Water is to be pumped from the large tank shown in Fig. P5.119 with an exit velocity of \(6 \mathrm{~m} / \mathrm{s}\). It was determined that the original pump (pump 1) that supplies \(1 \mathrm{~kW}\) of power to the water did not produce the desired velocity. Hence, it is proposed that an additional pump (pump 2) be installed as indicated to increase the flowrate to the desired value. How much power must pump 2 add to the water? The head loss for this flow is \(h_{L}=250 Q^{2}\), where \(h_{L}\) is in \(\mathrm{m}\) when \(Q\) is in \(\mathrm{m}^{3} / \mathrm{s}\).

Figure P5.119

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V = 6 m/s Nozzle area = 0.01 m Pump Pipe area = 0.02 m Pump #1 2 m 1 #2