Hey there. I am a Pakistani Chemical Engineering student...and trust me, I take help from Indian lectures most of the time. To be honest, they are a GREAT help. The teachers are expert in their profession. They teach you from the basics to the very minute details. I must say, they are spreading knowledge and for this, they will be rewarded with good. Thank You Sir. Stay Blessed.
The book used was Incropera and Dewitt, Fundamentals of heat and mass transfer 6th Edition. This book seems to be quite out of date as the 7th edition uses a more updated correlation for the external flow over the tubes, you may want to check both editions to compare.
If there would be no pressure losses required to force the flow through the heat exchanger, would there still be some power loss for a recuperated gas engine as compared with the nonrecuperated gas engine ?
I believe that the diameter used in the calculation of the Shell side Reynolds number should not be the same as the tube diameter (19 mm). The diameter should be, D = Aeff / Perimeter. Aeff is the area of the inside shell minus the area of the tubes, and Perimeter will be πDs( Ds is the inside diameter of the shell)
The cold fluid certainly is not water, because it has phase transition between Tci and Tco, so Cp(Tav) can not be an approximate value. If it is assumed as water, taking the phase change into account, q=h(Tco)-h(Tci)=7195 kW, not 998.4kW!. What are the cold and hot fluids? It would be much more beneficial if you provide this information
Hi, i need to design Heat exchanger for my project. The hot stream temperature is 75 and 25 for inlet and outlet respectively. heat duty is 4.018 kW. now i'm planning to use water to cool down the substance. But i don't have any info on its properties. I've read that maybe 21-22 deg C is an ideal for the inlet. but i can't determine the outlet because i don't have the flowrate of the water. so how am i supposed to determine the flowrate or the outlet temperature? Please help, i'm stuck because i can't even calculate the LMTD
Sounds like you are using water, cooling water. Generally cooling water. Temperature rise is limited to 115 degrees F. This prevents salts from dropping out. Consult Kern if available.
I think you must not to introduce a correction factor for DELTA (T ln) because in your example you are working with a 1:1 heat exchanger (single pass in tubes single pass in sheel). This is a very close situation to the one that is present in double pipes exchangers, were Delta (T) ln is a valid measurement of temperature diference. That graph is made for single pass in shell and a pair number of passes in tubes. FT=1
Ok im not an expert, i just want to know how did you get the Kf to calculate heat transfer coefficient, and what it means please I can't find it in the book.
Very useful.Thank you very much Sir. Please let me know about your e mail addresses so that I could communicate with you in case of a practical design which I expect to do .