Question #268314

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


1
Expert's answer
2021-11-18T15:08:55-0500

m=3050  kg/minp1=83400  Pap2=300000  PaD1=16  cmD2=10.16  cmm=3050 \; kg/min \\ p_1 = 83400 \; Pa \\ p_2 = 300000 \; Pa \\ D_1 = 16 \; cm \\ D_2 = 10.16 \; cm

Mass flow rate

v1=mρA1=30501000×3.144(0.16)2=151.77  m/min=2.529  m/sv2=mρA2=30501000×3.144(0.1016)2=376.48  m/min=6.274  m/sv_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=v22v122g+p2p1ρgH=(6.274)2(2.529)22×9.81+300000834001000×9.81H=39.3636.39519.62+2166009810H=1.68+2.20=3.88  mH = \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=ρQgHW=1000×A1v1×9.81×3.88W=1000×(π4×(0.16)2×2.529)×9.81×3.88W=1934.4  WW = 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


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