Page: CENTRI FUGAL PUMP
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Work Done: • In case of centrifugal pump, work is done by impeller on the water.

• Velocity triangles are drawn at inlet and outlet of the impeller in the same way as that of turbine, so as to obtain an expression for the work done by impeller on water.

• The water enters the impeller radically at inlet for bet efficiency of the pump, which means the absolute velocity of water at inlet makes an angle of 90 degree with the direction of motion of the impeller at inlet.

Hence a = $90_o$ and $V_{w1}$ = 0

Let N = Speed of impeller in r.p.m

$D_1$ = Diameter of impeller at inlet,

$u_1$ = Tangential velocity of impeller at inlet.

$= \frac{\pi D_1 N}{60}$

$D_2$ = Diameter of impeller at outlet,

$u_2$ = Tangential velocity of impeller at outlet.

$= \frac{\pi D_2 N}{60}$

$V_1$ = Absolute velocity of water at inlet.

$V_{r1}$ = Relative velocity of water at inlet.

a = Angle made by (v1) at inlet with direction of motion of wave.

θ = Angle made by relative velocity (vr1) at inlet.

As mentioned above, a = 90 degree, $V_{w1}$ = 0.

$\therefore$ work done by water on the runner per second per unit weight of the water striking per second is given by,

$\frac{1}{g}$ $[v_{w1} u1 - v_{w2} u2]$

$\therefore$ work done by the impeller on the water per second per unit weight of water striking per second.

$= [\frac{1}{g} (v_{w1} u1 - v_{w2} u2)]$

$= \frac{1}{g} [v_{w1} u2 - v_{w1} u1 ]$

$= \frac{1}{g} v_{w1} u_2$ ------(1)

work done by impeller on water per second,

$= \frac{w}{g} \times v_{w2} u_2$

where, w = weight of water

= p x g x q

Q = Volume of water

Q = $\pi D_1 B_1 \times vf_1$

and

Q = $\pi D_2 B_2 \times vf_2$

where B1 and B2 are width of impeller at inlet and outlet and $v_{f1}$ and $v_{f2}$ are velocities of flow at inlet and outlet.

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 modified 2 days ago by Abhishek Tiwari ★ 50 written 9 weeks ago by RB ♦♦ 100