Question 1.2: A centrifugal pump delivers 2.5 m³/s under a head of 14 m an......
A centrifugal pump delivers 2.5 m³/s under a head of 14 m and running at a speed of 2010 rpm. The impeller diameter of the pump is 125 mm. If a 104 mm diameter impeller is fitted and the pump runs at a speed of 2210 rpm, what is the volume rate? Determine also the new pump head.
First of all, let us assume that dynamic similarity exists between the two pumps. Equating the flow coefficients, we get [Eq. (1.3)]
\prod{}_1 = N^{-1}D^{-3}ρ^0Q = \frac{Q} {ND^3} (1.3)
\frac{Q_1}{ N_1D^3_1} = \frac{Q_2} {N_2D^3_2} \text{or} \frac{2.5} {2010 × (0.125)^3} = \frac{Q_2} {2210 × (0.104)^3}
Solving the above equation, the volume flow rate of the second pump is
Q_2= \frac{2.5 × 2210 ×(0.104)^3} {2010 × (0.125)^3} = 1.58 m³/s
Now, equating head coefficients for both cases gives [Eq. (1.9)]
\psi = H/ (U^2 /g )= gH/ (π^2N^2D^2) (1.9)
gH_1/N^2_1D^2_1 = gH_2/N_2^2D_2^2
Substituting the given values,
\frac{9.81 × 14} {(2010 × 125)^2} = \frac{9.81 × H_2}{ (2210 × 104)^2}
Therefore, H_2 = 11.72 m of water.