# Heat transfer of transfer tubes thought experiment [closed]

I need some help understanding the thought process behind this exercise I have set before me:

I've made a very poorly drawn graphic to encapsulate what's happening here for reference, as instructed in (a):

From here, I'm told to assume that the only mechanism for energy exchange between the outside of pipe and the environment is radiative, so apparently no conduction.. I guess? And, using this, I'm told to relate the temperatures of the environment and the inner and outer surfaces of the pipe are related by the equation given in the question.

The temperature of inner and outer surfaces of the pipe. The outer surface of the pipe must refer to $T_2$, but does the temperature of the inner surface of the pipe refer to $T_{env}$ or $T_1$?

Moving past that, here is my thought process:

• It can be shown that the heat transfer through between concentric cylinders (a pipe) can be shown using this graphic, given in the answer key to the exercise: Where $\frac{dQ_c}{dt}$ is the power from conduction from the inside to the outside of the pipe.
• I can be convinced that, if the interaction between the outside of the pipe and the environment is radiative, that the net energy transfer will be $$P_{net} = \sigma \epsilon 2 \pi r_2 l(T_2^4-T_{env}^4)$$

However, what trips me up is, to tie the two equations in and simplify to an answer, my lecturer states: "In the steady state, the energy transferred from the inside of the pipe to the outside (by thermal conduction) must be equal to the net energy transferred from the outside of the pipe to the environment (by radiation). Why is this an implication? Is that because the steady-state implies the pipe and its environment, along with the liquid will be in thermal equilibrium? Which means no net power output is possible? It's a confusing statement for me to grapple with, I feel. So, my main question is twofold: first, why is that the interaction between the outside of the pipe and the environment is only radiative is at all a reasonable assumption, and why can I make that deduction my lecturer made given the conditions here? Because he uses the following deduction to solve

$$\frac{dQ_c}{dt} = P_{net}$$

## closed as too broad by JMac, John Rennie, Jon Custer, heather, David Z♦Oct 27 '17 at 19:55

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

One of the problems is your picture, which shows the pipe only partially full. If fluid is being forced through the pipe, the pipe will be running 100% full. The assumption is that the flow rate is so high that, over the length of the pipe, the temperature of the fluid in the pipe does not have opportunity to change. So the temperature at the inner diameter of the pipe is $T_1$.

You have heat flowing by conduction through the wall of the pipe at a rate $\dot{Q}$, from the temperature $T_1$ at the inside wall to $T_2$ at the outside. The same heat then flows from the outside wall of the pipe by radiation to the surrounding environment. Why is it the same heat? Because it has nowhere else to go. So it flows by radiation at a rate $\dot{Q}$ from the temperature $T_2$ on the outside pipe surface to the environment at $T_{env}$.

This is like an electrical problem involving two resistances in series, with a current flow rate analogous to the heat flow rate $\dot{Q}$ and the overall voltage drop analogous to the overall temperature difference $T_1-T_{env}$. From the information you have been given, you should be able to determine the heat flow rate $\dot{Q}$ and the outside surface temperature $T_2$, given the inside fluid temperature $T_1$ and the environment temperature $T_{env}$.

• Big +1 for the circuit analogy. I find it makes situations like this a lot easier to conceptualize (as long as you're familiar with simple DC circuits). – JMac Oct 23 '17 at 19:22
• Thanks for your answer, couple things: a) why does the fluid and the pipe not have time for thermal contact? It's flowing very quickly but it still should be in thermal contact with the pipe, albeit constantly being in contact with different sections of it. b) So am I correctly understanding you when I say $T_{env}$ is actually the temperature outside the pipe, and $T_2$ is the temperature on the outside of the pipe? c) Am I correct in saying in the steady-state, the energy change from conduction from the inside of the pipe to the outside is the only thing source possible tor radiate to $env$? – sangstar Oct 24 '17 at 13:42
• (a) The fluid in the pipe does have thermal contact. It is just flowing so fast that it doesn't have enough time for its temperature to change significantly. The heat balance equation for the fluid from the inlet to the outlet is given by: $$\dot{m}C\Delta T=-\dot{Q}L$$where $\Delta T$ is the change in fluid temperature from inlet to outlet, L is the length of the pipe, $\dot{m}$ is the mass flow rate, and C is the heat capacity. If we solve for $\Delta T$, we obtain $$\Delta T=-\frac{\dot{Q}L}{\dot{m}C}$$ – Chet Miller Oct 24 '17 at 15:55
• Note that, as the mass flow rate of the fluid becomes very large, the steady state temperature change of the fluid between pipe inlet and outlet approaches zero. (b) Yes. It is the temperature out in the surrounding room. (c) Yes. This heat is supplied by the tiny change in the fluid temperature that we determined in item (a) – Chet Miller Oct 24 '17 at 15:58