EGR 334 Thermodynamics Chapter 9: Sections 5-6

EGR 334 ThermodynamicsChapter 9: Sections 5-6Lecture 35: Gas Turbine modeling with the Brayton Cycle

Quiz Today?

Today’s main concepts:• Be able to recognize Dual and Brayton Cycles• Understand what system may be modeled using

Brayton Cycle.• Be able to perform a 1st Law analysis of the Brayton

Cycle and determine its thermal efficiency.• Be able to explain how regeneration may be applied to

a Brayton Cycle model.

Reading Assignment:

Homework Assignment: Read Chapter 9, Sections 7-8

Problems from Chap 9: 42, 47, 55

3

OK….Quick Matching Quiza) Carnot b) Rankine c) Otto d) Diesel

p

v

..

..

4 1

3

1’

2 2’

AD

BC

4

Today you get to add two more cycles to your cycle repertoire.

Dual Cycle Brayton cycle.

Used as a hybrid cycle which includes elements of both the Otto and Diesel cycles. Used to model internal combustion engines

Used as a model for gas turbines (such as jet engines).

5Sec 9.4 : Air-Standard Duel Cycle

Neither the Otto or Diesel cycle describe the actual P-v diagrams of an engineHeat addition occurs in two steps• 2 – 3 : Constant volume heat addition• 3 – 4 : Constant pressure heat addition (first part of power stroke)

Process 1 – 2 : Isentropic compression

Process 2 – 3 : Constant volume heat transfer

Process 5 – 1 : Constant volume heat rejection

Process 3 – 4 : Constant pressure heat transfer

Process 4 – 5 : Isentropic expansion

2

323 PPTT

To set state 3: Use ideal gas law with V3 = V2.

1

212

r

r

PPPP and

6Dual Cycle analysis

2334 hhmQ

544545 uumUW

211212 uumUW

34 45 12

23 34

cycle

in

W W W WQ Q Q

Sec 9.4 : Air-Standard Duel Cycle

232323 uumUQ

34 4 3W p v v

155151 uumUQ

5 151

23 34 3 2 4 3

1 1u uQ

Q Q u u h h

process 1-2: s1 = s2

process 2-3: v2 = v3

process 3-4: p3 = p4


process 5-1: v5 = v1

7

Example (9.38): The pressure and temperature at the beginning of compression in an air-standard dual cycle are 14 psi, 520°R. The compression ratio is 15 and the heat addition per unit mass is 800 Btu/lbm. At the end of the constant volume heat addition process the pressure is 1200 psi. Determine,(a) Wcycle, in BTU/lb.(b) Qout, in BTU/lb.(c) The thermal efficiency.(d) The cut off ratio

State 1 2 3 4 5\

T (R) 520p (psi) 14 120

01200

u (Btu/lb) h (Btu/lb)vr

pr

State 1 2 3 4 5\

T (R)p (psi)u (Btu/lb)h (Btu/lb)vr

Pr

8

Example (9.38): State 1 2 3 4 5\

T (R) 520p (psi) 14 120

01200

u (Btu/lb) 88.62h (Btu/lb)vr 158.5

8Pr 1.214

7Identify State PropertiesState 1: p1 = 14 psi, T1 = 520 RState 2: s2 = s1 v2 = v1/rState 3: v3 = v2 and p3 = 1200 psiState 4: p4 = p3 = 1200 psiState 5: s5 =s4 and v5 = v1

Use Table A22E to fill in many of the other properties.

compression ratio, r = 15

Qin= Q23 + Q34 = 800 Btu

Given Information:

Qout = - Q51

State 1 2 3 4 5\

T (R) 520p (psi) 14 120

01200

u (Btu/lb)h (Btu/lb)vr

Pr

9

Example (9.38):

State 2: use r to find v2 and since 1-2 is isentropicfind vr2

572.101558.1581

2 rvv r

r

22 1

1

51.56114 594.26

1.2147r

r

pp p psi psip

State 1 2 3 4 5\

T (R) 520 1468.8

p (psi) 14 594.26

1200

1200

u (Btu/lb) 88.62 260.26

h (Btu/lb) 124.27

361.53

vr 158.58

10.572

pr 1.2147

51.561

State 1: given T = 520 Rlook up u, h, vr, and pr

then use Table A22E to look up T2, pr2, u2, and h2:

Pressure p2, can then be calculated using

State 1 2 3 4 5\

T (R) 520 p (psi) 14 120

01200

u (Btu/lb) 88.62h (Btu/lb) 124.2

7

vr 158.58

pr 1.2147

10

Example (9.38):

33 2

2

pT T

p

pv RT

State 1 2 3 4 5\

T (R) 520 1468.8

2966

p (psi) 14 594.26

1200

1200

u (Btu/lb) 88.62 260.26

577.4

h (Btu/lb) 124.27

361.53

780.7

vr 158.58

10.572

pr 1.2147

51.561

12001468.8594.26

R

State 3: given v3 = v2 andp3 = 1200 psi, use idealgas law:

then use Table A22E to look up u3 and h3:2965.97 R

State 1 2 3 4 5\

T (R) 520 1468.8

p (psi) 14 594.26

1200

1200

u (Btu/lb) 88.62 260.26

h (Btu/lb) 124.27

361.53

vr 158.58

10.572

pr 1.2147

51.561

11

Example (9.38):State 4: Knowing p4=p3

and the heat in:Qin= 800 Btu/lbuse the 1st Law:

4 3 2 3inQh u u hm

800 577.4 260.26 780.7 1263.56 / mBtu lb

Use Table A-22E to find T4 ,u4, pr4,and v4r

State 1 2 3 4 5\

T (R) 520 1468.8

2966 4577.6

p (psi) 14 594.26

1200

1200

u (Btu/lb) 88.62 260.26

577.4

949.7

h (Btu/lb) 124.27

361.53

780.7

1263.6

vr 158.58

10.572

0.2848

pr 1.2147

51.561

5961.6

4 2 24 24( )m u u Q W

23 34 23 343 2 4 3

Q Q W Wu u u um m m m

3 2 4 3 4 3( )inQu u u u p v vm

O

3 2 4 3 4 3 3 2 4 3( )inQ u u p v v u u u u h hm

State 1 2 3 4 5\

T (R) 520 1468.8

2966

p (psi) 14 594.26

1200

1200

u (Btu/lb) 88.62 260.26

577.4

h (Btu/lb) 124.27

361.53

780.7

vr 158.58

10.572

pr 1.2147

51.561

12

Example (9.38):State 5:

Use Table A-22E to look up T5, u5, h5, and pr5 and then find p5:

4

3

4

3

2

1

4

3

2

1

4

32

2

15

4

2

2

5

4

5

TTr

TT

VV

VV

VV

VVV

VVV

VV

VV

VV

Replace V’s using ideal gas. 7187.9

6.4577296615

4

3

4

5 TTr

VV

55 4

4

9.7187 0.2848 2.768r rV

v vV

State 1 2 3 4 5\

T (R) 520 1468.8 2966 4577.6

2299

p (psi) 14 594.26 1200 1200 61.44u (Btu/lb) 88.62 260.26 577.4 949.7 431.0h (Btu/lb) 124.27 361.53 780.7 1263.

6601.4

8vr 158.58 10.572 0.284

82.768

pr 1.2147 51.561 5961.6

305.24

process 4-5 is also isentropic

55 4

4

305.241200 61.445961.6

r

r

pp p psi

p

State 1 2 3 4 5\

T (R) 520 1468.8 2966 4577.6

p (psi) 14 594.26 1200 1200u (Btu/lb) 88.62 260.26 577.4 949.7h (Btu/lb) 124.27 361.53 780.7 1263.

6vr 158.58 10.572 0.284

8pr 1.2147 51.561 5961.

6

13

Example (9.38):(a) Wcycle, in Btu/lb.(b) Qout, in Btu/lb.(c) The thermal eff. (d) The cut off ratio

23 34 51cycle in outW Q Q Q Q Qm m m m m

515 1 431.0 88.62 342.4 /out

mQ Q u u Btu lbm m

800 342.4 457.6 /cyclem

WBtu lb

m

State 1 2 3 4 5\

T (R) 520 1468.8 2966 4577.6

2299


6601.4

8vr 158.58 10.572 0.284

82.768

pr 1.2147 51.561 5961.6

305.24

14

Example (9.38):(a) Wcycle, in Btu/lb.(b) Qout, in Btu/lb.(c) Thermal efficiency(d) The cut off ratio

cycle

in

WQ

457.6 0.572800

cycle

in

WQ

4

3

4577.6 1.5432965.9c

VrV

State 1 2 3 4 5\

T (R) 520 1468.8 2966 4577.6

2299


6601.4

8vr 158.58 10.572 0.284

82.768

pr 1.2147 51.561 5961.6

305.24

5 1

3 2 4 3

1u u

u u h h

1 out

in

QQ

Cut off ratio: from ideal gas equation at constant pressure: pV mRT

3 4

3 4

V VmRT p T

15Sec 9.5 : Modeling Gas Turbine Power Plants

Air-Standard analysis of Gas Turbine Power plants.Gas power plants are lighter and more compact than vapor power plants.

Used in aircraft propulsion & marine power plants.

16

Jet engine:Suck (intake)Squeeze (compressor)Bang/Burn (combustion)Blow (turbine/exhaust)

Air-Standard analysis:Working fluid is airHeat transfer from an external source (assumes there is no reaction)

Process 1 – 2 : Isentropic compression of air (compressor).Process 2 – 3 : Constant pressure heat transfer to the air from an external source (combustion) Process 3 – 4 : Isentropic expansion (through turbine)Process 4 – 1 : Completes cycle by a constant volume pressure in which heat is rejected from the air

Heat Ex

Sec 9.5 : Modeling Gas Turbine Power Plants

17Gas Turbine Analysis

121212 hhmHWW C

23

1243

23

1234

hhhhhh

QWW

QW

in

cycle

232323 hhmHQQ in

Sec 9.5 : Modeling Gas Turbine Power Plants

433434 hhmHWW T

14`441 hhmHQQ out

43

12

23

1234

hhhh

QWW

WWbwrT

C

For a gas turbine, the back work ratio is much larger than that in a steam cycle since vair>>vliquid

bwr for a gas turbine power cycle is typically 40-80% vs. 1-2% for a steam power cycle.





18Gas Turbine AnalysisGiven T1 & T3 use table to find h1 & h3 .

Find state 2.

Find state 4.

For Cold-Air Standard analysis: 1

2 2

1 1

k kT pT p

1 1

4 4 1

3 3 2

k k k kT p pT p p

For state 2. For state 4.

22 1

1r r

pp pp

Sec 9.3 : Air-Standard Diesel Cycle

44 3

3r r

pp pp

Compressor pressure ratio:

2

1

pp

19Gas Turbine AnalysisEffect of Compressor pressure on efficiency.

23

1243

hhhhhh

QW

in

cycle

23

1243

TTcTTcTTc

P

PP

2

1

23

14

2

1

23

14 11111

TT

TTTT

TT

TTTT

with2

3

1

4

TT

TT

12 1

11 k kp p

Max T3 is approximately 1700 K

Sec 9.3 : Air-Standard Diesel Cycle

20

Example : Air enters the compressor of an ideal cold air-standard Brayton cycle at 500°R with an energy input of 3.4x106 Btu/hr. The compression ratio is 14 and the max T is 3000°R. For k=1.4 calculate (a) The thermal efficiency(b) The back work ratio.(c) The net power developed.

State 1 2 3 4T (R) 500 300

0h (BTU/lb)

21

Example : Air enters the compressor of an ideal cold air-standard Brayton cycle at 500°R with an energy input of 3.4x106 BTU/hr. The compression ratio is 14 and the max T is 3000°R. For k=1.4 calculate (a) The thermal efficiency(b) The back work ratio.(c) The net power developed.

State 1 2 3 4T (R) 500 3000

Since we are given k=1.4, use a cold-air standard analysis.Temperatures for states 1 and 3 are given.

1

1.4 1 1.422 1

1

500 14 1062.76k k

pT T Rp

1 1.4 1 1.4

14 3

2

13000 1411.4214

k kpT T Rp

For state 2.

For state 4.

22


State 1 2 3 4T (R) 500 1063 3000 1411

2

11TT

529.010635001

T

C

WWbwr

43

12

43

12

43

12

TTTT

TTcTTc

TTmTTm

P

P

354.0

141130005001063

23


State 1 2 3 4T (R) 500 1063 3000 1411

12431243 TTTTcmhhhhmWWW PCTCycle

But need the mass flow rate. 2323 TTcmhhmQ Pin

6

3 2

(3.4 10 / ) 7078 /(0.248 / )(3000 1063)

inm

P

Q Btu hrm lb hrc T T Btu lbm R R

7078 / 0.248 / 3000 1411 1063 500Cycle m mW lb hr Btu lb R R

61.80 10 /CycleW Btu hr

24

Example (9.43): The rate of heat addition to an air-standard Brayton cycle is 3.4x109 BTU/hr. The pressure ratio for the cycle is 14 and the minimum and maximum temperatures are 520°R and 3000°R, respectively. Determine(a) The thermal efficiency(b) The net power developed. State 1 2 3 4

T (R) 520 3000

Pr h (Btu/lb)

25

State 1 2 3 4T (R) 520 1092 3000 1573pr 1.214

717.01 941.4 67.24

h (Btu/lb) 124.27

264.12

790.68

388.63

Example (9.43): The rate of heat addition to an air-standard Brayton cycle is 3.4x109 BTU/hr. The pressure ratio for the cycle is 14 and the minimum and maximum temperatures are 520°R and 3000°R, respectively. Determine(a) The thermal efficiency(b) The net power developed.

State 1 2 3 4T (R) 520 3000pr 1.214

7941.4

h (Btu/lb) 124.27

790.68Temperatures for states 1 and 3 are given. Relative

pressure and enthalpy values from Table A-22EFind state 2.

Find state 4.

22 1

1r r

pp pp

44 3

3r r

pp pp

0058.17142147.1

24.671414.941

26


23

1243

hhhhhh

QW

in

cycle

12.26468.790

27.12412.26463.38868.790

State 1 2 3 4T (R) 520 1092 3000 1573pr 1.214

717.01 941.4 67.24

h (Btu/lb) 124.27

264.12

790.68

388.63

498.0

27


State 1 2 3 4T (R) 520 1092 3000 1573pr 1.214

717.01 941.4 67.24

h (Btu/lb) 124.27

264.12

790.68

388.63 2143 hhhhmWWW CTCycle

But need the mass flow rate. 23 hhmQin

9

6

3 2

3.4 10 / 6.46 10 /790.68 264.12 /

inm

m

Q Btu hrm lb hrh h Btu lb

61.80 10 / 790.68 388.63 264.12 124.27 /Cycle m mW lb hr Btu lb

91.69 10 /CycleW Btu hr

28

End of Slides for Lecture 35

Documents

EGR 334 Thermodynamics Chapter 9: Sections 5-6