(adsbygoogle = window.adsbygoogle || []).push({}); 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[itex]\rho[/itex]V^{2}+[itex]\gamma[/itex]z=constant

Conservation of the mass

V_{1}A_{1}=V_{2}A_{2}

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=[itex]\frac{dVol}{dt}[/itex]=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

[itex]\gamma[/itex]z(t)=1/2*V(t)^{2}*[itex]\rho[/itex]

V(t)=[itex]\sqrt{\frac{2z(t)\gamma}{\rho}}[/itex] (EQ2)

By combining EQ1 and EQ2

20z(t)*z'(t)=[itex]\sqrt{\frac{2z(t)\gamma}{\rho}}[/itex]

[itex]\sqrt{z}[/itex] dz = [itex]\frac{\sqrt{\frac{2\gamma}{\rho}}}{20}[/itex] dt

By integrate

-2z^{3/2}/3 = [itex]\frac{\sqrt{\frac{2\gamma}{\rho}}}{20}[/itex]*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

I'm probably out of line, please help me.

PytLove

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