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

x=0.8

Step-by-step 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=(m_1-m_2)/(m_1)\\x=1-(m_2)/(m_1)\\x=1-(v_1)/(v_2)\\x=1-(0.9936)/(0.4972)\\x=0.8`