# Heat Exchange between Fluids (and similar questions)

1. Jun 28, 2010

### OsirisGuy

First let me apologize for any spelling mistakes as I am on a mobile device. Second I do have a few questions regarding thermodynamics for a project I would like to begin.

1) is there a way to calculate a volume of water in either the amount of time and the amount of energy required to make it boil?

2) I will paint a picture for this one. I have one copper tubing inside another. The outer tube has water flowing opposite to the fluid in the inner tube. The goal here is to have the water cool the inner fluid to a certain temperature.

I understand these may be slightly vague and I know I did not give hard numbers so formulas should work just fine (for now lol).

I have tried searching google and this forum for some answers but I do not know if what I'm looking for, and what I've found has been correct.

Thanks for the help everyone (and if this is in the wrong forum can a mod please move it to the correct area, thanks).

2. Jun 28, 2010

### alxm

Hi Osirisguy
1) If you're asking "Is there a way to calculate the amount of time/energy it takes to bring a volume of water to a boil", the answer would be yes. Typically you'd just take the heat capacity of water (Cp) and multiply by the difference in temperature. For water that's about 4.18 kJ/kg*C, so to heat water from 20 C to 100 C, that's 4.18*80 kJ/kg.

If your flow is x kg/s then you need: x*4.18*80 kJ/s (= kW) of power.

(If you want to be more precise you could take into account that Cp is not constant between 20 C and 100 C)

2) That'd be a counterflow heat exchanger of sorts. This is a classical engineering problem, but there's no way I could possibly summarize all the knowledge needed to do such a calculation properly in a single post. A standard textbook on the topic, such as Coulson and Richardson's "Chemical Engineering", vol 1 "Fluid Flow, Heat Transfer and Mass Transfer" should teach you what you need to know.

Unlike the above there isn't really a simple formula you can just plug your values into. It depends on the temperature range, flow rates, pipe diameters, thicknesses and material, etc.

3. Jun 28, 2010

### OsirisGuy

1) The water being boiled is standing so there is no flow (assuming the flow you are talking about is because of my second question on the counterflow heat exchanger). I didn't think it'd be as easy as finding the product. A specific example I'd be looking at:

I have 20 gallons of water to be boiled and I want to find an appropriate propane burner (in BTU).
20 gallons = 75.71 Liters = 75.71 kg. I will also go ahead and think that my tap water is slightly colder than room temperature so 15$$\circ$$C and to bring it to a boil at 100$$\circ$$C, $$\Delta$$T = 85 C.

So the energy required;
E = 75.71kg * 4.18 kJ/kg*C * 85 C
= 26,899.76 kJ
= 25,496.0531 BTU (1.060 kJ = 1.054 BTU).

This brings up another question in the back of my head and that is would a 25,000 BTU burner still be able to bring the 20 gallons (75.71 L) to a boil?

As for the "time", I was just curious as to see how much time it would then take to get that 20 gal (75.71 L) from 15 C to boil (100 C).

2) One of the reasons I did not give specific values for this question is because I can change some of my variables (e.g., copper tubing thickness, tube length, water (cooling) input temperature...).

A counterflow heat exchanger is exactly what I am describing. Unfortunately my "textbook" from physics was a CD and is very haphazardly organized (with a few overlooked mistakes). I am glad to know that there is a way to find such a calculation and that what I've been calling it is correct (so I can Google that term some more). I figured this calculation wouldn't be as easy as Q = mc*DeltaT.

I'm trying to remember some of the equations I've used. I remember using, between two materials an equation like:
Tf = (mwcwTw+mbcbTb)/(mwcw+mbcb)

where subscript "w" is for water and "b" is for my secondary liquid.

Thanks for both reading :D and the help!