# Homework Help: Time to drain water through a pipe

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1. Apr 18, 2015

### Bruneleski

1. The problem statement, all variables and given/known data
A closed and elevated vertical cylindrical tank with diameter 2m contains water to a depth of 0.8m.
A circular hole is made at the bottom of the tank with diameter 0.2m.As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure of 5000Pa at the surface of the water.
Ignore viscosity.How much time does it take for all the water to drain from the tank?

2. Relevant equations
Bernoulli's equation
Continuity equation
3. The attempt at a solution
I have no idea how to approach a problem like this in general but here is what i thought:
I can calculate water flow through a small hole using
$$dV/dt=A\upsilon$$
Now
$$dt=dV/A\upsilon$$
and integrate both sides to get t.
Unfortunately, I get nowhere.Units do not match and so on.I don't know if this is even conceptually right.
Now a friend of mine says i should watch out for dh/dt rate, but I can't see how is that useful

2. Apr 18, 2015

### Staff: Mentor

That approach is fine, but you have to find v as function of volume or (better) height.
Think of energy conservation. To flow out with speed v, the water needs energy, and its volume goes from a region with high pressure to one with lower pressure.

3. Apr 18, 2015

### Bruneleski

I can get that using Bernoulli.

p1+ρgy1+½ρ v12=p2+ρgy2+½ρ v22

½ρ v12 is zero because v1<<v1
ρgy2 is zero because potential energy at the bottom of the tank is 0
pressure p1 at the water surface is constant 5000Pa (it says so in text).
pressure at the bottom(hole) is assumed zero(it says so in text).
so we pull out water velocity at bottom(hole)

v=v2= sqrt(2p1/ρ+2gy1)

we plug that and take integral

t=∫(dV/Av) from initial volume V to V0=0 when water is completely drained
but because Av is constant we can take it outside the integral and integrate dV and get
t=-½V2/Av with units completely off (which obviously originates from the V^2)

4. Apr 18, 2015

### Staff: Mentor

Using your notation, if y1 represents the depth of water in the tank at any time, and AT represents the cross sectional area of the tank, what is the volume of water in the tank at any time? According to the continuity equation, the rate of change of this volume has to be equal to minus the velocity out the bottom hole times the cross sectional area of the bottom hole. You should obtain a differential equation for dy1/dt in terms of y1.

Chet

5. Apr 18, 2015

### Staff: Mentor

Integrating dV' would give V, in the same way as integrating dt' gives t on the other side of the equation.
v is not constant.

6. Apr 18, 2015

### Bruneleski

@chet
Woah, I treated the y1 as the constant all the time instead of variable of time.
Volume of water at any time is dy1(which changes over time) times the area of a cylinder.
I substituted that and got easy differential equation.

Thanks guys.

7. Apr 18, 2015

### rude man

EDIT: I didn't see your and mfb's further responses. I'll leave my post just in case. I'm surprised you got an 'easy' ODE. Mine required separation of variables.

Good start, except v is always +. Note that v = v(h=0). h=0 at hole. I'm using v for hole velocity. You also have dV/dt = A1dh/dt.
This is right. Remember y1 = y1(t).
v is not constant! v is velocity of hole water which starts max. and ends up zero when the tank is empty.
WARNING: are you confusing A, the area of the tank with A, the area of the hole? I think so.
You have v(y1). You also have v as a function of dy1/dt.
So form an ODE with y1 as dependent variable and t as independent. Solve for y1(t). Find the constant of integration (the ODE is separable) , set y1 = 0 and solve for t=T.