The overall efficiency of the pump is 70% and coefficient of friction =0.015 in the formula $4flv^2 /2gD$

Mumbai University > Electronics and telecommunication > Sem 7 > Applied Hydraulics

Marks: 08

Years: DEC 2014

Given:

Discharge $Q=0.05 m^3/s $

Height $Hs = h_s+ h_d =20m $

Dia of pipe $ D_s =D_d=15 cm = 0.15 m$

Length $L_s + L_d = L = 100 m $

Overall efficiency $N_o =70\% = 0.70 $

Coefficient of friction $f=0.015$

Velocity of water in pipe $V_s = V_d = V= \dfrac {\text{Discharge}}{\text {Area of pipe}} \\ = \dfrac {0.05}{\pi/4 (0.15)^2}\\ =2.82 m/s $

Frictional head loss in pipe

$(hf_s+hf_d ) = \dfrac {4fLV^2}{2gd}= \dfrac {4\times 0.015\times 100\times 2.82^2}{2\times 9.81\times 0.15} \\ =16.21m $

Using equation we get manometric head as

$H_m = (h_s +h_d)+ (hf_s+hf_d) + \dfrac {V_d^2}{2g} \\ =20+16.21+\dfrac {2.82^2}{2\times 9.81} \\ = 36.61m $

Overall efficiency is given by equation

$N_o = \dfrac {(\dfrac {WH_m}{1000})}{S.P} = \dfrac {pg\times Q\times H_m}{1000\times S.P} \\ S.P = \dfrac {pg\times Q\times H_m}{1000\times N_o} = \dfrac {1000\times 9.81\times 0.05\times 36.61}{1000\times 0.70} \\ =25.65 Kw$

S.P is the power required to drive the centrifugal pump.