Question

For the air wind tunnel shown Honeycomb Flow Straightener Anti-Turbulence Screens Silencer Fan Diffuser Test Section Inlet Assuming a circular cross section, the atmospheric pressure is 101.325 kPa, density of air is 1.2 kg/m?, and absolute viscosity of air is 1.825*10-S Ns/m? (assume there are no losses in the wind tunnel) Contraction ratio: 7 Diffuser length: 3 m Test section diameter = test section length = 70 cm a) When a static-pitot tube placed in the test section with a manometer (manometer fluid is water with a density of 1000 kg/m?), the height differential between the static and pitot measurement part is 12.4 cm Calculate the velocity and static pressure at the test section and b) If we measure pressures on the two surfaces of an airfoil using two pressure tabs where velocity is 0 m/s and 60 m's on two airfoil surfaces (assume inviscid and steady flow) respectively, at the test section of the wind tunnel, the pressure coefficients (C) calculated on these two surfaces are respectively c) If the required power for the fan is 5633 Watts, calculate the diameter at the exit of the diffuser and the diffuser expansion angle. d) If a flat plate is to be placed in the test section, what would be the maximum length of the flat plate to keep the boundary layer laminar throughout the flat plate

Answer 1

Pa = 101.325kle to fa = 1.2 kg/m² M = 1,825 105 None с 12.4 h = a h 12.4 Go :-) h = 103.20 m of water velocity = Jagh VQx9.41X/63.2 Velno - 44.99 m/s P * 2,3 + V? +2 P2 Ps 3:22 0 Pressun= fgh body 1000 X 4.11% 103.2 1012.392. klo

عج ) -1 تا 60 = ست Sg 2g و 4 کھرب 311, 1012 تعدا + و را F : پول كل اع = ا = ۲۷ 24 و ته P = 2024 34 kla P مع 20 %4 : 44 و2 و3 واط . با202 IP و یا 25 24 ha

areae of exit can be calculated by using formula

power = density*area*velocity^2