Heat Equation Derivation: Cylindrical Coordinates
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Heat Equation Derivation: Cylindrical Coordinates

September 11, 2019


In this screencast, I want to do a derivation
of the heat diffusion equation in cylindrical coordinates. And the reason you might come
across this equation, or the reason you might find it useful, is that you could be faced
with a problem where you need to calculate or derive the temperature profile for a cylindrical
geometry. So as an example, you might have a copper wire with an electrical current flowing
through it, and then there’s some convective heat transfer on the exterior, and the temperature
of the interior of this is greater than the temperature at the exterior of it. Another
example you might come across would be the temperature distribution across the walls
of a pipe. So you might have hot water flowing through the center of the pipe, and the exterior
of the pipe is cooler than the interior, and in both cases you can use this equation to
derive those temperature profiles. While it may not look like it, all this equation is
is an energy balance around a differential element. And what it says is that the rate
at which thermal energy is being stored within a small volume is equal to the rate at which
energy is entering, minus the rate at which energy – thermal energy – is leaving, plus
any rate of thermal energy generation. So thermal energy could be generated, for example,
as an electric current or maybe a chemical or a nuclear reaction, for example. Three
axes that we need in cylindrical geometry: The first would be the axial dimension, which
we call z in this case, so z is the length along this rod. The second is the radial direction,
or the distance from the center of the rod, and the third is this angular direction, which
we call phi, which gives us the angular position from some starting point. So on the right
what I’ve drawn is a geometric depiction of a differential control volume. Vertically
in this case, I’ve got my z axis, in the radial direction, I’ve got this variable r, and in
the angular direction, we use a variable phi. So you can see the three of those, we see
the variable r, phi, and z in this differential equation, which we’ll derive. What I’ve drawn
is a differential control element, or differential control volume. it’s characterized by the
dimensions, say, dz in the vertical dimension, and dr in the radial direction, and a distance
rd(phi) in the angular direction. Now we’ve got r times some change in angle, this r is
in front for dimensional consistency. Let’s examine the net flow of energy into this control
volume. We’ll think about the three different dimensions. And in the vertical direction,
we can call it qz entering the control volume from the bottom. Energy flowing in from the
back, in the radial direction, we’ll call q dot r, and there’s energy flowing in in the angular
direction, or the direction of increasing phi, into this face on the left, and we’ll
call that q dot phi. So that was energy coming in from the three faces, now what I’m showing
is energy leaving. I’ve got energy leaving in the radial direction on this front face,
so leaving in the direction of increasing r, leaving in the axial direction, or the
direction of increasing z, this top face right here, and then on this back face, leaving
in the angular direction which I call q dot phi +dphi. So those are the way energies can be
conducted into or out of the control volume, and the third term we need to consider is
the rate of thermal energy generation. The rate of thermal energy generation is this
term q dot multiplied by the volume of this control element. So we’ve got q dot times
a volume, or we could say q dot times, let’s say we want dr times dz, the vertical dimension,
multiplied by r dphi. And this is analogous to a rectangular geometry, where we say the
length times the width times the height, so in this case r dphi is the width, we’ve got
the height of it, dz, and the length of it, dr. And finally, the rate at which thermal
energy is being stored is proportional to the heat capacity and the rate the temperature
is changing. So we take the mass, or the density times this volume, dr times rdphi *dz. Density
times a volume is a mass, multiplied by the heat capacity, times the rate at which temperature
is changing. So here I’ve written it all out. I’ve got the energy stored term, the energy
coming in, the energy coming out through the other three faces, and the energy being generated.
One thing we need to figure out is, how does the heat leaving at q(r+dr) compare to the
heat entering at q(r). So how does the heat leaving from this face, this front face, compare
to the amount of heat coming in from this back face. We can do it for all three dimensions,
the r, the phi, and the z dimension, by way of Taylor series expansions, which I’ve done
down here. So with these expansions, we can plug in values now for q(r+dr), and the phi
and z components of it. When I make this substitution, what we’ll find is that these q(r)’s will
drop out because of the negative sign, and the q(phi)’s just as well, and the q(z)’s
will also drop out because of the negative sign in the expression. So here’s what we’re
left with after making the substitution. On the left is the energy storage term, on the
right, the last term is the energy generation term, and these three terms now in the middle
represent the net rate at which energy is entering the control volume. What we need
to do now is show how the flow of energy in the r, phi, and z directions, q(r), q(phi)
and q(z), how can we relate those to the temperature? And we can do that by looking at Fourier’s
Law. Let’s examine the flow of energy in the radial direction first. What I’ve written
is, q(r) is equal to -kdT/dr times r*d(phi)*d(z). And what I’m writing is simply, the total
flow of energy, or the net flow of energy, in the r direction is equal to the flux, which
is -k*dT/dr, the flux times this cross-sectional area, or I should say the area of this face,
which is r*d(phi), this length, times d(z), or this height, but the area in the back or
the area in the front. And I could do something analogous for the angular direction, this
phi direction, and in the z direction. So in the phi direction, I’m talking about the
area drdz, so this area here, and this area in the back of it, and in the z or axial direction,
I’m talking about the area on the bottom and the top of our control volume. Note the presence
of r in the denominator of the phi direction, and the presence of that r is to make the
equation dimensionally homogeneous, so we’re talking about a differential change in the
distance in this angular direction, so we’ve got r*d(phi). So let’s make the substitution,
we’ll put in for q(r), q(phi), and q(z). So here’s what we’re left with after making the
substitution: we’ve got the energy stored term, the three net energy transport into
and out of the control volume now in terms of temperature, and finally the energy generation
term. And finally the last thing to do is to divide by this dr times r d(phi) d(z), divide
both sides of the equation. It’ll fall out in the energy generation term, it’ll fall
out in the energy storage term, and we’ll see some combination of it in the
net energy coming in minus energy leaving. And so here you have it, the heat diffusion
equation in cylindrical coordinates.

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  1. I disliked the video because you had an error and you seemed to skip a bit of math at the end, which would have revealed the error. In the video you show that q_(Phi+d Phi)=q_Phi + partial of q_Phi wrt Phi, *d Phi, but the end should be r*d Phi, since as you stated we are dealing with a differential distance, and d Phi would only be a arc length…Then you do not explain in any way why the r for the radius term stays in but the den. r comes out in the Phi term.

  2. 3D solution (programming done in MATLAB) of discretized momentum, energy and mass balances for an arbitrary gas phase reaction (pressure variations must be taken into account) taking place in a tubular reactor with baffles and equipped with a heat exchanger, please.

  3. The second part of your statement is correct but the first isn't. The differential distance is already included in the definition of q_Phi and it doesn't make sense to compute the taylor series expansion with r*d Phi. Think about it, if you that were the case we wouldn't be able to write the taylor series expansion of an unknown function because we wouldn't know the factors to the differential we're multiplying. It's a great video, the only thing missing the one you pointed out.

  4. at 6:45 you substitute from fourier's law but you pull the "r" that is with the d(phi) into the parenthesis for the first term (d/dr term) but that is not the case for the other two terms…why is that?

  5. Cengel's 5th ed:
    "After lengthy manipulation, we obtain [Conduction Equation in cylindrical coordinates]"
    In my opinion, the author of the book clearly had no respect for the reader's intellectual capacity. This is pretty straight forward, even though I could not think of it myself at first.

  6. Hello

    How are you doing brother ?
    Why don't you drive for
    ? Heat Equation : spherical Coordinates

    Would you mind do a video explaining the derivation
    Please guys give me like in order to let (LearnChemE) to see my comment
    ..and do a video for this type of an equation

    best regards

  7. Awesome video, had the exact concepts that I was looking for! Been working on a problem involving heat conduction in spherical coordinates, and was able to take the method here and apply it to my specific problem. Thanks for the clear explanation, really appreciate the effort that went into this video!

  8. In the second term of the final equation, why is it written (1/r)(d/dr)(kr dT/dr) instead of just (d/dr)(k dT/dr)? Why not eliminate the r from that term?

  9. please could you draw us an elemental volume for a sphere,like you drew here for cylindrical coordinates,i am having a hard time understanding the "rsin(theta)d(pHi)" dimension in the derivation for heat conduction in spherical coordinates,other videos dont have a clear explaination for it.

  10. Thank you so much…this too tough for me to understand during the class discussion…now u just save me from being dead..😂

  11. I have a question: i the general equation you have 1/r d/dr multiplied by (kr dt/dr). shouldn't the r's cancel out?

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