EGR 334 Thermodynamics Chapter 9: Sections 5-6 Lecture 35: Gas Turbine modeling with the Brayton Cycle Quiz Today?

EGR 334 ThermodynamicsChapter 9: Sections 5-6

Lecture 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 P

PTT

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

1

212

r

r

P

PPP and

6

Dual Cycle analysis

2334 hhmQ

544545 uumUW

211212 uumUW

34 45 12

23 34

cycle

in

W W W W

Q 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) 520

p (psi) 14 1200

1200

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) 520

p (psi) 14 1200

1200

u (Btu/lb) 88.62

h (Btu/lb)

vr 158.58

Pr 1.2147

Identify State Properties

State 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) 520

p (psi) 14 1200

1200

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

572.1015

58.15812

r

vv rr

22 1

1

51.56114 594.26

1.2147r

r

pp p psi psi

p

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 1200

1200

u (Btu/lb) 88.62

h (Btu/lb) 124.27

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

12001468.8

594.26R

State 3: given v3 = v2 and

p3 = 1200 psi, use ideal

gas 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/lb

use 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 u

m 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

T

Tr

T

T

V

V

V

V

V

V

V

VV

V

VV

V

V

V

V

V

V

Replace V’s using ideal gas. 7187.9

6.4577

296615

4

3

4

5 T

Tr

V

V

55 4

4

9.7187 0.2848 2.768r r

Vv v

V

State 1 2 3 4 5\

T (R) 520 1468.8 2966 4577.6

2299

p (psi) 14 594.26 1200 1200 61.44

u (Btu/lb) 88.62 260.26 577.4 949.7 431.0

h (Btu/lb) 124.27 361.53 780.7 1263.6

601.48

vr 158.58 10.572 0.2848

2.768

pr 1.2147 51.561 5961.6

305.24

process 4-5 is also isentropic

55 4

4

305.241200 61.44

5961.6r

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 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

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 Q

m m m m m

515 1 431.0 88.62 342.4 /out

m

Q Qu u Btu lb

m 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

p (psi) 14 594.26 1200 1200 61.44

u (Btu/lb) 88.62 260.26 577.4 949.7 431.0

h (Btu/lb) 124.27 361.53 780.7 1263.6

601.48

vr 158.58 10.572 0.2848

2.768

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

W

Q

457.60.572

800cycle

in

W

Q

4

3

4577.61.543

2965.9c

Vr

V

State 1 2 3 4 5\

T (R) 520 1468.8 2966 4577.6

2299

p (psi) 14 594.26 1200 1200 61.44

u (Btu/lb) 88.62 260.26 577.4 949.7 431.0

h (Btu/lb) 124.27 361.53 780.7 1263.6

601.48

vr 158.58 10.572 0.2848

2.768

pr 1.2147 51.561 5961.6

305.24

5 1

3 2 4 3

1u u

u u h h

1 out

in

Q

Q

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

3 4

3 4

V VmR

T 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

17

Gas Turbine Analysis

121212 hhmHWW C

23

1243

23

1234

hh

hhhh

Q

WW

Q

W

in

cycle

232323 hhmHQQ in

Sec 9.5 : Modeling Gas Turbine Power Plants

433434 hhmHWW T

14`441 hhmHQQ out

43

12

23

1234

hh

hh

Q

WW

W

Wbwr

T

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.





18


Given T1 & T3 use table to find h1 & h3 .

Find state 2.

Find state 4.

For Cold-Air Standard analysis:

12 2

1 1

k kT p

T p

1 1

4 4 1

3 3 2

k k k kT p p

T p p

For state 2. For state 4.

22 1

1r r

pp p

p

Sec 9.3 : Air-Standard Diesel Cycle

44 3

3r r

pp p

p

Compressor pressure ratio:

2

1

p

p

19


Effect of Compressor pressure on efficiency.

23

1243

hh

hhhh

Q

W

in

cycle

23

1243

TTc

TTcTTc

P

PP

2

1

23

14

2

1

23

14 11

111

T

T

TT

TT

T

T

TT

TT

with

2

3

1

4

T

T

T

T

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 4

T (R) 500 3000

h (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


State 1 2 3 4

T (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 R

p

1 1.4 1 1.4

14 3

2

13000 1411.42

14

k kp

T T Rp

For state 2.

For state 4.

22



State 1 2 3 4

T (R) 500 1063 3000 1411

2

11T

T 529.0

1063

5001

T

C

W

Wbwr

43

12

43

12

43

12

TT

TT

TTc

TTc

TTm

TTm

P

P

354.0

14113000

5001063

23



State 1 2 3 4

T (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)in

mP

Q Btu hrm lb hr

c 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 4

T (R) 520 1092 3000 1573

pr 1.2147

17.01 941.4 67.24

h (Btu/lb) 124.27

264.12

790.68

388.63



State 1 2 3 4

T (R) 520 3000

pr 1.2147

941.4

h (Btu/lb) 124.27

790.68Temperatures for states 1 and 3 are given. Relative

pressure and enthalpy values from Table A-22E

Find state 2.

Find state 4.

22 1

1r r

pp p

p

44 3

3r r

pp p

p

0058.17142147.1

24.6714

14.941

26



23

1243

hh

hhhh

Q

W

in

cycle

12.26468.790

27.12412.26463.38868.790

State 1 2 3 4

T (R) 520 1092 3000 1573

pr 1.2147

17.01 941.4 67.24

h (Btu/lb) 124.27

264.12

790.68

388.63

498.0

27



State 1 2 3 4

T (R) 520 1092 3000 1573

pr 1.2147

17.01 941.4 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 /in

mm

Q Btu hrm lb hr

h 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

EGR 334 Thermodynamics Chapter 9: Sections 5-6 Lecture 35: Gas Turbine modeling with the Brayton Cycle Quiz Today?

Documents

pressure p

p psi14594

actual p

dual cycle brayton cycle

psi state

brayton cycle model

hybrid cycle

cycle repertoire