Question

Prob-2 If the flow of 0.10 m/s of water is to be maintained in the system shown, what power must be added to the water by the pump? The pipe is made of steel and is 20 cm in diameter. Use table 10.4 and AS for ks ? and ? Elevation = 13 m Elevation = 10 m Water T-10°C 40 m 40 m

please show clear steps solution.

Answer 1

Ans) Apply Bernoulli equation between point 1 and 2 located at water surface elevation of lower and upper reservoir respectively,

P1/$\gamma$ +
V1
/ 2g + Z1 + Hp = P2/$\gamma$ +
V22
/ 2g + Z2 + Hf

Since,both point 1 and 2 are open to atmosphere, pressure is only atmospheric,hence gauge pressure, P1 = P2 = 0

Velocity at surface is negligible so, V1 = V2 = 0

Elevation, Z1 = 10 m and Z2 = 13 m

Hp is pump head

Hf = Head loss due to friction  

Hence, above equation reduces to,

0 + 0 + 10 + Hp = 0 + 0 + 13 + Hf

=> Hp = 3 + Hf..........................................(1)

Also, Hf = f L $V^2$ / (2 g D)

where, f = Darcy friction factor

L = Pipe length = 40 + 40 = 80 m

V = Flow velocity = Q / A

D = Pipe diameter = 20 cm or 0.20 m

Flow rate (Q) = 0.10 $m^3$ /s

Pipe area = ($\pi$/4) ($0.2^2$) = 0.0314 $m^2$

=> Velocity (V) = 0.10 / 0.0314 = 3.18 m/s

To calculate Darcy friction factor (f) , determine Reynold number (Re)

Re = V D / $\nu$

where, $\nu$ = Kinematic viscosity of water = $10^{-6}$ $m^2$/s

=> Re = 3.18 x 0.20 / $10^{-6}$

=> Re = 636000

According to table 10.4, roughness of steel pipe (ks) = 0.046 mm

=> Relative roughness (ks/D) = 0.0046 mm / 200 mm = 0.000023

According to Moody diagram, for Re = 636000 and ks/D = 0.000023, Darcy friction factor (f) = 0.015

=> Hf = 0.015(80)($3.18^{2}$) / (2 x 9.81 x 0.20)

=> Hf = 3.09 m

Putting values in equation 1,

Hp = 3 + 3.09 = 6.09 m

Therefore, required pump power (P) = $\rho$ g Q Hp

=> P = 1000 x 9.81 x 0.10 x 6.09

=> P = 5974 W or 5.974 kW

Hence, power required by pump to be added to water is 5.974 kW