Question

1. [20 pts] An indraft wind tunnel pulls air into an inlet, through a contraction, the test section, and the diffuser, as shown below. The test section has a circular cross section and it has a diameter of 0.5 m. Assume the wind tunnel is at sea level and the temperature and pressure of the atmosphere are given by standard atmospheric conditions. (a) If the test section is designed to have a velocity of 200 m/s, determine the Mach number, temperature and pressure in the test section. Is the pressure in the test section greater than or less than the surrounding atmospheric pressure? (b) Determine the mass flow rate through the wind tunnel. (c) The contraction between the inlet and the test section has an area ratio of 5. Determine the velocity, Mach number, temperature and pressure at the inlet of the wind tunnel. (d) The diffuser section downstream of the test section has area ratio of 3. At the exit of the diffuser is the fan that drives the wind tunnel. The air exiting the fan must be at atmospheric pressure, since the flow there is subsonic. Determine the pressure change across the fan. Inlet rcontraction diffuser ir exit Pe=Pa test section ----- 个个个个 KR ------

Answer 1

Sol D. Us Contraction sta Test section d₂ = 0.5m A₂ = I (0.5² = .0. 19625 In wird tunne our flow is considered to be incompressibe & inviscid from contraction to test section arra is decreasing And from test section to fan area is increasing We are assuming sea level conditions at inlet Po= 101325Pa T. - 288 k s- 1.22 kg/m3

Thise conditions are like we are getting air from reservoir whose stognation pressure conditions are those of sealand (a) vsung bernoulli theorem to get pressure in test section Po= Pe + I SV 101325 = Pe + +(1.22) (200) Pt=76925 Pa From enthalpy egh, we will find out. static temperation -To - Tat v e Toronto 288 = Tt + (200² 2x1005 2 x 1005 Tz - DIRES 268.099 K

So = M = Vt VRT My = 200 V1.4x287 x 268.00 My = 200 328.14 = 0.609 is less than Pressure in test section the atmospheric pressure b. mass flow rate - SAV - 1.22 x 0.196 X200 = 47.824 kglon Mass flow rate will be constant through out wind tunnel

(C) Giren Ai - 5 At Ai=5x 0.196 - 0.98 m² Using continuity eg" S.Ai Vi = SE AVE Since flow is incompressible Si= si Vi = At Vt - 1x200 = 40m/s Ai For inct conditions using bernoulli egas. Po-Pi + 1 svie 101325 = P + 4 (1-22) (40) Pů = 100349 Pa

Again, using enthalpy egh Toe Ti + Vi 20p 288 = Ti + (40) 2(1005) Ti= 287.20K Mach No. Mi - Vi = VERTi Maret vex 27x281.3 V 1.4x 287 x 287.20 Mi = 0.11 (d) Grisen Ad = 3 At Using conternity equ Ad Vd = At Ve vd = A + XVt = 200 - 66.6 m/s 3

ho Using Bernoulli for diffeuses section ede bel Por Potts Vol Po=101325 - 4x 1.22 x (66.62 Po=98619.3 Pa nou ath Pressure change across fan Pr = Paton - Pe Pq = 101325-98619.3 Plan = 2705.69 Pa 9059 W