tutorial gasturbine solution

Jabatan Kejuruteraan Loji dan AutomotifFakulti Kejuruteraan Mekanikal dan PembuatanUniversiti Tun Hussein Onn Malaysia

SolutionTutorial 2/ ws

BDA 3043

Tutorial 2 – SolutionGas turbinesWinardi Sani

¶ Air enters the compressor of an ideal air standard Brayton cycle at 100 kPa, 300K, with a volumetric flow rate of 5 m3/s. The compressor pressure ratio is 10. Theturbine inlet temperature is 1400 K.

1

2 3

4

Condenser

Compressor

Combustor

Turbine

QinQin

Qout

Win

Wout

QoutWin

Wout

v s

1

2 3

4

1

2isentropicisobar

4

3

Tp

= 10

0 kP

a

1400

100300

p

Using air-standard assumptions: cp = f(T ).

(a) The thermal efficiency of the cycle

• Referring to the diagrams above:

ηth,B =|wout| − |win||qin|

=|h4 − h3| − |h2 − h1|

|h3 − h2|(1)

The value of each enthalpy is taken from the air property table.

• Change of states:1 – 2: Isentropic compressionState 1 : T1 = 300 K, p1 = 100 kPa=⇒ h1 = 300.19 kJ/kg, pr1 = 1.3860

pr2 = pr1 ∗p2

p1, isentropic relation

= 1.3860 ∗ 10 = 13.860 (2)

1


BDA 3043


At the air properties table, read the required values on pr2 = 13.860.Interpolation is needed.

h2 = 575.59 +[586.04− 575.5914.38− 13.50

]∗ (13.86− 13.50)

= 579.87 kJ/kg (3)

T2 = 570 +[

580− 57014.38− 13.50

]∗ (13.86− 13.50)

= 574.09 K (4)

3 – 4: Isentropic power output:State 3 : T3 = 1400 K, pr3 = 450.5, h3 = 1515.42 kJ/kg

Ideal-gas properties of air

T h pr u vr so

K [kJ/kg] - [kJ/kg] - [kJ/kg K]

200 199.97 0.3363 142.56 1707.0 1.29559

300 300.19 1.3860 214.07 621.2 1.70203

540 544.35 11.10 389.34 139.7 2.29906

570 575.59 13.50 411.97 121.2 2.35531

580 586.04 14.38 419.55 155.7 2.37348

780 800.03 43.35 576.12 51.64 2.69013

800 821.95 47.75 592.30 48.08 2.71787

1400 1515.42 450.5 1113.52 8.919 3.36200

pr4 = pr3 ∗p4

p3= pr3 ∗

1p3p4

= pr3 ∗1p2p1

= 450.5 ∗ 110

= 45.05

Interpolation is needed here.

h4 = 800.03 +[821.95− 800.0347.75− 43.35

]∗ (45.05− 43.35)

= 808.5 kJ/kg (5)

T4 = 780 +[

800− 78047.75− 43.35

]∗ (45.05− 43.35)

= 787.73 K (6)

• The value of the enthalpy at all states has been calculated:

|h4 − h3| = |808.5− 1515.42| = 706.92 kJ/kg

|h2 − h1| = |579.87− 300.19| = 279.68 kJ/kg

|h3 − h2| = |1515.42− 579.87| = 935.55 kJ/kg (7)

2



BDA 3043

=⇒ ηth,B =|h4 − h3| − |h2 − h1|

|h3 − h2|

=706.92− 279.68

935.55=

427.24935.55

=⇒ ηth,B = 45.7% (8)

(b) The back work ratio

rbw =|wc||wt|

=|w12||w34|

=|h2 − h1||h4 − h3|

=279.68706.92

= 0.40 = 40%

40% of the turbine work is used to drive the compressor.

(c) The net power developed, in kW

Wnet = m ∗ [|wt| − |wc|]|wt| − |wc| = 427.24 kJ/kg

m =pV

RT=p1V

RT1

=100 kPa ∗ 5 m3/s0.2870 kJ

kg K ∗ 300 K∗ N/m2

Pa∗ J

Nm

= 5.81 kg/s

=⇒ Wnet = 5.81kgs∗ 427.24

kJkg

=⇒ Wnet = 2, 481.1 kW (9)

Using cold air-standard assumptions: κ = 1.4, cp = 1.007 kJ/kg.


ηth,B = 1− 1rp(κ−1)/κ

With rp =p2

p1= 10 and

κ− 1κ

= 0.286

ηth,B = 1− 1100.286 = 1− 1

1.93= 48.2% (10)


rbw =|h2 − h1||h4 − h3|

=cp(T2 − T1)cp(T3 − T4)

=(T2T1− 1)

(T3T4− 1)

∗ T1

T4(11)

3


BDA 3043


T2

T1=

[p2

p1

]κ−1κ

isentropic relation

T3

T4=

[p3

p4

]κ−1κ

(12)

With p1 = p4 and p2 = p3, see p – v graph→ T2T1

= T3T4

, so eq. (11) becomes:

rbw =T1

T4(13)

T4 can be calculated using eq. (12)

κ− 1κ

=1.4− 1

1.4= 0.286

p3

p4=p2

p1= rp = 10

→[p3

p4

]κ−1κ

= 100.286 = 1.93

=⇒ T4 =T3

1.93=

14001.93

= 725.39 K (14)


Wnet = m ∗ [|wt| − |wc|]|wt| − |wc| = (h3 − h4)− (h2 − h1)

(h3 − h4) = cp(T3 − T4)

= 1.007kJ

kg K(1400− 725.39) K

=⇒ h3 − h4 = 679.33 kJ/kg (15)

(h2 − h1) = cp(T2 − T1)

T2 =T3

T4∗ T1 (16)

=1400

725.39∗ 300

→ T2 = 579 K

h2 − h1 = 1.007kJ

kg K(579− 300) K

=⇒ h2 − h1 = 280.95 kJ/kg (17)

|wt| − |wc| = (h3 − h4)− (h2 − h1)

= 679.33− 280.95 = 398.38 kJ/kg

(18)

4



BDA 3043

m =pV

RT=p1V

RT1

=10 kPa ∗ 5 m3/s

0, 2870 kJkg K ∗ 300 K

∗ N/m2

Pa∗ J

Nm

= 5.81 kg/s

=⇒ Wnet = 5.81kgs∗ 398.38

kJkg

=⇒ Wnet = 2, 314.6 kW (19)

Comparison between the results using the air-standard and the coldair-standard analysis

Parameter Air-standard analysis Cold air-standard analysis

T2 574.09 K 579 K

T4 787.73 K 725.39 K

η 45.7 % 48.2 %

Wnet 2,481.1 kW 2,314.6 kW

Isentropicrelationship

p2

p1=pr2pr1

(use air table)T2

T1=

[p2

p1

]κ−1κ

cp Temperature-dependent 1.007 kJ/kg K

κ - 1.4

5


BDA 3043


· Reconsider the question 1, but include in the analysis that the turbine and com-pressor each have an isentropic efficiency of 80%.

Qin

WcQout

Wtturbineworkoutputmaximum

compressorworkinputminimum

s

1

3

T

1400

p = 100 kPa

42

isentropic

300

Qin

Wt

QoutWc

s

31400

p = 100 kPa

4

4s22s

1

irreversibilitiesWith

T

isen

trop

ic

actu

al

300isen

trop

ic

actu

al

turb

ine

com

pres

sor

The question will be solved using the cold air-standard assumptions meaning,κ = 1.4, cp = 1.007 kJ/kg 6= f(T ).

To make better understanding about the isentropic process, the following deriva-tions regarding the isentropic process are given. The index, s, indicating theisentropic process is omitted to avoid confusing and the formulas below are as-sociated with the left side of the graph above.Thermal efficiency:

ηth =|wt| − |wc||qin|

(20)

=|h4 − h3| − |h2 − h1|

|h3 − h2|(21)

The relationship between enthalpy and temperature of ideal gas is given follow-ing:

|h4 − h3| = cp(T4 − T3) = cp(T3 − T4)

|h2 − h1| = cp(T2 − T1)

|h3 − h2| = cp(T3 − T2)

Substituting these relations into eq. (21), it yields:

ηth =(T3 − T4)− (T2 − T1)

T3 − T2=

(T3 − T2)− (T4 − T1)T3 − T2

= 1− T4 − T1

T3 − T2(22)

The second term in the right side of the equation is written in other form:

T4 − T1

T3 − T2=

(T4T1− 1)

(T3T2− 1)

T1

T2(23)

6



BDA 3043

Change of states 1 – 2 and 3 – 4 experience isentropic processes (see lecturenote). The isentropic relation:

T1

T2=

[p1

p2

]κ−1κ

(24)

T4

T3=

[p4

p3

]κ−1κ

(25)

Change of states 2 – 3 and 4 – 1 are isobaric processes meaning p2 = p3 andp4 = p1. Rewriting the eq. (24) and (25):

T1

T2=

[p1

p2

]κ−1κ

T4

T3=

[p1

p2

]κ−1κ

Consequently:

T1

T2=T4

T3=⇒ T4

T1=T3

T2(26)

And the eq. (23) becomes:

T4 − T1

T3 − T2=

(T4T1− 1)

(T3T2− 1)

T1

T2= 1 ∗ T1

T2

=T1

T2(27)

The eq. (22) now has a simple form:

ηth = 1− T1

T2(28)

T1

T2=

1T2T1

=1[

p2

p1

]κ−1κ

With the pressure ratio: rp = p2p1

, the eq. (28) becomes:

ηth = 1− 1

rpκ−1κ

(29)

Isentropic process means there is no losses along the path during the changeof state, for example from state 1 to 2 or state 3 to 4. If any irreversiblity or lossduring the process to be considered, it increases the entropy and causes:

7


BDA 3043


• On the turbine:The actual work output is less than the isentropic work output (see the rightside of the previous graphs.). The ratio between the actual and the isen-tropic work output is defined as the turbine efficiency, ηt:

ηt =Actual Turbine Work

Isentropic Turbine Work(30)

=wt

wt,s< 1

The isentropic work output:

wt,s = w3-4s see T – s diagram

wt,s = h4s − h3 (31)

And the actual work output:

wt = w34 see T – s diagram

wt = h4 − h3 (32)

ηt =h4 − h3

h4s − h3(33)

=h3 − h4

h3 − h4s(34)

• On the compressor:The isentropic work output is less than theactual work input (see the rightside of the previous graphs.). The ratio between the isentropic and the ac-tual work output is defined as the compressor efficiency, ηc:

ηc =Isentropic Compressor Work

Actual Compressor Work(35)

=wc,s

wc< 1

The isentropic work input:

wc,s = w1-2s see T – s diagram

wc,s = h2s − h1 (36)

And the actual work input:

wc = w12 see T – s diagram

wc = h2 − h1 (37)

ηc =h2s − h1

h2 − h1(38)

8



BDA 3043

The thermal efficiency of the cycle


(39)

Turbine work output:

wt = ηt ∗ wt,s (40)

wt,s = h4s − h3

= cp(T4s − T3)

= cpT3

[T4s

T3− 1

](41)

3 – 4s is an isentropic process, so the relation between T and p is given as follows:

T4s

T3=

[p4s

p3

]κ−1κ

(42)

Along the isobar lines: p4s = p1 and p3 = p2, so:

T4s

T3=

[p1

p2

]κ−1κ

=[

1rp

]κ−1κ

, with rp =p2

p1

→ T4s

T3=

1rp(κ−1)/κ

(43)

rp, cp, κ and T3 are given, so wts and T4s can be calculated.

Compressor work input:

wc = wc,s/ηc (44)

wc,s = h2s − h1

= cp(T2s − T1)

= cpT1

[T2s

T1− 1

](45)

1 – 2s is an isentropic process, so the relation between T and p is given as follows:

T2s

T1=

[p2s

p1

]κ−1κ

(46)

9


BDA 3043


Along the isobar lines: p2s = p2 and p1 = p4s, so:

T2s

T1=

[p2

p1

]κ−1κ

=[rp

]κ−1κ

, with rp =p2

p1

→ T2s

T1= rp

(κ−1)/κ (47)

Or

→ T1

T2s=

1rp(κ−1)/κ

(48)

Comparison with eq. (43)

→ T4s

T3=

T1

T2ssee eq. (26) (49)

rp, cp, κ and T3 are given, so wcs and T2s can be calculated.

Heat input:

qin = q23

= h3 − h2 (50)

The first law of T/D on the compressor:

h2 = h1 + wc

= cpT1 + wc,s/ηc (51)

qin can be now calculated. Hence the thermal efficiency of the cycle is deter-mined.The first law of T/D on the turbine:

h4 = h3 + wt

= cpT3 + wt,s ∗ ηt (52)

10



BDA 3043



(53)

Turbine work output:

wt = ηt ∗ wt,s (54)

wt,s = cpT3

[T4s

T3− 1

]T4s

T3=

1rp(κ−1)/κ

rp = 10, and (κ− 1)/κ = 0.286

→ T4s

T3=

1100.286 = 0.518 (55)

cp = 1.007 kJ/kg and T3 = 1400K

wt,s = 1.007kJ

kg K∗ 1400 K ∗ (0.518− 1)] = −679.52

kJkg

→ wt = −0.80 ∗ 679.52kJkg

→ wt = −543.62kJkg

(56)

Compressor work input:

wc = wc,s/ηc (57)

wc,s = cpT1

[T2s

T1− 1

]→ T1

T2s=

1rp(κ−1)/κ

→ T4s

T3=

T1

T2s= 0.518

With T1 = 300 K

wc,s = 1.007kJ

kg K∗ 300 K ∗

[1

0.518− 1

]= 281.105

kJkg

wc = 281.105kJkg

/0.80

→ wc = 351.381 kJ/kg (58)

Heat input:

qin = q23

= h3 − h2 (59)

11


BDA 3043


The first law of T/D on the compressor:

h2 = h1 + wc

= cpT1 + wc

= 1.007kJ

kg K∗ 300 K + 351.381 kJ/kg

→ h2 = 653.48 kJ/kg (60)

h3 = cpT3

= 1.007kJ

kg K∗ 1400 K

→ h3 = 1409.8 kJ/kg (61)

qin = (1409.8− 653.48) kJ/kg

= 756.32 kJ/kg (62)

The thermal efficiency of the gas turbine is then:


=| − 543.62| − |351.381|

|756.32|=

192.24756.32

=⇒ ηth = 0.254 = 25.4% (63)


rbw =|wc||wt|

=351.381543.62

→ rbw = 0.65 = 65% (64)


Wnet = m ∗ [|wt| − |wc|]

|wt| − |wc| = 192.24 kJ/kg

The mass flow rate is determined by using the equation of state for idealgas:

m =pV

RT=p1V

RT1

=10 kPa ∗ 5 m3/s

0, 2870 kJkg K ∗ 300 K

∗ N/m2

Pa∗ J

Nm

= 5.81 kg/s

=⇒ Wnet = 5.81kgs∗ 192.24

kJkg

=⇒ Wnet = 1, 116.89 kW (65)

12



BDA 3043

¸ A regenerator is incorporated in the cycle of the question 1. Determine:

(a) The thermal efficiency of the cycle with the regenerator effectiveness of80%



s

2

4

5

1

3

regen

erat

ion actual

max

p1

p2

com

busti

on

3’T

1400

300

6

Turbine

Combustor

2

3

4 5

6

1

Regenerator

Comp.

exhaust gas

air

Assumptions:

a) Steady state

b) The compressor and turbine processes are isentropic.

c) Kin. and potential energy effects are negligible.

d) The working fluid is air modeled as an ideal gas with the constant specificheat, cp = 1.007 kJ/kg, and specific heat ratio κ = 1.4 (cold air-standardassumptions).

Solution

(a) Thermal efficiency:


(66)

wt = h5 − h4 (67)

wc = h2 − h1 (68)

qin = h4 − h3 (69)

• Turbine work output

wt = h5 − h4

= cpT4(T5

T4− 1) (70)

T5

T4=

1rp(κ−1)/κ

13


BDA 3043


Given: rp = p2p1

= 10 and T4 = 1400 K , (κ− 1)/κ = 0.286

T5

T4=

1100.286 = 0.518

→ wt = 1.007kJkg∗ 1400 K ∗ (0.518− 1)

=⇒ wt = −679.52kJkg

(71)

• Compressor work input

wc = h2 − h1

wc = cpT1[T2

T1− 1]

→ T1

T2=

1rp(κ−1)/κ

→ T1

T2=T5

T4= 0.518 (72)

With T1 = 300 K

=⇒ wc = 1.007kJ

kg K∗ 300 K ∗

[1

0.518− 1

]= 281.105

kJkg

Heat input:

qin = q34

= h4 − h3 (73)

The regenerator effectiveness is defined as follows (see the left-side graphabove):

ηreg =h3 − h2

h3′ − h2

h3 = ηreg(h3′ − h2) + h2

With T3′ = T5:

h3 = ηreg ∗ cp(T5 − T2) + cpT2

T2 and T5 are calculated using eq. (72)

T2 =T1

0.518=

3000.518

K

→ T2 = 579.15 K (74)

T5 = T4 ∗ 0.518 = 1400 ∗ 0.518 K

→ T5 = 725.2 K (75)

h3 = 0.80 ∗ 1.007 ∗ (725.2− 579.15) + 1.007 ∗ 579.15

→ h3 = 700.86kJkg

(76)

14



BDA 3043

h4 = cpT4

h4 = 1.007 ∗ 1400kJkg

→ h4 = 1409.8kJkg

(77)

qin = 1409.8− 700.86

→ qin = 708.94kJkg

(78)

The thermal efficiency:


=679.52− 281.105

708.94= 0.562

=⇒ ηth = 56.2% (79)

15


BDA 3043


¹ Air enters the compressor at 100 kPa, 300 K and is compressed to 1 MPa. Thetemperature at the first turbine stage is 1400 K. The expansion takes place isen-tropically in two stages, with reheat to 1400 K between the stages at a constantpressure of 300 kPa. A regenerator having an effectiveness of 100% is also incor-porated in the cycle. Determine the thermal efficiency.

Combustor

1

2

3

45 6 7

8Regenerator

Turbine 1 Turbine 2

Reheater

Loa

d

Comp.

64

heat

inpu

t

753

s

T

rege

nera

tion

1400

300 1

2

p2=3

00 kP

a

p3=1

MPa

p1=1

00 kP

a

Assumptions:

a) Steady state




Solution

• Thermal efficiency:


wt = wt1 + wt2 = (h5 − h4) + (h7 − h6)

wc = h2 − h1

qin = qcombust. + qreheater = (h4 − h3) + (h6 − h5)

• Compressor work input (isentropic process)

wc = h2 − h1

wc = cpT1[T2

T1− 1]

→ T1

T2=

1rp(κ−1)/κ

16



BDA 3043

Given: rp = p2p1

= 1000100 = 10 and T1 = 300 K , (κ− 1)/κ = 0.286

→ T2 = T1 ∗ rp(κ−1)κ = 579.59 kg K

=⇒ wc = 1.007kJ

kg K∗ 300 K ∗

[1

0.518− 1

]= 281.105

kJkg

• Turbine work output (isentropic process)

wt = wt1 + wt2 = (h5 − h4) + (h7 − h6)

= cpT4(T5

T4− 1) + cpT6(

T7

T6− 1) (80)

T5

T4=

[p2

p3

](κ−1)/κ

=[

3001000

]0.286

= 0.709

T7

T6=

[p1

p2

](κ−1)/κ

=[100300

]0.286

= 0.730

cpT4(T5

T4− 1) = 1.007

kJkg∗ 1400 K ∗ (0.709− 1) = −410.25

kJkg

cpT6(T7

T6− 1) = 1.007

kJkg∗ 1400 K ∗ (0.730− 1) = −380.65

kJkg

=⇒ wt = −790.9kJkg

(81)

Heat input:

qin = (h4 − h3) + (h6 − h5)

= cp ∗ (T4 − T3 + T6 − T5) = cp ∗ (T4 + T6 − (T3 + T5))

Referring to the graph, we see T4 = T6 = 1400 K and T3 = T7 (100% effective-ness of the regenerator).

T7

T6= 0.730 → T7 = 0.730 ∗ T6 = 1, 022 K = T3

T5

T4= 0.709 → T5 = 0.709 ∗ T4 = 992.60 K

→ qin = 1.007 ∗ (2 ∗ 1, 400− (1, 022 + 992.60))

=⇒ qin = 790.9 kJ/kg (82)

The thermal efficiency of the power plant:


=790.9− 281.105

790.9= 0.645

=⇒ ηth = 64.5% (83)

17


BDA 3043


º Air is compressed from 100 kPa, 300 K to 1 MPa in a two-stage compressorwith intercooling between stages. The intercooler pressure is 300 kPa. The air iscooled back to 300 K in the intercooler before entering the second compressor.Each compressor stage is isentropic. For steady-state operation and negligiblechanges in kinetic and potential energy from the inlet to exit, determine:

(a) The temperature at the exit of the second compressor stage

(b) The total compressor work input per unit mass flow

(c) Repeat for a single stage of compression from the given inlet state to thefinal pressure

(d) How much percent is the power input per unit mass flow rate can be savedusing the two-stage compression?

2

1

34

Comp.1

Intercooler

Comp.2

2s

2s

1s

sv

1s

1

W

W

T

p

s = const.

2

p2=300 kPa

1

2

intercooling

p3=1 MPa

p1=100 kPa

c,2

c,1100

300

1000

300

Assumptions:

a) Steady state




Solution

(a) The temperature at the exit of the second compressor stage

• 1 – 1s : isentropic compression

→ T1

T1s=

1rp(κ−1)/κ

Given: rp = p2p1

= 300100 = 3 and T1 = 300 K , (κ− 1)/κ = 0.286

→ T1s = T1 ∗ rp(κ−1)κ = 300 ∗ 30.286 = 300 ∗ 1.369 = 410.75 K

• 2s – 2: isentropic compression

→ T2s

T2=

1rp(κ−1)/κ

18



BDA 3043

Given: rp = p2p2s

= 1000300 = 3.33 and T2s = 300 K , (κ− 1)/κ = 0.286

T2 = T2s ∗ rp(κ−1)κ = 300 ∗ 3.330.286 = 300 ∗ 1.411

Temperature at the second compressor stage:

→ T2 = 423.32 K

(b) The total compressor work input per unit mass flow

wc = wc,1 + wc,2

wc,1 = w1−1s = h1s − h1 = cp(T1s − T1)

= cpT1(T1s

T1− 1)

T1s/T1 has been calculated at the item (a).

→ wc,1 = 1.007 ∗ 300 ∗ (1.369− 1) = 111.53 kJ/kg

Work input for the second compressor, with T2s = T1 = 300 K

wc,2 = cpT2s(T2

T2s− 1)

In the item (a), the value of T2/T2s = 1.411

wc,2 = 1.007 ∗ 300 ∗ (1.411− 1) = 124.16 kJ/kg

→ wc,2 = 124.16 kJ/kg

=⇒ wc = 111.53 + 124.16 = 235.7 kJ/kg

(c) Repeat for a single stage of compression from the given inlet state to thefinal pressure.

• Isentropic compression from state 1 to 2 (single stage)

wc = w12 = h2 − h1 = cpT1(T2

T1− 1)

T2

T1=

[p2

p1

](κ−1)/κ

=[1000100

]0.286

= 1.93

→ wc = 1.007 ∗ 300 ∗ (1.93− 1)

=⇒ wc = 280.95 kJ/kg (84)

(d) How much percent is the power input per unit mass flow rate can be savedusing the two-stage compression?

savings =wc,single − wc,two stage

wc,single

=280.95− 235.7

280.95= 0.16

=⇒ savings = 16% (85)

19


BDA 3043


» A regenerative gas turbine with intercooling and reheat operate at steady state.Air enters the compressor at 100 kPa, 300 K with a mass flow rate of 5.807 kg/s.The pressure ratio across the two-stage compressor is 10. The pressure ratioacross the two-stage turbine is also 10. The intercooler and reheater each op-erates at 300 kPa. At the inlets to the turbine stages, the temperature is 1400 K.The temperature at the inlet to the second compressor stage is 300 K. The isen-tropic efficiency of each compressor and turbine stage is 80%. The regenerativeeffectiveness is 80%. Determine:

(a) The thermal efficiency of the cycle(b) The back work ratio(c) The net power developed, in kW

2

3 41

Combustor5

67 98

Comp.1

Intercooler

Comp.2

Regenerator

Turbine 1 Turbine 2

Reheater

3

6 8

4s

7s5

22s

4

1

99s

T

s

p=300 kPap=1 M

Pa

p1=100 kPa

5’7

1400

300

Assumptions:

a) Steady stateb) The compressor and turbine processes are not isentropic.c) Kin. and potential energy effects are negligible. No pressure drop for flow

through the heat exchangers.d) The working fluid is air modeled as an ideal gas with the constant specific

heat, cp = 1.007 kJ/kg, and specific heat ratio κ = 1.4 (cold air-standardassumptions).

20



BDA 3043

Solution



wt = wt1 + wt2 = w67 + w89

wc = wc1 + wc2 = w12 + w34

qin = qcombust. + qreheater = q56 + q78

• The total compressor work input per unit mass flow. ηc1 = ηc2 = ηc = 0.8.

wc = wc1 + wc2 = w12 + w34

w12 = wc1,s/ηc

wc1,s = w1−2s = h2s − h1

Change of state 1 – 2s is an isentropic compression

wc1,s = cpT1(T2s

T1− 1) and

T2s

T1=

[p2s

p1

]κ−1κ

Given:p2s

p1=

300100

= 3 and T1 = 300 K, and κ−1κ = 0.286

T2s

T1= 30.286 = 1.369 → T2s = 1.369 ∗ 300 = 410.7 K

wc1,s = 1.007 ∗ 300 ∗ (1.369− 1) = 111.526 kJ/kg

wc1 = w12 = 111.526/0.8 kJ/kg

=⇒ wc1 = 139.41 kJ/kg (86)

wc2 = w34 = wc2,s/ηc

Change of state 3 – 4s is an isentropic compression

wc2,s = cpT3(T4s

T3− 1) and

T4s

T3=

[p4s

p3

]κ−1κ

Given:p4s

p3=

1000300

= 3.333 and T3 = T1 = 300 K, and κ−1κ = 0.286

T4s

T3= 3.3330.286 = 1.411 → T4s = 1.411 ∗ 300 = 423.3 K

wc2,s = 1.007 ∗ 300 ∗ (1.411− 1) = 124.17 kJ/kg

wc2 = w34 = 124.17/0.8 kJ/kg

=⇒ wc2 = 155.21 kJ/kg (87)

The total compressor work input:

=⇒ wc = 294.62 kJ/kg (88)

21


BDA 3043


• The total turbine work output per unit mass flow. ηt1 = ηt2 = ηt = 0.8.

wt = wt1 + wt2 = w67 + w89

w67 = wt1,s ∗ ηtwt1,s = w6−7s = h7s − h6

Change of state 6 – 7s is an isentropic expansion

wt1,s = cpT6(T7s

T6− 1) and

T7s

T6=

[p7s

p6

]κ−1κ

Given:p7s

p6=

3001000

= 0.3 and T6 = 1400 K, and κ−1κ = 0.286

T7s

T6= 0.30.286 = 0.709 → T7s = 0.709 ∗ 1400 = 992.2 K

wt1,s = 1.007 ∗ 1400 ∗ (0.709− 1) = −410.69 kJ/kg

wt1 = −410.69 ∗ 0.8 kJ/kg

=⇒ wt1 = w67 = −328.55 kJ/kg (89)

Work output of the second turbine:

wt2 = w89 = wt2,s ∗ ηt

Change of state 8 – 9s is an isentropic expansion

wt2,s = cpT8(T9s

T8− 1) and

T9s

T8=

[p9s

p8

]κ−1κ

Given:p9s

p8=

100300

= 0.333 and T8 = 1400 K, and κ−1κ = 0.286

T9s

T8= 0.3330.286 = 0.73 → T9s = 0.73 ∗ 1400 = 1022.2 K

wt2,s = 1.007 ∗ 1400 ∗ (0.73− 1) = −380.42 kJ/kg

wt2 = −380.42 ∗ 0.8 kJ/kg

=⇒ wt2 = w89 = −304.34 kJ/kg (90)

The total turbine output:

=⇒ wt = −632.89 kJ/kg (91)

• The total heat input per unit mass flow. Regeneerator effectiveness,ηreg = 0.8.

qin = qcombust. + qreheater = q56 + q78

22



BDA 3043

The regenerator effectiveness, ηreg.

ηreg =h5 − h4

h5′ − h4

h5 = ηreg(h5′ − h4) + h4

With T5′ = T9, it follows→ h5′ = h9

h5 = ηreg ∗ (h9 − h4) + h4 (92)

h9 is calculated using the 1. law of T/D on the second turbine:

h9 = wt2 + h8 = wt2 + cpT8

= −304.34 + 1.007 ∗ 1400 = 1, 105.46 kJ/kg (93)

h4 is calculated using the 1. law of T/D on the second compressor:

h4 = wc2 + h3 = wc2 + cpT3

= 155.21 + 1.007 ∗ 300 = 457.31 kJ/kg (94)

Eq. (92) can be now calculated:

h5 = 0.8 ∗ (1, 105.46− 457.31) + 457.31 = 975.83 kJ/kg (95)

(96)

Heat input supplied by the combustor:

q56 = h6 − h5

= cpT6 − h5 = 1.007 ∗ 1, 400− 975.83

= 433.97 kJ/kg (97)

(98)

Heat input supplied by the reheater:

q78 = h8 − h7

h7 is calculated using the 1. law of T/D on the first turbine:

h7 = wt1 + h6

q78 = h8 − (wt1 + h6) = −wt1= 328.55 kJ/kg (99)

The total heat input per unit mass flow:

qin = 433.97 + 328.55

= 762.52 kJ/kg (100)

The thermal efficiency of the power plant:

ηth =632.89− 294.62

762.52=

338.27762.52

=⇒ ηth = 44.4% (101)

23


BDA 3043


¼ A regenerative gas turbine power plant is shown in Fig. 1. Air enters the com-pressor at 1 bar, 27 °C with a mass flow rate of 0.562 kg/s and is compressed to 4bar. The isentropic efficiency of the compressor is 80%, and the regenerator ef-fectiveness is 90%. All the power developed by the high-pressure turbine is usedto run the compressor. The low-pressure turbine provides the net power output.Each turbine has an isentropic efficiency of 87% and the temperature at the inletto the high-pressure turbine is 1200 K. Determine:

(a) The net power developed, in kW

(b) The thermal efficiency of the cycle

(c) The temperature of the air at state 2, 3, 5, 6, and 7, in K.

Combustor

1

2

3

4

6

7

5

Regenerator

Comp.1 HPT

LPT

1 bar, 27 Co

1 bar

Figure 1: Regenerative gas turbine power plant

Solution

(a) The net power developed, in kW

1

22s

3

3’

p=4 bar

5

64s

4

5s

s

T 7

p1=1 bar

300.15

1200

T – s diagram

24



BDA 3043

• The net power output per unit mass flow rate:

wnet = |wt| − |wc|wt = wt1 + wt2 = w45 + w56

wc = w12

with |wt1| = |wc|

→ wnet = |wt2| = |w56| (102)

• Turbine work output = Compressor work input

−wt = wc

−(h5 − h4) = h2 − h1

h5 = h4 − (h2 − h1) (103)

h4 and h1 can be directly calculated since T1 and T4 are both given. h2

is determined through the compressor efficiency and the relationship of anisentropic compression.

h4 = cpT4 = 1.007 ∗ 1200 = 1, 208.4 kJ/kg

ηc =h2s − h1

h2 − h1

→ h2 = (h2s − h1)/ηc + h1 (104)

h2s − h1 = cpT1(T2s

T1− 1)

→ T2s

T1=

[p2s

p1

]κ−1κ

=[41

]0.286

= 1.49 (105)

h2s − h1 = 1.007 ∗ 300.15 ∗ (1.49− 1) = 148.1 kJ/kg

h1 = cpT1 = 1.007 ∗ 300.15 = 302.251 kJ/kg

→ h2 = 148.1/0.8 + 302.25 = 487.375 kJ/kg

→ T2 = 487.375/1.007 = 484 K

h5 = h4 − (h2 − h1)

→ h5 = 1, 208.4− (487.375− 302.251) = 1, 023.28 kJ/kg (106)

→ T5 = h5/cp = 1016.2 K

• Calculation of the intermediate pressure

ηt1 =h4 − h5

h4 − h4s, with ηt1 = ηt2 = ηt = 0.87

→ h4s = h4 − (h4 − h5)/ηt= 1, 208.4− (1, 208.4− 1, 023.28)/0.87 = 995.61 kJ/kg

→ T4s = h4s/cp = 995.61/1.007 = 988.69 K (107)

25


BDA 3043


Isentropic expansion of the process 4 – 4s:

T4s

T4=

[p4s

p4

]κ−1κ

→ p4s

p4=

[T4s

T4

] κκ−1

=[988.691200

]3.5

= 0.508

p4s = p4 ∗ 0.508 = 4 ∗ 0.508 = 2.031 bar (108)

• Compressor work input

wc = h2 − h1

→ wc = 487.375− 302.251 = 185.12 kJ/kg (109)

Calculation of h6:

ηt2 =h5 − h6

h5 − h5s

h6 = h5 − ηt ∗ (h5 − h5s)

h5s − h5 = cpT5(T5s

T5− 1)

T5s

T5=

[p5s

p5

]κ−1κ

=[

12.031

]0.286

= 0.817

→ h5s − h5 = 1.007 ∗ 1, 016.2 ∗ (0.817− 1) = −187.665 kJ/kg

→ h6 = 1, 023.28− 0.87 ∗ 187.665 = 860.0 kJ/kg

→ T6 = h6/cp = 860.0/1.007 = 854 K (110)

• The net power output per unit mass flow rate:

→ wnet = h6 − h5 = 860.0− 1, 023.28

wnet = −163.3 kJ/kg (minus sign means work output)

Wnet = mwnet

= 0.562 kg/s ∗ 163.3 kJ/kg

=⇒ Wnet = 91.775 kW (111)

(b) The thermal efficiency of the cycle

• Regenerative effectiveness:

ηreg =h3 − h2

h3′ − h2

h3 = ηreg(h3′ − h2) + h2

26



BDA 3043

With T3′ = T6 → h3′ = h6

h3 = ηreg(h6 − h2) + h2

= 0.90 ∗ (860.0− 487.375) + 487.375

→ h3 = 822.738 kJ/kg

→ T3 = h3/cp = 822.738/1.007 = 817.0 K

(112)

• Heat input:

qin = q34 = h4 − h3

= 1, 208.4− 822.738 = 385.663 (113)

• The thermal efficiency of the cycle:

ηth =wt2qin

=163.3

385.663= 0.423

=⇒ ηth = 42.3% (114)

(c) The temperature of the air at state 2, 3, 5, 6, and 7, in K.Energy balance on the regenerator (thermal contact of a heat exchanger):

h7 = h2 + h6 − h3

= 487.375 + 854− 822.738 = 518.637 kJ/kg

→ T7 = h7/cp = 515.03 K (115)

State Temp. [K]

T1 300.15

T2 484.00

T3 817.00

T4 1,208.40

T5 1016.20

T6 854.00

T7 515.03

27

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