Question

Honeycomb Flow Straightener - Anti-Turbulence Screens Silencer Fan Test Section Diffuser Iniet Assuming a circular cross section, the atmospheric pressure is 101.325 kPa, density of air is 1.2 kg/m², and absolute viscosity of air is 1.825*10% 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 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 (Cp) calculated on these two surfaces are and 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

Given all a) sman VT? Vi 1 2 air 1 ) Vi = Jagna wateret 2 - Path = 101.825 kea Sair: 2 kg CR = 7 D.L=8m 7-825X105 Ns/m2 dia) test seetiaus L7 = 70 cm 100rg im B De = 12.4cm . PT 9 Vi= fagn & haasm - H TE h = 12:4 (10.00 = 10320193 em 29 (103 20193)xiom P2=59h 20240967 44.999 m/s 1.289.81 x 103-2094 = 45m/s 1214.979 kq = 14 ms? m² - is stagnation point Pepee 1.2149 kpa V2-0. noul cp- pressure P-Pool & P2 P1= 1.2149 lepa Llovo ² - Pitl. 2149 lepa Po, Borto + free stream properties Po press at which cp to be measured. P2= 101.3257112149 as u = omp & U60 m/s V-0 4 Stagnation point → B = 102.549 kpa Pe 102.54 kpa ep- (102689 - 101.3282102 es Bello (eople 112149x109 *** 2 C45) 2 40199 0.6 x 452 at v bomps. P=9qh at 1-0 By Bernouis eq? ansb. Pat Vg² = P2 +U2² P +60² 39 29 39 2 39 ag ९१ 29 Pg 102-SUXIOS 602 870-183.18 = e.epaqe prepg 39 1,2x9.81 2x9.81 coefficient? Pi= path sep = 0.99 Xlos 2 to = 8527.02 Px – theepers 439 8527.02 eleostesports 100380.07 = 100* 380 kpa - 0.777 as op- p-po - (1000380-101.325) 103 I love? x 12 x 452 :: At Vooms + cp=-0.747 ansb. 56 33 ulatt diameter lexit # x 7ocm = 4qoem - 4.90 usian Power fan = diffuser expa angreso 210.ch tano=201 7 0.7 e 3m 2.1 ortant (017 0-340990 di d2 Fom 490cm as ARS C. Re= SUD Le = .=0.ofre for lamina& flow 112 x 45 x toxió? = 2074.23X10 et 1.825X165 Ns/m2 20.747105 are Les toxió? x 10.07) X (2074.23X10) - 101490.41 as for plate (flat) Re for laminar < 5x10s but Rey 5x105 test & turbulent flow in section n Scanned with CamScanner

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