Answer:
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
[tex]\dfrac{T_2}{T_1}=r^{\dfrac{\gamma-1}{\gamma}}[/tex]
Now by putting the values
[tex]\dfrac{T_2}{T_1}=r^{\dfrac{\gamma-1}{\gamma}}[/tex]
[tex]\dfrac{T_2}{300}=10^{\dfrac{1.4-1}{1.4}}[/tex]
T₂=1.93 x 300
T₂=579 K
[tex]\dfrac{T_3}{T_4}=r^{\dfrac{\gamma-1}{\gamma}}[/tex]
Now by putting the values
[tex]\dfrac{T_3}{T_4}=r^{\dfrac{\gamma-1}{\gamma}}[/tex]
[tex]\dfrac{1400}{T_4}=10^{\dfrac{1.4-1}{1.4}}[/tex]
T₄=726.32 K
a)
We know that efficiency ,η
[tex]\eta=1-\dfrac{Q_r}{Q_a}[/tex]
We know that heat addition
Q a= Cp ( T₃-T₂)
Heat rejection
Qr= Cp ( T₄-T₁)
[tex]\eta=1-\dfrac{Q_r}{Q_a}[/tex]
[tex]\eta=1-\dfrac{T_4-T_1}{T_3-T_2}[/tex]
[tex]\eta=1-\dfrac{726.32-300}{1400-579}[/tex]
η = 0.48
b)
Back work ratio
[tex]bwr=\dfrac{W_{com}}{W_{tur}}[/tex]
[tex]bwr=\dfrac{T_2-T_1}{T_3-T_4}[/tex]
[tex]bwr=\dfrac{579-300}{1400-726.32}[/tex]
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
