I am playing with the Idea that I should make a my own heath exchanger to generate warm water.

But as a hobbyist I would like to calculate the potential of my design and here is the question for me, how do I do this?

I need to calculate the heat transfer of a pipe passing trough liquid.

For example: I have a boiler containing 100liters of water, in this boiler there is a spiral with length of 10 meters with liquid of 50°C. How do you calculate the heath transfer from the liquid in the pipe to the water inside the boiler ?

Lets forget the fact that my boiler water is loosing heat to it's surrounding.

Additional info: the spiral pipe will be made out 1/2" (Ø21.3mm) x sched. 5 (1.65mm)

And the liquid speed is 1m³/h

  • $\begingroup$ You can do this, but the heat exchanger coefficients are secrets guarded by heat exchanger manufacturers... Any decent textbook on heat transfer will take you as far as you can go... I like and still use Heat Transfer by Simonson... $\endgroup$ – Solar Mike Feb 22 '19 at 15:04


The starting point is the general heat exchanger equation below.

$$\dot{q} = U A \Delta T_{LM} $$

Here, $\dot{q}$ is heat flow (J/s = W), $U$ is the overall heat transfer coefficient of the system (W/m$^2$ $^o$C), $A$ is the tube area (m$^2$), and $\Delta T_{LM}$ is the log mean temperature through the exchanger ($^o$C).

The next equation is the energy balance on the fluid.

$$ \dot{q} = \dot{m}\tilde{C}_p \Delta T $$

Here, $\dot{m}$ is the mass flow (kg/s) and $\tilde{C}_p$ is the specific heat of the fluid in the pipe (J/kg $^o$C).


Reasonable values for the overall heat transfer coefficient $U$ can be obtained using charts or correlation tables in a range of textbooks on heat transfer equipment or from on-line sources. Exact values are only obtained by measurements on the final working system.

The value of $A$ is obtained by geometry along the tube. For a tube of length $L$ and external radius $r$, the external tube area is $2\pi r L$.

The specific heat and flow rate of the fluid in the tube are supposedly known.

Assign the temperature of the external fluid as $T_\infty$ and the inlet temperature as $T_i$. You end with one equation and one unknown, the outlet temperature $T_o$.

$$ U A \frac{\left(T_i - T_o\right)}{\ln\left(\frac{T_i - T_\infty}{T_o - T_\infty} \right)} = \dot{m}\tilde{C}_p \left(T_o - T_i\right)$$

Use this to estimate the output temperature of the fluid through the tube. Alternatively, use this to determine what tube length you need to get a desired temperature.


Recast the above as a dimensionless equation with $\Theta = (T_o - T_\infty)/(T_i - T_\infty)$ and $\beta = U/\dot{m}\tilde{C}_p$.

$$ \Theta = \exp(\beta\ A)$$

Here is a plot of $\Theta$ versus $A$ for values of $\beta$ of 1 (lower right black), 3, and 10 (upper left red).

Heat Exchanger Temperature Profile Plot

Estimates with Variations (Uncertainties)

Variations (uncertainties) in $\beta$, $A$, and $\Theta$ can related through linear uncertainty propagation.

$$ \left(\frac{\Delta \Theta}{\Theta}\right)^2 = \beta^2 \Delta^2 A + A^2 \Delta^2 \beta $$

By example, when you know the area precisely, $\Delta A = 0$. When you have a 10% uncertainty in $\beta$, this translates to an estimate of the relative uncertainty in $\Theta$ as

$$ \left(\frac{\Delta \Theta}{\Theta}\right) = 0.10\ \beta A $$

In plain English, when you know the area of the exchanger, the relative uncertainty in the output temperature increases linearly with the area and effective heat transfer function. Larger heat exchangers will have greater uncertainty in their end temperature, as will exchangers with low flow or high heat transfer coefficient.

| improve this answer | |
  • $\begingroup$ I used this approach and sized a parallel plate heat exchanger... When I contacted the company I was told "not bad you got to within 20%..." and no they would not give me their working values... $\endgroup$ – Solar Mike Feb 23 '19 at 19:46
  • $\begingroup$ @SolarMike I've added a note about uncertainty estimates. $\endgroup$ – Jeffrey J Weimer Feb 24 '19 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.