Optimizing DIY Air-to-Air Counterflow Heat Exchanger Size with Experimental Data

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SUMMARY

This discussion focuses on optimizing the size of a DIY air-to-air counterflow heat exchanger based on experimental data. The user conducted an experiment revealing that a 5°C temperature difference results in a 1.3°C cooling of the warm air outlet, indicating a direct correlation between temperature change and heat exchanger length. The formula derived, y = x(L/(c+L)), allows for predicting temperature changes based on varying lengths of the heat exchanger. The conversation emphasizes the importance of understanding flow rates and the linear relationship of temperature changes within the heat exchanger.

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  • Understanding of heat exchanger principles, specifically counterflow designs.
  • Familiarity with basic thermodynamics, particularly temperature differentials.
  • Knowledge of fluid dynamics, especially airflow rates and their impact on heat transfer.
  • Ability to apply mathematical formulas to practical engineering problems.
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  • Research the impact of varying airflow rates on heat exchanger efficiency.
  • Explore advanced heat exchanger design techniques, including materials and surface area optimization.
  • Learn about the mathematical modeling of heat transfer in counterflow systems.
  • Investigate the effects of humidity on heat exchanger performance and efficiency.
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This discussion is beneficial for DIY enthusiasts, engineers, and anyone involved in designing or optimizing air-to-air heat exchangers, particularly in residential or small-scale applications.

Xalt
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Hi
I am planning to make a DIY heat exchanger. It will be an air-to-air counterflow heatexchanger with equal air flow. I made a small test heat exchanger so I would be able to estimate the required size of the heat exchanger. I did a small experiment with the small exchanger. The outcome is that with a 5 degree (Celsius) temperature difference between the cold air inlet and the warm air inlet, the warm air outlet cools down 1.3 degrees. Since I have equal airflow this means the cold air would be warmed up 1.3 degrees. I have a 9 square cm heat exchange surface (6 cm long x 1.5 cm broad). The question now is: what happens if I would double the length of the heat exchanger? Any help will be appreciated!
 
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Xalt said:
Since I have equal airflow this means the cold air would be warmed up 1.3 degrees.
Approximately, if the humidity is not too different.

With a temperature difference of x between the inlets, a temperature change of y and the length L of your heat exchanger:
Temperature change inside should be approximately linear within the length of the heat exchanger, that leads me to ##cy=L(x-y)## where c is a parameter that depends on the flow rate and other parameters that can stay constant here.

Using your values, I get c=6cm*(3.7K/1.3K)=17cm. That value roughly corresponds to a "typical length" in your problem, but independent of its interpretation we can plug it in the first equation and solve for y for different values of L:
$$y=x\frac{L}{c+L}$$
With L=12cm and the same airflow, I get a temperature change of 2.1 K.
 
I think that should be y=(x/2)*(L/(c+L))

You would need y ---> x/2 as L ---> infinity, no?
 
mikeph said:
I think that should be y=(x/2)*(L/(c+L))

You would need y ---> x/2 as L ---> infinity, no?

Thank you both for your replies. With a counter-flow heat exchanger, I think y--> x as L --> infinity; with a parallel flow heat exchanger the limit is indeed x/2
 
Xalt said:
Thank you both for your replies. With a counter-flow heat exchanger, I think y--> x as L --> infinity; with a parallel flow heat exchanger the limit is indeed x/2
Right.
 

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