Answer :
Answer:
[tex]P_2=4091\ KPa[/tex]
Explanation:
Given that
T₁ = 290 K
P₁ = 100 KPa
Power P =5.5 KW
mass flow rate
[tex]\dot{m}= 0.01\ kg/s[/tex]
Lets take the exit temperature = T₂
We know that
[tex]P=\dot{m}\ C_p (T_2-T_1)[/tex]
[tex]5.5=0.01\times 1.005(T_2-290})\\T_2=\dfrac{5.5}{0.01\times 1.005}+290\ K\\\\T_2=837.26\ K[/tex]
If we assume that process inside the compressor is adiabatic then we can say that
[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{0.285}[/tex]
[tex]\dfrac{837.26}{290}=\left(\dfrac{P_2}{100}\right)^{0.285}\\2.88=\left(\dfrac{P_2}{100}\right)^{0.285}\\[/tex]
[tex]2.88^{\frac{1}{0.285}}=\dfrac{P_2}{100}[/tex]
[tex]P_2=40.91\times 100 \ KPa[/tex]
[tex]P_2=4091\ KPa[/tex]
That is why the exit pressure will be 4091 KPa.