Question

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max A small round object is tested in a 0.75-m diameter wind tunnel. The pressure is uniform across sections 1 and 2. The upstream pressure is 30 mm H2O (gage), the downstream pressure is 15 mm H20 (gage), and the mean air speed is 12.5 m/s. The velocity profile at section 2 is linear; it varies from zero at the tunnel centerline to a maximum at the tunnel wall. V Calculate a) the mass flow rate in the wind tunnel, [5 pts] b) the maximum velocity at section 2, [5 pts] c) the drag of the object and its supporting vane. [15 pts] Neglect viscous resistance at the tunnel wall. 1 2

Answer 1

assume

density of air = 1.225 kg/m^3

density of water = 1000 kg/m^3

a. mass flow rate of air = 6.765 kg/s

b. maximum velocity at section 2 = 18.75 m/s

c. drag of the object and its supporting vane = 54.44 N

Gn Page No :) Umaa v 0 0.75 m = 15 mm 130 (gage) unalream presani, hi 30 mm ho (gage) mean air relocity, Va = 12.5ms down abream preoane, ha Velocity profile raries from zero & at the tunnel centerline maximum eat the tunnel nall Neglect viacous reaistance at the tunnel mal to a

tunnel 2 A = 1/4 D 1/4 x 0.75 air m in = ing Sex X20 Yd Solution Page No: 2 (a) Man flow rate is the drinal m = SAVO = 0.4417865 m2 density of Calandard Condition Saiz ora 1.22587/m2 Vo = 12.5 m/s m = 1.225.% 0.44 17865 12.5 6. 765 kg)s Applying continuity equation at section 2 Em S Sair Ug dag Az Since the relocity profile varies linearly at section 2 dag Qor.do ń = S Sair o dag Ag R R R R m Soot * Vamaata R m = Ug = Vamaya R (Sair = constant) SA s Vaman? Xano da 0 SAir S 0 *r* si.) - 0

Page No: 3 R S re do R Se alle R3 -0 0 R3 3 3 3 ③ in ① 3 Xan R. X 3 in = Saio x Vamax R m = Sar xamax XOD. 3 in = 6.765 kg/s R D/2 = - 0.75 - 2 0.375 m Soir 1.225 18/03 ( standard condition) from ean mx 3 & max 2 6.765 x 3 1.225 x 0.3752 XD Shox R x 20 Va, max 7 An = 1000kg / m3 (exhandand condition = 18.75 m/s Concernation momentum principle betreen ④ 40 P, A timu,+ FR = PA + S U Sair Ua dag - FR- Reaction force P = lgh, Chi = 30 mm 30x10 m) of the object IN = 1000 X9.81 * 307103 of inahero density) = 294.300 Nm2 P2 Cha 15*3.) = 1000 x 9.0x 15 x16 = 147950 N/m 2 u, 18.5m/s g = 9.81 mis In ghz -1 mm -3

Page No 14 R S "o Jair "a dag = San S u² dag 0 २ = Paio S (Verea Xaordo 0 R 2 R R = Jair X V2 max t2 s 2² xr do R2 s 83 dr on la 0 S asa 4 S Sair ug dag = Sain x V 2 2 2mox x 80 x R4 Ag R? 2 Pair & Vernos art R 4 = 1.225* 18.75 X20% 0.3762 4 = 95.131 N P, е 2943007 0.44 17 8 65 130,0178 N - 147.1 5 0 % 0.4417865 65,00 89 egno Apply all natuer in 130,0178 + 6.765 x12.5 + FP = 66.0089 + 96.131 FR = 65.0089 196.131 - 130,0178 - 66.7651*12+5) FR = -54.44 N on the object is equal and the reaction force acting on the object Fdney FR - (54.44) opposite to = 54.44 N