Question

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

(a) the thermal efficiency of the cycle,

(b) the back work ratio,

(c) the net power developed, in kW.

Answer 1

Given that

P₁= 100 KPa

T₁=300 K

Volume flow rate Q= 5 m³/s

T₃=1400 K

r= 10

We know that for air ,heat capacity ratio is 1.4 and specific heat constant pressure is 1.005 KJ/kg.k

γ= 1.4

Cp=1.005 KJ/kg.k

For Brayton cycle

`(T_2)/(T_1)=r^{(\gamma-1)/(\gamma)}`

Now by putting the values

`(T_2)/(T_1)=r^{(\gamma-1)/(\gamma)}`

`(T_2)/(300)=10^{(1.4-1)/(1.4)}`

T₂=1.93 x 300

T₂=579 K

`(T_3)/(T_4)=r^{(\gamma-1)/(\gamma)}`

Now by putting the values

`(T_3)/(T_4)=r^{(\gamma-1)/(\gamma)}`

`(1400)/(T_4)=10^{(1.4-1)/(1.4)}`

T₄=726.32 K

a)

We know that efficiency ,η

`\eta=1-(Q_r)/(Q_a)`

We know that heat addition

Q a= Cp ( T₃-T₂)

Heat rejection

Qr= Cp ( T₄-T₁)

`\eta=1-(Q_r)/(Q_a)`

`\eta=1-(T_4-T_1)/(T_3-T_2)`

`\eta=1-(726.32-300)/(1400-579)`

η = 0.48

b)

Back work ratio

`bwr=(W_(com))/(W_(tur))`

`bwr=(T_2-T_1)/(T_3-T_4)`

`bwr=(579-300)/(1400-726.32)`

bwr=0.414

c)

Net work = Net heat

Net heat = Qa- Qr

Qr= Cp ( T₄-T₁)

Q a= Cp ( T₃-T₂)

Net heat = 1.005 (1400- 579 - 726.32+ 300 ) KJ/kg

Net heat = 396.65 KJ/kg

We know that

P V = m R T

P = ρ R T

By putting the values

P = ρ R T

100 = ρ x 0.287 x 300

ρ =1.16 kg/m³

mass flow rate m =ρ Q

m = 1.16 x 5 = 5.80 kg/s

Net power P = m . Net heat

P= 2303.42 KW