- #1
swiper122
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- Homework Statement
- A barrel of cylinder shape, with radius 6 cm, is filled with water 10 cm high. You open a pipe at the bottom with area of 1 cm^2. How much time is needed for water to drop 5 cm in height?
- Relevant Equations
- Bernoulli equation: p + (⍴ *v^2)/2 + ⍴gh = constant.
Volumetric flow rate = Q= dV/dt = A*v
v1- velocity of water at the top of the barrel
v2- velocity of water at the the pipe (bottom of cylinder)
p + (⍴ *(v1)^2)/2 + ⍴gh=p + (⍴ *(v2)^2)/2 + ⍴gh
atmosferic pressure cancels out
(⍴ *(v1)^2)/2 + ⍴gh = (⍴ *(v2)^2)/2 + ⍴gh
density doesn't affect the result(cause its in every term)
(v1)^2/2 + gh = (v2)^2/2 + gh
(finaly i can set the height of the pipe to zero)
(v1)^2/2 + gh = (v2)^2/2
(i am left with this equation and 2 unknowns)
Q= dV/dt = A*v
(Now, I am not sure how to use this equation, since i think Volumetric flow is not constant, because velocity is not constant (potential energy is decreasing as height lowers with time)
I feel like i need to integrate but i don't know how to set it up. The result is supposedly 63 s. Thank you
v2- velocity of water at the the pipe (bottom of cylinder)
p + (⍴ *(v1)^2)/2 + ⍴gh=p + (⍴ *(v2)^2)/2 + ⍴gh
atmosferic pressure cancels out
(⍴ *(v1)^2)/2 + ⍴gh = (⍴ *(v2)^2)/2 + ⍴gh
density doesn't affect the result(cause its in every term)
(v1)^2/2 + gh = (v2)^2/2 + gh
(finaly i can set the height of the pipe to zero)
(v1)^2/2 + gh = (v2)^2/2
(i am left with this equation and 2 unknowns)
Q= dV/dt = A*v
(Now, I am not sure how to use this equation, since i think Volumetric flow is not constant, because velocity is not constant (potential energy is decreasing as height lowers with time)
I feel like i need to integrate but i don't know how to set it up. The result is supposedly 63 s. Thank you
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