Answer to Question #268314 in Molecular Physics | Thermodynamics for dandan

Question

a centrifugal pump operating under steady flow conditions delivers, 3,050 kg/min of water from initial pressure of 83, 400 Pa to final pressure of 300,000 Pa. the diameter of the inlet pipe to the pump is 16cm and the diameter of discharge pipe is 10.16cm. what is the work

Answer 1

$m=3050 \; kg/min \\ p_1 = 83400 \; Pa \\ p_2 = 300000 \; Pa \\ D_1 = 16 \; cm \\ D_2 = 10.16 \; cm$

Mass flow rate

$v_1 = \frac{m}{ρA_1} = \frac{3050}{1000 \times \frac{3.14}{4}(0.16)^2} = 151.77 \; m/min = 2.529 \;m/s \\ v_2= \frac{m}{ρA_2} = \frac{3050}{1000 \times \frac{3.14}{4}(0.1016)^2} = 376.48 \; m/min = 6.274 \; m/s$

Heat difference

$H = \frac{v^2_2 -v^2_1}{2g} + \frac{p_2 -p_1}{ρg} \\ H = \frac{(6.274)^2 -(2.529)^2}{2 \times 9.81} + \frac{300000-83400}{1000 \times 9.81}\\ H = \frac{39.363 -6.395}{19.62} + \frac{216600}{9810} \\ H = 1.68 + 2.20 = 3.88 \;m$

The work done

$W = ρQgH \\ W = 1000 \times A_1v_1 \times 9.81 \times 3.88 \\ W = 1000 \times (\frac{\pi}{4} \times (0.16)^2 \times 2.529) \times 9.81 \times 3.88 \\ W = 1934.4 \;W$