Question 9.2.2: A centrifugal pump, in which water enters radially, delivers......
A centrifugal pump, in which water enters radially, delivers water to a height of 165 m. The impeller has a diameter of 360 mm and width 180 mm at inlet and the corresponding dimensions at the outlet are 720 mm and 90 mm respectively; its rotational speed is 1200 rpm. The blades are curved backwards at 30° to the tangent at exit and the discharge is 0.389 m³/s. Determine (a) theoretical head developed, (b) manometric efficiency, (c) pressure rise across the impeller assuming losses to be 12% of velocity head at exit, (d) pressure rise and the loss of head in the volute casing, (e) the vane angle at inlet, and (f) power required to drive the pump assuming an overall efficiency of 70%. What would be the corresponding mechanical efficiency?
Refer Fig. 9.2.2 for nomenclature and Fig.9.2.4(a) for velocity triangle at outlet.
Given:
\begin{matrix} H_m=165 m & D_1 =0.36 m & D_2=0.72 m & \beta _2=30° \\B_1=0.18 m & B_2=0.09 \ m&Q=0.389 & N=1200 \ \mathrm{rpm}\end{matrix}
∴ u_2 = π D_2 N/60 = 45.24 m/s \text{and } u_1 = u_2/2 = 22.62 m/s
Further Q = π D_1 B_1 C_{f1} = π D_2 B_2 C_{f2}
But D_1 B_1 = D_2 B_2, (check from given data)
∴ C_{f1} = C_{f2} = 0.389/(π × 0.72 × 0.09) = 1.91 m/s
Assumption. (i) Radial entry at inlet and (ii) 100% volumetric efficiency, since no information is given.
a. Theoretical head developed is Euler head = H_e = C_{w2} u_2/g. C_{w2} is obtained from velocity triangle at outlet.
From Fig. 9.2.4(a),
C_{w2} = u_2 – C_{f2}/\tan β_2 = 45.24 – 1.91/\tan 30 = 41.93 m/s
∴ H_e = 41.93 × 45.24/9.81 = 193.4 m
b. Manometric efficiency = Manometric Head/ Euler Head = 165/193.4 = 85.3%
c. Applying modified Bernoulli equation at points inlet (1) and outlet (2) of the pump and neglecting the change in potential energy, we get
Pressure rise in impeller =H_e+\frac{C_1^2 -c_2^2}{2g}-h_{\mathrm{loss}} (A)
But, C_1 =C_{f1}=1.91 \rightarrow C_1^2/(2g)=0.186 m \text{and } \frac{C_2^2}{2g}=\frac{C^2_{f2}+C^2_{w2}}{2g}=89.79 m
→ h_{\mathrm{loss}}= 0.12 \frac{C^2_2}{2g} 0.12 \times 89.79 =10.77 m
Substituting the different values in the Eq. (A), we get Pressure rise in impeller = 193.4 + (0.1862 – 89.79) – 10.77 = 93.02 m
d. H_m = pressure rise in impeller + pressure rise in volute casing. This gives Pressure rise in volute casing = 165 – 93.02 = 71.98 m
Loss of head in pump = loss in impeller + loss in volute casing
This gives, loss of head in volute casing
= H_e – H_m – h_{\mathrm{loss}} = 193.4 – 165 – 10.77 = 17.63 m
e. Refer Fig. 9.1.4 for inlet velocity triangle. From this triangle,
\tan β_1 = C_{f1}/u_1 = 1.91/22.62 → β_1 = 4.8°
f. Power required = w Q H_m/η_o
= (9.81 × 1000) × 0.389 × 165 / 0.7 = 899.5 × 1000 W = 899.5 kW
η_o = η_{\mathrm{mano}} × η_{\nu} × η_{\mathrm{mech}} → η_{\mathrm{mech}} = 0.7/0.853 = 82%
