Question

Question 3 Show that the pressure head rise in the impeller of a centrifugal pump when frictional and other losses in the impeler are neglected, is given by (a) where Vni is the velocity of flow at the inlet, while at the outlet of the backward curved blades, u2, m2, and ßz are the blade velocity, velocity of flow and blade angle, respectively. (4 marks) A centrifugal pump rotating at 1450 rpm discharges 0.015 m'ls of water. Its impeller has internal and external diameters of 200 mm and 400 mm, respectively. Width of the impeller at the inlet is 15 mm and at the outlet 8 mm. If the blade angle at outlet is 35°, find the pressure head rise in the impeller neglect all losses. (b) (4 marks) How much will it impact the pressure head rise if the impeller blades were straight, instead of curved backward? (e) (2 marks)

Answer 1

$\frac{P_1}{\rho g}+\frac{V^2_1}{2g}+\frac{V_{w2}u_2}{2g}=\frac{P_2}{\rho g}+\frac{V^2_2}{2g}$

$\frac{V_{w2}u_2}{g}=H_m+\frac{V^2_2}{2g}-\frac{V^2_1}{2g}$..........................(1)

V2 Outlet rl V., Inlet w l

$tan \beta_2= \frac{V_{f2}}{u_2-V_{w2}}$

$u_2-V_{w2}= V_{f2} \ cot \beta_2$

$V_{w2}= u_2-V_{f2} \ cot \beta_2$

$V^2_2=V^2_{w2}+V^2_{f2}$

$V^2_2=(u_2-V_{f2} \ cot \beta_2)^2+V^2_{f2}$

$H_m=\frac{V^2_1}{2g}+\frac{u_2(u_2-V_{f2} \ cot \beta_2)}{g}-\frac{[(u_2-V_{f2} \ cot \beta_2)^2+V^2_{f2}]}{2g}$

$H_m=\frac{V^2_1}{2g}+\frac{2u^2_2-2u_2 \ V_{f2} \ cot \beta_2-u^2_2+2V_{f2} \ u_2 \ cot \beta_2-V^2_{f2} \ cot^2 \beta_2-V^2_{f2}}{2g}$

$H_m=\frac{V^2_1}{2g}+\frac{u^2_2-V^2_{f2} \ cot^2 \beta_2-V^2_{f2}}{2g}$

$H_m=\frac{V^2_1}{2g}+\frac{u^2_2-V^2_{f2} (1+\ cot^2 \beta_2)}{2g}$

$H_m=\frac{1}{2g}{[V^2_1+u^2_2-V^2_{f2} (1+\ cot^2 \beta_2)]}$   $(Because \ cosec^2\beta=1+cot^2\beta)$

$H_m=\frac{1}{2g}{[V^2_1+u^2_2-V^2_{f2} cosec^2 \beta_2]}$

If there is no loss of energy in impeller

$\frac{P_2-P_1}{\rho g}=\frac{1}{2g}{[V^2_1+u^2_2-V^2_{f2} cosec^2 \beta_2]}$

$u_2=\frac{\pi D N}{60}=\frac{\pi (0.4)(1450)}{60}=30.368 \ m/s$

$\\tan \beta_2= \frac{V_{f2}}{u_2-V_{w2}}\\ \\ \\ tan 35\degree=\frac{1.492}{30.368-V_{w2}}$

$Q=\pi D_2 B_2V_{f2}$

$V_{f2}=\frac{Q}{\pi D_2 B_2}=\frac{0.015}{\pi (0.4)(8 \times 10^{-3})}=1.492 \ m/s$

$V_1=V_{f1}=\frac{Q}{\pi D_1 B_1}=\frac{0.015}{\pi (0.2)(15 \times 10^{-3})}=1.591 \ m/s$

$\frac{P_2-P_1}{\rho g}=\frac{1}{2 \times 9.81}{[1.591^2+30.368^2-1.492^2 cosec^2 35]}$

$\frac{P_2-P_1}{\rho g}=46.788 \ m$

$\Delta P=46.788 \times 10^3 \times 9.81=459kPa$

$\\ For\ backward\ curved \ vane \ the \ value \ of \ \beta_2 < 90 \degree \ while \ for \ straight\ impeller\ blades\ \beta_2=90 \degree$

Forward curved vane > 90 P2- 90° 6 Radial vane Backward curved vane Discharge (Q)

$H_{m-back}=\frac{1}{2g}{[V^2_1+u^2_2-V^2_{f2} cosec^2 \beta_2]}$$(\mathit{For\ backward\ curved \ blade})$

$H_{m-straight}=\frac{1}{2g}{[V^2_1+u^2_2-V^2_{f2}]}$$(\mathit{For\ straight \ blade\ as \ cosec90\degree=1})$

$\frac{P_2-P_1}{\rho g}=\frac{1}{2 \times 9.81}{[1.591^2+30.368^2-1.492^2 cosec^2 90]}$

${\color{Red} Hence \ head\ rise \ increases \ with\ straight\ blade}$

$\% \ increase \ in\ head=\frac{47.019-46.788}{46.788}$