Answer:
The answers to the question are
(A) 27.86 m/s
(B) 41.71 m/s
(C) 64.48 m/s
Step-by-step explanation:
(A) From steam tables we
h1 = 3064.2 kJ/kg and h2 = 2676.2 kJ/kg
With speed, C2 = (2×(h1-h2))^0.5
The speed= 27.86 m/s
(B) Where nitrogen is modelled as an ideal gas we have
T1/T2 =(P1/P2)^(Y-1)/Y
Therefore, Y = 3.15 and Cp = Y×R/(Y-1) where R = 0.297 kJ/(kg K, Cp = 0.435
h1 = Cp×T1 = 249.33 kJ/kg
h2 = Cp×T2 = 162.33 kJ/kg
We have 10 moles therefore speed, C2 = (2×(h1-h2))^0.5 = (2×(249.33 kJ/kg - 162.33 kJ/kg)×10)^0.5 = 41.71 m/s
(C) We have R = R'/m = 8314.5 kJ/(g-mol K)/(28 g/mol) = 0.297 kJ/(kg K) and Cp = 7/2×R = 1.0395 kJ/(kg K)
h1 = Cp×T1 = 1.0395 kJ/(kg K) × 573.15 K = 595.79 kJ/(kg) and h2 = Cp×T2 = 387.89 kJ/(kg)
Therefore we have for 10 moles C2 = (2×(595.79 kJ/(kg)-387.89 kJ/(kg))×10)^0.5 = 64.48 m/s