# CL: Multiple modes of heat transfer

Previously, we introduced heat transfer problems in cylindrical coordinates, and looked at a very simple example. As it turns out, problems in cylindrical coordinates are slightly more interesting than they let on. Let's take a look...

Let's look at our simple pipe again. Where the inside diameter (ID) is 2mm and the outside diameter (OD) is 2.5mm. Let's now include convective heat transfer, so the inside fluid is at Ti = 300K and the outside fluid at To = 350K. What is the rate of heat flow per length of pipe if the thermal conductivity of the pipe is k = 10 W/mK, and the inside and outside convective heat transfer coefficients are 35 W/msK (inside) and 70 W/m2K (outside)?

We have already derived the governing equation:

$\displaystyle{q = \frac{\Delta T}{R}}$

Noting that for conduction through a cylinder, qcond was

$\displaystyle{q_{1\to 2} = \frac{2\pi k L}{\ln(r_2/r_1)}(T_1-T_2)}$

so that R for conduction is given by

$\displaystyle{R_{cyl} = \frac{\ln(r_2/r_1)}{2\pi kL}}$

Recalling that the resistance due to convection is given by

$\displaystyle{R_{conv} = \frac{1}{hA} = \frac{1}{2\pi rLh}}$

(which r do we use now?!)

For this problem our thermal circuit is simply three resistors in series, so our total resistance, RT is

RT = Rconv(i) + Rcyl + Rconv(o)

so that

$\displaystyle{R_{tot} = \frac{1}{2\pi r_1Lh} + \frac{\ln(r_2/r_1)}{2\pi kL} + \frac{1}{2\pi r_2Lh} }$

From here it is simply plugging in numbers!

Similarly, if we had both convection and radiation leaving the exterior of the pipe, we would have a thermal circuit that would look like:

and we would simply sum the total resistance as

RT = Rconv(i) + Rcyl + Reff

where

$\displaystyle{R_{eff} = \frac{1}{1/R_{rad} + 1/R_{conv}}}$

##### OUTCOME:

Calculate the resistance and magnitude of heat flow in systems in which multiple modes of heat transfer are present

##### TEST YOURSELF

What happens if we increase the outer radius of the pipe? (or, equivalently, we add insulation to the exterior of the pipe)