Question

A storage tank holds methane at 120 K, with a quality of 25 %, and it warms up by 5°C per hour due to a failure in the refrigeration system. How long time will it take before the methane becomes single phase and what is the pressure then?

Answer 1

One of the fundamental pillars to solve this problem is the use of thermodynamic tables to be able to find the values of the specific volume of saturated liquid and evaporation. We will be guided by the table B.7.1 'Saturated Methane' from which we will obtain the properties of this gas at the given temperature. Later considering the isobaric process we will calculate with that volume the properties in state two. Finally we will calculate the times through the differences of the temperatures and reasons of change of heat.

Table B.7.1: Saturated Methane

$T_1 = 120K$

$p_1 = 191.6kPa$

$v_f = 0.002439m^3/kg$

$v_{fg} = 0.30367 m^3/kg$

Calculate the specific volume of the methane at state 1

$v_1 = v_f+x_1v_{fg}$

$v_1 = 0.002439+ (0.25)(0.30367)$

$v_1 = 0.0783m^3/kg$

Assume the tank is rigid, specific volume remains constant

$v_2 = v_1$

$v_2 = 0.0783m^3/kg$

Now from the same table we can obtain the properties,

At $v_g = 0.0783m^3/kg$

$T_2 = 145K$

$p_2 = 823.7kPa$

We can calculate the time taken for the methane to become a single phase

$t = \frac{T_2-T_1}{\dot{T}}$

Here

$T_1 =$ Initial temperature of Methane

$\dot{T} =$ Warming rate

Replacing

$t = \frac{(145-273)-(120-273)}{5}$

$t = \frac{25}{5}$

$t = 5hr$

Therefore the time taken for the methane to become a single phase is 5hr