Develop an expression for the ratio of the rate of heat transfer to water at 40°C from a thin flat strip of width ?D/2 and length L at zero angle of attack and a tube of the same length and diameter D in cross-flow with its axis normal to the water flow in the Reynolds number range between 50 and 1000. Assume both surfaces are at 90°C. for air flowing over the same two surfaces in the Reynolds number range between 40,000 and 200,000. Neglect radiation.

Question

GIVEN
• Air flowing over a thin flat strip at zero angle of attack or a tube in crossflow
• Air temperature (T?) = 40°C
• Tube diameter = D
• Strip width = ?D/2
• Tube and strip length = L
• Reynolds number : 40,000 < Re < 200,000
FIND
• The ratio of the heat transfer from the strip and that from the cylinder. (qs/qt)
ASSUMPTIONS
• Radiative heat transfer is negligible
• Steady state for both cases
• The tube and strip temperatures (Ts) are 90°C
PROPERTIES AND CONSTANTS
Kinematic viscosity (?) = 17.6 × 10–6 m2/s
Prandtl number (Pr) = 0.71 at 90°C Prs = 0.71

Answer 1

This solution follows the same procedure as the solution to



For the tube, from



but Pr = Prs. The heat transfer rate from the strip will be 165% of that from the tube with the same

Reynolds number.