Fluid Mechanics Problem: Evacuating a Tank with Vacuum Pump for Ideal Gas Air

[TSenior]

Homework Statement

6. A tank filled with air of initial density ρo=1.2 kg/m3 and initial pressure Po=100 kPa is to be evacuated by a vacuum pump. The tank volume is 1 m3 and the pump evacuates 0.0013 m3 of air per second regardless of the pressure. The process is isothermal and for an ideal gas such as air, this means that P/ρ = Po/ρo = constant. Derive an equation that shows how the pressure inside the tank, P, changes with time, t and then use it to calculate the time needed to reduce the tank pressure to 1 kPa.

Homework Equations

m1+m2-m3-m4= dM/dt

Mass Flow Rate = Rho x V x A
Mass inside control volume = Rho(Volume)

The Attempt at a Solution

The answer is (t = 3,542 s) but I am unsure of the method to get there.

 

[Valkarie]

I am thinking that the equation would look something like this, P/ρ = Po/ρo P = (Po/ρo) x ρ m1 + m2 - m3 - m4 = dM/dt (ρoV + (dM/dt))/t = P However I am not sure how to isolate t, and even if this is the right equation. Any help would be appreciated!

 

[Mene]

I would first start by defining the variables in the problem. The tank is filled with air of initial density ρo=1.2 kg/m3 and initial pressure Po=100 kPa. The tank volume is 1 m3 and the pump evacuates 0.0013 m3 of air per second regardless of the pressure. The process is isothermal, meaning that the temperature remains constant throughout the process. We can use the ideal gas law, PV = nRT, to relate the pressure, volume, and temperature of the air inside the tank.

Next, I would use the given information to set up a mass balance equation. The mass flow rate into the tank is equal to the mass flow rate out of the tank, so we can write:

m1 + m2 = m3 + m4

Where m1 and m2 are the initial and final masses of air in the tank, and m3 and m4 are the masses of air being evacuated by the pump. We can express these masses in terms of density, volume, and mass flow rate using the equations given in the problem statement.

m1 = ρoV

m2 = ρV

m3 = ρoV - 0.0013t

m4 = ρV - 0.0013t

Substituting these into the mass balance equation, we get:

ρoV + ρV = (ρoV - 0.0013t) + (ρV - 0.0013t)

Simplifying and rearranging, we get:

ρoV + ρV = ρoV + ρV - 0.0026t

0.0026t = ρoV + ρV - ρoV - ρV

0.0026t = 0

This equation is not possible, so we must have made a mistake in our assumptions. Looking back at the problem statement, we see that we are given the initial density and pressure, but not the initial volume. Therefore, we cannot solve for the time needed to reduce the tank pressure to 1 kPa.

To complete this problem, we would need to know the initial volume of the tank. We could then use the ideal gas law to solve for the initial mass of air in the tank and use that to set up the mass balance equation. From there, we