Heat Exchange between Fluids (and similar questions)

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SUMMARY

This discussion focuses on calculating the energy required to boil water and the design of a counterflow heat exchanger. The specific heat capacity of water is established at 4.18 kJ/kg°C, which is used to determine the energy needed to heat 20 gallons (75.71 liters) of water from 15°C to 100°C, resulting in approximately 25,496 BTU. The conversation also highlights the complexity of designing a counterflow heat exchanger, referencing Coulson and Richardson's "Chemical Engineering" as a key resource for understanding the necessary calculations and variables involved.

PREREQUISITES
  • Understanding of specific heat capacity (Cp) and its application in thermodynamics
  • Familiarity with counterflow heat exchanger design principles
  • Basic knowledge of energy conversion between kJ and BTU
  • Proficiency in using thermodynamic equations for heat transfer calculations
NEXT STEPS
  • Study the principles of heat transfer in counterflow heat exchangers
  • Learn to apply the equation Q = mcΔT for various materials
  • Research the design specifications for propane burners and their BTU ratings
  • Explore advanced thermodynamics textbooks, particularly Coulson and Richardson's "Chemical Engineering"
USEFUL FOR

Engineers, thermodynamics students, and professionals involved in heat transfer applications, particularly those designing heating systems or studying fluid dynamics.

OsirisGuy
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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).
 
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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.
 
Thanks for that reply alxm, helped quite a bit. Some follow up questions/comments.

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\circC and to bring it to a boil at 100\circC, \DeltaT = 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!
 

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