Question

a pump that delivers 4.5 kw to the water (v=0.898x10^-6 m^2/s ; p=99.1kg/m^3) is used to fill a cylindrical tank as shown, calculate the flow rate and the depth of the water in the cylindrical tank 5 minutes after the pump is started. Assume the sum of the loss coefficients of the fittings=12.8

B 2 m d=100 mm smooth pipe Pipe length = 50 m 12 m 19 A

Answer 1

Let we have following data,

Power input, P = 4.5 kW = 4500 W

Density, $\rho$ = 991 kg/m³

(Note - Mass density of water can be taken as approx 1000 kg/m³ But in problem statement it is indicated as 99.1 kg/m³ which is considered as typo error and taken as 991 kg/m³)

Total head if water, H_t against which pump have to work is given by

H_t = H + Head loss in fittings = 12 m +12.8 m = 24.8 m

Power input, P = Q x H_t x g x $\rho$ , Where, Q = flow rate of water

Therefore from above equation we have

Q = 4500 / (24.8 x 9.81 x 991) = 0.01866 m³/s

.

Now lets find out depth of water in tank after 5 minutes

Time, t = 5 min = 5 x 60 s = 300 sec

Volume, V of water delivered in 300 sec = Q x t = 0.01866 x 300 = 5.598 m³

Area, A of tank = (
7
/4) x D² = (
7
/4) x 2² = 3.141 m²

We know that, Volume, V = A x tank height(h) filled by water

Therefore,

tank height, h = V / A = 5.598 / 3.141 = 1.782 m

Hence after 5 min tank height filled by pump will be 1.782 m