Heat transfer and radiation question help

Watch
Lion1996
Badges: 9
Rep:
?
#1
Report Thread starter 5 months ago
#1
Hi I have been asked to solve this. he power of an electronic component is dissipated through a heat sink by convection
and radiation into surroundings at 20C. The electronic component is powered by a
voltage of 10 V and the heat sink has an effective area of 25 mm2
. Using the bisection
method, determine the surface temperature of the heat sink knowing that the heat sink
is made in silica.
The properties of silica are:
✓ Resistance 𝑅 = 3 k
✓ Emissivity 𝜀 = 0.8
✓ Convective heat transfer coefficient ℎ = 17 W m2 ⁄ ∙ K
Schematic:
Assumptions:
• Steady-state conditions.
• Constant properties.
• Heat sink thickness is negligible.
• The total rate of heat transfer through the heat sink is equal to the power P
dissipated by the component, where 𝑃 = 𝑈 ∙ 𝐼 with 𝑈 = 𝑅 ∙ 𝐼.
Methodology to be implemented: bisection method
i. Express 𝑄𝑇𝑜𝑡𝑎𝑙 = 𝑄𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 + 𝑄𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 as equation 𝑓(𝑇𝑆
) = 0 where 𝑓 is a
function of unknown 𝑇𝑆
. Function 𝑓(𝑇𝑆
) will be a function of degree 4 of the
form: 𝑓(𝑇𝑆
) = 𝑎 ∙ 𝑇𝑠
4 + 𝑏 ∙ 𝑇𝑆 + 𝑐
where constants a, b and c must be determined.
ii. Calculate the derivative d𝑓(𝑇𝑆
)⁄d𝑇𝑆
.
iii. Study the sign of the derivative to sketch the graph of 𝑓(𝑇𝑆
).
iv. Observe that it exists 2 solutions. One of the 2 solutions should be clearly
improbable and therefore be dismissed.
v. Use the bisection method to approximate the valid solution to 2 decimal placeName:  Screenshot_20210416-235724_Drive.jpg
Views: 12
Size:  135.4 KB
Last edited by Lion1996; 5 months ago
0
reply
Callicious
Badges: 22
Rep:
?
#2
Report 5 months ago
#2
See https://en.wikipedia.org/wiki/Bisection_method for information on how to use the bisection method if that's what you're stuck with. What are you stuck with, exactly?

Also, please for the love of all that is good send a photograph of the question.
0
reply
Lion1996
Badges: 9
Rep:
?
#3
Report Thread starter 5 months ago
#3
(Original post by Callicious)
See https://en.wikipedia.org/wiki/Bisection_method for information on how to use the bisection method if that's what you're stuck with. What are you stuck with, exactly?

Also, please for the love of all that is good send a photograph of the question.
Just the formula to use and how to calculate it.

I have attached the question.
Attached files
0
reply
Callicious
Badges: 22
Rep:
?
#4
Report 5 months ago
#4
(Original post by Lion1996)
Just the formula to use and how to calculate it.

I have attached the question.
Thanks for the img. The bisection method itself (at least to me) seems iterative: there isn't some formula for the final answer, you just follow it until you're satisfied with the convergence to a solution. I'm sure it'd be great to code with (and I've used a similar method before for root-finding for practice) but I can't see it's use practically.

Anyhow, if you just want to determine the surface temperature of the heat sink, you'll just need to balance out the equation of state for it. In this context, that's satisfying

P = \dot{Q}_{rad} + \dot{Q}_{conv}

where your \dot{Q}_{rad} will be given by

\dot{Q}_{rad}\propto{T^4}

with T the temperature of the surface of the heat sink. I've introduced dots because I'm talking about a rate of something, i.e. the amount of heat released per unit time from the heat sink, but anyhow, that's a bit pedantic I know...

The exact constant of proportionality depends on if the area A is just one side or both ("effective" area... is that effective over both sides, or just one? The terminology seems a bit ambiguous. I'd say "effective area of the top surface" or whatever else...).

If you assume A is just the total area of the heatsink, then you get something like

\dot{Q}_{rad} = \epsilon\times{A\sigma{T^4}}

where all the constants I've used are as typical for this. That's your radiation component. Convection you can get via https://en.wikipedia.org/wiki/Heat_transfer_coefficient . In a steady state, it's just a matter of rearranging it all to solve for T... so I'm not sure where this root finding bisection nonsense comes in. Seems like they want you to use it just to see that you can use it, in which case what's the point? It's math, not physics, anyway I digress... Note that for convection, you'll be considering temperature difference.

If you do intend to go through the bisection nonsense they've dished out at you, here's my guide (from what I can understand from the Wiki page on it.)

i) get f. Q_{tot} is just the power of the resistor. The rest you can solve using what I've said above. Just rearrange it so that you have something like P + etcetc = 0 = f(T_s).

ii) ...

iii) Just consider the eq and sketch the graph. You should see where the derivative is positive, negative, etc. Anyhow, sketch the graph.

iv) The graph should have two roots (as they say, apparently) and one of them will be unphysical, i.e. it might lie in the domain T<0 or otherwise might just not be realistic for your setup. This one is one you can ignore.

v) Look at the "valid" root and the domain that it occupies. You might know that it lies somewhere above 0, but below 10,000. Use the bisection method and narrow the domain down.

If you know that the value switches from positive to negative (for f) in this range, where does that happen? Is it between 0 and 5000, or 5000 and 10000?

If the former, you look at the former domain. Split it again. 0 to 2500, 2500 to 5000.

Which part flips? First or second? Focus the one that does. Split it. Do this again.

See the wiki page if you're stuck with it, it actually does a good job. You just need to keep going until you find the point where the value of f(T_s) flips, which will let you find the solution you want.
0
reply
Lion1996
Badges: 9
Rep:
?
#5
Report Thread starter 5 months ago
#5
(Original post by Callicious)
Thanks for the img. The bisection method itself (at least to me) seems iterative: there isn't some formula for the final answer, you just follow it until you're satisfied with the convergence to a solution. I'm sure it'd be great to code with (and I've used a similar method before for root-finding for practice) but I can't see it's use practically.

Anyhow, if you just want to determine the surface temperature of the heat sink, you'll just need to balance out the equation of state for it. In this context, that's satisfying

P = \dot{Q}_{rad} + \dot{Q}_{conv}

where your \dot{Q}_{rad} will be given by

\dot{Q}_{rad}\propto{T^4}

with T the temperature of the surface of the heat sink. I've introduced dots because I'm talking about a rate of something, i.e. the amount of heat released per unit time from the heat sink, but anyhow, that's a bit pedantic I know...

The exact constant of proportionality depends on if the area A is just one side or both ("effective" area... is that effective over both sides, or just one? The terminology seems a bit ambiguous. I'd say "effective area of the top surface" or whatever else...).

If you assume A is just the total area of the heatsink, then you get something like

\dot{Q}_{rad} = \epsilon\times{A\sigma{T^4}}

where all the constants I've used are as typical for this. That's your radiation component. Convection you can get via https://en.wikipedia.org/wiki/Heat_transfer_coefficient . In a steady state, it's just a matter of rearranging it all to solve for T... so I'm not sure where this root finding bisection nonsense comes in. Seems like they want you to use it just to see that you can use it, in which case what's the point? It's math, not physics, anyway I digress... Note that for convection, you'll be considering temperature difference.

If you do intend to go through the bisection nonsense they've dished out at you, here's my guide (from what I can understand from the Wiki page on it.)

i) get f. Q_{tot} is just the power of the resistor. The rest you can solve using what I've said above. Just rearrange it so that you have something like P + etcetc = 0 = f(T_s).

ii) ...

iii) Just consider the eq and sketch the graph. You should see where the derivative is positive, negative, etc. Anyhow, sketch the graph.

iv) The graph should have two roots (as they say, apparently) and one of them will be unphysical, i.e. it might lie in the domain T<0 or otherwise might just not be realistic for your setup. This one is one you can ignore.

v) Look at the "valid" root and the domain that it occupies. You might know that it lies somewhere above 0, but below 10,000. Use the bisection method and narrow the domain down.

If you know that the value switches from positive to negative (for f) in this range, where does that happen? Is it between 0 and 5000, or 5000 and 10000?

If the former, you look at the former domain. Split it again. 0 to 2500, 2500 to 5000.

Which part flips? First or second? Focus the one that does. Split it. Do this again.

See the wiki page if you're stuck with it, it actually does a good job. You just need to keep going until you find the point where the value of f(T_s) flips, which will let you find the solution you want.
Thanks for the help. Just wondering how could you code it via Matlab please.
0
reply
Callicious
Badges: 22
Rep:
?
#6
Report 5 months ago
#6
(Original post by Lion1996)
Thanks for the help. Just wondering how could you code it via Matlab please.
You're doing this as a coding assignment? That makes it way easier!

The wiki page has some pseudocode you can base it on, but this also exists https://www.mathworks.com/matlabcent...d-root-finding
0
reply
Lion1996
Badges: 9
Rep:
?
#7
Report Thread starter 5 months ago
#7
(Original post by Callicious)
You're doing this as a coding assignment? That makes it way easier!

The wiki page has some pseudocode you can base it on, but this also exists https://www.mathworks.com/matlabcent...d-root-finding
Thanks for the help. do you know how to do this question please
Attached files
0
reply
Callicious
Badges: 22
Rep:
?
#8
Report 5 months ago
#8
(Original post by Lion1996)
Thanks for the help. do you know how to do this question please
I can take a look but I recommend making a new thread about it so that others can do so too

Also I don't know what level this is, but it certainly isn't GCSE or maybe not even AS/A... could you mark it as "AS/A" or "Undergrad" dependent on which one it is?
Last edited by Callicious; 5 months ago
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

What support do you need with your UCAS application?

I need help researching unis (5)
9.09%
I need help researching courses (5)
9.09%
I need help with filling out the application form (3)
5.45%
I need help with my personal statement (24)
43.64%
I need help with understanding how to make my application stand out (13)
23.64%
I need help with something else (let us know in the thread!) (2)
3.64%
I'm feeling confident about my application and don't need any help at the moment (3)
5.45%

Watched Threads

View All