Question

An engineer is measuring the lift and drag on an aerofoil section mounted in a square wind tunnel. The wind tunnel is 0.5 m high and 0.5 m deep into the screen). The upstream air velocity is uniform at 25 m/s. The downstream velocity is uniform at 28.75 m/s in the lower half of the wind tunnel. The downstream velocity in the upper half is uniform. The vertical component of velocity is zero at the beginning and end of the test section. The test section is 1 m long. The engineer measures the pressure distribution in the tunnel along the upper and lower walls and finds Pupper = 90 - 50.-20.(1 - 2) (Pa, gauge) Power = 90 - 50x + 20.0(1 - 2) (Pa, gauge) where x is the distance in metres measured from the beginning of the test section. The air density is constant at 1.2 kg/m? Find the lift and drag forces acting on the aerofoil. Neglect shear on the walls of the wind tunnel. Beginning End of Test section Test section of 0.25 m 25 m/s 0.25 m 28.75 m/s 1 m

Answer 1

Answer :- given data wind tunnel high = 0.5m width 0.5m nu ii potjob Uo (upstream Velocity) = 25 miseen Vertical velocity o milseed test length (L) = 1m.. of Shear stress=0 stol &- at m Papper = 90-500- 200 (1-x) [pa, gauge] - Single 20x Pilower = 90- 50x+20x (1-x) [pa, gauge] fett - 2 uxoso sair = 1.2 kg/m3 test tes 3A Beginning end 2 pupper 0,25/m é v=2 15 misec of 28.75msee wer rso = uper stream velouty at enol 2 = lower stream velocity at and y lett force Flast wo Due to uper & lower pressure difference voltood of

feeft = Corda CCCCCC AP = Plower - Pupper AP = go-soutzon (1-2) - (90 - 500 - zon(1-2) pe ДР : чох (1-x) Pa dA = bxda (1-xx bda 0.5m) flett = yox 0.5 x 2(1-x) da 0 (L=im) fleft = 20 [ ] feept = 20 [ 1 - 1 / 3 feeft = 20 N flest = 3.33N Drag force by apply transport theorem tot i =inlet e cii Reporolds =exit pH = 25misee or te no 125misel 6 = 28.75misee 0.25 Be Control Volume

momentum form of Reynold theorm :- For SS S ÚC Treni dih) Velocity is uniform Cross section across the fo=SS SV²dA halos lahan to = -8 to Ai + 9V²A, 9²ke a Aiz Inlet Area = 0.5X0.5 = 4 m2 Ag = Alz= outlet Aread uper Streum = 0.25x0.5 = 12 m2 outlet Area lower stream = 0.25x0.5= 1 m2 fo = -1,2 X25)*4 + 1.2x625)*£ + 1.28728.45) xk fo = 1.[-6253* + (09)? + (20,75)*7 to, = 12 + 1 [ (28.75)² – ( 2 592 ] If = 30.234 N7 Note : Drag force force applied direction of flow left force » perpendicular of flow. in the to motion