heat transfer area

[Aramis]

hi to one of you, i wonder if u can tell me if there is any way to calculate the heat transfer area that i need to cool down wort to fermentation temperature, Hope you can help me
best regards

[Alex T]

indeed there is, but your question doesn't have enough information. is it for wort flowing through a heat exchanger? if you actually ring any supplier of heat exchangers and ask for a quote they will run a simulation for you. otherwise, you need to consult the heat transfer textbooks. give some more info (qty or flowrate, start temp, final temp, SG, etc), and we might be able to help.

[Aramis]

heat exchange trasfer area

iTxs for ur answer, yes wort is flowing trough a heat exchanger, I know that suppliers of heat exchangers can give that value, but I really like to know how to calculate that. 500 L/h, 95°C, 20°C 1.045, what else would you need. hope you can help me
best regards

[Moonlight]

Mass/time x temp=Mass/time x temp. That is for equilibrium at 100% efficient heat transfer.
You can always slow everything down and make any heat exchanger do almost any job. It is difficult to get closer than a 10 F difference between the optimal cooled wort and the incoming chilled water. The relationship is exponential so it is a bit challenging using normal math.
Sorry, I cannot really help you with a simple number relating to sq. footage, each plate style has a different efficiency, and the flow rate determines turbulence which is a huge part of the equation-more turbulent, better heat exchange. Ask your HX supplier as they do it every day.

[headlessbob]

iTxs for ur answer, yes wort is flowing trough a heat exchanger, I know that suppliers of heat exchangers can give that value, but I really like to know how to calculate that. 500 L/h, 95°C, 20°C 1.045, what else would you need. hope you can help me
best regards

I can give you a basic run-down:
Q=G x Cp x (T2-T1)
Q(wort) = Q(chill water) = Q(through exchanger wall)
A=Q / [U x (log mean temp difference)]

Where Q is heat, G is the mass flow rate, Cp is heat capacity of liquid, T2-T1 is temperature change of liquid in question. Heat in to chill water = heat out of wort = heat transferred through metal in exchanger (assuming no losses to surroundings); the second equation calculates the exchanger area (A) based on the thermal resistance of the metal between the two liquids (U), the 'mean log temp change', and the heat transfered between the two liquids (from first equation). The 'mean log temp difference' is a bit tricky to explain in plaintext where I can't draw a diagram, but its a calculation of the average temperature of the each liquid as they pass through the exchanger in opposite directions... see http://en.wikipedia.org/wiki/Log_mea...ure_difference

You'd basically start by calculating the heat transferred to the water from the wort from equation 1. From that, you can calculate the temp change of the water as it passes through the exchanger by equation 1 as well, depending on water inlet temp and flowrate. These temperature changes are used to calculate the log mean temp difference, which is used to calculate the exchanger area required with euqation 2.

Area "required" is hard to nail down, some of the variables you could consider is your chilling water inlet temp and flowrate, as well as how hot do you want the water to exit the exchanger at? 500 L/hr seems a fairly slow flowrate for wort in a production brewery, unless your entire batch size is 500 L. You should be able to look up the heat capacity of wort (water is 4.2 kJ/kmol.K I think) as well as the heat transfer coeff (U) value... however that will depend on the heat exchanger design as how turbulent the liquid flow is near the exchange surfaces.

Its a complicated system certainly, causing students all kinds of problems... wikipedia has a decent description of heat transfer. Or even better, grab a chemical engineering textbook on heat transfer from the library. In the end, the exchanger supplier can do all the calculations you need.
Cheers