Question

Air at a temperature of 300 K flows over one side of a flat plate of width 1 m at a velocity of 20 m/s. The plate has a constant surface temperature of 350 K. Assume Re(x,c)=5x10^5.

a) What is the velocity boundary layer thickness at the end of the plate if L=0.25 m? What if L=1 m?

b) Calculate the drag on the plate if L=0.25 m. What is the drag if L=1 m?

c) Find the heat transfer rate from the plate to the air if L=0.25 m. What is the heat transfer rate if L=1 m.

d) A small diameter wire is placed at the leading edge of the flat plate. This wire trips the flow, causing it to be turbulent over the entire length of the plate. Repeat the calculations performed in parts a), b), and c) for this condition.

Answer 1

Given: Temperature of flowing air To = 300 K Temperature of plate Tp = 350 K Velocity of air flow of flat plate Vair = 20 m Width of plate is w = 1 m Calculate Reynolds number at L = 0.25 m and L=1 m Properties of air at film temperati To +T * 300+ 350 =325 K re T = 2 Kinematic viscosity v=1.807x10-3 n Prandtl number Pr = 0.701 Density of air p=1.086 kg Thermal conductivity k = 0.028 W /mK

Rez-035 = Vair L - 20x0.25 <= 2.76x10 - = Rea VaivL20x1 V 1.807x10-5 -5 = 11.06x103 At L=0.25 m, Reynolds number is less that critical value Rec = 5x10°, so flow is laminar. At L=1 m, Reynolds number is greater that critical value Rec = 5x10°, so flow is turbulent. (a) Calculate velocity boundary layer thickness at L=0.25 m and L=1 m. 5L 5x0.25 OL=0.25 = 7 - = 2.37x10- m Re, V2.76x103 81=0.25 = 2.37 mm

0.38L 0.38x1 7 = 0.023 m | = = 0.382 Ref (11.06x10°)* Oza1 = 23 mm (b) Calculate drag on the plate if L=0.25 m and L=1 m. At L=0.25 m, flow is laminar. (0,)-08=1.33 Re*% =1.33x(2.76x109) * = 0.0025 At L=1 m, flow is turbulent. Check length of plate where flow becomes critical. Rec = 510* = Px

20x. 5x10ʻ =- 1.807x10-5 x = 0.451 m 7. – 9.451 – 0,451 5, mixed boundary layer condition exists. Re So, for mixed boundary layer condition: (Cr), = 0.074 Re-*_1742 (Cr),4 = 0.074(11.06x10%)*__1742 11.06x105 (Cr)4 = 0.003 Drag force for L=0.25 m length of plate.

Ep =(Cs)-ox *324,73 Fp = x= 0.0025x2x1.086x(0.25x1)x 20° F) = 0.135 N Drag force for L=1 m length of plate. Fr= C, ).mx5p4,Vä, Fp = 0.003x5X1.086x(1x1)x202 Fp = 0.6516 N

(c) Calculate heat transfer from the plate at L=0.25 m and L=1 m. For plate length L=0.25 m, flow is laminar. Average Convective heat transfer coefficient for laminar flow. NUZ=0.25 = 0.664 Re%2 Pr%3 NU L-028 = 0.664(2.76x10°) (0.701) Nu 2-0.25 = 309.9 H2+0.25 0.25 – 309.9 0.028 hZ+0.25 = 34.71 W/m2K

For plate length L=1 m, mixed boundary layer condition exists. Average convective heat transfer coefficient for mixed boundary layer. Nu a = (0.037 Re%–871)x Pa% Niu, 4 = (0.037(11.06x10*)% – 871)x0.701% Nu 2-1 = 1474.18 1,42 = 1474.18 0.028 = 1474.18 |974 = 41.22 W m2 Kb

Heat transfer for plate length L=0.25 m. 22-0.25 = H20.25A,(T. -T.) Ql+0.25 = 34.71(0.25x1)(350 – 300) Ql=0.25 = 433.875 W Heat transfer for plate length L=1 m. Q==h44, (Tp-T.) 2z= = 41.22(1x1)(350 – 300) Qz=1 = 2061 W (d) As per question turbulent boundary layer exist over entire plate length. Velocity boundary layer at L=0.25 m, corresponding to Reynolds number for turbulent flow Re = 5x105.

0.38L 0.38x0.25 F = 0.0068 m turb ) L=0.25 11-025 Re's (5x109) (Sterb )2-0.25 = 6.8 mm Velocity boundary layer at L=1m, Reynolds number will be same as before Rez-1 = 11.06x10". 0.38L 0.38x1 SL= 17 Reza Rey (uroa (11.06 xlos y=0.023 m 8ja = 23 mm Calculate drag force for turbulent flow over entire plate. At L=0.25 m, flow is turbulent. (C)2-025 = 0.074 Re-** = 0.074x(5x10º)** = 0.0053 At L=l m, flow is fully turbulent. Reynolds number will be same as before

(C:)., = 0.074 Re * = 0.074x(11.06x109)*** = 0.0045 Drag force at x=0.25 m length of plate. (Fo],028 =(,)x xz p4,Vidal (Fo )1028 = 0.0053x5x1.086x (0.25x1)x 202 (Fb ),-0.25 = 0.287 N

r=1 (F) = (Cs)->p4,V (FD),sa = 0.0045*5*1.086x(1x1)x202 (FD)== 0.977 N = Convective heat transfer coefficient for turbulent flow(as per question). At x=0.25 m NU -0.25 = 0.037 Re%s Pr% NU L-025 = 0.037(5x109)*x0.701% Nu 2-0.25 =1191.11 hq=0.25 0.25 = 1191.11

ha=0.25 +0.25 =1191.11 0.028 hx=0.25 = 133.404 W/ /m²K At x=1 m, Reynolds number will be same as before Rez=1 =11.06x109. Nu 4 = 0.037 Re%s Pro Nu -4 = 0.037(11.06x109)% *0.701% Nu 1-1 = 2247.91 ht=2×1 – 2247.91 hx-x1 – 2247.91 ht=1 = 62.94 W/ ./m²K

Heat transfer for plate length L=0.25 m. Q2-0.25 = HZ-0.25A, (T) –T.) Q2-0.25 = 133.404x(0.25x1)(350 – 300) QL=0.25 = 1667.55 W Heat transfer for plate length L=1 m. 2z= = h,=4p(T. -T.) Qla1 = 62.94x(1x1)(350–300) Q2-1 = 3147 W