Air Kinematic Viscosity V 1 57 10 4 Ft Sec Enters A Square Duct Through A Do 0 9 Ft Opening As Shown In The Fi 1 (131.86 KiB) Viewed 36 times
Air (kinematic viscosity, v = 1.57 × 10-4 ft²/sec) enters a square duct through a do = 0.9-ft opening as shown in the figure below. Because the boundary layer displacement thickness increases in the direction of the flow, it is necessary to increase the cross- sectional size of the duct if a constant U = 2.0 ft/sec velocity is to be maintained outside the boundary layer (i.e., in the inviscid core). (a) Determine an expression for the duct size, d, as a function of x if U is to remain constant. Round coefficients to 3 significant digits. (b) Use the expression to determine d for x = 1, 3, 5, 8 and 10 ft. Assume laminar flow. (Your expression can also be plotted from 0≤x≤ 10 ft as will be shown after you determine the expression and duct sizes for the given x-values.) U d(x) U -X
(a) d = (i where n = (b) For x = 1 ft d = For x = 3 ft d = For x = 5 ft d = For x = 8 ft d = For x = 10 ft_d= M MO + i ft ft ft ft ft •x") ft,
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