Water flows through a duct of square cross section as shown in Fig. P4.62 with a

Question

Water flows through a duct of square cross section as shown in Fig. P4.62 with a constant, uniform velocity of V = 19 m/s. Consider fluid particles that lie along line A-B at time t = 0. The position of these particles when t = 0.22 s is denoted by line A'-B'. Use the volume of fluid in the region between lines A-B and A'-B' to determine the flowrate in the duct.

Figure P4.62

Water flows through a duct of square cross section as shown in Fig. P4.62 with a constant, uniform velocity of V = 19 m/s. Consider fluid particles that lie along line A-B at time t = 0. The position of these particles when t = 0.22 s is denoted by line A'-B'. Use the volume of fluid in the region between lines A-B and A'-B' to determine the flowrate in the duct. Figure P4.62 Q =

Explanation / Answer

Since V is constant in time and space,

Length = (V)*(t) = 19 m/s * 0.22 s = 4.18m

Thus the volume comes out to be (0.5)^2 * 4.18 = 1.045 m^3

Q = Vab/t = 1.045/0.19 = 5.5

Hence the value of Q will be 5.5 regardless of which line we consider since flow is constant

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