Question 16.13: A centrifugal pump is placed at height 3 m above the sump wa......
A centrifugal pump is placed at height 3 m above the sump water level, and the outlet of the delivery pipe is 8 m above the sump level. The vane angle at outlet is 45°. The velocity of flow through the impeller is constant at 1.6 m/s. Determine the pressure heads at inlet and outlet to the wheel. Assume velocity of water in the pipe is equal to the velocity of flow through the impeller. Neglect losses.
Given: H_{s} = 3m, H_{s} + H_{d} = H = 8 m, Φ = 45°, V_{f} = V_{f_{1}} = 1.6 m/s, and V_{s} = V_{d} = V_{f} = 1.6 m/s
All the losses are neglected.
To find the pressure heads at inlet, i.e. in CC level, we apply Bernoulli’s equation at AA (sump level) and at CC, i.e. centre or inlet [Figure 16.16 (a)],
Z_{A}+{\frac{p_{A}}{\rho g}}+{\frac{V_{A}^{2}}{2g}}=Z_{\mathrm{inlet}}+{\frac{p_{\mathrm{inlet}}}{\rho g}}+{\frac{V_{\mathrm{inlet}}^{2}}{2g}}+\mathrm{Losses} (i)
Here, Z_{A} = 0 as datum level,
\frac{p_{A}}{\rho g} = Atmospheric pressure head
Velocity at sump level = 0
Losses = 0
Hence, Eq. (i) may be written as
or 10.3=3+{\frac{p_{\mathrm{inlet}}}{\rho g}}+0.13
Thus, Pressure head at inlet = \frac{p_{\mathrm{inlet}}}{\rho g}
= 10.3 – 3.13 = 7.17 m of water absolute
= -3.13 m (Vacuum)
Considering outlet diagram in Figure 16.16(b), we get
or V_{w_{1}}=u_{1}-1.6\;\mathrm{cot}\phi=u_{1}-1.6\;\times1=(u_{1}-1.6)\;\mathrm{m/s} (ii)
Now, Energy head at outlet-energy head at inlet = \frac{V_{w_{1}}u_{1}}{g} as loss is neglected
Again, Energy head at outlet = Energy head at delivery point D
=8+{\frac{(1.6)^{2}}{2\times9.81}}=8.18\,{\mathrm{m\ of~water}}Thus, {\frac{V_{\mathrm{w_{1}}}u_{1}}{g}}\,=\,8.18\,\operatorname{m}\quad{\mathrm{or}}\quad{\frac{(u_{1}-1.6)u_{1}}{9.81}}=8.18\,{\mathrm{m}}
or u_{1}^{2}-1.6u_{1}-80.245=0 (iii)
Solving the quadratic equation [Eq. (iii)] in u_{1}, we get
u_{1}={\frac{1.6\pm{\sqrt{\left(-1.6\right)^{2}-4\times1\times\left(-80.245\right)}}}{2\times1}}\quad{\mathrm{or}}\quad u_{1}={\frac{1.6\pm17.987}{2}}Hence, u_{1} = 9.793 m/s
Therefore, V_{w_{1}} = 9.793 – 1.6 = 7.183 m/s
From the outlet velocity diagram in Figure 16.16(b), we have
We have {\frac{V_{1}^{2}}{2g}}={\frac{(7.36)^{2}}{2\times9.81}}=2.761\,{\mathrm{m}}
Again, Energy head at outlet-Energy head at inlet = {\frac{V_{w_{1}}u_{1}}{g}}\,=\,8.18\,\,{\mathrm{m}}
\left[{\frac{p_{\mathrm{outlet}}}{\rho g}}+{\frac{V_{\mathrm{1}}^{2}}{2g}}\right]-\left[{\frac{p_{\mathrm{inlet}}}{\rho g}}+{\frac{V^{2}}{2g}}\right]=8.18~\mathrm{m}or {\Bigg[}{\frac{p_{\mathrm{outlet}}}{\rho g}}+2.761{\Bigg]}-[-3.13+0.13]=8.18
or {\frac{\rho_{\mathrm{outlet}}}{\rho g}}=8.18-5.76=2.42\operatorname{m \ of~water}
