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

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Dâu 6 months 2021-07-28T05:22:11+00:00 1 Answers 36 views 0

Answers ( )

  1. Nho
    0
    2021-07-28T05:23:26+00:00
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    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) + \frac{V1^{2}-V2^{2} }{2}  + g(z1-z2)]

    0= [Qcv + m[(h1-h2) + \frac{V1^{2}-V2^{2} }{2} ]

    0=( h1-h2) + ( \frac{V1^{2}-V2^{2} }{2} )

    0= (3445.3-3051.1)+ \frac{100^{2} - V2^{2}}{2}

    0= 394.2 + \frac{100^{2} - V2^{2}}{2}

    -394.2= \frac{100^{2} - V2^{2}}{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

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