Centrifugal pump head, discharge and power

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:

$$u_2=\frac{\pi D_2N}{60}=\frac{\pi\times0.4\times1450}{60}=30.37\ \text{m/s}$$

        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.