA and U.S. Dept. of Energy.)

No Slide Title

Transcript No Slide Title

Gas Power Cycles

9-1 Basic Considerations in the Analysis of Power Cycles

General classes of engines

Single vs. double action engines

Indicator diagrams

9-2 The Caront Cycle and Its Value in Engineering

9-3 Air Standard Assumptions

9-4 An Overview of Reciprocating Engines

Indicator Diagrams

Indicator Diagrams

Work = MEP x A x Displacement.

The spark ignition engine The Otto cycle

(

(

)

)

1

9-6 Diesel Cycle:

The compression ignition engine - the Diesel cycle

Eliminates pre-ignition of the fuel-air mixture when compression ratio is high.

All heat transfer to a constant mass system.

Key parameters for

the Diesel cycle

Efficiency Comparisons (Approximate)

r

•

Air-standard Diesel Cycle Expansion ratio Pressure ratio Cut of ratio Compression ignition

Comparison of the Otto and Diesel cycles

Gas Power Systems - 3

The Dual cycle

Q

Gas Power Systems - 4

Air-standard Carnot cycle

The closed, air standard cycle for the gas turbine.

The open, actual cycle for the gas turbine.

Air Standard Brayton Cycle

Air Standard Brayton Cycle

Overview

Internally irreversible Brayton cycles

Real turbine performance

The Carnot Efficiency for the Cycle

One stage of inter-cooling with turbine and compressor inefficiencies.

887

1295

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