Answer:
a)r= 7.27
b)[tex]\eta =0.54[/tex]
c)P₂=1449.21 KPa
d)W=540 KJ/kg
Explanation:
Given that
T₁= 290 K
P₁=90 KPa
Qa = 1000 KJ/kg
T₃=2050 K
For air
specific heat at constant volume Cv
Cv=0.71 KJ/kg.k
γ = 1.4
Qa= Cv ( T₃ - T₂)
Bu putting the values
1000 = 0.71 x ( 2050 - T₂)
T₂ = 641.54 K
[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]
r=Compression ratio
By putting the values
[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]
[tex]\dfrac{641.54}{290}=r^{1.4 -1}[/tex]
r= 7.27
[tex]\dfrac{P_2}{P_1}=r^{\gamma}[/tex]
By putting the values
[tex]\dfrac{P_2}{P_1}=r^{\gamma}[/tex]
[tex]\dfrac{P_2}{90}=7.27^{\gamma}[/tex]
P₂=1449.21 KPa
This is maximum pressure
The efficiency η
[tex]\eta =1-\dfrac{1}{r^{\gamma -1}}[/tex]
[tex]\eta =1-\dfrac{1}{7.27^{1.4-1}}[/tex]
[tex]\eta =0.54[/tex]
We also know that
[tex]\eta=\dfrac{W}{Q_a}[/tex]
[tex]0.54=\dfrac{W}{1000}[/tex]
W=540 KJ/kg
This is the net work output.
