Thermodynamics: Mass Flow Rate

[parth6512]

Homework Statement

A cylindrical tank with a diameter, d, is being filled with a liquid having a density,p. There is an inlet at the top of the tank and an outlet at the bottom of the tank. The mass flowrate at the inlet is given by m(in) and the mass flowrate at the outlet is given by m(out). The outlet mass flowrate is proportional to the height, h, of the liquid in the tank such that m(out)=k*h where k is the proportionality constant. Using conservation of mass determine the height of the liquid level,h, as a function of time,t, and other variables p,m(in),d, and k. m(in)>m(out)

Homework Equations

m(total)=m(in)-m(out)
m=p*volume*area
area=(pi/4)*d^2

The Attempt at a Solution

I used the formula d(p*area(tank)*h)/dt = p*v(in)*area(in)-p*((2*g*h)^.5)*area(out). Is this any leap forward?

 

[KataKoniK]

Hello,

Yes, your attempt at a solution is a good start. You have correctly used the conservation of mass principle to set up an equation that relates the change in mass in the tank to the mass flowrates at the inlet and outlet. However, there are a few things that can be improved in your solution.

Firstly, the formula you have written is not fully correct. The left side of the equation should be d(m(t))/dt, not d(p*area(tank)*h)/dt. This is because we are interested in the change in mass in the tank over time, not the change in volume.

Secondly, you have not taken into account the fact that the outlet mass flowrate is proportional to the height of the liquid in the tank. This means that m(out) = k*h, not just k. Therefore, your equation should be d(m(t))/dt = p*v(in)*area(in)-p*((2*g*h)^.5)*k*h.

Finally, to solve this equation, you will need to use the fact that the mass flowrate at the inlet, m(in), is greater than the mass flowrate at the outlet, m(out). This means that m(in) > k*h, which can be rearranged to h < m(in)/k. This will give you an upper limit for the height of the liquid in the tank.

I hope this helps. Let me know if you have any further questions.