Temperature of fluid flowing through pipe

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SUMMARY

The discussion centers on the mathematical relationship for temperature change in fluids flowing through pipes, specifically in contexts such as exhaust pipes and radiator hoses. Key parameters influencing temperature change include cross-sectional diameter, viscosity, volumetric flow rate, specific heat, length of the pipe, density, and thermal conductivity. Participants recommend consulting "Transport Phenomena" by Bird, Stewart, and Lightfoot for detailed equations and methodologies. The complexity of heat transfer in these systems necessitates a thorough understanding of both fluid dynamics and thermodynamics.

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  • Understanding of fluid dynamics principles
  • Knowledge of thermodynamic properties such as specific heat and thermal conductivity
  • Familiarity with heat transfer concepts, including conduction and convection
  • Basic mathematical skills for applying equations related to heat transfer
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  • Study "Fanno flow" analysis for stationary pipes to understand flow exit velocity
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Engineers, thermodynamics students, and professionals involved in automotive engineering or fluid mechanics who are looking to optimize heat transfer in piping systems.

  • #91
Jay_ said:
Okay. One more thing. How do we obtain the constants for the outside of the pipe (of standard air)? You mentioned BSL section on submerged objects.

We can find the thermal conductivity using this equation :
http://bouteloup.pierre.free.fr/lica/phythe/don/air/air_k_plot.pdf
You have actual data for air in Perry's Handbook.

BTW, I should mention that we made a slight oversight in our approach so far. For evaluating the physical properties (see BSL), we should be using the arithmetic average of the wall temperature and the bulk gas temperature (inside the pipe), and the arithmetic average of the wall temperature and the ambient air for the outside of the pipe. Sorry. I forgot that.
From BSL would I be using equation 14.4-7 and 14.4-8 to find the Nusselt number (since it is a cylinder), and then find the value of 'h'? In doing so what would be the value of the characteristic length L as we go to find h from Nu?
This is not strictly Kosher, since, the correlation is for gas flow perpendicular to the cylinder, and in our case, it is parallel to the cylinder. I mentioned earlier that, if we do this, we will get an upper bound to the Nu and h.

Doing heat transfer to a submerged body is much more uncertain than for internal flow. For one thing, even in the turbulent flow region, the turbulent boundary layer thickness is growing with distance along the body, and the heat transfer coefficient is decreasing with distance. For a sphere or a cylinder in cross flow, this is not important, but for a tailpipe (where the air flow is axial), it can be. It might be useful to calculate the local heat transfer coefficient on the tailpipe by treating it as a flat plate (valid if the boundary layer thickness is small compared to the pipe radius), but the correlation in BSL is for a sharp edged entry, while, in our case, there is a muffler at the entrance (which certainly does not provide a sharp edged entrance).

If I were you, I would do 4 scouting calculations, and compare the results:

1. Assume a sharp edged entrance for the flat plate situation and with the Re evaluated at the distance x behind the muffler that the thermocouple is situated (as the characteristic length). This would give an upper bound to h.

2. Calculate the h from the tool box equation

3. Calculate the h using the cross flow correlation for a cylinder. This would give an upper bound for h

4. Calculate the h using the internal flow calculation, with v taken as the car speed and the characteristic length taken as D. Since, for internal flow, the boundary layer thickness does not grow, this would also give an upper bound for h.

Compare these 4 results, and see what you get. At worst, choose the one from 1, 3, 4 that gives the lowest upper bound, or choose 2. I would run some numbers to see.

Chet
 
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  • #92
You have actual data for air in Perry's Handbook.

But how is the upper bound for h be useful? I need a value to put into the equation.

I want to just go with the toolbox equation. But the toolbox equation has no connection to temperature. Heat transfer coefficient depends on temperature too right, because it depends on the thermal conductivity and that changes with temperature?

TW, I should mention that we made a slight oversight in our approach so far. For evaluating the physical properties (see BSL), we should be using the arithmetic average of the wall temperature and the bulk gas temperature (inside the pipe), and the arithmetic average of the wall temperature and the ambient air for the outside of the pipe.

Okay. So that means when I am finding the viscosity at a given temperature for the gas, I should fin the average of the gas temperature and the wall temperature inside and look for the pure gas viscosity values at that temperature?

Same for the thermal conductivity and specific heat? Example if I got gas temperature as 582 C, and I get the outside wall temperature as 150 C (and we are assuming this is the same as the inside wall temperature), then, I need to find the values of the pure gas constants at 366 C? And then use these values in the mixing rules correct?

------------

Also the if the mole fractions are 0.734 (N2), 0.125 (CO2), 0.141 (H2O) the specific heat at a given temperature of this mixture would just be

Specific_heat_mix = 0.734(Specific_heat_N2) + 0.125(Specific_heat_CO2) + 0.141(Specific_heat_H2O)

Correct? I am sorted with viscosity, and thermal conductivity, and specific heat for both sides of the exhaust (that is inside wall and outside wall), I don't see the temperature relation to the toolbox equation which is why I don't feel good using it.
 
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  • #93
Jay_ said:
But how is the upper bound for h be useful? I need a value to put into the equation.
You choose the one that gives the least upper bound.
I want to just go with the toolbox equation. But the toolbox equation has no connection to temperature. Heat transfer coefficient depends on temperature too right, because it depends on the thermal conductivity and that changes with temperature?
Yes. That's what I don't like about it. But, it's better than nothing.


Okay. So that means when I am finding the viscosity at a given temperature for the gas, I should fin the average of the gas temperature and the wall temperature inside and look for the pure gas viscosity values at that temperature?

Same for the thermal conductivity and specific heat? Example if I got gas temperature as 582 C, and I get the outside wall temperature as 150 C (and we are assuming this is the same as the inside wall temperature), then, I need to find the values of the pure gas constants at 366 C? And then use these values in the mixing rules correct?

yes.

------------
Also the if the mole fractions are 0.734 (N2), 0.125 (CO2), 0.141 (H2O) the specific heat at a given temperature of this mixture would just be

Specific_heat_mix = 0.734(Specific_heat_N2) + 0.125(Specific_heat_CO2) + 0.141(Specific_heat_H2O)
yes
Correct? I am sorted with viscosity, and thermal conductivity, and specific heat for both sides of the exhaust (that is inside wall and outside wall), I don't see the temperature relation to the toolbox equation which is why I don't feel good using it.
Yes. It doesn't have a length scale either.

P.s. Don't forget to use the properties of air on the outside of the pipe, not the exhaust gas, when you use the correlations in bsl.

Chet
 
  • #94
Don't forget to use the properties of air on the outside of the pipe, not the exhaust gas, when you use the correlations in bsl.

That's is where I am stuck. I need one calculation because I want to make it into code. So is there any text that explains what the value of h would be if air flow was parallel (and not perpendicular)?

This link below maybe? It doesn't give me an equation though.

http://deepblue.lib.umich.edu/bitst...98/707_2005_Article_BF01410516.pdf?sequence=1

I found an online powerpoint presentation too :

http://www.kostic.niu.edu/352/Cen4Ed/Heat_4e_Chap07_lecture.ppt

1. Does to top most equation in slide 22 look like the one I should use?

2/Again, in slide 23 they have so many equations and for the Nusselt number. And for the Reynolds number of 6717 we got (4000 < Re < 40,000) it says we should use Nu = 0.193*(Re^0.618)*(Pr^(1/3))

I calculated the Nusselt number using that and the Pr = 0.68, I get Nu = 39.38

Isn't this significantly different from the value of Nu = 24 we got earlier using Nu = 0.004*Re*(Pr^(1/3)).

Is slide 23 better or is BSL more accurate?
 
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  • #95
Jay_ said:
That's is where I am stuck. I need one calculation because I want to make it into code. So is there any text that explains what the value of h would be if air flow was parallel (and not perpendicular)?

I don't think you are going to find it. The presence of the hot muffler before the tailpipe screws everything up. If it weren't for that, I would use the correlation for heat transfer in flow along a flat plate. As I said, this could be used for the tailpipe if the BL thickness is small compared to the tailpipe radius. It would be a function of the placement of the thermocouple relative to the entrance to the tailpipe.

What we're doing here is going to require a judgement call. That's why I've been urging you to do calculations for the various cases that I mentioned. After looking at the results from all these upper bound cases, we are going to have to decide which equation, or modification thereof, we feel will best quantify the heat transfer coefficient for the flow over the outside of the tailpipe. But, right now, we don't have any basis for making such a judgement call because we don't have any results for comparison of these various cases.

This link below maybe? It doesn't give me an equation though.

http://deepblue.lib.umich.edu/bitst...98/707_2005_Article_BF01410516.pdf?sequence=1

I found an online powerpoint presentation too :

http://www.kostic.niu.edu/352/Cen4Ed/Heat_4e_Chap07_lecture.ppt

1. Does to top most equation in slide 22 look like the one I should use?
This equation is for external flow across a cylinder. This is one of the cases we are want to look at.
2/Again, in slide 23 they have so many equations and for the Nusselt number. And for the Reynolds number of 6717 we got (4000 < Re < 40,000) it says we should use Nu = 0.193*(Re^0.618)*(Pr^(1/3))

I calculated the Nusselt number using that and the Pr = 0.68, I get Nu = 39.38

Isn't this significantly different from the value of Nu = 24 we got earlier using Nu = 0.004*Re*(Pr^(1/3)).

Is slide 23 better or is BSL more accurate?
The equation on slide 23 is for flow across a cylinder. The equation in BSL is for axial flow inside a cylinder. The equation in BSL works well for internal flow, and the equation on slide 23 works well for external flow across a cylinder. They have nothing to do with one another.

The equation that I think will work best for your situation is on slide 11, which applies to flow over a flat plate (for the reasons that I mentioned above). But, before making the judgement call, i think we should look at the results for the other cases I mentioned.

Chet
 
  • #96
The equation on slide 23 is for flow across a cylinder.

So couldn't we use this to find hout?

1. Assume a sharp edged entrance for the flat plate situation and with the Re evaluated at the distance x behind the muffler that the thermocouple is situated (as the characteristic length). This would give an upper bound to h.

But I don't know the value of x as of now. Let me assume it to be 10 cm. Now we started with an assumption of 650 C gas temperature. And my outer wall temperature was 250 C, ambient is 35 C, so I should take the average : 142.5 C. I should do all calculations using this, right?

From BSL,
Rx = vρx/μ =>

v = 15 m/s
ρ (@ 142.5 C, air) = 0.847 (SI)
μ (@ 142.5 C, air) = 2.3922E-5
x = 0.1

=> Rex = 53011 (pretty high?!)

Pr (@ 142.5 C, air) = 0.695

Nu = 2sqrt(37/1260)*(Re^(1/2))*(Pr^(1/3)) = 69.962

k (@ 142.5 C, air) = 0.0346
h = Nu*k/x => hout = 24.207

2. Calculate the h from the tool box equation

Let me calculate it at 15 m/s. So, h2_out = 34.18

3. Calculate the h using the cross flow correlation for a cylinder. This would give an upper bound for h

Which equation?

4. Calculate the h using the internal flow calculation, with v taken as the car speed and the characteristic length taken as D. Since, for internal flow, the boundary layer thickness does not grow, this would also give an upper bound for h.

In this don't I have to start from Re = 4W/(mu*pi*D)? What would the flow rate W be?

Just as how we had a few steps for the internal flow, couldn't we just have equations for a certain value for the external flow also? I am getting really confused. Could you post the calculation you expected from slide 11? I don't know what x is or what to take its value as.
 
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  • #97
If my calculation in the second section of my previous post where I got h = 24.207 seems right, I think I will go with that. But it looks wrong because it is nowhere near 34.

I showed my professor some calculations the other day, but I want to finish the model soon. This has taken way longer than expected.

I really appreciate all your help. This would have been impossible otherwise. I just want to know one reliable way to calculate
 
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  • #98
Jay_ said:
So couldn't we use this to find hout?

Yes. This is approach #3.
But I don't know the value of x as of now. Let me assume it to be 10 cm. Now we started with an assumption of 650 C gas temperature. And my outer wall temperature was 250 C, ambient is 35 C, so I should take the average : 142.5 C. I should do all calculations using this, right?

From BSL,
Rx = vρx/μ =>

v = 15 m/s
ρ (@ 142.5 C, air) = 0.847 (SI)
μ (@ 142.5 C, air) = 2.3922E-5
x = 0.1

=> Rex = 53011 (pretty high?!)

Pr (@ 142.5 C, air) = 0.695

Nu = 2sqrt(37/1260)*(Re^(1/2))*(Pr^(1/3)) = 69.962

k (@ 142.5 C, air) = 0.0346
h = Nu*k/x => hout = 24.207
Nicely done, if the "arithmetic" is correct.

Let me calculate it at 15 m/s. So, h2_out = 34.18
So, do you see what I'm saying. That's why I don't have a lot of confidence in that toolbox equation.
Which equation?
You use either the equation in BSL or the equation in the Powerpoint presentation. They should give very similar results. You calculate the Reynolds number using 15 m/s, and the characteristic length for flow across a cylinder, namely, the tube diameter.
In this don't I have to start from Re = 4W/(mu*pi*D)?
No. This equation for Re is specific to flow inside a tube. You use the general equation for the Reynolds number, with 15 m/s, characteristic length equal to the diameter, and the density and viscosity evaluated at 142.5 C. We are just using the "axial flow inside the tube" result to get an upper bound to the "axial flow outside the tube" result.

Just as how we had a few steps for the internal flow, couldn't we just have equations for a certain value for the external flow also? I am getting really confused. Could you post the calculation you expected from slide 11? I don't know what x is or what to take its value as.
You already did this correctly above under item 1; assuming 10 cm was a pretty good idea and approximation. So please, don't despair. You're doing a great job. For the outside flow, you don't need to iterate on the temperature, since the wall temperature and the bulk flow temperature (ambient air) are already known, and the film temperature is established.

So far, you've shown that the toolbox equation gives too high a value for h in your situation. Now let's see what items 3 and 4 predict for the upper bound.

Chet
 
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  • #99
Jay_ said:
If my calculation in the second section of my previous post where I got h = 24.207 seems right, I think I will go with that. But it looks wrong because it is nowhere near 34.

I showed my professor some calculations the other day, but I want to finish the model soon. This has taken way longer than expected.

I really appreciate all your help. This would have been impossible otherwise. I just want to know one reliable way to calculate
You've made great progress. This is what I was hoping for. We're almost there.

Chet
 
  • #100
Rx = vρx/μ =>

v = 15 m/s
ρ (@ 142.5 C, air) = 0.847 (SI)
μ (@ 142.5 C, air) = 2.3922E-5
x = 0.1

=> Rex = 53011
=> Pr (@ 142.5 C, air) = 0.695
------------------------------

Silde 11

Nu = 0.0296*(Re^0.8)*(Pr^(1/3)) = 157.8

k (@ 142.5 C, air) = 0.0346
x = 0.1 m

h = Nu*k/x
h_slide_11_out = 54.6

------------------------------------------

Slide 12 (1)


Nu = 0.037*(Re^0.8)*(Pr^(1/3)) = 197.25
k (@ 142.5 C, air) = 0.0346
x = 0.1 m

h = Nu*k/x
h_slide_12.1_out = 68.25

I think this is not the equation because it has L and not x
-----------------------------------------

So do I take the value as the one obtained from slide 11, viz. h_slide_11_out = 54.6

Doing that, and using :

T_{gas} = \frac{(T_{otp} - T_{amb})*54.6}{22}+T_{inp}

If I have a wall temperature of 250 C (same as Totp and Tinp), and Tamb = 35 C, I get Tgas = 783.6 C Sound right?

Or do I have to work based on the mean values? So that would give me Tgas_boundary = (142.5 - 35)*54.6/22 + 142.5 = 409.3 C for the boundary.

And so (Tgas + Twall)/2 = 409.3 => Tgas = 676.1 C
 
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  • #101
Jay_ said:
Rx = vρx/μ =>

v = 15 m/s
ρ (@ 142.5 C, air) = 0.847 (SI)
μ (@ 142.5 C, air) = 2.3922E-5
x = 0.1

=> Rex = 53011
=> Pr (@ 142.5 C, air) = 0.695
------------------------------

Silde 11

Nu = 0.0296*(Re^0.8)*(Pr^(1/3)) = 157.8

k (@ 142.5 C, air) = 0.0346
x = 0.1 m

h = Nu*k/x
h_slide_11_out = 54.6
I'm a little confused. I thought in post 96 you got the Nu and h for flow along a flat plate. This result should match that one, but it doesn't. Where did you get the constants for post #96.

Also, in this calculation, you determined a Re of 56000, and yet you used the Nu equation applicable to Re>500000. What gives. You should have used the other equation from slide 11. Maybe that will match better.

Note also that, the lower the value of h2, the closer your measurement of the wall temperature will come to the hot exhaust temperature, and the more accurate your calculation will be.

------------------------------------------
Slide 12 (1)


Nu = 0.037*(Re^0.8)*(Pr^(1/3)) = 197.25
k (@ 142.5 C, air) = 0.0346
x = 0.1 m

h = Nu*k/x
h_slide_12.1_out = 68.25

I think this is not the equation because it has L and not x
Right. This gives the average value of the heat transfer coefficient from x = 0 to x = L, and you want the local value where your temperature measurement is situated. Here again, you used the wrong equation for Nu, applicable to Re > 500000.

-----------------------------------------
So do I take the value as the one obtained from slide 11, viz. h_slide_11_out = 54.6

We haven't decided yet what we want to use to get h2.
Doing that, and using :

T_{gas} = \frac{(T_{otp} - T_{amb})*54.6}{22}+T_{inp}

If I have a wall temperature of 250 C (same as Totp and Tinp), and Tamb = 35 C, I get Tgas = 783.6 C Sound right?
Methodologically correct.
Or do I have to work based on the mean values? So that would give me Tgas_boundary = (142.5 - 35)*54.6/22 + 142.5 = 409.3 C for the boundary.

And so (Tgas + Twall)/2 = 409.3 => Tgas = 676.1 C
Methodologically incorrect.

Chet
 
  • #102
I'm a little confused. I thought in post 96 you got the Nu and h for flow along a flat plate. This result should match that one, but it doesn't. Where did you get the constants for post #96.

I used the same constants for all of them. I got them in various sites:

http://www.lmnoeng.com/Flow/GasViscosity.php

I used different equations for Nu though.

The one I did in post # 96 is the equation from BSL (equation 14.4-2). I didn't notice the Reynolds number limits in the slides. Sorry for that, I just went by the word "Turbulent".

Now using the first equation in slide 11.

Rex = 53011
Pr (@ 142.5 C, air) = 0.695

Nu = 0.332(Re^0.5)*(Pr^(1/3)) = 67.71

h = Nu*k/x = 67.71*0.0346/0.1 => h = 23.43

The value seems a little small
 
  • #103
Jay_ said:
I used the same constants for all of them. I got them in various sites:

http://www.lmnoeng.com/Flow/GasViscosity.php

I used different equations for Nu though.

The one I did in post # 96 is the equation from BSL (equation 14.4-2). I didn't notice the Reynolds number limits in the slides. Sorry for that, I just went by the word "Turbulent".

Now using the first equation in slide 11.

Rex = 53011
Pr (@ 142.5 C, air) = 0.695

Nu = 0.332(Re^0.5)*(Pr^(1/3)) = 67.71

h = Nu*k/x = 67.71*0.0346/0.1 => h = 23.43

The value seems a little small
Much better.
 
  • #104
So what are we using for the air ultimately?
 
  • #105
So far the leading candidate is the one you just did, but you haven't done flow over a cylinder yet. Let's see what that gives.

Chet
 
  • #106
Would that be equation number 14.4-7?

Using

Num = (0.4(Re^(1/2)) + 0.06(Re^(2/3)))*Pr^(0.4) (I neglect the viscosity factor, assuming it to be unity)

Given the Reynold number values Re = 53011 and Pr = 0.695, I get

Num = 152.82

k = 0.0346 (air @ 142.5)
x = 0.1

h = Nu*k/x => h = 52.88

Which one do we use?
 
  • #107
We use the one giving the lowest upper bound, namely #3. From this, it looks like, at least for 35 mph, the ratio of h2 to h1 is close to 1.0. I wonder how much this ratio would vary with speed. Maybe if it doesn't change much with speed, you can use 1.0 for all speeds. This would certainly simplify things for you greatly.

Chet
 
  • #108
We use the one giving the lowest upper bound, namely #3.

You mean the one in post #102, where I got a value of h = 23.43? I think this is rather surprising. I would have expected the value to be higher. Another thing I want to ask: Is our assumption of considering it to be a flat plate wrong? The reason I ask is because the pipe is nothing like a flat plate, its more like a cylinder moving along its height-axis.
 
  • #109
I'll explain later the connection between axial external flow along a cylinder, and flow at the entrance of a flat plate. But, right now, I'm out with my grandchildren. For now, consider the case in which the cylinder diameter is very large (like say a mile). How important would the curvature be then?

Why don't you do the calculation for the case of axial flow along a cylinder using the info in the paper you referenced? I predict you will get about the same result.

Chet
 
  • #110
That is so sweet. Have a nice time with your grand-kids :-)

Okay. I assume you are talking about the equation in slide 22?

Nucyl = 0.3 + (A/B)*C

where, A = 0.62*(Re^(1/2))*(Pr^(1/3))
B = (1 + (0.4/Pr)^(2/3))^(1/4)
C = (1 + (Re/282000)^(5/8))^(4/5)

Using Re = 53011, and Pr = 0.695

Nucyl = 141.4022

k = 0.0346, x = 0.1

h = Nu*k/x = 48.93

When you are free could you let me know if this is all correct? This value is far from the other 'h'. But it seems more correct because I would get a gas temperature that is double the value and that seems right. Should I go with this one?
 
  • #111
No. That equation is for flow across a cylinder, rather than axial. I found a reference that has axial flow, but I don't have my computer with me; it's bookmarked.

Chet
 
  • #112
Post me the link. I need to complete this without much more delay. :-/
 
  • #113
Jay_ said:
Post me the link. I need to complete this without much more delay. :-/
Hi Jay. Unfortunately, I misspoke. I did not bookmark the link, and now I can't find it again. Sorry.

Back to heat transfer in flow over a flat plate, and its relationship to heat transfer in axial flow over a cylinder. Imagine in the case of a cylinder that you increase its radius until it is infinite. Then you end up with a flat plate. So you know that a flat plate is going to be a limiting case of a cylinder. All the "action" in heat transfer and momentum transfer in axial gas flow over a cylinder occurs in very close proximity to the surface; this is the so-called boundary layer region. Outside the boundary layer, the temperature and velocity variation in the air are unaware that the cylinder is there. If the boundary layer is very thin compared to the radius of the cylinder, the curvature of the cylinder is unimportant in determining the heat transfer through the boundary layer, and the system can be treated as a flat plate.

Even though the car is moving through stagnant air, if the car velocity is constant, it doesn't matter whether you regard the car as moving and the air stagnant, or the car stationary and the air blowing backwards over the exhaust. This is just a change in inertial reference frame. But, it makes it much easier to analyze what is happening. So if air is blowing backwards axially along the tailpipe, the leading edge of the tailpipe is at the muffler. The boundary layer builds up in thickness from the muffler (leading edge) toward the back of the tailpipe, in proportion to the square root of distance backward. Since heat has to be conducted through the boundary layer to the fresh air in the free stream, this causes the heat transfer coefficient to decrease with distance measured backwards from the muffler. This can now be recognized as something very similar to what happens in heat transfer to a gas flowing along a flat plate, starting at the leading edge.

The question of the validity of using the correlation for flow over a flat plate to describe flow and heat transfer over a cylinder hinges on the thickness of the thermal boundary layer compared to the radius of the cylinder. If this ratio is low compared to unity, then it is valid to use the flat plate approximation. Let's see how this plays out in our situation. The equation for the heat flux through the boundary layer is given by:
q= k\frac{ΔT}{δ}
where δ is the boundary layer thickness and ΔT is the wall temperature minus the free-stream air temperature. This is the equation we used in freshman physics for heat conduction through a slab of thickness δ. The only question is, "what is the boundary layer thickness δ?" Now, from our heat transfer correlation, we also know that:
q = h ΔT
If we combine these two equations, we get:
h=\frac{k}{δ}
Now, the local Nusselt number for flow over a flat plate from our correlation is given by:
Nu_x=\frac{hx}{k}
where x is the distance from the leading edge of the plate (i.e., in our situation, the distance measured backwards from the muffler). If we combine these two equations, we obtain:
δ=\frac{x}{Nu_x}
So, for x = 10 cm, we calculate δ≈0.5 cm. This compares with a cylinder radius of 3.5 cm, so the ratio of the boundary layer thickness to the cylinder radius is only ≈ 0.15. This validates the use of the flat plate approximation for our situation.

There are lots of factors that can affect the accuracy of the heat transfer coefficient we calculate, and the uncertainties are large (over and above neglecting the curvature of the tailpipe surface). The biggest one is the effect of the muffler. Its presence prevents the existence of a sharp leading edge to the tailpipe. There is also heat from the muffler entering the free-stream air that blows backwards over the tailpipe, and this affects the temperature difference ΔT. There can also be a circulation zone (stagnation zone) that forms in the wake of the muffler which will affect both the air flow and the heat transfer.

To model the heat transfer coefficient very accurately, one would have to construct a computational fluid dynamics (CFD) model of the flow and heat transfer that includes air flow over both the muffler and the exhaust pipe. This is well beyond the scope of what your assignment calls for. Therefore, we will have to live with the inaccuracies, and do the best we can with the correlations we have (using our best judgement). We have gotten about as much as we can out of this approach. That's why I was suggesting earlier some calibration experiments which measure both the exhaust gas temperature out of the tailpipe (or in the muffler) and the surface temperature of the exhaust pipe.

Chet
 
  • #114
Thanks Chet. So ultimately, we are using the value (and equation) in post # 102?

I did not bookmark the link, and now I can't find it again. Sorry.

It might be in your browser history. If you could, dig it out of there. I find it hard to accept that value of the gas being very close to the value of the outside wall.
 
  • #115
Jay_ said:
Thanks Chet. So ultimately, we are using the value (and equation) in post # 102?
Yes.

I find it hard to accept that value of the gas being very close to the value of the outside wall.
On what do you base this? I'm having trouble understanding why the calculated h2 seems low to you.

I might also point out that the lower the value of h2, the better off you are. If h2 were zero, for example, the wall temperature measurement would give you the exact exhaust gas temperature.

Chet
 
  • #116
Okay. Let me use what we have done.
 
  • #117
Hey Chet. I hope you are still watching this thread because I may have questions all of a sudden. Can I use the formulas for the heat capacities in Perry's book table 2-194?

For the gases, N2, CO2 and H2O, we have a neat formula with the temperature variable and the maximum uncertainty is 3% (on that of Nitrogen). I have a question though. The unit is in cal/deg. mol if I need it in J/deg. kg which is the S.I. unit.

So for this conversion I multiply by 4.184 (convert calories to joules) and divide by the molecular weight (convert moles to grams in the denominator) times a 1000 (make it kilograms in the denominator) right?
 
  • #118
Jay_ said:
Hey Chet. I hope you are still watching this thread because I may have questions all of a sudden. Can I use the formulas for the heat capacities in Perry's book table 2-194?
I don't have Perry's Handbook in front of me, but, if the table heading is something like heat capacities for ideal gas region, then yes, these are the formulas you want.
For the gases, N2, CO2 and H2O, we have a neat formula with the temperature variable and the maximum uncertainty is 3% (on that of Nitrogen). I have a question though. The unit is in cal/deg. mol if I need it in J/deg. kg which is the S.I. unit.

So for this conversion I multiply by 4.184 (convert calories to joules) and divide by the molecular weight (convert moles to grams in the denominator) times a 1000 (make it kilograms in the denominator) right?
Yes. Another way to check is to see if it gives you the expected values when used to calculate the Prantdl number.

Chet
 
  • #119
I checked them directly, its right. The equations from Perry have to be multiplied by 4.2 (or 4.184, whichever) and divided by molecular weight to get it in S.I. I don't see why, but I get the answer without having to multiply by 1000. I guess its because they use the "small calorie".

And I am trying to find functions, instead of tables because functions can use temperature as the input to an equation. That way its continuous and not discrete, and any value of temperature can be used.

So I need the following as functions of temperature:

1. Specific heat capacities for N2, CO2 and H2O (CHECK)
2. Thermal conductivity for N2, CO2 and H2O (I have two equations for N2, CO2 which I am yet to verify).
3. Dynamic viscosity for N2, CO2 and H2O (this is giving me a hard time)
4. Dynamic viscosity for air (CHECK)
5. Thermal conductivity for air (CHECK)

If you come across anything (equation, as a function of temperature) do let me know.
 
  • #120
Jay_ said:
I checked them directly, its right. The equations from Perry have to be multiplied by 4.2 (or 4.184, whichever) and divided by molecular weight to get it in S.I. I don't see why, but I get the answer without having to multiply by 1000. I guess its because they use the "small calorie".

Are you expressing Cp in units of kJ/(kg-K)? If so, you, of course, don't need the factor of 1000.
And I am trying to find functions, instead of tables because functions can use temperature as the input to an equation. That way its continuous and not discrete, and any value of temperature can be used.

So I need the following as functions of temperature:

1. Specific heat capacities for N2, CO2 and H2O (CHECK)
2. Thermal conductivity for N2, CO2 and H2O (I have two equations for N2, CO2 which I am yet to verify).
3. Dynamic viscosity for N2, CO2 and H2O (this is giving me a hard time)
Make a semilog graph of viscosity as a function of 1/T. It should come close to a straight line (that you can fit with an equation).
4. Dynamic viscosity for air (CHECK)
5. Thermal conductivity for air (CHECK)

If you come across anything (equation, as a function of temperature) do let me know.
You can always put the data in as discrete values in a table that you can interpolate.
 

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