# Heat exchanger calculations

1. Nov 10, 2014

### Xalt

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!

2. Nov 10, 2014

### Staff: Mentor

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.

3. Nov 10, 2014

### mikeph

I think that should be y=(x/2)*(L/(c+L))

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

4. Nov 10, 2014

### Xalt

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

5. Nov 10, 2014

Right.