Using Control Volumes to Solve Heat Transfer Problems Intro: In heat transfer problems we are interested in understanding a couple of things. For example, we may be interested in the temperature gradient in a given material so we know what the temperature is at different points in the object. We may also be interested in understanding the amount of heat flow there is within an object, or how long it will take to for the object to reach steady state given a constant input of energy. Admittedly, we can use the differential model to solve for such problems using the general equation: = ! + However, as useful as this method may be in solving many temperature gradient problems, we are unable to solve a more complex variety of problems. These more complex problems include a varying surface area in which the heat flow is allowed to transfer through:

You should be able to tell that the differential equation is unable to incorporate for this change in surface area in calculating the temperature gradient for there is no area term in the differential equation. And as small as this may be, the changing area changes the what the temperature gradient is. Not to mention, although a bit more tedious to perform, the control volume method is able to solve problems that the differential temperature gradient can. The Method: In essence, the control volume method is a simple energy balance equation where: = !" !"# + !"#"$%&'(# !"##

(Control volume equation) 1. qin, qout refer to the heat flow or the heat entering or leaving a specific volume either through conduction, convection or radiation in a given time. The units are watts or joules per second. 2. The qgneration and qloss refer to the heat energy being created within the specific volume specified. This can be found either by the amount of energy a chemical reaction releases or by the amount of energy a chemical reaction needs to proceed forward. 3. The energy accumulation term refers to the amount of energy accumulated in the specific region in the given amount of time. This can vary from problem to problem, but an example would be:

a. mass * heat capacity at constant pressure* dT/dt. In practice: Understanding the control volume method comes from practice so let us look at the figure below. The figure below is at steady state.

Immediately, we should write down the control volume equation: = !" !"# + !"#"$%&'(# !"## We should then proceed to fill in the terms we know. We first begin with the energy accumulation term, but because the problem states that the figure is at steady state we know that there is no energy accumulation. With that, our equation becomes: 0 = !! !"# + !"#"$%&'(# !"## Likewise, our figure is not generating nor using any heat so we can take out those heat flow terms, giving us: 0 = !" !"# We see that in the figure, qin is through conduction and the qout is through convection and conduction. We can then rewrite out equation as: 0 = !" !"#$%!&'"# !" ! !"# !"#$%!&'"# !" !!! !"# !"#$%!&'"# !"# !"#$%!&'"# For the conduction terms we can use Fouriers conduction transfer equation to describe the heat flow, and for the convection we use Fouriers convection transfer equation to describe the heat flow. 1. !" !"#$%!&'"# !" !"#$% ! = !!!!"#! !"#$% !"!" ! 2. !"# !"#$%!&'"# !" !"#$% !!! = !!!"#$! !"#$% !"!" !!! 3. !"# !"#$%!&'"# = !"#$% !" !"#(!"# )

Where, k is the thermal conductivity constant for the material, A is the area, and h is the thermal convection constant for the surrounding fluid. Substituting this into our control volume equation we get: 0 = !!!"#$! !"#$% ! !!!"#$! !"#$% !!! 2!"#$% !" !"!(!"# ) Using algebra to rearrange this, we can simplify the equation to: !!!"#$! !"#$% ! !!!"#$! !"#$% !!! = 2!"#$% !" !"#(!"# ) We then rewrite the Area of each respective part by saying that: 1. Athrough metal = thickness of the metal * width of the metal 2. Ametal air = width of the metal *delta z Doing that, we can simplify the above equation to: ( ) ! ( ) !!! = 2 (!"# ) We can then divide everything by delta z and width and use algebra to rearrange the equation to look like: !!! () ! = 2(!"# ) By taking the limit as z goes to 0, we get: () !! = 2(!"# ) To solve the rest of the problem we would simply use normal ODE techniques.