I have a question, which seems simple, but for the life of me I cannot figure out. So, let's say I have a very thin cylinder (consider an interface), between two fluid with different temperatures. The fluid inside the pipe is flowing with a mass rate of $ \dot{m}=\rho U S $, where $ \rho $ is the density of the fluid, $ U $ is the velocity and $ S $ the section of the cylinder.

Naively I could say that the rate of heat transfer, $Q$, between the two fluid is $ Q=hA(T_o-T_i) $, where $h$ is the heat transfer coefficient, $A$ the "contact area between the two fluid" (or the lateral surface of the cylinder), and $ T_o, T_i $ are respectively, the temperature outside and inside the cylinder.

Since I am interested in the heat added per unit mass, naively (again) I would define it as $$q=\frac{Q}{\dot{m}}=\frac{hA(T_o-T_i)}{\rho U S }=\frac{2L}{r}\frac{h(T_o-T_i)}{\rho U}$$

where $L$ is the cylinder length and $r$ its radius.
From this equation one could conclude that, if we increase the cylinder radius, the heat per unit mass will decrease, even if $Q$ is proportional to $r$.
If $r$ increases, the volume of the cylinder increases, so even if we add more heat, it will be "distributed" on a larger volume.
But this reasoning breaks down if we increase $L$. In this case, we have both $q$ and $Q$ are directly proportional to the length. Even if, we should also see the effect of the volume increment if we increment $L$.

Have I been too naive in the use of the equations? Or is there something that I haven't caught ?


1 Answer 1


Heat transfer coefficient is used for heat transfer between fluid and the wall surface, not conduction through the wall. For a steady state, you would need 2 heat transfer coefficients $h_i$, $h_e$ and also wall conductivity $\lambda$. $$\dot{Q} = \left(T_{fi}-T_{wi}\right)\cdot A_i\cdot h_i = \left(T_{wi}-T_{we}\right)\cdot \frac{2\pi \lambda L}{\ln\left(\frac{r_e}{r_i}\right)} = \left(T_{we}-T_{fe}\right)\cdot A_e\cdot h_e$$ Increasing length is a little bit different than increasing radius, because fluid temperature will probably change along the length (in some situations this can be negligible). For inlet and outlet fluid temperatures $T_{fi0}$ and $T_{fi1}$ and thermal capacity $cp$: $$\dot{Q} = \left(T_{fi0}-T_{fi1}\right)\cdot \dot{m}\cdot cp$$

I am not sure what the question is, but differences in sensitivities to $r$ and $L$ are easily explained by $r$ influencing heat transfer area $\propto r$ but also cross-section area $\propto r^2$, so the net effect for $q\propto \frac{1}{r}$. On the other hand, $L$ influences just the heat transfer area, so $q\propto L$.

Edit: More accurate dependence on length

If you are interested in dependence on pipe length, it is better to use a more precise model, where a very short section of pipe with length $dx$ transfers small amount of heat $d\dot{Q}$ into the environment with constant temperature $T_e$ ($U$ is overall heat transfer coefficient per unit length of the pipe), which also changes temperature of the fluid in a pipe by $dT$:

$$d\dot{Q} = \left(T(x)-T_e\right)\cdot U = -dT\cdot \dot{m}\cdot cp$$

This is a simple differential equation, which you can directly integrate:

$$\int\limits_{T_{i0}}^{T(x)}\frac{1}{T(x)-T_e} dx = -\frac{U}{\dot{m}\cdot cp}\cdot \int\limits_0^x dx$$

This leads to a internal fluid temperature profile along the pipe $T(x)$:

$$T(x) = T_e+\left(T_{i0}-T_e\right)\cdot \exp\left(-\frac{U}{\dot{m}\cdot cp}\cdot x\right)$$

Here is an example: enter image description here

  • $\begingroup$ Thanks for the comment! My question is related to physical intuition. We have a temperature difference between the pipe and the fluid. Consider a very long pipe, infinite if you like. From the formula seems that the heat added by unit mass would be infinite. But since $q$ is a quantity "by unit mass" I would have expected not to have this strong dependence on $L$. That is why I was asking if the equation that I derived is correct. $\endgroup$
    – GMV871
    Sep 5, 2022 at 7:17

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