Question 9.2.3: A centrifugal pump lifts water under a static head of 36 m o......
A centrifugal pump lifts water under a static head of 36 m of water of which 4 m is suction lift. Suction and delivery pipes are both 150 mm in diameter. The head loss in suction pipe is 1.8 m and in delivery pipe 7 m. The impeller is 380 mm in diameter and 25 mm wide at mouth and revolves at 1200 rpm. Its exit blade angle is 35°. If the manometric efficiency of the pump is 82%, determine (a) the discharge through the pump, and (b) the pressure at the suction and delivery branches of the pump.
Refer section 9.2.1 for various heads used in the following solution.
Assumption. Radial entry at the pump inlet.
Given:
\begin{matrix} H=36 m & h_{fs} =1.8 m & h_{fd}=7 m\rightarrow H_m=36 +1.8 +7 =44.8 m \\ D_s=D_d=0.15 m & D_2 =0.38 m & N=1200\rightarrow u_2=23.87 m/s \end{matrix}
B_2 = 0.025 m \ \ β_2 = 35° \ \ h_s = 4 m → h_d = H – h_s = 32 m
η_{\mathrm{mano}} = 0.82 → H_e = C_{w2} u_2/g = Hm/ η_{\mathrm{mano}} \text{or} C_{w2} = 22.45 m/s
a. Q = (π D_2 B_2) C_{f2}, \text{where} C_{f2} = (u_2– C_{w2}) \tan β_2 = (23.87 – 22.45) \tan 35 = 0.99 m/s This gives
Q = π × 0.38 × 0.025 × 0.99 = 0.0295 m³/s
Also Q = πD^2_s V_s/4, where V_s is velocity of water in suction pipe. This gives, Vs = Vd = 1.67 m/s, since pipe diameter is same for suction and discharge pipes.
∴ {V_s^2}/{2g}={V_d^2}/{2g}=0.142 \ m
=-\left\lgroup h_s+h_{fs}{V^2_s}/{2g}\right\rgroup =-5.942 m=10 -5.942 =4.058 \ m
=(4.058 \times 9.81 \times 1000 )/1000 =39.8 \ kN/m^2
=\left\lgroup h_d+h_{fd} {V^2_d}/{2g}\right\rgroup =39.14 \ m =3.84 bar (gauge)