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In a Nut Shell: External torque drives the impeller of a centrifugal pump resulting in a change

of angular momentum of the fluid across the impeller and is key to the analysis of centrifugal

pumps.

Centrifugal pumps produce rotary motion of an impeller pushing fluid across vanes.

Consequently cylindrical coordinates are handy in the analysis of fluid motion.

Recall r = r e_r+ z k where r is the position vector to a fluid particle

Now V = V_r e_r + V_t e_t + V_z k Note: V_r is the radial component of absolute velocity

where V is the absolute velocity of fluid particle in cylindrical components

So the cross product is: r x V = (˗ z V_t ) e_r + ( z V_r ˗ r V_z ) e_t + ( r V_t ) k

For pumps and turbines the important term is the z-component since it relates to axial torque.

˗ ∫ (r x V) ρ V . n dS + ∫ (r x V) ρ V . n dS = Σ T (Use z-component of the

cs in cs out angular momentum only.)

and from conservation of mass: ∫ ρ V . n dS = ρ Q = ρ A₁V_r1 = ρ A₂ V_r2

Thus for steady, uniform flow the principle of angular impulse and angular momentum for

pumps simplifies to: let 1 denote the entrance and 2 denote the exit on the vanes of the impeller

ρ Q [ r₂ V_t2 ˗ r₁ V_t1 ] = T_z

where ρ Q is the mass flow rate crossing the control surface (It will be the same

entering and leaving the control surface.) ρ Q = dm/dt

r₂ V_t2 where r₂ is the radial distance on the vane at the exit and V_t2 is the tangential

component of the absolute fluid velocity of the fluid particle at the exit

r₁ V_t1 where r₁ is the radial distance on the vane at the entrance and V_t1 is the tangential

component of the absolute fluid velocity of the fluid particle at the entrance

T_z is the net torque driving the pump Click here to continue discussion.