Gas Turbine Power Plant

The Brayton Cycle (closed)

Product of
air fuel Ec Combustion
2 mixture
QA EP
S≠C 2’
3

Start-up Wc Wሖ 𝑐 Combustor TP IP
BP
motor BPT
BPc P=C
Compressor Gas Turbine
ḿ

S≠C S=C Generator

P=C
4’ S≠C
1
Heat Sink
4
where, QR
E𝑐 → energy charged

E𝑐 = mf HHV or Vf HHV

PV diagram TS diagram
P
T

3
QA P=C

2 2’ P=C

2 2’ S≠C
S=C

S≠C S=C S≠C

4 4’
3
S=C

S≠C P=C

P=C
4’ 1
QR
4
S
V

processes:

1-2: isentropic process → compression

1-2’: irreversible adiabatic compression
2-3:
isobaric process → heat addition
2’-3:
3-4: isentropic process → expansion
3-4’: irreversible adiabatic expansion
4-1:
isobaric process → heat rejection
4’-1 1
 fuel
2 mf
mg = ma + mf
2’
ma Combustor
S≠C S≠C
P=C 3

Compressor Gas Turbine

S=C S=C
4’ S≠C
1
4

Energy analysis:

for compressor,

ḿ
2 h2
S≠C
2’

Compressor

S=C

h1 1 Wc

considering for ideal compression ; Ein = Eout

m′ h1 + Wc = m′ h2

Wc = m′ (h2 − h1 ) ; h = cp T

Wc = m′ (cp T2 − cp T1 )

Wc = m′ cp (T2 − T1 ) → ideal compression work

for actual compression,

Wሖ c = m′ cp (T2 ′ − T1 ) → actual compression work

2
for nc ,

nc → compression efficiency
Wc Wc
nc = × 100% ; Wሖ c : × 100%
Wሖ c nc

also,

m′cp (T2 − T1 )
Ƞc = × 100%
m′cp (T2 − T1 )

or,
(T2 − T1 )
Ƞc = × 100%
(T2 ′ − T1 )

but,
T2 − T1
T2 ′ = + T1
nc

for Ƞmc ,

Ƞmc → mechanical efficiency of compressor

Wሖ c
Ƞmc = × 100%
BPc

Wሖ c
BPc =
Ƞmc

for combustor,
QA

2’

h2 Combustor
3 h3
2

ḿ
for ideal process heat addition,

m′ h2 + Q A = m′ h3

Q A = m′ (h3 − h2 ) ; h = cp T

3
Q A = m′ (cp T3 − cp T2 )

so,

Q A = m′ cp (T3 − T2 ) → ideal heat added

for actual heat addition,

Q′A = m′ cp (T3 − T2 ′) actual heat added

for combustor efficiency,

Ƞcomb → combustor efficiency

𝑄𝐴
Ƞcomb = × 100%
Ec

Ec = mf HHV ; kJ/kg
or
vf HHV ; kJ/m3

for Gas Turbine,

h3
3

ḿ
Gas Turbine TP

4’
4

h4

for ideal Gas turbine expansion,

m′ h3 = TP + m′ h4

TP = m′ (h3 − h4 ) ; h = cp T

TP = m′ (cp T3 − cp T4 )

TP = m′ cp (T3 − T4 ) → turbine theoretical power 4

for actual expansion process → process 4 − 4′,

IP = m′ cp (T3 − T4 ′) → turbine internal power

for ȠT ,

ȠT → turbine expansion efficiency

IP
ȠT = × 100%
TP

m′ cp (T3 − T4 ′)
= ′ × 100%
m cp (T3 − T4 )

but,
T4 ′ = T3 − ηT (T3 − T4 )

for ηmT ,

ηmT → mechanical efficiency of turbine

BPT
ηm T = × 100%
IP

BPT = (IP) ηmT

for net brake power,

BP → BP input to generator

BP = BPT − BPC

for generator efficiency,

𝐸𝑃
𝜂𝑔 = 𝑋 100%
𝐵𝑃

EP = (BP)ηg

5
for heat sink,

𝑄𝑅

ℎ4
1 Heat Sink 4
ℎ1 4′

ḿ
for ideal heat rejection,

m′ h4 = Q R + m′ h1

Q R = m′ (h4 − h1 ) ; h = cp T

Q R = m′ cP (T4 ′ − T1 )

Q R = m′ cp (T4 − T1 ) ; ideal heat rejection

for actual heat rejection,

Q R = m′ cp (T4 ′ − T1 ) ; actual heat rejection

for ideal cycle only,

Wnet
ec = × 100% → ideal cycle thermal efficiency
QA

where,
Wnet = Q A − Q R

or,

Wnet = TP − WC

for actual cycle thermal efficiency,

ሖ net
W
é c = × 100% → actual cycle thermal efficiency
QሖA

ሖ net = Qሖ A − Qሖ R
W

6
or,

ሖ net = IP − W
W ሖC

considering 𝑟𝑝 ,

rp → compression pressure ratio

Pmax P2
rp = =
Pmin P1

NOTE: if no pressure drop across combustor and heat sink

∴ P2 = P3 and P1 = P4

then,

P2 P3
rp = =
P1 P4

for maximum Wnet ,

T2 = √T1 T3

otherwise,
consider isentropic process 1-2; PVK = C
k−1
T2 P2 k
=( )
T1 P1
or,
k−1
P2 k
T2 = T1 ( )
P1

7
 T2 k−1
= (rp ) k
T1
k−1
T2 = T1 (rp ) k

considering isentropic process 3-4 ; PVk = C

k−1
T3 P3 k
=( )
T4 P4
or,
T3 k−1
= (rp ) k
T4

k−1
T3 = T4 (rp ) k

also,

Wnet QA − QR
ec = × 100% → ec = × 100%
QA QA

|QR | m′cp (T4−T1 )

ec = [1 − ] × 100 % → ec = [1 − ] × 100 %
QA m′cp (T3−T2 )

in terms of temperature,

(T4 − T1 )
ec = [1 − ] × 100 %
(T3 − T2 )

8
 substituting T2 and T3 ,

(T4 − T1 )
𝑒𝑐 = 1 − k−1 k−1
× 100 %
(T4 (rp ) k − T1 (rp ) k )
[ ]

simplifying,

(T4 − T1 )
𝑒𝑐 = [1 − k−1
] × 100 %
(T4 − T1 )(rp ) k

finally,

1
𝑒𝑐 = [1 − 𝑘−1
] × 100 %
(𝑟𝑝 ) 𝑘
× 100 %

SAMPLE PROBLEM

1. A gas electric turbine plant is to generate 80 000 kW. The unit receives air at
100 kPaa and 295 °K. The maximum temperature is 1200°K and the maximum
pressure is 400 kPaa .Other data are as follows: for air and fuel mixture, 𝐶𝑃 =
1.0047 kJ⁄kg − k , k=1.5; compression adiabatic eff., 90%; compressor
mech’l eff., 85%; turbine internal eff., 90%; turbine mech’l eff., 80%;
generator eff., 95%; combustor eff., 95%.
Determine:
a) EC
b) cycle thermal efficiency
c) over-all station thermal eff.