Water is being stored in a tank such that the water level is 60m above the groun

Question

Water is being stored in a tank such that the water level is 60m above the ground. The tank is connected by a long pipe to a house on the ground below. Assume a faucet in the house is at ground level. The radius of the tank is 5.0m and the radius of the faucet opening is 0.010m.

*What is the pressure at the house faucet when it is closed?

*If the faucet were opened fully, what would be the speed of the water as it came out?

*How fast would the level of the water in the tank above go down?

Thank you!

Explanation / Answer

The hieght of the water level h = 60m

(a) The pressure at the house faucet

            P = P0 + gh

               = 101kPa + (10^3 kg/m^3)(9.8m/s^2)(60)

               = 689 kPa

(b) From bernoulli's equation

           P = P0 + 1/2v^2

          P -P0 = 1/2 v^2

therefore the velocity

           v = 2(P-P0)/   

              =  2( 607.6*10^3)/10^3

              = 34.859 or 35m/s

(c) From volume flow rate

          R = Av

          A1v1 = A2v2

therefore the speed of water leaves tank

          v1 = (A2/A1) v2

               = (0.01^2/5^2) (35)

                = 1.4*10^-4 m/s

           or 0.14 mm/s

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