# Heat added per unit mass in a cylindrical pipe

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 ?

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:

• 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. Sep 5, 2022 at 7:17