A body of weight 300N is lying rough
1 answer:
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Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 , = 1.8
T1 = (1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02= = (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) = = 135.7*
J/Kg.
Answer:
attached below
Explanation:
