Question

A rigid, insulated tank contains steam at 3 MPa, 400 Celsius. A valve on the tank is opened, allowing steam to escape. The overall process is irreversible, but the steam remaining inside the tank goes through a reversible adiabatic expansion. Determine the fraction of steam that has escaped when the final state inside is saturated vapor.

Answer 1

Answer:

x=0.8

Explanation:

Defining our values we have

$p_1=3Mpa\\T_1=400\° c\\q_{1-2}=0$

Where $p_1$ is the pressure of the super heated steam

$T_1$ is the Temperature of the super heated steam

and $q_{1-2}$ is the pressure in an adiabatic process.

State 1

$v_1 = 0.09936m^3/Kg\\s_1=6.9213kJ/kg.K$

State 2

$s_2=s_1=6.9212$

Note that here in state 2 the process is Reversible.

At the value where $T_2 = 141\°c$ we have

$v_2=(v_g)_{T_2}=0.4972m^3/kg$

Through this values we can calculate the Fraction of steam,

$x=\frac{m_1-m_2}{m_1}\\x=1-\frac{m_2}{m_1}\\x=1-\frac{v_1}{v_2}\\x=1-\frac{0.9936}{0.4972}\\x=0.8$