Pump Performance: 22-Mec-A6 Fluid Machinery, December 2013, Q3

Question 3: Pump Performance (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. The radial-flow impeller geometry, speed and duty are as tabulated below.

Given data
D₁ / D₂	0.130 / 0.300 m	h₁ / h₂	0.020 / 0.010 m
β₁ / β₂	20° / 25°	Speed N	1750 rev/min
Flow Q	0.030 m³/s	Head H	35 m

Find. Blade, radial and whirl velocities at both stations; the shaft torque and power; and the hydraulic power and efficiency.

Inlet: radial water velocity V_r1 = 3.67 m/s, small pre-whirl V_w1 = 1.82 m/s.

Outlet: whirl V_w2 = 20.66 m/s drives the head and torque.

Approach. Blade speed $U=\omega r$; radial velocity from continuity over the cylindrical flow area $\pi D h$; whirl from the blade-angle triangle $\tan\beta=V_r/(U-V_w)$; torque from Euler’s moment-of-momentum, power $T\omega$; efficiency = water power / shaft power.

(a) Blade velocities. $\omega=2\pi\cdot 1750/60=183.3\ \text{rad/s}$: $$U_1=\omega r_1=183.3\cdot 0.065=11.9\ \text{m/s},\quad U_2=183.3\cdot 0.150=27.5\ \text{m/s}$$
(b) Radial velocities. Continuity over $A=\pi D h$: $$V_{r1}=\frac{Q}{\pi D_1 h_1}=\frac{0.030}{\pi(0.130)(0.020)}=3.67\ \text{m/s},\quad V_{r2}=\frac{0.030}{\pi(0.300)(0.010)}=3.18\ \text{m/s}$$
(c) Whirl velocities. From $\tan\beta=V_r/(U-V_w)$, i.e. $V_w=U-V_r/\tan\beta$: $$V_{w1}=11.9-\frac{3.67}{\tan20^\circ}=1.8\ \text{m/s},\quad V_{w2}=27.5-\frac{3.18}{\tan25^\circ}=20.7\ \text{m/s}$$
(d) Torque and power. Euler moment of momentum $$T=\rho Q(r_2V_{w2}-r_1V_{w1})=1000\cdot 0.030\,(0.150\cdot 20.7-0.065\cdot 1.8)=\boxed{89.4\ \text{N}\,\text{m}}$$$$P_{shaft}=T\omega=89.4\cdot 183.3=\boxed{16.4\ \text{kW}}$$
(e) Hydraulic power and efficiency. $$P_{hyd}=\rho g Q H=1000\cdot 9.81\cdot 0.030\cdot 35=10.3\ \text{kW}$$$$\eta=\frac{P_{hyd}}{P_{shaft}}=\frac{10.3}{16.4}=\boxed{62.9\%}$$

Question 3 results
Quantity	Inlet	Outlet
Blade velocity U	11.9 m/s	27.5 m/s
Radial velocity V_r	3.67 m/s	3.18 m/s
Whirl velocity V_w	1.8 m/s	20.7 m/s
Torque	89.4 N·m
Shaft power / hydraulic power	16.4 kW / 10.3 kW
Pump efficiency	62.9%

Topic: Centrifugal pump velocity triangles (Euler)

Key relations: $U=\omega r$; continuity $Q=\pi D h\,V_r$; $\tan\beta=V_r/(U-V_w)$; Euler torque $T=\rho Q(r_2V_{w2}-r_1V_{w1})$, $P=T\omega$; $\eta=\rho gQH/P$.

Why this works: All the shaft work enters the water as a change in moment of momentum, set by the whirl (tangential) velocities at inlet and outlet. The blade angles fix the relative-velocity direction, from which the whirl follows once the radial velocity is known from continuity. The gap between shaft power and the useful $\rho gQH$ is the hydraulic + volumetric + mechanical loss.

Common pitfall: Using the flow area as $\tfrac{\pi}{4}D^2$ (a pipe) instead of the impeller periphery $\pi D h$ badly mis-states the radial velocity; also, whirl is measured relative to the blade, not the casing.

Source: Fox & McDonald, Turbomachinery chapter (Euler turbomachine equation); Dixon & Hall §7 (centrifugal pumps).

Question 3 of 8: Pump Performance

Question 3: Pump Performance (10 marks)