Question 16.11: A centrifugal pump runs at 960 rpm. The outlet vane angle is......
A centrifugal pump runs at 960 rpm. The outlet vane angle is 45° and velocity of flow is 2.4 m/s at outlet. The discharge through the pump is 0.18 m³/s, when the pump works against a total head 18 m. If the manometric efficiency is 78%, determine the diameter and width of the impeller at outlet.
Given: N = 960 rpm, Φ = 45°, V_{f_{1}}= 2.4 m/s, Q = 0.18 m³/s, H_{m} = 18 m, and \eta_{mano} = 78% = 0.78
Considering outlet triangle diagram of Figure 16.14, we have
\tan\phi={\frac{V_{f_{1}}}{u_{1}-V_{w_{1}}}}\qquad{\mathrm{or}}\qquad\tan45^{\circ}={\frac{2.4}{u_{1}-V_{w_{1}}}}Thus, 1={\frac{2.4}{u_{1}-V_{\mathrm{w_{1}}}}}
V_{w_{1}} = u_{1} – 2.4
Using the equation of manometric efficiency, we have
\eta_{\mathrm{mano}}={\frac{g H_{m}}{V_{\mathrm{w}_{1}}u_{1}}}\qquad{\mathrm{or}}\qquad0.78={\frac{9.81\times18}{V_{\mathrm{w}_{1}}u_{1}}}or V_{w_{1}}u_{1}={\frac{9.81\times18}{0.78}}=226.38 (ii)
Substituting the value of V_{w_{1}} from Eq. (i) in Eq. (ii), we get
(u_{1}-2.4)u_{1}=226.38or u_{1}^{2}-2.4u_{1}-226.38=0 (iii)
Equation (iii) is a quadratic in u_{1}. Hence solving for u_{1}, we have
u_{1}={\frac{2.4\pm{\sqrt{\left(-2.4\right)^{2}-4\times1\times\left(-226.38\right)}}}{2\times1}}or u_{1}={\frac{2.4\pm35.85}{2}}
Considering the plus sign, we get
u_{1}={\frac{2.4+35.85}{2}}=19.125\,\mathrm{m/s}Now, diameter D_{1} of impeller at outlet, is obtained by equation
u_{1}={\frac{\pi D_{1}N}{60}}or 19.125={\frac{\pi D_{1}\times960}{60}}=50.265\ D_{1}
Thus, Diameter of impeller at outlet = D_{1}= 0.3805 m = 38.05 cm
Again, we have the discharge equation at outlet as
Hence, Width of impeller at outlet = B_{1}= 0.06277 m = 6.277 cm
