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Prove u= u_f + xu_g and h= h_f + xh_fg

  1. Mar 24, 2015 #1

    I'm learning thermodynamics using the book titled "Thermodynamics An Engineering Approach (5th edition)".
    In page 130, the derivation of v_avg = v_f + xv_fg is shown but how about deriving u_avg= u_f + xu_g and h_avg= h_f + xh_fg ?

    The derivation of the derivation of v_avg = v_f + xv_fg is like this:
    Consider a tank filled with saturated liquid-vapor mixture. Then the total volume of them, V, is V_f + V_g, where V_f is volume occupied by saturated liquid while V_g is volume occupied by saturated vapor.

    x= m_f/m_total

    V= mv ---> m_total v_avg= m_f*v_f + m_g*v_g
    also, we know
    m_f= m_total- m_g ---> m_total* v_avg = (m_total-m_g)v_f+ m_g*v_g
    dividing by m_total, we have
    v_avg= (1-x) v_f + xv_g

    and since x= m_f/m_total, we rewrite the equation...
    v_avg= v_f + xv_fg
    where v_fg= v_g - v_f and the subscript "avg" is usually dropped for simplicity.

    The book says these analysis can be repeated for internal energy, u and enthalpy, h....Could anybody tell me how?

  2. jcsd
  3. Mar 24, 2015 #2


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    Have you tried substituting hf and hfg for vf and vfg in the derivation of vavg?

    You don't have to re-invent the wheel, mathematically speaking, each time you want to calculate the average of a certain physical property.
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