TITLE.PM5

684 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-6.pm5

or, T 22 = T 1 T 3 or T 2 = TT 13

or, T 2 = 303 1073× = 570.2 K

Now, Wturbine – Wcompressor = m&f × C × η

or, 100 = m&f × 45000 ×^1
41
32

− −

−

L

N

M

O

Q

P

TT

TT

= m&f × 45000^1

570.2 303

1073 570

− −

−

L

N

M

O

Q

.2P

= m&f × 21085.9

or, m&f =
100
21085.9 = 4.74 × 10

–3 kg/s. (Ans.)

Again, Wturbine – Wcompressor = 100 kW

(mmTT m&&af+−−××−=)( 34 ) &a 1 ()TT 21100

or, (m&a+ 0.00474)(1073 – 570.2) – m&a(570.2 – 303) = 100

or, (m&a+ 0.00474) × 502.8 – 267.2 m&a = 100

or, 502.8 m&a + 2.383 – 267.2 m&a = 100
or, 235.6 m&a = 97.617
∴ m&a = 0.414 kg/s. (Ans.)
Example 13.37. In a gas turbine plant working on Brayton cycle, the air at inlet is 27°C,
0.1 MPa. The pressure ratio is 6.25 and the maximum temperature is 800°C. The turbine and
compressor efficiencies are each 80%. Find compressor work, turbine work, heat supplied, cycle
efficiency and turbine exhaust temperature. Mass of air may be considered as 1 kg. Draw T-s
diagram. (N.U.)
Solution. Refer to Fig. 13.56.

T

s

2

1

2 ′

3

4

4 ′

T-s diagram

Fig. 13.56

Given : T 1 = 27 + 273 = 300 K ; p 1 = 0.1 MPa ; rp = 6.25, T 3 = 800 + 273 = 1073 K ;

ηcomp. = ηturbine = 0.8.