Hussaini Muhammad Your approach to calculating the wall heat flux appears to be correct. The formula you used, which is wall heat flux equals heat flux divided by the area, is appropriate. However, it seems there might have been a minor error in your calculation. Let's go through the calculation step by step to ensure accuracy:

Given data: Heat flux (Q) = 172.87 W Height of the wall (h) = 200 mm = 0.2 m Outer diameter (D) = 22 mm = 0.022 m Inner diameter (d) = 20.2 mm = 0.0202 m

Area of the wall (A) = π * (D/2)^2 - π * (d/2)^2 (assuming the wall is cylindrical) = π * (0.011^2 - 0.0101^2) = π * (0.000121 - 0.00010201) = π * 0.00001899 = 0.059566 mm^2 = 0.000059566 m^2

Now, apply the formula:

Wall heat flux = Q / A = 172.87 W / 0.000059566 m^2 = 2,902,340.97 W/m^2

So, the wall heat flux would be approximately 2,902,340.97 W/m^2, which is significantly higher than your previous calculation. Double-check your calculations using the corrected area value, and make sure all units are consistent (in this case, meters). If you encounter any further issues or require additional assistance, please feel free to ask.

Wow!, I even thought the value of wall heat flux I found is so large. Actually, in the paper, 2-D simulation were run, so the wall is rectangular in shape and you know an area of rectangle is length and breadth. But the experimental paper used 3-D wall which may be considered as cylindrical in shape. And I found that an area of a cylinder is the sum of the area of bottom and top circles and the area of the rectangle. So, now I think may be the answer can be found using the area of rectangle of a cylinder (an open cylinder or tube) or the area of an open-top cylinder as you can see below. Actually, that is why I was confused. And the author(s) of the paper did not tell the kind of area they used. Thank you so for your kind response.

Qamar seems to have used the wrong area for heat transfer: he assumes that you are heating the ends of the cylinder. The heat transfer area is the wall area of the cylinder. Assuming that the cylinder is open at each end, the area is PI x D x L = 1.269*10^-2 m2. The heat flux is 173 / Area = 13.63 kW / m2. The value of 2.9 MW/m2 quoted by Qamar would be massive indeed.

Jack Broughton. But I'm heating the bottom end of the cylinder. it a vertical thermosyphon. First part (the bottom part is the heating section 200mm, section part adiabatic section 100mm and top part is the cooling section 200mm. so, it is a vertical cylinder divided into three parts. Please look at the image of the area of cylinder I uploaded and recommend the right one for me.

So, as I now understand this is a section of a heat pipe (or thermosyphon). It appears that you are heating the bottom section and the model is what you have called "open-top cylinder). Thus the heat flux is the heat flow divided by this area.