Water is being stored in a tank such that the water level is 62 m above the grou

Question

Water is being stored in a tank such that the water level is 62 m 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 5.2 m and the radius of the faucet opening is 0.010 m

a) What is the pressure at the house faucet when it is closed?
b) If the faucet were opened fully, what would be the speed of the water as it came out?
c) How fast would the level of the water in the tank above go down?

Explanation / Answer

The hieght of the water level h = 62m

(a) The pressure at the house faucet

            P = P0 + gh

               = 101kPa + (10^3 kg/m^3)(9.8m/s^2)(62)

               = 708.6 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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