Question

Problem #3 A student performed a flow visualization experiment for the flow over an airfoil as shown below. Smoke was generated from a pipe upstream of the airfoil with holes spaced 1 cm apart. The free stream velocity is 2 m/s. Estimate the velocity at point A, then estimate the gauge pressure at the same point. Assume standard conditions. U = 2m/s Problem #4 For the same airfoil above the wake velocity distribution is measured downstream of the airfoil as shown below. Calculate the drag force acting on the airfoil. Assume all pressures atmospheric At the exit the velocity is given by uly) = v) (0.5U (1 + lyl/8). 0<lyl <8 Tyl 28 10.50

Answer 1

*. ribanc 43 as moving the Line Spacing huicm - 0.01m free stream velocity UX = 2 m/s moving the fluid line initially from Straight to zero ..ang vat point A. So the kinetic energy of Initiat condition is reduced due to Kinetic height Change - - - - CKEJA - CKEDx - mgh Hace usz - Uge - ghi Ця - це* - 24h -V22-(2x9:8140.01) LA = 1.95 m/s velocity at A UA - 1.95 m/s pressure at A = PA PA + Į PUA? = Px + 1 e Ua? p=1-225 kg/m3 Pa = 1 atm = 101.325 kPa PA- Pa = < P (Ux2 - 442) (PA- Pc) - 4x1.225 Ⓡ (22-1-952) (A-PX): 0.12 N/m2 Gauge pressure at A PGA = PA - atmospher PGA = PA - PA CS Scanned with PGA = 0.12 N/m2 CamScanner PGAPA - atmospheric pressure

→ Drag force is cre the Velocity Variation Ō to Torte is created due created at exit due to boundary layer → So consider a small strip dy in exit region mass flow through strip m exx ucy) dy - Drag forte is due to momentum change Drag D - momentum at inlet - Momentum at exit so elemental drag force dF = Pa ucy) x dy * ( u-ucys) df = Par ucy [u- ucy?] dy so total drag force i ex [Luxurys - ucy?] dy DE Px flux 0:54 (1 + 19!) - 0-25 4*(1 + 92)*]dig Do Perut 12 + zugu J ay – Pave (v + say'dy - Paya [ 24 * 4* Hotel so gutt 2 9 ) dry - Pqe [40] - popie x [ 9 to get the cs sem [ 40 - 20-29]

- Per a [20-20] D = PxUz • (-5) D = Px U²0 3 Scanned with CamScanner