Question 4 of 8: Pump Design

Question 4: Pump Design (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Preliminary geometry and duty of a centrifugal water pump:

Find. Ideal shaft power and torque, the blade speeds and radial velocities at both stations, the outlet whirl implied by the duty head, and the two blade angles taken from velocity diagrams drawn to scale.

Approach. Power from P = ρgQH and torque from P = 2πNτ (both page 21–22); kinematics from V_B = ωr and continuity through the cylindrical periphery; then Euler's pump equation with zero inlet whirl for the outlet tangential velocity, which closes both triangles and delivers the blade angles.

1. Part (a) — ideal power. Under ideal (frictionless) conditions all the shaft power ends up as water power, so page 21's P = ρgQH applies directly: $$P=1000\times 9.81\times 0.25\times 40=98\,100\ \text{W}=\boxed{98.1\ \text{kW}}$$ A real pump of this duty would draw perhaps 120 kW; the question deliberately strips the losses out so that the kinematics can be isolated.
2. Part (b) — shaft torque. Page 22 gives P = 2πNτ with N in rev/s, which is the same as P = ωτ. The angular speed is $$\omega=\frac{2\pi N}{60}=\frac{2\pi(1720)}{60}=180.1\ \text{rad/s}$$ so $$\tau=\frac{P}{\omega}=\frac{98\,100}{180.1}=\boxed{544.6\ \text{N}\!\cdot\!\text{m}}$$
3. Part (c) — blade tangential velocities. The blade speed at any radius is simply ωr: $$V_{B1}=180.1\times 0.100=\boxed{18.01\ \text{m/s}},\qquad V_{B2}=180.1\times 0.180=\boxed{32.42\ \text{m/s}}$$
4. Part (d) — radial (through-flow) velocities. Neglecting blade thickness, the whole flow crosses a cylindrical surface of area 2πrb at each station: $$V_{1R}=\frac{Q}{2\pi r_1b_1}=\frac{0.25}{2\pi(0.100)(0.050)}=\frac{0.25}{0.03142} =\boxed{7.96\ \text{m/s}}$$ $$V_{2R}=\frac{Q}{2\pi r_2b_2}=\frac{0.25}{2\pi(0.180)(0.030)}=\frac{0.25}{0.03393} =\boxed{7.37\ \text{m/s}}$$ The passage width is tapered from 50 mm to 30 mm precisely so that the radial velocity stays nearly constant as the radius grows; a designer aims for this because a decelerating radial component inside the impeller promotes separation. Note the area must be the cylindrical periphery 2πrb, never a disc area.
5. Part (e) — outlet tangential velocity from the duty head. Euler's pump equation with no inlet whirl (V_1T = 0, the pure radial inlet the question specifies) reduces to $$gH=V_{B2}V_{2T}\qquad\Longrightarrow\qquad V_{2T}=\frac{gH}{V_{B2}}=\frac{9.81\times 40}{32.42}=\boxed{12.10\ \text{m/s}}$$ As a check on parts (a), (b) and (e) together, the page-22 hydraulic torque τ = ρQ(r₂V_2T − r₁V_1T) gives 1000(0.25)(0.180 × 12.10) = 544.6 N·m, identical to part (b) — the two routes through the problem agree exactly.
6. Part (f) — close the triangles and read the blade angles. The blade angle is the angle of the relative velocity to the tangential direction, so at inlet, where the absolute velocity is purely radial, $$\beta_1=\tan^{-1}\!\left(\frac{V_{1R}}{V_{B1}}\right)=\tan^{-1}\!\left(\frac{7.96}{18.01}\right) =\boxed{23.8^{\circ}}$$ and at outlet the relative tangential component is what the blade fails to impart, V_B2 − V_2T = 32.42 − 12.10 = 20.32 m/s, giving $$\beta_2=\tan^{-1}\!\left(\frac{V_{2R}}{V_{B2}-V_{2T}}\right) =\tan^{-1}\!\left(\frac{7.37}{20.32}\right)=\boxed{19.9^{\circ}}$$ For completeness the diagram also fixes the absolute discharge: V₂ = √(12.10² + 7.37²) = 14.17 m/s at α₂ = 31.3° to the tangential, and the relative velocities are 19.69 m/s at inlet and 21.61 m/s at outlet. Both blade angles are well under 90°, so the impeller is backward-curved — the standard choice, because it gives a falling head-capacity characteristic and a brake power that does not run away at high flow.