Question 3.18: Figure 3.15 shows a schematic of a centrifugal pump. The flu...

Part (a)

The control volume is chosen to be the annular region between sections 1 and 2 where the flow passes through the pump blades (see Fig. 3.15). The flow is steady and assumed incompressible. The contribution of pressure to the torque about axis O is zero since the pressure forces at 1 and 2 act radially through O. Equation (3.59) becomes

\sum{M_0} = \frac{\partial}{\partial t}\left[\int_{CV}{(r \times V)\rho  d^\circ \mathcal{V}} \right] + \int_{CS}{(r \times V)\rho(V \cdot n)  dA}             (3.59)

\sum{M_o} = T_o = (r_2 \times V_2)\dot{m}_{out} – (r_1 \times V_1)\dot{m}_{in}             (1)

where steady flow continuity tells us that

The cross product r × V is found to be clockwise about O at both sections:

r_2 \times V_2 = r_2V_{t2} \sin 90^\circ k = r_2V_{t2}k      clockwise

r_1 \times V_1 = r_1V_{t1}k      clockwise

Equation (1) thus becomes the desired formula for torque:

T_o = \rho Q (r_2V_{t2} – r_1V_{t1})k     clockwise Ans. (a) (2a)

This relation is called Eulers’ turbine formula . In an idealized pump, the inlet and outlet tangential velocities would match the blade rotational speeds V_{t1} = \omega r_1  and  V_{t2} = \omega r_2. Then the formula for torque supplied becomes

T_o = \rho Q \omega (r_2^2 – r^2_1)       clockwise               (2b)

Part (b)

Convert ω to 600(2π/60) = 62.8 rad/s. The normal velocities are not needed here but follow from the flow rate

V_{n1} = \frac{Q}{2\pi r_1 b} = \frac{2.5  m^3/s}{2\pi (0.2  m) (0.15  m)} = 13.3 m/s

V_{n2} = \frac{Q}{2\pi r_2 b} = \frac{2.5}{2\pi (0.5)(0.15)} = 5.3 m/s

For the idealized inlet and outlet, tangential velocity equals tip speed:

V_{t1} = \omega r_1 = (62.8 rad/s)(0.2 m) = 12.6 m/s

V_{t2} = \omega r_2 = 62.8(0.5) = 31.4 m/s

Equation (2a) predicts the required torque to be

T_o = (1000  kg/m^3)(2.5  m^3/s) [(0.5  m)(31.4  m/s) – (0.2  m)(12.6  m/s)] = 33,000 (kg \cdot m^2)/s^2 = 33,000  N \cdot m                  Ans.

The power required is

P = ωT_o = (62.8 rad/s)(33,000 N · m) = 2,070,000 (N · m)/s = 2.07 MW (2780 hp)                      Ans.

In actual practice the tangential velocities are considerably less than the impeller-tip speeds, and the design power requirements for this pump may be only 1 MW or less.