Power Plant Engineering

298 POWER PLANT ENGINEERING

p 2 =1.01 × 6 = 6.06 bar

Rp =^2

1

P

p

= 6

Pressure at point 4 = 6.06 – 0.15 = 5.91 bar

Applying isentropic law to the process 1 – 2

T 2 ′ = T 1 (RP)(γ – 1)/γ = 288(6)0.286 = 480 K

ηc =^21

21

(T T )

(T T )

′−

−

But T 2 = T 1 + ηc (T 2 ′ – T 1 ) = 288 + 0.8(480 – 288) = 528 K

p 3 = 6.06 – 0.15 = 5.91 bar

and p 4 = 1.01 + 0.15 = 1.16 bar

Applying isentropic law to the process 4 – 5′

T 5 ′ = (1)/^4

3

4

T

P

P

γ− γ







= 0.286

(700 273)

5.91

1.16

+







= 612 K

ηt =^45

45

(T T )

(T T )′

−

−

or, T 5 = T 4 – ηt(T 4 – T 5 ′)

= 973 – 0.85(973 – 612) = 666 K

The effectiveness of the regenerator is given by

ε =^45

45

(T T )

(T T )

−

−

T 3 = T 2 + 0.75 (T 5 – T 2 ) = 528 + 0.75(666 – 528) = 631.5 kW

Wc = Cp(T 2 – T 1 ) = 1 × (528 – 288) = 240 kJ/kg

Wt = Cp(T 4 – T 5 ) = 1 × (973 – 666) = 307 kJ/kg

Wn = Wt – Wc = 307 – 240 = 67 kJ/kg

QS = Cp(T 4 – T 3 ) = 1 × (973 – 631.5) = 341.5 kJ/kg

ηth =

W

Q

n

s

=

67

341.5

= 0.196 = 19.6%.

Example 2. In a constant pressure open cycle gas turbine air enters at 1 bar and 20°C and
leaves the compressor at 5 bar. Using the following data; Temperature of gases entering the turbine
= 680°C, pressure loss in the combustion chamber = 0.1 bar, ηcompressor = 85%, ηturbine = 80%, ηcombustion
= 85%, γ = 1.4 and cp = 1.024 kJ/kgK for air and gas, find:

(1) The quantity of air circulation if the plant develops 1065 kW.