Question 16.9: Prove that for a centrifugal pump running at speed N rpm giv......
Prove that for a centrifugal pump running at speed N rpm giving discharge Q, manometric head H_{m} is expressed in the form:
H_{m} = AN²+ BNQ + CQ², where A, B, and C are constants.
Manometric head (H_{m}) is the head against which pump has to work. Hence, considering no loss was to occur, the manometric head with usual notations of Figure 16.12 is:
H_{m}={\frac{V_{\mathrm{m}_{1}}u_{1}}{g}}But some loss occurs in the impeller and in the casing. Hence, manometric head is:
H_{m}={\frac{V_{w_{1}}u_{1}}{g}} – Loss of heads in the impeller and casing
or H_{m}=\frac{V_{w_{1}}u_{1}}{g}-K\frac{V_{1}^{2}}{2g}=\frac{V_{w_{1}}u_{1}}{g}-K(V_{w_{1}}^{2}+V_{f_{1}}^{2})
or H_{m}=\frac{V_{w_{1}}}{g}-\frac{K}{2g}(V_{w_{1}}^{2}+V_{f_{1}}^{2}) (i)
Peripheral velocity u_{1}={\frac{\pi D_{1}N}{60}}=K_{1}N
where K_{1} is a constant.
\mathrm{Discharge}\ Q_{1}=\pi B_{1}D_{1}V_{f_{1}}or V_{f_{1}}={\frac{1}{\pi B_{1}D_{1}}}Q=K_{2}Q
where K_{2} is a constant
from outlet diagram of centrifugal pump in Figure 16.12, we have
V_{w_{1}}=K_{1}N-K_{2}Q\;\mathrm{cot}\phior V_{{w}_{1}}=K_{1}N-(K_{2}\cot\phi)Q
or V_{w_{1}}=K_{1}N-K_{3}Q,{\mathrm{~where~}}K_{3}{\mathrm{~is~a~constant}}.
Substituting the values of V_{w_{1}} and u_{1} in Eq. (i), we have
{H}_{m}=\frac{(K_{1}N-K_{3}Q)\times K_{1}N}{g}-\frac{K}{2g}[(K_{1}N-K_{3}Q)^{2}+(K_{2}Q^{2})]or { H}_{m}=\frac{(K_{1}^{2}N^{2}-K_{1}K_{3}N Q)}{g}-\frac{1}{2g}[(K K_{1}N-K K_{3}Q)^{2}+(K K_{2}Q^{2})]
Further simplifying by grouping the constants of N², NQ and Q², it is written as:
H_{m} = AN²+ BNQ + CQ², where A, B and C are constants.
Hence shown.
