Answer to Question #234651 in Molecular Physics | Thermodynamics for Peter

Question

A rigid vessel contains 5.0 kg of wet steam at 0.4 MPa. After the addition of 9585 KJ, the steam has a pressure of 2.0 MPa and a temperature of 700 degrees Celsius. Determine the initial energy and the specific volume of the steam?

Answer 1

Solution;

Given;

$m=5.0kg$

$Q_A=9585kJ$

$P_1=0.4MPa$

Vessel is rigid, meaning;

$V_1=V_2$

$P_2=2.0MPa$

$T_2=700°c=973K$

Heat added;

$Q_A=m(h_2-h_1)$

From steam tables,at 700°c and 2.0MPa;

$h_2=3923.81kJ/kg$

Therefore;

$9585=5×(3923.81-h_1)$

$h_1=\frac{19619.05-9585}{5}$

$h_1=2006.81kJ/kg$

From steam tables ,at 0.4MPa and h₁=2006.81kJ/kg;

$h_f=604.37kJ/kg$

$h_g=2737.68kJ/kg$

$v_f=0.001084m^3/kg$

$v_g=0.4624m^3/kg$

$u_g=2552.74kJ/kg$

$u_f=603.93kJ/kg$

$h_1=h_{f_1}+xh_{g_1}$ =$h_f+x(h_g-h_f)$

$2006.81=604.37+x(2737.68-604.37)$

$x=0.657$

Take;

$v_1=v_f+x(v_g-v_g)$

$v_1=0.001084+0.657(0.4624-0.001084)$

Hence,the specific volume is;

$v_1=0.3043m^3/kg$

To calculate internal energy;

$u=u_f+x(u_g-u_f)$

$u=603.93+0.657(2552.74-603.93)$

Hence the internal energy is;

$u=1884.29kJ/kg$