# Fluid mechanics flowrate problem

1. Feb 13, 2012

### PythagoreLove

1. The problem statement, all variables and given/known data
A long water trough of triangular cross section is formed from two planks as is show in Fig P3.66. A gap of 0.1 in. remains at the junction of the two planks. If the water depth initially was 2 ft, how long a time does it take for the water depth to reduce to 1 ft?

http://alkaspace.com/is.php?i=133522&img=Photo_du_643845.jpg

2. Relevant equations
Bernoulli equation
p+1/2$\rho$V2+$\gamma$z=constant

Conservation of the mass
V1A1=V2A2

3. The attempt at a solution

Q=water flow
V=Velocity of the water
A=Area
Vol=Volume of water
L=Length of the planks
z=height of the water level

A=0.1*L
Vol=L*z(t)2*tan(45)=z(t)2*L

Q=$\frac{dVol}{dt}$=2z(t)*z'(t)*L

Q=V(t)*A=-0.1*L*V(t)

0.1*L*V(t)=2z(t)*z'(t)*L

0.1*V(t)=2z(t)*z'(t)

V(t)=20z(t)*z'(t) (EQ1)

With the bernoulli equation we can find the velocity of the water in fonction of the height of the water level

$\gamma$z(t)=1/2*V(t)2*$\rho$

V(t)=$\sqrt{\frac{2z(t)\gamma}{\rho}}$ (EQ2)

By combining EQ1 and EQ2
20z(t)*z'(t)=$\sqrt{\frac{2z(t)\gamma}{\rho}}$

$\sqrt{z}$ dz = $\frac{\sqrt{\frac{2\gamma}{\rho}}}{20}$ dt

By integrate

-2z3/2/3 = $\frac{\sqrt{\frac{2\gamma}{\rho}}}{20}$*t+C

When t=0, z=2 so C=1,89

if I put z=1, t=3.04s and the answer is 36.5s

PytLove

2. Feb 14, 2012

### LawrenceC

I get your answer of 36.5 seconds. But what is the 3.04s?

3. Feb 14, 2012

### PythagoreLove

I am sorry my sentence was wrong I meant:

when I put z=1, my answer is 3.04s. But the answer in the book is 36.5s.

I have no idea if my error is at the beginning of the problem or at the end. Please help.

PytLov

4. Feb 15, 2012

### LawrenceC

What are your units of gamma and rho? Are your terms in your equations uniform in units?

My solution is similar to yours. First I found an expression for the volume of the trough.

V = h^2

I use unit length to unclutter the equations. Then I take the derivative of the above expression.

I used Bernoulli's eqn. to get another expression for the rate of change of volume.
It is based on unit length as well.

I equate the derivatives. It is a variable separable differential equation. One variable is height, the other is time.

Then I integrate using definite integrals. The time integral goes from 0 to T. The height integral goes from 2 to 1. You wind up with an expression for T and it works out to be 36.5 seconds.

5. Feb 15, 2012

### PythagoreLove

I found my error, I used 0.1*L for the area of the falling water but I needed to put it in feet. Thank you very much, you have no idea how much time I passed on that problem.

PytLov

6. Feb 15, 2012

### LawrenceC

Good for you!