Time taken for heat transfer

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  • #1
ken93

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Ques: I would like to calculate the time taken where the internal pipe's temperature would reach 150 degree Celsius too. A heat source will be constantly feeding on outside pipe where the temperature will be maintaining at 150 degree Celsius. Feel free to make any assumptions if there is parameters that i have missed out.

Assumptions:
Nominal pipe size = 5 inch
carbon steel thermal conductivity, k = 52 w/mk
Weight = 21.77 kg/m
Length = 0.5 m
Temperature outside pipe = 150 degree Celsius
Temperature inside pipe = 60 degree Celsius
Power source = 40 kw
Specific heat of carbon = 0.49 kj / (kgk)
 

Answers and Replies

  • #2
davenn
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Ques: I would like to calculate the time taken where the internal pipe's temperature would reach 150 degree Celsius too. A heat source will be constantly feeding on outside pipe where the temperature will be maintaining at 150 degree Celsius. Feel free to make any assumptions if there is parameters that i have missed out.

Assumptions:
Nominal pipe size = 5 inch
carbon steel thermal conductivity, k = 52 w/mk
Weight = 21.77 kg/m
Length = 0.5 m
Temperature outside pipe = 150 degree Celsius
Temperature inside pipe = 60 degree Celsius
Power source = 40 kw
Specific heat of carbon = 0.49 kj / (kgk)

well since you have tagged you post "A" for post graduate (it assumes you have a university degree), what have you done so far to find some answers ?[Mentors' note: the thread tag has been changed to 'B' since this post was made]
 
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  • #3
ken93
well since you have tagged you post "A" for post graduate (it assumes you have a university degree), what have you done so far to find some answers ?
I have tried to solve it but the answer i had is not logical. That is why im just checking if i could get any help. Thank you
 
  • #4
sophiecentaur
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I have tried to solve it but the answer i had is not logical. That is why im just checking if i could get any help. Thank you
What calculation did you use and what was the result? What was 'illogical' about your result?
 
  • #5
Nidum
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Let's see if we can arrive at a complete problem statement .

The outside wall temperature of the pipe is maintained at 150 C . The inside wall temperature is initially 60 C .

You want to know how long it will take for the inside wall temperature to reach 150 C .

What are the actual dimensions of the pipe ? Inside diameter ? Outside diameter ? Length of heated section1 ?

Is there anything taking heat away from the inside surface ? Any fluid flow ? Contact with other components ?

Is this an academic exercise or a real technical problem ? If the latter then please tell us more about it .

Note (1) : May not actually be needed for the calculations but useful to know anyway .
 
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  • #6
ken93
What calculation did you use and what was the result? What was 'illogical' about your result?
The fact is, im not even sure if the way i attempted on this question is even correct. Please, do enlighten if you know how. This is the way i approached it.
Q= (2*PI*31*162*1.1)/(5*3.412*(ln(5.563/5.345))
= 52535 W/hr
t = 52535/ 40000
t = 1.3 hr
 
  • #7
ken93
Let's see if we can arrive at a complete problem statement .

The outside wall temperature of the pipe is maintained at 150 C . The inside wall temperature is initially 60 C .

You want to know how long it will take for the inside wall temperature to reach 150 C .

What are the actual dimensions of the pipe ? Inside diameter ? Outside diameter ? Length of heated section1 ?

Is there anything taking heat away from the inside surface ? Any fluid flow ? Contact with other components ?

Is this an academic exercise or a real technical problem ? If the latter then please tell us more about it .

Note (1) : May not actually be needed for the calculations but useful to know anyway .
YES, i want to know how long it will take for the inside wall temperature to reach 150 C. This my ultimate answer im looking for.
OD = 5.563 in
ID = 5.345 in
Length of heated = 0.5 m
Assume the pipe is empty.
This is just a project im working on. Do let me know if you need any other extra information and i will check if i can get it. Your help is much appreciated. Thank you in advance. Cheers
 
  • #8
sophiecentaur
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The fact is, im not even sure if the way i attempted on this question is even correct. Please, do enlighten if you know how. This is the way i approached it.
Q= (2*PI*31*162*1.1)/(5*3.412*(ln(5.563/5.345))
= 52535 W/hr
t = 52535/ 40000
t = 1.3 hr
Formulae with just magic numbers in them really don't get across a lot of information. It helps if you show what the formula is actually doing before slotting in the values. Some derivation or a reference is useful. For instance, it will be an exponential change so it 'never actually gets there' so you need to decide what temperature is 'near enough'.
Also, you have the units W/hr in the formula. What does that mean? Watts are Power so what's watts per hour??
Can you step back a bit and describe how you are getting that result, i.e. the reasoning behind it? I don't think you have specified enough parameters to work it out. That's where the 'workings' would help to get us somewhere,
 
  • #9
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What makes you think that if you maintain a constant heat flux on the outside surface of the pipe, the outside surface temperature will remain constant at 150 C?
 
  • #10
jbriggs444
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The outside wall temperature of the pipe is maintained at 150 C . The inside wall temperature is initially 60 C .

You want to know how long it will take for the inside wall temperature to reach 150 C .
That's an easy question to answer: Never.

Temperature will never equilibriate in a finite time. It will just get close enough not to matter.
 
  • #11
ken93
Formulae with just magic numbers in them really don't get across a lot of information. It helps if you show what the formula is actually doing before slotting in the values. Some derivation or a reference is useful. For instance, it will be an exponential change so it 'never actually gets there' so you need to decide what temperature is 'near enough'.
Also, you have the units W/hr in the formula. What does that mean? Watts are Power so what's watts per hour??
Can you step back a bit and describe how you are getting that result, i.e. the reasoning behind it? I don't think you have specified enough parameters to work it out. That's where the 'workings' would help to get us somewhere,
0106ph-chromalox-eq7.png




This is the formula i gotten. Its called supplemental heating. So im a lil confused if this particular formula is actually correct for me to tackle this question. Okay, lets say i've decided the near enough temperature is 140C (Assume), is there any formula you can show me that i can achieve the time needed for it to reach that temperature?
 
  • #12
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This is the formula i gotten. Its called supplemental heating. So im a lil confused if this particular formula is actually correct for me to tackle this question. Okay, lets say i've decided the near enough temperature is 140C (Assume), is there any formula you can show me that i can achieve the time needed for it to reach that temperature?
Can you please show your derivation of this equation?

During the time that the pipe is heating up, there will be a radial temperature profile within the pipe that changes with time. This, of course, needs to be taken into account in determining the time required for the far (inner) boundary to reach a prescribed temperature. As I asked before, are you trying to solve this assuming that the rate of heat flow at the outer radius is constant (in which case, the temperature at the outer radius will be increasing with time) or assuming that the outer radius is held at a constant temperature of 150 C? Either way, I can tell you how to actually solve this, provided you are not rude enough to fail to respond to my questions again.

The solution to this problem depends on the solution to the transient heat conduction equation. I will refrain from providing more information at this time.
 
  • #13
ken93
Can you please show your derivation of this equation?

During the time that the pipe is heating up, there will be a radial temperature profile within the pipe that changes with time. This, of course, needs to be taken into account in determining the time required for the far (inner) boundary to reach a prescribed temperature. As I asked before, are you trying to solve this assuming that the rate of heat flow at the outer radius is constant (in which case, the temperature at the outer radius will be increasing with time) or assuming that the outer radius is held at a constant temperature of 150 C? Either way, I can tell you how to actually solve this, provided you are not rude enough to fail to respond to my questions again.

The solution to this problem depends on the solution to the transient heat conduction equation. I will refrain from providing more information at this time.
Im sincerely sorry if i have offended you that i did not reply you in the first place.

Back to the question, i did not derived the equation myself. it just an equation i found online which i THINK it works in this situation. Maybe there is some misunderstanding in my question. What i meant was a constant heat source of 150C will be applying to the outer surface of the pipe. So, in real world scenario, the pipe outer's temperature would increase overtime. Therefore, in this case i would want to know the solution assuming that the rate of heat flow at the outer radius is constant. It would be great if you could show me how it works. Thanks a lot
 
  • #14
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The thickness of the wall is only 0.218" , the average diameter is 5.454", and the average radius is 2.727". So the wall thickness is less than 10% of the average radius, and we can neglect the curvature of the wall. We can do this by slitting the pipe along its length, and then laying the pipe wall flat. This leaves us with a flat slab ##h=0.218(2.54)=0.554\ cm## thick, with cross section ##\pi (5.454)(2.54)=W=43.5## cm wide and ##L=50## cm long. If Q is the magnitude of the power source, the heat flux at the outer surface of the pipe (x = h) is ##\frac{Q}{WL}##, and the heat flux at the inner surface of the pipe (x = 0) is zero (the inner surface is assumed to be insulated).

The solution for the temperature distribution within the slab is a function of time t and distance above the inner surface x. In the present problem, this temperature distribution will feature a short term transient that dies out after a short amount of time, and a long time asymptotic temperature distribution that grows linearly with time. We will focus here primarily of the long time asymptote.

The average temperature of the slab (averaged over the volume of the slab) and any time will be given by:
$$\bar{T}=T_i+\frac{Qt}{\rho C LWh}$$, where Q is the total heating rate, t is the time since heating began, ##T_i## is the initial slab temperature (60 C), ##\rho## is the density of the steel, and C is the heat capacity of the steel. At long times, in addition to the average temperature, there will also be a superimposed temperature variation with location x through the slab, ##\hat{T}(x)##. So the overall long term temperature distribution will be given by $$T=\bar{T}+\hat{T}(x)=T_i+\frac{Qt}{\rho C LWh}+\hat{T}(x)\tag{1}$$

More generally, the temperature variations within the slab are described by the transient heat conduction equation:$$\rho C \frac{\partial T}{\partial t}=k\frac{\partial^2T}{\partial x^2}\tag{2}$$
If we substitute Eqn. 1 into Eqn. 2, we obtain:$$\frac{Q}{LWh}=k\frac{d^2\hat{T}}{d x^2}\tag{3}$$If we integrate Eqn. 3 once with respect to x, we obtain:$$k\frac{d\hat{T}}{dx}=\frac{Q}{LWh}x\tag{4}$$where use has been made here of the zero heat flow boundary condition at x = 0. If we next integrate Eqn. 4 with respect to x, we obtain: $$\hat{T}=\hat{T}(0)+\frac{Q}{kLWh}\frac{x^2}{2}\tag{5}$$where ##\hat{T}(0)## is the constant of integration. The value of this constant of integration can be established by noting that, since ##\bar{T}## is the average temperature of the slab, the average value of ##\hat{T}## must be equal to zero. This leads to the result that:
$$\hat{T}(0)=-\frac{Qh}{6kLW}\tag{6}$$

So, our solution for the long-term temperature distribution is given by: $$T=T_i+\frac{Qt}{\rho C LWh}-\frac{Qh}{6kLW}+\frac{Q}{kLWh}\frac{x^2}{2}$$
So, the temperature at the inner surface of the pipe is: $$T(0)=T_i+\frac{Q}{\rho C LWh}\left(t-\frac{1}{6}\frac{h^2}{\alpha}\right)\tag{7}$$where ##\alpha## is the thermal diffusivity ##\left(\frac{k}{\rho C}\right)##. And the temperature at the outer surface of the pipe is: $$T(h)=T_i+\frac{Q}{\rho C LWh}\left(t+\frac{1}{3}\frac{h^2}{\alpha}\right)\tag{8}$$The difference in temperature between the inner and outer surface will be:
$$T(h)-T(0)=\frac{Qh}{2kLW}\tag{9}$$
 
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  • #15
ImNotOliver
There is another way you can calculate the problem. A method that Thomas Edison would have used. Attach a thermometer to the pipe, start your stop watch and begin heating.
 
  • #16
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There is another way you can calculate the problem. A method that Thomas Edison would have used. Attach a thermometer to the pipe, start your stop watch and begin heating.
I don't need to purchase or set up any equipment, get it safety checked, and spend time doing the experiments (and analyzing the experimental results). What is your estimate of the time and cost it would take for TAE to do this in today's dollars with modern equipment?
 
  • #17
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Just a quick check, do you actually have some of this pipe? Only reason I ask, 5 inch pipe is kind of unusual, especially your thin-wall schedule 5. I don't think I've ever actually seen pipe that size. Maybe it is used in food industry or something I'm not familiar with?
 
  • #18
ken93
The thickness of the wall is only 0.218" , the average diameter is 5.454", and the average radius is 2.727". So the wall thickness is less than 10% of the average radius, and we can neglect the curvature of the wall. We can do this by slitting the pipe along its length, and then laying the pipe wall flat. This leaves us with a flat slab ##h=0.218(2.54)=0.554\ cm## thick, with cross section ##\pi (5.454)(2.54)=W=43.5## cm wide and ##L=50## cm long. If Q is the magnitude of the power source, the heat flux at the outer surface of the pipe (x = h) is ##\frac{Q}{WL}##, and the heat flux at the inner surface of the pipe (x = 0) is zero (the inner surface is assumed to be insulated).

The solution for the temperature distribution within the slab is a function of time t and distance above the inner surface x. In the present problem, this temperature distribution will feature a short term transient that dies out after a short amount of time, and a long time asymptotic temperature distribution that grows linearly with time. We will focus here primarily of the long time asymptote.

The average temperature of the slab (averaged over the volume of the slab) and any time will be given by:
$$\bar{T}=T_i+\frac{Qt}{\rho C LWh}$$, where Q is the total heating rate, t is the time since heating began, ##T_i## is the initial slab temperature (60 C), ##\rho## is the density of the steel, and C is the heat capacity of the steel. At long times, in addition to the average temperature, there will also be a superimposed temperature variation with location x through the slab, ##\hat{T}(x)##. So the overall long term temperature distribution will be given by $$T=\bar{T}+\hat{T}(x)=T_i+\frac{Qt}{\rho C LWh}+\hat{T}(x)\tag{1}$$

More generally, the temperature variations within the slab are described by the transient heat conduction equation:$$\rho C \frac{\partial T}{\partial t}=k\frac{\partial^2T}{\partial x^2}\tag{2}$$
If we substitute Eqn. 1 into Eqn. 2, we obtain:$$\frac{Q}{LWh}=k\frac{d^2\hat{T}}{d x^2}\tag{3}$$If we integrate Eqn. 3 once with respect to x, we obtain:$$k\frac{d\hat{T}}{dx}=\frac{Q}{LWh}x\tag{4}$$where use has been made here of the zero heat flow boundary condition at x = 0. If we next integrate Eqn. 4 with respect to x, we obtain: $$\hat{T}=\hat{T}(0)+\frac{Q}{kLWh}\frac{x^2}{2}\tag{5}$$where ##\hat{T}(0)## is the constant of integration. The value of this constant of integration can be established by noting that, since ##\bar{T}## is the average temperature of the slab, the average value of ##\hat{T}## must be equal to zero. This leads to the result that:
$$\hat{T}(0)=-\frac{Qh}{6kLW}\tag{6}$$

So, our solution for the long-term temperature distribution is given by: $$T=T_i+\frac{Qt}{\rho C LWh}-\frac{Qh}{6kLW}+\frac{Q}{kLWh}\frac{x^2}{2}$$
So, the temperature at the inner surface of the pipe is: $$T(0)=T_i+\frac{Q}{\rho C LWh}\left(t-\frac{1}{6}\frac{h^2}{\alpha}\right)$$where ##\alpha## is the thermal diffusivity ##\left(\frac{k}{\rho C}\right)##. And the temperature at the outer surface of the pipe is: $$T(h)=T_i+\frac{Q}{\rho C LWh}\left(t+\frac{1}{3}\frac{h^2}{\alpha}\right)$$The difference in temperature between the inner and outer surface will be:
$$T(h)-T(0)=\frac{Qh}{2kLW}$$
I havent had the time to review and study it yet. However, thank you so much sir for your assistance. Really appreciate it. Thank you. Have a good day sir.
 
  • #19
ken93
Just a quick check, do you actually have some of this pipe? Only reason I ask, 5 inch pipe is kind of unusual, especially your thin-wall schedule 5. I don't think I've ever actually seen pipe that size. Maybe it is used in food industry or something I'm not familiar with?
Oh it's just an example im giving. I dont actually have an 5" pipe.
 
  • #20
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There is another way you can calculate the problem. A method that Thomas Edison would have used. Attach a thermometer to the pipe, start your stop watch and begin heating.
I don't need to purchase or set up any equipment, get it safety checked, and spend time doing the experiments (and analyzing the experimental results). What is your estimate of the time and cost it would take for TAE to do this in today's dollars with modern equipment?
Might not cost the OP anything, and may not take much time. 150 C (~ 300 F) is in cooking temperature ranges, a thermostatically controlled frying pan, or a frying pan in a 150 C oven should get you close enough for what they seem to be after. OP seems to be looking for rough numbers (140 C out of 150 C good enough). So a scrap of 1/4" sheet steel at room temperature, in a frying pan with a bit of cooking oil (for good heat transfer to the 'pipe') heated and stabilized to ~ 150 C, measured with a infrared thermometer (doesn't everyone own one? If not they should! They are very handy).
 
  • #21
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Might not cost the OP anything, and may not take much time. 150 C (~ 300 F) is in cooking temperature ranges, a thermostatically controlled frying pan, or a frying pan in a 150 C oven should get you close enough for what they seem to be after. OP seems to be looking for rough numbers (140 C out of 150 C good enough). So a scrap of 1/4" sheet steel at room temperature, in a frying pan with a bit of cooking oil (for good heat transfer to the 'pipe') heated and stabilized to ~ 150 C, measured with a infrared thermometer (doesn't everyone own one? If not they should! They are very handy).
And you really feel that that would satisfy the OP for his pipe example? And you really feel that that would be easier to do that than to just calculate $$t=WC\Delta T/Q$$where W is the mass of the pipe, C is the pipe heat capacity, ##\Delta T## is the temperature increase, and Q is the heating rate?
 
  • #22
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I know from your previous postings that he calculation is easier for you.

But I don't know the OP well enough to know which is easier for them. It was just a suggestion, OP can do with it what they wish. I think doing both might be the most illuminating.
 
  • #23
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I know from your previous postings that he calculation is easier for you.

But I don't know the OP well enough to know which is easier for them. It was just a suggestion, OP can do with it what they wish. I think doing both might be the most illuminating.
Does the equation I gave seem complicated to you?
 
  • #24
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Does the equation I gave seem complicated to you?
I'm not sure it matters, I'm not the OP.

I don't find the equation complicated, but I am maybe a little confused about its application to the OPs condition - I see the mass of the pipe, but I don't see where the thickness of the pipe's walls comes into play.

Intuitively, it seems to me that it would take longer to reach 140 C on the inside of a thick walled pipe than it would with a thin walled pipe, both being heated to 150 C on the outside. Am I missing that? Is it part of "Q"? That wasn't obvious to me, you used the term "heating rate" earlier, I thought maybe that was the rate of heat applied from the source? But maybe that's the key - heat rate through the steel?

So I guess from the view of the application of the equation, and understanding where some real-life conditions might affect the expected result from a calculated one, maybe it does seem somewhat complicated. And apparently the OP didn't absorb it immediately, they've been gone for ~ 24 hours.
 
  • #25
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I'm not sure it matters, I'm not the OP.

I don't find the equation complicated, but I am maybe a little confused about its application to the OPs condition - I see the mass of the pipe, but I don't see where the thickness of the pipe's walls comes into play.

Intuitively, it seems to me that it would take longer to reach 140 C on the inside of a thick walled pipe than it would with a thin walled pipe, both being heated to 150 C on the outside. Am I missing that? Is it part of "Q"? That wasn't obvious to me, you used the term "heating rate" earlier, I thought maybe that was the rate of heat applied from the source? But maybe that's the key - heat rate through the steel?

So I guess from the view of the application of the equation, and understanding where some real-life conditions might affect the expected result from a calculated one, maybe it does seem somewhat complicated. And apparently the OP didn't absorb it immediately, they've been gone for ~ 24 hours.
I guess what you say here has some validity. But, in the analysis I presented, I also provided simple algebraic equations for calculating the temperatures of the inner and outer pipe surfaces as functions of time (Eqns. 7 and 8). And I also provided a simple algebraic equation for calculating the difference in temperature between the outside surface and the inside surface. Would you find these equations difficult to apply?
 

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