Respuesta :
Answer:
a) [tex]T_2=569.35 K[/tex]
b)Work done per kg of air=196.84 KJ/Kg
Explanation:
Given: [tex]\gamma =1.4[/tex] for air.
[tex]P_1=90 KPa ,T_=22^\circ C,P_2=900 KPa[/tex]
We know that
[tex]\dfrac{T_2}{T_1}=\left (\frac{P_2}{P_1}\right )^{\dfrac{{\gamma-1}}{\gamma}}[/tex]
So [tex]\dfrac{T_2}{295}=\left (\frac{900}{90}\right )^{\dfrac{{1.4-1}}{1.4}}[/tex]
[tex]T_2=569.35 K[/tex]
(a) [tex]T_2=569.35 K[/tex]
(b)Work for adiabatic process
W=[tex]\frac{P_1V_1-P_2V_2}{\gamma -1}[/tex]
We know that PV=mRT for ideal gas.
W=[tex]mR\frac{T_1-T_2}{\gamma -1}[/tex]
Now by putting values
work per kg of air=[tex]0.287\times \frac{295-569.35}{1.4 -1}[/tex]
Work w=-196.84 KJ/Kg (Negative sign indicate work given to input.)
So work done per kg of air=196.84 KJ/Kg
