Power Plant Engineering

304 POWER PLANT ENGINEERING

(4) The thermal efficiencies of the actual and ideal cycle

(5) The required airflow rate.

Solution. T 1 = 273 + 20 = 293 K

T 3 = 273 + 850 = 1123 K

T 2 a = T 1

(1)/

2

1

P

P

γ− γ







= 293.(4.28)0.2857 = 443.9 K

Similarly T 4 a = 0.2857

1123

(4.28)

= 741.25 K

Now ηcompressor =

21 1

21

(T T )

(T T )

a −

−

ηturbine =^34

34

(T T )

(T T )a

−

−

T 2 =^121

compressor

T(T T)+−a

η

=

293 (443.9 293)

0.8

+−

= 481.6 K

T 4 = T 3 – ηturbine (T 3 – T 4 a)

= 1123 – 0.8(1123 – 741.25) = 817.6 K

(2) and (3) specific work of compressor = Cp (T 2 – T 1 )

= 1.005(481.6 – 293) = 189.54 kJ/kg

Specific work of turbine = 1.005 (T 3 – T 4 )

= 1.005(1123 – 817.6) = 306.93 kJ/kg

Net work = 306.93 – 189.54 = 117.4 kJ/kg

(4) Thermal efficiency (ηt) of ideal cycle,

ηt = (1)

2

1

11

P

P

γ− γ

−







= 1 – 0.66 = 34%

Thermal efficiency of actual cycle,

ηt =

(Heat supplied-Heat rejected)

Heat supplied

=

32 41

32

{C (T T ) C (T T )}

{C ( T T )}

pp

p

−− −

− = 1 –

41

32

(T T )

(T T )

−

−

= 1 –

(817.6 293)

(1123 481.6)

−

−

= 1 – 0.818 = 18.20%

(5) Air flow rate =

3600

net work output in kJ/kg

kg/kW-hr.

I

φ

2a^2

1

4

3

4a

Fig. 9.35