Recent content by lycee

Of course it's obvious when you look at it logically.Thanks for the heads up.I can know find the LMTD and ultimately find the required area

If you refer to the first post from Mitch1 all the details of the first part are there.The second part of the question is asking for the area required to perform this duty when the overall heat coefficient is 54 Wm^-2 K^-1

I'm new to this forum and not sure yet how to write down the symbols, but heat transfer= Phi = U A (LMTD).So finding A would be phi/U(LMTD).Finding the LMTD is proving difficult as I'm not sure of the outlet temp of the feed water.

Should be Wm^-2 K^-1

On the same question: The overall heat transfer coefficient based on outside area of the tubes is given as 54 Wm K .Determine the area of heat transfer required to perform this duty