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1 January, 19:59

Steam at 700 bar and 600 oC is withdrawn from a steam line and adiabatically expanded to 10 bar at a rate of 2 kg/min. What is the temperature of the steam that was expanded, and what is the rate of entropy generation in this process

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  1. 1 January, 23:10
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    Final temperature of the steam = 304.29 K = 31.14°C

    Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T = 0 J/K.

    But 0.01 J/k. s was obtained though, which is approximately 0.

    Explanation:

    For an adiabatic system, the Pressure and temperature are related thus

    P¹⁻ʸ Tʸ = constant

    where γ = ratio of specific heats. For steam, γ = 1.33

    P₁¹⁻ʸ T₁ʸ = P₂¹⁻ʸ T₂ʸ

    P₁ = 700 bar

    P₂ = 10 bar

    T₁ = 600°C = 873.15 K

    T₂ = ?

    (700⁻⁰•³³) (873.15¹•³³) = (10⁻⁰•³³) (T₂¹•³³)

    T₂ = 304.29 K = 31.14°C

    b) Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T

    To prove this

    Entropy of the process

    dQ - dW = dU

    dQ = dU + dW

    dU = mCv dT

    dW = PdV

    dQ = TdS

    TdS = mCv dT + PdV

    dS = (mCv dT/T) + (PdV/T) =

    PV = mRT; P/T = mR/V

    dS = (mCv dT/T) + (mRdV/V)

    On integrating, we obtain

    ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

    To obtain V₁ and V₂, we use PV = mRT

    V/m = specific volume

    Pv = RT

    R for steam = 461.52 J/kg. K

    For V₁

    P = 700 bar = 700 * 10⁵ Pa, T = 873.15 K

    v₁ = (461.52 * 873.15) / (700 * 10⁵) = 0.00570 m³/kg

    For V₂

    P = 10 bar = 10 * 10⁵ Pa, T = 304.29 K

    v₂ = (461.52 * 304.29) / (10 * 10⁵) = 0.143 m³/kg

    ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

    m = 2 kg/min = 0.0333 kg/s, Cv = 1410.8 J/kg. K

    ΔS = [0.03333 * 1410.8 * In (304.29/873.15) ] + (0.03333 * 461.52 * In (0.143/0.00570)

    ΔS = - 49.56 + 49.57 = 0.01 J/K. s ≈ 0 J/K. s
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