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

1. Mar 24, 2015

### Curiosity 1

Hello.

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?

Thanks.

2. Mar 24, 2015

### SteamKing

Staff Emeritus
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.