Question 9.2.1: The impeller of a centrifugal pump has an outer diameter of ......
The impeller of a centrifugal pump has an outer diameter of 250 mm and an affective area of 0.017 m². The blades are bent backwards so that the direction of outlet relative velocity make an angle of 148° with the tangent drawn in the direction of impeller rotation, the diameters of suction and delivery pipes are 150 mm and 100 mm respectively. The pump delivers 0.013 m³/s at 1450 rpm when the gauge pressure on the suction and delivery pipes close to the pump show heads of 4.6 m below and 18.0 m above atmosphere respectively. The head losses in the suction and delivery pipes are 2.0 m and 2.9 m respectively. The motor, driving the pump delivers 8.67 kW. Assuming that water enters the pump without shock and whirl, determine (i) the solution efficiency and (ii) the overall efficiency of the pump.
Given:
\begin{matrix}D_2=0.25 \ m & A_f=0.017 m^2 & \beta _2=180-148 =32º & D_s = 0.15 m\\D_d=0.1 m &N = 1450 \ \mathrm{rpm} &Q=0.031 m^3/s &h_{fs}=2.0 \ m\\h_{fd}=2.9 m & P=8.67 \ \mathrm{kW} & p_s/w=-4.6 m &p_d/w=18.0 m \\ C_{w1}=0 \end{matrix}
For determining manometric efficiency, we calculate manometric and Euler heads.
Manometric head is given by Eq. 9.2.3 for which velocity of flow in suction and discharge pipe is obtained using continuity equation. This gives:
H_m=H_d+(-H_s)=\left\lgroup\frac{p_d}{w}-\frac{p_s}{w} \right\rgroup +\frac{V_d^2}{2g} -\frac{V^2_s}{2g} (9.2.3)
V_s=\frac{4Q}{\pi D^2_s}=1.754 m/s V_d=\frac{4Q}{\pi D^2_d} =3.947 m/s ,\text{ and }
H_m=\left\lgroup\frac{p_d}{w}-\frac{p_s}{w} \right\rgroup + \left\lgroup\frac{V^2_d}{2g} -\frac{V^2-s}{2g} \right\rgroup =\frac{(18+4.6)+\left(3.947^2 -1.754^2\right) }{2\times 9.81} =23.3 m (A)
To determine Euler head, see Fig. 9.2.4 for backward curved vane.
From given data, we get
u_2 = π D_2 N/60 = 18.98 m/s \text{and } C_{f2} = Q/A_f = 1.82 m/s
From outlet velocity diagram:
C_{w2} = u_2– Cf_2 \cot β_2 = 16.07 m/s \text{and } H_e = C_{w2} u_2/g = 31.1 m
∴ η_{man} = H_m/H_e = 74.71 %
η_o = w Q H_m/P = 9.81 × 0.031 × 23.23/8.67 = 81.48%
