My understanding is that thermal conductivity is responsible for the time until something will heat up. For example, if you hold a copper bar over a candle it will quickly become hot at the other end. If you instead hold a brass bar it will take a longer time, but the end result will be the same.

If this is correct (is it?) then why does thermal conductivity matter in a double pipe heat exchanger? It will take some more time for the pipe to heat up and start transferring heat, but once that happens it's all the same, is it not?



Since the rest of the answer is too long I will start with an example.

Imagine you could build a heat exchanger from wooden pipes. If you run through the liquid you would observe very little efficiency (i.e. the hot liquid would remain hot and the cold cold). There would be too little heat exchange, because the wood would not allow heat to pass through the material at any significant rate.

combined conduction and convection

You have to remember that the problem of heat transfer between the hotter fluid (h) and the cooler (c) is a combined conductivity and convection problem.

enter image description here

Figure Conducting wall with convective heat transfer (source: MIT thermodynamics)


  • $T_2$ is the hotter fluid
  • $T_1$ is the cooler fluid
  • $\delta_1$, $\delta_2$ are the convective zones. I.e. the zone where the temperature in the fluid will change. This zone will change with the velocity and the viscosity and other geometric parameters

In those problem sometimes, the concept of thermal Resistance is introduced. Thermal resistance shows what is the energy per area for a degree of temperature difference. It is important to note that as the Thermal resistance increases, less heat energy is transmitted per temperature degree difference $\Delta T$ (or $\Delta T_{LMTD}$ to connect to the previous question)

In the above case, the thermal resistance is: $$R= \frac{1}{h_1A} + \frac{L}{A \color{red}{k}} + \frac{1}{h_2A}$$


  • L is the thickness of the wall (pipe)
  • A is the exchange area
  • $h_1$ and $h_2$ the heat convectivity coefficient
  • $\color{red}{k}$ heat conductivity coefficient

As you can see the thermal resistance is inversely proportional to the thermal conductivity. This means that when thermal conductivity of the pipe is high, then the thermal resistance is low.

Steady state of a heat exchanger

The problem with heat exchanger is that they come into too many flavors. Usually you encounter the parallel and the counterflow (but there are several others). Below is the example of those types of heat exchangers. In this image you can see the temperature difference

enter image description here

figure: Temperature difference along the length of the heat exchanger for different types of flows

In that image you can see immediately the problem. If you have a heat exchanger with a parallel flow you will have different temperature difference at different section of the heat exchanger (thus different heat transfer rates because the $\Delta T$ varies). (The same problem exists with the counterflow to a lesser extent.)

However, this is the temperature distribution of the steady state.

Transient state (short parenthesis)

The transient state is (in the simplest case) when the heat exchanger starts its operation. Then the fluid that passes can be assumed to have a uniform distribution (i.e. does not change with location).

However, the temperature distribution will change with time, and it will approach the steady state solution. It is obvious that during the transient state the heat transfer will be greater because the temperature difference will be greater compared to the steady state. So the bottom line is that, at the steady state there is the least heat transfer compared to all the transient states that led to it.

Putting everything together

From the previous paragraphs, I was trying to highlight that the steady state will have a temperature distribution along the length of the heat exchanger that should remain constant (wrt to time).

Therefore for a infinitesimally small portion along the length of the heat exchanger (with area A) the temperature difference can be assumed to remain constant wrt time. In that portion, the heat transfer rate can be assumed to be:

$$\dot{Q} = \frac{\Delta T}{R}$$

However as we've seen before as the thermal conductivity (k) increases, then thermal resistance R decreases, and as R decreases the heat transfer rate increases. That means that:

for increasing thermal conductivity (k), there is more heat transfer for the same temperature difference.

Since in the steady state the temperature difference is constant, the heat exchange will be dominated by the thermal resistance, and therefore it would be affected by changes in the thermal conductivity.

Why is the effect not pronounced in metals

The reason the heat conductivity effect is as not pronounced in metals is that because Thermal Resistance R is determined by the equation:

$$R= \frac{1}{A}\left(\frac{1}{h_1} + \frac{L}{\color{red}{k}} + \frac{1}{h_2}\right)$$

Then, the lowest of the $h_1$, $h_2$ and $\frac{k}{L}$ will have the dominating effect on the heat conductivity. Since usually $h_1$, $h_2$ coefficients are significantly lower than $\frac{k}{L}$ (unless L becomes too great), any change of thermal conductivity is not affecting affecting significantly the thermal resistance. (this is exactly the point mart's second paragraph is trying to raise).


In the steady-state case you describe, the thermal conductivity determines how much heat (kW) a material will trasnmit per unit surface for a given temperature difference (K). In heat exchanger applications, the thermal conductivity is one of three heat transfers in series: primary medium to pipe, transfer through pipe, pipe to secondary medium.

To my best knowledge, thermal conductivity is secondary consideration in most double pipe applications because the pipe wall will be very thin compared to the rest. Actual heat transfer through most HX is mostly determined by flow conditions.

  • $\begingroup$ Also known as "skin" affect ; much more important than thermal conductivity of the tubes. Refineries generally use Admiralty brass tubes, not for conductivity , but for corrosion resistance. $\endgroup$ Aug 7 at 15:49

I didn't realize how confusing the terms could be for a heat exchanger until I started typing it out:

I will refer to the hot side as:

  • the side where fluid leaves cooler than it enters
  • the side that ends up being cooled
  • the side that heat flows out of

I will refer to the cold side as:

  • the side where fluid leaves hotter than it enters
  • the side that is being heated
  • the side that heats flows into

The final temperature of something is all about the equilibrium between the heat flow in and out of a system. There is always heat flowing into and out of the system

Remember the heat exchanger's purpose is to exchange heat between a hot fluid (thereby cooling it down) and a cold fluid (thereby heating it up).

That means the cold fluid (on the cold side) is constantly removing heat from the hot fluid (on the hot side) exchanger interface.

So what happens if the cold fluid is removing heat faster than it can pass through the walls of the pipe from the hot fluid?

The pipe wall on the cold side will drop in temperature until the temperature gradient is large enough to increase heat flow and establish a new equilibrium. The pipe wall on the cold side will always be at a cooler temperature than the wall on the hot side side, and the cold fluid will at most reach the temperature of the wall on the cold side.

So if your goal is to raise the cold fluid to a given temperature, you need run a hotter fluid than you otherwise would (or run more of it relative to the cold fluid).

If your goal is to lower the hot fluid to a given temperature, you need run a colder fluid than you otherwise would (or run more of it relative to the hot fluid).

  • $\begingroup$ So how would you explain what thermal conductivity of a material tells me? Would it be correct to say that, given the same temperature difference, a material with a higher conductivity will "release" more energy per second compared to a material with lower conductivity? $\endgroup$ Aug 6 at 22:50
  • $\begingroup$ @user1477107 Well how quickly heat flows depends on the gradient and the low thermal conductivity material will require a higher gradient in order to produce the same heat flow. Like friction in a gearbox. All the friction does is does is make you have to turn the input shaft harder to get the same torque on the output shaft, right?. But oh wait, it turns out that is super important. Why would you want to turn (or heat or cool) any harder than you need to? $\endgroup$
    – DKNguyen
    Aug 6 at 22:51

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.