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ivancsc1996
 one year ago
Please help! This problem has been bugging me for a while now. I am not to good with variable mass problems.
An enclosed cylindrical tank with a radius R and a height h is half filled with water. The air on the top part of tank is maintained at a pressure P. A hole of diameter d is opened. Derive an expression for the upward thrust excerted by the escaping water.
ivancsc1996
 one year ago
Please help! This problem has been bugging me for a while now. I am not to good with variable mass problems. An enclosed cylindrical tank with a radius R and a height h is half filled with water. The air on the top part of tank is maintained at a pressure P. A hole of diameter d is opened. Derive an expression for the upward thrust excerted by the escaping water.

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rajat97
 one year ago
Best ResponseYou've already chosen the best response.0to find the speed of the escaping water stream, we can use the Bernoulli's equation. Bernoulli's equation says \[P_{1} + \rho gh + 1/2(\rho v_{1}^{2}) = P_{2} + 1/2(\rho v_{2}^{2})\] where, P1 is the pressure at the top surface of the liquid rho is the density of the liquid g is the acceleration due to gravity h is the height of the liquid or distance between the top surface and the bottom surface of the liquid v1 is the speed of the top surface of the liquid P2 is the pressure at the bottom surface of the liquid and v2 is the speed of the bottom surface of the liquid here in this question, we have P1 = P (as the pressure of the air maintained above the liquid is P) h is h/2 and P2 is the pressure at the bottom now the pressure at the bottom is an important thing the pressure at the bottom is the pressure exerted by the atmosphere on the bottom of the liquid as the bottom surface of the liquid is exposed to the atmosphere through a small hole. so, P2 = P0 (and P0=atmospheric pressure) i think now you can solve the question Hint:if you do not get what to do with the speeds of the liquid surfaces v1 and v2, just use the equation of continuity that relates v1 and v2 using the areas of cross section and get the equation converted to only one variable 'v2' now that you have got the speed v2 of the ejecting liquid, you can apply law of conservation of momentum on the system(cylinder and water) density of liquid x area of crosssection x speed of the stream of liquid = rate of flow of mass out of the liquid and rate of flow of liquid x speed of the liquid stream = the upthrust exerted by the liquid you can use the physical definition of force to derive 'rate of flow of liquid x speed of the liquid stream = the upthrust exerted by the liquid' the physical definition of force says that force exerted is change in momentum/time or more precisely, dp/dt where dp is infinitesimal small change in momentum in infinitesimal small time dt the change in momentum of the tank will be equal to the change in momentum of the water coming out because both were initially at rest (you can get it from the equation of conservation of momentum) if you have any query feel free to ask because this answer dose not seem to be very clear as i'm in a hurry sorry but hope this helps you:)

ivancsc1996
 one year ago
Best ResponseYou've already chosen the best response.0Thanks @rajat97. My problem comes from the fact that as the water flows out the height of water changes and therefore Bernoullis equation becomes a differential equation I don't know how to solve

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0oh! okay. i'll try it once again
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