I fucked up sci.
I have literally no idea how to use differential equations. Its been years.
Pic related, an old exam problem the likes of which I'm gonna need to be able to solve for my final.
I'm in panic mode, somebody please try and explain this shit to me... use small words.
>>7669342
Just review the dynamic sections in your heat transfer textbook. You're looking for the heat transfer PDE in cylindrical coordinates. In 1.a set all differentials with respect to time equal to zero.
This problem is too textbook tier for an exam imo. I'm assuming your writing closed book then? Your prof is a fag.
JUST
>>7669366
What textbook are you using?
>>7669365
Yeah closed book and I know questions are recycled, I should have failed the midterm because my conceptual understanding honestly is shit but I worked through old midterms and enough of the questions got recycled with slightly changed numbers that I scraped a C.
>>7669392
Welty is extremely shit beyond its fluid sections imo. Cengel is my personal preference -I recommend it if you have to retake the module, though Bergman is probably fine, unfortunately I don't have a copy.
In any case start by finding and writing down the general energy balance including the energy generation term). For Welty it's on page 220:
[math] \nabla . k \nabla T +\dot{q} + \Phi = \rho C_v \frac{DT}{Dt}[/math]
Then notice that "after a time" when the system is at steady state (and because the rod is a solid so there is no bulk motion) the material derivative term -which changes with time- should be zero [math] \frac{DT}{dt} = 0[/math].
[math] \Phi [/math] is the viscous energy dissipation term which is zero for this problem.
Do this then after you've done that assume that the temperature profile along the length of the cylinder is constant (write this assumption down, it's a good assumption if the rod is long enough due to the symmetry of the rod, the external convection at r = R_max = 50mm is your boundary condition, this is what causes a non-constant axial temperature profile at steady state)
I'll help you work through this problem, but I'm not solving it for you there would be no point.
>>7669456
OK so this is what I have. How do i go about putting this in terms of r?
>>7669555
Basically I have
0= /nabla * (k /nabla T) + q
I download welty if theres a specific section you want me to look at or if you want to refer to equations by number?
Don't download Welty unless you feel you need to textbook.
To get to cylindrical coordinates. Assume k is relatively constant with temperature so the equation becomes:
[math] k \nabla^2 T + \dot{q} = 0 [/math]
For cylindrical coords, using our prior assumption that [math] \frac{\partial^2 T}{\partial z^2} = 0 [/math], further assuming that due to symmetry
[math]\frac{1}{r^2} \frac{\partial^2 T}{\partial \theta^2} = 0 [/math]
so from (Laplacian of cylindrical coordinates; I'm not sure if your professor expects you to memorize this or not)
[math]\nabla^2 T = \left[ \frac{\partial^2 T}{\partial r^2} + \frac{1}{r} \frac{\partial T}{\partial r} +\frac{1}{r^2} \frac{\partial^2 T}{\partial \theta^2} + \frac{\partial^2 T}{\partial z^2} \right][/math]
the equation becomes
[math] k ( \frac{\partial^2 T}{\partial r^2} + \frac{1}{r} \frac{\partial T}{\partial r}) + \dot{q} = 0 [/math]
which is a simple(1 dimensional) linear second order ODE (with k constant and [math] \dot{q}[/math] invariant with r)
[math] ( \frac{\partial^2 T}{\partial r^2} + \frac{1}{r} \frac{\partial T}{\partial r}) = - \frac{\dot{q}}{k } [/math]
In ODE form:
[math] ( \frac{d^2 T}{d r^2} + \frac{1}{r} \frac{d T}{d r}) = - \frac{\dot{q}}{k } [/math]
which is the answer to 1.a. You could've simply written this answer as long as you stated all the assumptions use the get there (NOTE: You get marks for stating correct assumptions, writing down your assumptions is very important).
Now find the general solution of this second order linear ODE, note that there is a simple trick to make equations like this separable. I will post the answer I found in my next post, but try it first.
>>7669631
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>ANSWERS-----------------------------------------------------------------
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The general solution to this equation can either be found by using a dummy variable then using the general solution to a first order ODE, however, a simple way is to notice that the LHS of this equation can be factored with the chain rule (which if you recall from your intro DE class is how we derived the formula for the general solution to a first order ODE anyway), so
[math] ( r \frac{d^2 T}{d r^2} + \frac{d T}{d r}) = - \frac{\dot{q}r}{k } [/math]
[math] \frac{d}{d r} ( r \frac{dT}{dr}) = - \frac{\dot{q } r}{k } [/math]
Integrate twice
[math]r \frac{dT}{dr} = - \frac{\dot{q} r^2}{2k } + C_1[/math]
[math]T(r) = - \frac{\dot{q} r^2}{4k } + C_1 ln(r) + C_2[/math]
Which answers 1.b. Notice at this point that my result differs from the form in the question statement. I either made a mistake or your prof. did. You might want to email him about this. I'm not sure how he found that form, but it's possible he didn't properly consider how the 1/r factor in cylindrical coords differ from the "normal" Cartesian coordinates. Also Welty Appendix A explains the coord systems in great detail, but your text probably has a similar section somewhere describing how to convert the equations from Cartesian to other coordinate systems.
For 1.c The boundary conditions are given by the T(r) function in the question at r = 0 and r = 50mm. You only need 2, but it is easier to find by C_1 by differentiatin T(r) once:
[math]\frac{dT(r)}{dr} = b r [/math]
So
[math]r \frac{dT}{dr} = - \frac{\dot{q} r^2}{2k} + C_1[/math]
[math]r br = - \frac{\dot{q} r^2}{2k} + C_1[/math]
[math]C_1 = r^2 \left( b + \frac{\dot{q}}{2k}\right)[/math]
1.d Calculate the constants (just need C_2) then calculate [math]\dot{q}[/math
>>7669654
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>ANSWERS [cont.]--------------------------------------------------------
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For 1.e recall that the formula for conduction is simply
[math] \frac{q_r}{A} = - k \frac{dT }{dr}[/math]
So you simply need to calculate [math]\frac{dT(r)}{dr}[/math] at both [math]T(r=0)[/math] and [math]T(r = 0.05 m)[/math] then calculate the surface area A(r) through which heat flows at both points
Note that the heat flux must be symmetrical, so the total heat flow at a point [math]r[/math] is will be the entire circumference of cylinder circle at r multiplied by one "unit length of the rod" which is to say 1 m, it's important to state your final answer with units of W/m (or W/(m^2 m) for flux per unit length).
Part II.a (a) The first PDE we derived with [math] \frac{\partial T}{dr}[/math] non-zero
II.b) The initial conditions are the same as the final steady state conditions of Part I.
c) A simple calculation just solve for [math] \frac{\partial T}{dr}[/math] with the new [math]\dot{q}[/math] value.
>>7669631
I'm kind of confused about how you go from this form of the laplacian to that one, but if I memorize the form you used both would be correct?
>>7669654
Ah, OP I found a subsection where Welty does this (should've read the section first).
Ignore that post sorry.
Ok so back up to the differential equation:
[math]\dot{q} + \frac{k}{r} \frac{d}{dr} \left( r \frac{dT}{dr}\right) = 0 [/math]
integrate then rearrange
[math]rk\frac{dT}{dr} + \dot{q} \frac{r^2}{2} + \frac{k}{r} = C_1[/math]
so
[math]k\frac{dT}{dr} + \dot{q} \frac{r}{2} + \frac{k}{r} = \frac{C_1}{r}[/math]
Welty has a rather convoluted explanation here, but it basically means that because the temperature profile is symmetrical then at the centre it must be true that [math]\frac{dT(r = 0)}{dr} = 0 [/math]
So at r = 0:
[math] 0 + \dot{q} \frac{(0)^2}{2} + \frac{k}{r} = C_1[/math]
So [math]C_1 = 0[/math]
and
[math]k\frac{dT}{dr} + \dot{q} \frac{r}{2} + \frac{k}{r} = 0[/math]
From which
[math]T(r) = -\frac{\dot{q} r^2}{4k} + C_2[/math]
Which is the same functional form as in the problem statement.
This is discussed in detail in Welty on page 231. It looks like your Prof. is drawing problems from the text. So it might be worth it for you to download Welty if you still have some time before your exam.
Never mind I follow. This is a really thorough explanation. I'm gonna need a while to digest it all but I appreciate you holding my hand.
>>7669701
Don't memorize an extra form both the same:
[math]\frac{1}{r} \left[ \frac{\partial}{\partial r} \left( r \frac{\partial f}{\partial r} \right)\right][/math]
Use the chain rule for the factor inside the square brackets:
[math]\frac{1}{r} \left[ \frac{\partial }{\partial r} \left( r \right)\frac{\partial f}{\partial r} + r \frac{\partial}{\partial r} \left(\frac{\partial f}{\partial r} \right)\right][/math]
[math]\frac{1}{r} \left[ \frac{\partial f }{\partial r} + r \frac{\partial^2 f}{\partial r^2}\right][/math]
[math]\frac{1}{r} \frac{\partial f }{\partial r} + \frac{\partial^2 f}{\partial r^2}[/math]
>>7669717
I have an appointment, I'll check on the thread tomorrow if you have more questions.
>>7669731
You're a hero, I'm going to brood over this for a while.
All those k/r terms shouldn't be there.
integrate then rearrange*
[math]rk\frac{dT}{dr} + \dot{q} \frac{r^2}{2} = C_1 [/math]
so
[math] k\frac{dT}{dr} + \dot{q} \frac{r}{2} = \frac{C_1}{r} [/math]
...
So at r = 0:
[math] 0 + \dot{q} \frac{(0)^2}{2} = C_1 [/math]
...
and
[math] k\frac{dT}{dr} + \dot{q} \frac{r}{2} = 0[/math]
>>7669342
>can't solve a problem, make people solve it for you
That won't do you any good in any career. I am pleased to know you will fail at some point in time.
>>7669342
So I did 2 the way you described (i think), I found it a bit strange that dT/dt can be expressed independent of r so I suspect I may have done something wrong, but you've cleared up a lot of this for me. I am very grateful.
Trying my hand at this one now. I'll post what I've done. Trying to remember how to solve a 2nd order DE.
>>7671404
I downloaded welty and page 470 gives me the general solution to it, but that won't help me on an exam if I cant get it myself
>>7671412
Would this method be valid? Just integrate both sides similar to how you showed me earlier? If I can avoid using hyperbolic functions/ solving from a characteristic equation I'd like to lol
>>7671412
You solve an ODE like that by setting ay''+by'+c and solving it as if it with the quadratic equation. You'll get two distinct real roots so you put those two roots as the exponent of the exponential function with a constant of integration in front of them.
This link explains how you change that to the hyperbolic functions http://math.stackexchange.com/questions/237466/odes-and-hyperbolic-functions.
>>7671518
Homogeneous second order ODEs follow a very simple set of rules. Just Google them and memorise it.
>>7671547
>>7671545
Thanks, I have to go for a couple hours but I glossed over that example quickly and it looks like it will help. I'll be back in a few hours and start over :(
>>7671574
I missed the h when writing sinh
>>7671575
Whenever you see "diffusion limiting" or "relative fast [and/or] instantaneous reaction" on a catalytic surface you can reduce the problem to that simple separable ODE.
Professors love asking that question because it's what was essentially first used to crack integrated circuit design.
Pretty cool how accurate those analytical solutions are. It's almost on the real data points.
>>7671588
That's good to know, I'll remember that.
For the second case where there is a first order reaction I would need to include an R term like >>7671404
>>7671518
though?
>>7673182
Or no, its still a diffusion limited case so the only thing that changes is that I cant say at the surface boundary Ca = 0, instead Ca is an unknown which is dependent on R? I think I'm starting to get it
>>7673182
The reaction term is only present if the reaction occurs [math] \it{throughout}[/math] the entire control volume since R_a(z) is a function of z that is zero everywhere except on the boundary condition on the catalytic surface.
The reaction term occurs when your component is reacting with another component in the fluid.
