Question

Water is pumped through a 60 m pipe of 0.3 m diameter from a lower reservoir to a higher reservoir whose surface is 10 m above the lower one. The sum of minor loss coefficients is KL = 14.5. When the pump adds 40 kW to the water the flowrate is 0.20 m3 /s. Determine the pipe roughnes

Answer 1

Answer:

$\epsilon = 0.028*0.3 = 0.0084$

Explanation:

$\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2$

where $P_1 = P_2 = 0$

V1 AND V2  =0

Z1 =0

$h_P = \frac{w_p}{\rho Q}$

$=\frac{40}{9.8*10^3*0.2} = 20.4 m$

$20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10$

we know thaT$V  =\frac{Q}{A}$

$V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec$

$20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10$

f  = 0.0560

$Re =\frac{\rho v D}{\mu}$

$Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5$

fro Re = 7.53*10^5 and f = 0.0560

$\frac{\epsilon}{D] = 0.028$

$\epsilon = 0.028*0.3 = 0.0084$