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PythagoreLove
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Homework Statement
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
Homework Equations
Bernoulli equation
p+1/2[itex]\rho[/itex]V2+[itex]\gamma[/itex]z=constant
Conservation of the mass
V1A1=V2A2
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
-2z3/2/3 = [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