Share
Water vapor enters an insulated nozzle operating at steady state with a velocity of 100 m/s and specific enthalpy of 3445.3 kJ/kg, and exits
Question
Water vapor enters an insulated nozzle operating at steady state with a velocity of 100 m/s and specific enthalpy of 3445.3 kJ/kg, and exits with specific enthalpy of 3051.1 kJ/kg. The velocity at the exit is most closely
(a) 104 m/s
(b) 636 m/s
(c) 888 m/s
(d) 894 m/s
in progress
0
Physics
6 months
2021-07-28T05:22:11+00:00
2021-07-28T05:22:11+00:00 1 Answers
36 views
0
Answers ( )
Answer:
(a) 104 m/s
Explanation:
enters with specific enthalpy ‘h1’ = 3445.3 kJ/kg
exits with specific enthalpy ‘h2’ 3051.1 kJ/kg
velocity ‘V1’= 100 m/s
In order to find velocity at the exit ‘V2’, we use this relation
0 = Qcv – Wcv + m[(h1-h2) +
+ g(z1-z2)]
0= [Qcv + m[(h1-h2) +
]
0=( h1-h2) + (
)
0= (3445.3-3051.1)+
0= 394.2 +
-394.2=
-788.4 = 10000-V2²
V2²= 10000+ 788.4
V2²= 10784.44
Taking square root on both sides
V2=103.84
V2≈ 104m/s
Therefore, The velocity at the exit is most closely 104m/s