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.

Question

GIVEN

• Water flowing over a thin flat strip at zero angle of attack or a tube in crossflow

• Water temperature (T?) = 40°C

• Tube diameter = D

• Strip width = ?D/2

• Tube and strip length = L

• Reynolds number: 50 < Re < 1000

FIND

• The ratio of the heat transfer from the strip and that from the cylinder. (qs/qt)

ASSUMPTIONS

• Steady state for both cases

• The tube and strip temperatures (Ts) are 90°C SKETCH

PROPERTIES AND CONSTANTS

for water at 40°C: Prandtl number (Pr) = 4.3 At the surface temperature of 90°C: Prs = 1.94

SKETCH

Answer 1

Note that the heat transfer area (?D) is the same in both cases. Thin Strip: The flow over the thin strip is laminar for the Reynolds number given. The Nusselt number is given by



Tube:

The Nusselt number for the tube is given by



Since the transfer areas and temperature differences are the same, the ratio of the rates of heat transfer

is equal to the ratio of the heat transfer coefficients. The heat transfer rate from the strip will be 64% of

that from the tube with the same Reynolds number.