--- title: "Centrifugal pump head, discharge and power" subject: "Fluid Machines" chapter: 4 key: "semester_5/fluid_machines:centrifugal-pump-head-power-numerical" source: https://www.bctengineer.com/q/semester_5/fluid_machines/centrifugal-pump-head-power-numerical site: BCT Engineering --- # Centrifugal pump head, discharge and power Fluid Machines · Chapter 4: Pump - Asked in: year not given, 8 marks - Video: [Velocity Triangles Diagram For Impeller of Centrifugal Pump | Fluid Mechanics | Shubham Kola](https://www.youtube.com/watch?v=SHjzpNpCjwE) — Shubham Kola - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/fluid_machines/centrifugal-pump-head-power-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-4 ## Question A centrifugal pump impeller has an outer diameter of 0.4 m and an outlet width of 0.03 m. The impeller runs at 1450 rpm and the vanes are curved backward at an outlet angle of 30 degrees to the tangent. The flow velocity at outlet is 2.8 m/s and the entry is without whirl. The manometric efficiency is 78 % and the overall efficiency is 70 %. Neglecting the thickness of vanes, calculate (a) the absolute velocity and its direction at outlet, (b) the theoretical (Euler) head and the manometric head, (c) the discharge, and (d) the power required to drive the pump. ## Short answer Compute blade speed u2=30.37 m/s, whirl velocity Vw2=25.52 m/s. (a) V2=25.67 m/s at 6.3° to tangent. (b) Euler head He=79.00 m, manometric head Hm=61.62 m. (c) Discharge Q=0.1056 m³/s. (d) Shaft power P=91.2 kW. ## Answer **Given:** $D_2=0.4$ m, $B_2=0.03$ m, $N=1450$ rpm, $\beta_2=30^\circ$, $V_{f2}=2.8$ m/s, $V_{w1}=0$, $\eta_{man}=0.78$, $\eta_o=0.70$. #### Outlet velocity triangle Blade speed: $$ u_2=\frac{\pi D_2N}{60}=\frac{\pi\times0.4\times1450}{60}=30.37\ \text{m/s} $$ ```text V2 Vf2 \ | \ | Vr2 (at beta2 = 30 deg to u2 direction) \a| / --------*-----> u2 (blade direction) |<-Vw2->|<-Vf2 cot b2->| ``` Whirl velocity: $$ V_{w2}=u_2-\frac{V_{f2}}{\tan\beta_2}=30.37-\frac{2.8}{\tan30^\circ}=30.37-4.85=25.52\ \text{m/s} $$ #### (a) Absolute velocity and direction $$ V_2=\sqrt{V_{w2}^2+V_{f2}^2}=\sqrt{25.52^2+2.8^2}=25.67\ \text{m/s},\qquad \tan\alpha_2=\frac{V_{f2}}{V_{w2}}\Rightarrow\alpha_2=6.3^\circ $$ #### (b) Heads Euler head (no whirl at inlet): $$ H_e=\frac{V_{w2}u_2}{g}=\frac{25.52\times30.37}{9.81}=79.00\ \text{m} $$ $$ H_m=\eta_{man}H_e=0.78\times79.00=61.62\ \text{m} $$ #### (c) Discharge $$ Q=\pi D_2B_2V_{f2}=\pi\times0.4\times0.03\times2.8=0.1056\ \text{m}^3/\text{s}=105.6\ \text{L/s} $$ #### (d) Power Water power: $$ P_w=\rho gQH_m=1000\times9.81\times0.1056\times61.62=63.81\ \text{kW} $$ $$ P_{shaft}=\frac{P_w}{\eta_o}=\frac{63.81}{0.70}=91.2\ \text{kW} $$ **Answer:** (a) $V_2=25.67$ m/s at $6.3^\circ$; (b) $H_e=79.00$ m, $H_m=61.62$ m; (c) $Q=0.1056$ m³/s; (d) 91.2 kW. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the whirl velocity at outlet found for a backward-curved vane with outlet angle beta2 measured from the tangent? A. Vw2 = u2 + Vf2 / tan(beta2) B. Vw2 = u2 - Vf2 / tan(beta2) C. Vw2 = Vf2 tan(beta2) D. Vw2 = u2 / tan(beta2) Answer: B. For backward vanes the flow component Vf2 cot(beta2) is subtracted from the blade speed: 30.37 - 4.85 = 25.52 m/s. (Key point: Whirl velocity at outlet) 2. With no whirl at inlet, how are the Euler head and manometric head related to the given data? A. He = Vw2 u2 / g and Hm = eta_man x He B. He = V2^2 / 2g and Hm = He / eta_man C. He = Vw2 u2 / g and Hm = He / eta_man D. He = u2^2 / g and Hm = eta_o x He Answer: A. He = 25.52 x 30.37 / 9.81 = 79.0 m, and Hm = 0.78 x 79.0 = 61.62 m. (Key point: Euler head and manometric head) 3. Which expression gives the discharge when vane thickness is neglected? A. Q = pi D2 B2 u2 B. Q = pi D2 B2 Vf2 C. Q = pi D2 B2 V2 D. Q = pi D2 B2 Vw2 Answer: B. Discharge is outlet area times the flow velocity: pi x 0.4 x 0.03 x 2.8 = 0.1056 m^3/s. (Key point: Discharge from outlet area) 4. How is the power required to drive the pump obtained? A. Water power rho g Q Hm divided by overall efficiency (about 91.2 kW) B. Water power multiplied by overall efficiency C. rho g Q He divided by manometric efficiency D. Water power rho g Q Hm itself (63.8 kW) Answer: A. Input shaft power = output water power / eta_o = 63.81 / 0.70 = 91.2 kW. 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