Answer:
The exit temperature is 293.74 K.
Explanation:
Given that
At inlet condition(1)
P =80 KPa
V=150 m/s
T=10 C
Exit area is 5 times the inlet area
Now
[tex]A_2=5A_1[/tex]
If consider that density of air is not changing from inlet to exit then by using continuity equation
[tex]A_1V_1=A_2V_2[/tex]
So [tex]A_1\times 150=5A_1V_2[/tex]
[tex]V_2=30 [/tex]m/s
Now from first law for open system
[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]
Here Q=0 and w=0
[tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}[/tex]
When air is treating as ideal gas
[tex]h=C_pT[/tex]
Noe by putting the values
[tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}[/tex]
[tex]1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}[/tex]
[tex]T_2=293.74K[/tex]
So the exit temperature is 293.74 K.