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rawimpact
Apr29-09, 10:44 PM
1. A large tank isopen to the atmosphere and is filled with water. A small hole is 16m below the surface of the water. If the flow rate of the water is out of the hole is 2.5e^3 m^3/min, calculate: a)the velocity of the water as it leaves the hole and b) the diameter of the hole.



2. Lets say the pool is 1 and the hole is 2, P = pressure and P is density, i was given the equation: P1 + Pv1^2 + 1/2PY1 = P2 + 1/2PV2^2 + PY2


3. I have no clue how to solve this equation, can someone help me out please?

rl.bhat
Apr29-09, 10:57 PM
Check the relevant formula given in 2.
Inside the pool velocity is zero. And outside the pool Y2 is zero.

rawimpact
Apr29-09, 11:28 PM
Check the relevant formula given in 2.
Inside the pool velocity is zero. And outside the pool Y2 is zero.

So the equation is:

P1 + 1/2PY1 = P2 + 1/2PV2^2

Correct?

Where do i go from here?

rl.bhat
Apr29-09, 11:33 PM
Again check the left hand side of the equation.

rawimpact
Apr29-09, 11:55 PM
Can you please explain the equation, i do not understand it so i really do not know what the velocities are

rl.bhat
Apr29-09, 11:58 PM
OK. At the opening what is the total pressure inside the tank?

rawimpact
Apr30-09, 12:00 AM
OK. At the opening what is the total pressure inside the tank?

Well since it is exposed to the atmosphere, isnt it atmospheric pressure? 1.01x10^5 Pa?

rl.bhat
Apr30-09, 12:04 AM
Well since it is exposed to the atmosphere, isnt it atmospheric pressure? 1.01x10^5 Pa?
Plus the pressure due to the water at the depth Y1

rawimpact
Apr30-09, 12:15 AM
Ok, i've figured it out, thank you for all of your help. I have another problem, should i continue that here or start another thread?

rl.bhat
Apr30-09, 12:38 AM
You can continue here.