Question 16.12: A centrifugal pump has the following dimensions: Inlet diame......

Given: D= 15 cm = 0.15 m, D_{1}= 30 cm = 0.3 m, B = B_{1}= 4.8 cm = 0.048 m, θ = 0.5 rad = \left\lgroup{\frac{0.5}{\pi}}\right\rgroup \times180^{\circ},\;\phi=0.3\;{\mathrm{rad}}\left\lgroup{\frac{0.3}{\pi}}\right\rgroup \times180^{\circ}, and ω = 100 rad

Water enters from inlet to outlet without shock, i.e. the energy loss in the pump is zero.

{\frac{D}{2}}=r={\frac{0.15}{2\ cm}}=0.075\,{\mathrm{m}}

{\frac{D_{1}}{2}}=r_{1}={\frac{30}{2}}cm =0.15\,{\mathrm{m}}

It is required to determine Q and H_{m}. The required variables, vane angles etc. are shown in Figure 16.15.

The discharge equation is written as

The tangential velocity of the impeller at inlet is

={\frac{2\pi N}{60}}\times{\frac{D}{2}}=\omega r=100\times0.075=7.5\,{\mathrm{m/s}}

The tangential velocity of the impeller at outlet is:

={\frac{2\pi N}{60}}\times{\frac{D_{1}}{2}}=\omega r_{1}=100\times0.15=15\,{\mathrm{m/s}}

From the inlet diagram of Figure 16.15, we have

or            V_{f}=u\ \mathrm{tan}\,\theta=7.5\times\tan{\left\lgroup{\frac{0.5}{\pi}}\right\rgroup }\times180^{\circ}=4.097\,{\mathrm{m/s}}

Thus,  Discharge Q = πDBV_{f}

= π × 0.15 × 0.048 × 4.097 = 0.09267 m³/s

The loss in the pump is zero due to shockless entry, hence the head developed is given by

H_{m}={\frac{V_{\mathrm{m_{1}}}u_{1}}{g}}        (i)

From the outlet velocity diagram, we have

or            V_{w_{1}}=15-3.2328V_{f_{1}}          (ii)

Discharge      Q=\pi D_{1}B_{1}V_{f_{1}}

or         0.09267 = π × 0.3 × 0.048 V_{f_{1}}
Hence,                V_{f_{1}}= 2.048 m/s
Substitute V_{f_{1}} = 2.048 m/s in Eq. (ii) to get V_{w_{1}} as
V_{w_{1}} = 15 – 3.2328 × 2.048 = 8.38 m/s

Substituting V_{w_{1}} in Eq. (i) to get H_{m} as:

H_{m}=\frac{V_{\mathrm{w_{1}}}u_{1}}{g}

={\frac{8.38\times15}{9.81}}=12.813\,{\mathrm{m}}