Question 16.16: A two-stage centrifugal pump with an impeller in series has ......

Given: D_{1}= 60 cm = 0.6 m, B_{1}= 3 cm = 0.03 m, Q = 0.08 m³/s, N = 600 rpm, Φ = 45° and \eta_{mano} = 80%

Now,              u_{1}={\frac{\pi D_{1}N}{60}}={\frac{\pi\times0.6\times600}{60}}=18.85\,{\mathrm{m/s}}

Again,          Q=\pi D_{1}B_{1}V_{f_{1}}

or            0.08=\pi\times0.6\times0.03\times V_{f_{1}}

Thus,          V_{f_{1}}=1.4147\;\mathrm{m/s}

From the velocity diagram at outlet of Figure 16.19, we have

or       1={\frac{18.85-V_{w_{1}}}{1.4147}}

or            1\times1.4147=18.85-V_{w_{1}}

Hence,        V_{w_{1}}=17.4353\ \mathrm{m/s}

Head generated by one impeller =  {\frac{V_{w_{1}}u_{1}}{g}}={\frac{17.4353\times18.85}{9.81}}=33.5\,\mathrm{m}

If H_{m}, is the manometric head developed by one impeller, we get

or        H_{m}=0.80\times33.5=26.8~{\mathrm{m}}

Since, as the pump is two-stage, we have
Total head generated H_{Total} = 2 × 26.8 m = 53.6 m