Engineering Mass flow rate as a function of time

[oggillia]

Homework Statement

A barrel of height H and base diameter D is initially filled with water at STP
(density=1000kg/m3). At the base there exist two circular openings of diameter d and 2d, both of which are initially stoppered.

Subsequently, the small opening of diameter d is unplugged, and the water is allowed to
drain. We can assume that the kinetic energy of the flowing water from the hole is approximately
equal to the potential energy of the column of water directly above it, i.e. ½mve
2= mgh(t) where h(t) is the height of water and it is of course a function of time since it decreases as the tank is draining. You may assume that the flow at the hole is normal and 1-dimensional. Develop expressions for the height of the water, h(t) and the mass-flow rate at the hole,
and mass flow rate as a function of time and given dimensions, i.e., H, D and d.

Relevant Equations

We know mass flow rate= density*velocity*Area since the flow is 1-D and normal.
Mass flow rate at exit= -(denisty)*(sqrt(2gh))*(pi/4)*(d^2)
The velocity at the exit is practically given to us in the problem (½mve 2= mgh(t)). Solve for v.
I integrated the conservation of mass formula to find h(t).
h(t) = H - [(d^4/D^4)*gt^2]/2

The only problem I'm having, is finding the mass flow rate as a function of time, m-dot(t).

At first, I inputted h(t), of which I solved for, into the mass flow rate formula.
So it looked something like this, m-dot(t) = -(density)*[sqrt(g*(H-(g*d^4*t^2/D^4)]*(pi/4)*d^2

But I don't think that's right? Any thoughts?

 

[pasmith]

Homework Statement:: A barrel of height H and base diameter D is initially filled with water at STP
(density=1000kg/m3). At the base there exist two circular openings of diameter d and 2d, both of which are initially stoppered.

Subsequently, the small opening of diameter d is unplugged, and the water is allowed to
drain. We can assume that the kinetic energy of the flowing water from the hole is approximately
equal to the potential energy of the column of water directly above it, i.e. ½mve
2= mgh(t) where h(t) is the height of water and it is of course a function of time since it decreases as the tank is draining. You may assume that the flow at the hole is normal and 1-dimensional. Develop expressions for the height of the water, h(t) and the mass-flow rate at the hole,
and mass flow rate as a function of time and given dimensions, i.e., H, D and d.
Relevant Equations:: We know mass flow rate= density*velocity*Area since the flow is 1-D and normal.
Mass flow rate at exit= -(denisty)*(sqrt(2gh))*(pi/4)*(d^2)
The velocity at the exit is practically given to us in the problem (½mve 2= mgh(t)). Solve for v.
I integrated the conservation of mass formula to find h(t).
The only problem I'm having, is finding the mass flow rate as a function of time, m-dot(t).

At first, I inputted h(t), of which I solved for, into the mass flow rate formula.
So it looked something like this, m-dot(t) = -(density)*[sqrt(g*(H-(g*d^4*t^2/D^4)]*(pi/4)*d^2

But I don't think that's right? Any thoughts?

I agree that <br /> \dot M = - \tfrac{1}4\pi \rho d^2 \sqrt{2gh}. I don't agree with your expression for \sqrt{2gh(t)}, but since you haven't explained how you arrived at it in sufficient detail I don't know where you went wrong.

What is the differential equation which governs h(t)? How did you solve it?

 

[oggillia]

Here's how I solved h(t).