im designing a heat exchanger and the value of the overall heat transfer coefficient for the fluids im using (cooling water and organic vapours) according to coulson and richardson vol.6 falls in the range 250 - 750 W/metre sq. K. After numerous iterations i established that 1011 W/metre sq. K is the optimum heat transfer coefficient. Is this correct or must it fall within this range . If so what can i do to make it closer to this range because im sure my calculations are accurate.
the standard formula for thermal transfer using U(thermal coefficient) is Q=Au ( T1 - T2) Q is the absorption of heat in btu T1 is the temp differential internal T2 is the temp differential external A is the area of the material as an example lets do a problem Q= 200in sq (.19) (22degrees)=836 Q=200 in sq (.30) (22degrees)=1320 as you can see from the ex's that the higher the thermal transfer the greater the btu absorption. Respectfully the reverse or rather the loss of btu is the same. If you have a piece that is in operation measure the temp differential use a selected surface area and you can estimate the U value or you can look in the ASHREA manuals for a really long list. Since you mentioned it the relationship of U-value to the resistance is a relative matter the higher the number the greater the transfer,inversely 1/U = the R value which by practical example is more like the resistance to transfer....Well I hope I got in there somwhere with some help...Have a good one from the E
You need to realize that calculating the overall heat transfer coefficient for any heat exchanger is as best an estimate and unless you have empirical data, it is very difficult to get the right answer. As an example: the heat transfer coefficient for natural gas versus cooling water can vary from 35 to 100 btus/(hr per square foot per deg F) depending on just the pressure on the natural gas side of the exchanger. (100 psi to 1000 psi) You might find some help if you have access to a Perrys Chemical Engineering Data Book.
However good your calculation, a simple thing like hard water can hit your coefficient for a six. In any case , you cannot increase the overall coeff by calculation and any number of iterations. If you ask me, i will suggest that you stick to the lower end of reported values, which will give some insurance. Organic vapours are tricky in that if any high boiling components are present they can form a fouling film quickly dropping the coefficient far below the designed values. You may mail me through my profile if you want a more detailed discussion.
You'll have to reduce the coefficient of your design by increasing the heat transfer area.
