Question

Consider the process of filling a tank with compressed air from a constant pressure supply The tank has a volume of 500 ft' and initially is at atmospheric pressure and 75°F. The tank is connected to a high pressure air line that contains a control valve. The valve is quickly opened when the filling process begins, and the flow rate of air through the valve and entering the tank is given by the following equation: 2 80 (AP)12 w where w is the mass flow rate in lbm/min and supply pressure and the air pressure in the tank in units of psi. Determine the time required to fill the tank to a pressure of 425 psia if the supply pressure is 450 psial and the process is isothermal. AP is the difference between the

Process Design & Control

Consider the process of filling a tank with compressed air from a constant pressure supply.

Answer 1

Temperature of the tank = 75 ^oF (534.65 ^oR)

Volume = 500 ft³

Initial Pressure= 14.7 psia

Using Ideal Gas Law to relate Pressure, Volume and Temperature inside the Tank

$\large P*V=n*R*T$

n= number moles of air present in the tank

n can be represented as mass of air (m) divided by it molecular weight (mw).

Molecular weight of air is 28.97

mw

.. m = mw * P * R *T

here mw, V, R and T are constant

dm mwV dP

Writing material balance for tank

In = Out + Accumulated

dm 80 * (AP)0.5-0 + dt

The supply of compressed air is at constant pressure of 450 psia

ДР = 450-P

dm -0+ dt 0.5 80 * (450-P)

Substituting dm/dt in above equation

$\large \therefore \frac{mw*V}{R*T}*\frac{dP}{dt} = 80*\left ( 450-P \right )^{0.5}$

80 R T dt dP V (450 - P)0.5m

Integrating the above equation within limits as

1. t=0 P =14.7 psia
2. t=t P=P

R T t dt dP 14.7 (450 - P).5mw* V

80*R* T [t 0] |-2 * (450-P)0.5. - 14.7mw V

80 *R T 0.5 2 * (450-P) + 41.7 = * t

In the above equation, we substitute P=425 psia and calculate time

Time = 1 min