The figure shows a section of a long tube that narrows near its open end to a di

Question

The figure shows a section of a long tube that narrows near its open end to a diameter of 1.0 mm. Water at 20 degrees Celsius flows out of the open end at 0.020 L/s. What is the gauge pressure at point P, where the diameter is 4.0 mm? [Hint: Be sure to account for everything that requires a pressure gradient...This problem integrates what you learned about viscous fluids with what you learned about ideal fluids.] The answer is not 8.2 x 10^5; it says ''In addition to the pressure change in the narrow part of the tube, you also need to include the additional pressure change that occurs when the tube widens.'' But I'm not sure. Thank you!

Explanation / Answer

volume rate flow , V/t = 0.02 L/s                                            = (0.02*10-3) m3/s                                            = 0.02*10-3 m3/s according to equation of continuity ,         volume rate flow , V/t = A1v1 = A2v2 ............. (1) diameter of the pipe at open end is                          d1 =1 mm = 1*10-3 m then radius of the pipe , r1 = d1/2 = 0.5*10-3 m crosssectional area A1 = r12                                   = 0.785*10-6 m2 diameter of the pipe at point 'P' is                          d2 =4 mm = 4*10-3 m then radius of the pipe , r2 = d2/2 = 2*10-3 m diameter of the pipe at point 'P' is                          d2 =4 mm = 4*10-3 m then radius of the pipe , r2 = d2/2 = 2*10-3 m diameter of the pipe at point 'P' is                          d2 =4 mm = 4*10-3 m then radius of the pipe , r2 = d2/2 = 2*10-3 m crosssectional area of the pipe at point 'P' is                               A2 = r22                                    = 12.56*10-6 m2                                    = 12.56*10-6 m2 ........................................................................... from eq (1),            V/t = A1v1         (0.02*10-3) = (0.785*10-6)(v1)               v1 = 25.47 m/s from eq (1),             V/t = A2v2            (0.02*10-3) = (12.56*10-6)(v1)                   v2 = 1.59 m/s ................................................................................ from eq (1),            V/t = A1v1         (0.02*10-3) = (0.785*10-6)(v1)               v1 = 25.47 m/s from eq (1),             V/t = A2v2            (0.02*10-3) = (12.56*10-6)(v1)                   v2 = 1.59 m/s ................................................................................ from eq (1),             V/t = A2v2            (0.02*10-3) = (12.56*10-6)(v1)                   v2 = 1.59 m/s ................................................................................ from Bernoulli's Equation ,               P1 + (1/2) v12 = P2 + (1/2) v22 Gauge Pressure is              P2 - P1 = (1/2) (v12 - v22) where , density of water = 1000 kg/m3 where , density of water = 1000 kg/m3 substitute the given data in above eq, we get           P2 - P1 = 3.23*105 Pa

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