Temperature of fluid flowing through pipe

[Chestermiller]

Check the "Dittus-Boelter" relation here in the very end of the document:

http://www.me.umn.edu/courses/old_me_course_pages/me3333/gallery/eq 3.pdf

I found another link (I can't locate it now), which had different equations for turbulent flow in a circular pipe for different range of Reynolds numbers.

Like I said, these correlations are based on experimental data. They are what the individual authors judged as the best fit to the data. They will all give very similar predictions. In the case of the viscosity factor, as I said, leaving that out is equivalent to assuming that the temperature difference between the bulk and the wall is small.

Like in example 1.4-2 I guess (for viscosity). But then what does the viscosity factor become?

The viscosity factor is the ratio of the viscosity at the tube wall temperature to the viscosity at the bulk gas temperature (to the 0.14 power).

In another chapter, they have the mixing rules for thermal conductivity. Of course, for heat capacity, the mixture rule is just mole fraction.

Also, there is example 1.4-1 in which they find the viscosity at various temperatures for CO₂. It looks like this uses other constants characteristic to the gas. How do we know these values for the gas in the exhaust?

These constants are molecular properties characteristic of the particular substance, and are independent of the temperature and pressure.

But shouldn't we have done this when we assumed the constants of the gas? We picked all the constants for air.

Yes, but I wanted you to see some results, even if they are not very accurate (yet). And, I wanted to bring you along gradually, so that all the complexity was not included at one time. I have lots of experience doing modelling, and I have found great value in starting simple and seeing results early on. I have found this approach to be very effective. My three basic principles for doing modelling are:
1. Start simple
2. Start simple
and
3. Start simple

Why? If you can't solve a simple version of your problem, then you certainly won't be able to solve it in full complexity. Plus, you get to see some results right away. Plus, you will get to see what each successive refinement makes in the final results.

I have literature from Volkswagen that shows the percentages in the exhaust emissions by mass are as follows :

N₂ = 71%
CO₂ = 14%
H₂O = 13%
CO, HC, NO_X = 2% (I assume all of this to be CO)

This is by mass. Now, by my calculations on 100 g of this exhaust gas. We have the following:

Moles of each gas

moles of N₂ = 71/28 ≈ 2.5357
moles of CO₂ = 14/44 ≈ 0.3182
moles of H₂O = 13/18 ≈ 0.7222
moles of CO = 2/28 ≈ 0.0714

Total moles : 3.6475

Mole fractions

mole fraction of N₂ = 2.5357/3.6475 ≈ 0.6952
mole fraction of CO₂ = 0.3182/3.6475 ≈ 0.0872
mole fraction of H₂O = 0.7222/3.6475 ≈ 0.1980
mole fraction of CO = 0.0714/3.6475 ≈ 0.0196

OK. This is good. Now, if you have data on the individual species (like N2 from toolbox), you don't need to use the estimation procedures in BSL for those species. Of course, this is better than using the estimation procedures. However, you will still need to apply the mixing rules.

Chet

 

[Chestermiller]

In post #70, you gave VW data on the composition of the exhaust gas. What does this data imply about the air:fuel ratio? I calculate about 9.6. What do you calculate?

Assuming complete combustion of octane, I get an air to fuel ratio of 15.1, and the following mole percentages in the exhaust:

N2 73.4
CO2 12.5
H20 14.1

Chet

 

[Jay_]

Like I said, these correlations are based on experimental data. They are what the individual authors judged as the best fit to the data. They will all give very similar predictions. In the case of the viscosity factor, as I said, leaving that out is equivalent to assuming that the temperature difference between the bulk and the wall is small.

Thats okay. Could you tell me what equation we should use here. I need to get done with this, I am sorry if I sound like I am not interested in knowing more.

The thing is this due by September. And I need to get started with the actual experiment of getting the data from a car. My professor has been asking me for the 'model' before he goes on to allow us to show him the experimental data, and I was hoping this discussion would come to an end soon. I certainly do appreciate all your help. This would have been impossible without you.

But I want to get to final the answer (like I typed in post # 70) in steps.

What does this data imply about the air:fuel ratio? I calculate about 9.6. What do you calculate?

I am not sure how to go from mole fractions to the air fuel ratio. Given that we assume a air:fuel ratio of 14.7:1 for the Toyota Corolla. I would like to work the other way, going from air:fuel ratio to mole fractions. Could you show me how you calculated the air:fuel ratio from the percentages?

I would like to show him some sort of code (MATLAB) of the model, by Wednesday - day after. We can certainly fine tune the model after that.

But I want to show him one complete iteration (with the gas constants, not the air constants) soon. I keep telling him I am almost done with it and I may have given him the impression that I am doing nothing. Could we get the constants of the gas considering 14.7 air:fuel ratio and finish the iterations?

I won't be able to type the MATLAB code until I know the exact process for each.

We need to get three constants : viscosity, Prandtl number and thermal conductivity. How do we get these knowing the gas mixture?

 

[Chestermiller]

I am not sure how to go from mole fractions to the air fuel ratio. Given that we assume a air:fuel ratio of 14.7:1 for the Toyota Corolla. I would like to work the other way, going from air:fuel ratio to mole fractions. Could you show me how you calculated the air:fuel ratio from the percentages?

Given the time constraints, let's skip the VW data and how to go from the percentages to the air:fuel ratio. Let's start out by assuming that octane is a surrogate for your fuel. The balanced chemical reaction equation for the complete combustion of octane is:

C₈H₁₈+12.5 O₂ = 8 CO₂ + 9 H₂O

The number of moles of N2 in the air that enters along with this amount of O2 is 12.5x79/21=47.0, where 79 and 21 are the mole percents of N2 and O2 in the air. So the total number of moles of air that are required for complete combustion of 1 mole of octane is 59.5. The molecular weight of air is 29, and the molecular weight of octane is 114. So the air:fuel ratio is 59.5x29/114 ~ 15.1 This is close enough to 14.7. So for every mole of fuel that enters, there are 47 moles of N2, 8 moles of CO2, and 9 moles of water as the exhaust. This gives the mole fractions I showed in my previous post.

I would like to show him some sort of code (MATLAB) of the model, by Wednesday - day after. We can certainly fine tune the model after that.

I don't know how to program in MATLAB, so I can't help you there. FORTRAN is what I always use.

But I want to show him one complete iteration (with the gas constants, not the air constants) soon. I keep telling him I am almost done with it and I may have given him the impression that I am doing nothing. Could we get the constants of the gas considering 14.7 air:fuel ratio and finish the iterations?

We need to get three constants : viscosity, Prandtl number and thermal conductivity. How do we get these knowing the gas mixture?

Regarding the Prantdl number, that's the heat capacity times the viscosity divided by the thermal conductivity. So, you really need to get those. You start out by getting the properties of the pure substances. You look up heat capacity vs temperature in a handbook for each species. You can get the viscosity and thermal conductivity of each species either from literature equations that were fit to data or using the estimation procedures in chapters 1 and 9 in BSL. Once you know the properties of the pure species, you use the mixing rules in chapters 1 and 9 to get the properties of the mixtures (except for heat capacity, which is weighted in terms of mole fraction).

Chet

 

[Jay_]

That's okay Chet. I am pretty good with MATLAB. I will show some calculations based on this post of yours soon, so that I will have something to show him. Let me then get your approval on a flowchart diagram before I get to the code. :-)

There is also one more thing. The final quantity we want to measure is Q (rate of energy) of the gas. So for that we need mass flow rate (which we calculate), Cp (specific heat capacity of gas, which we also calculate) and temperature of gas (which we again find out by iteration).

But what does ΔT represent? You mentioned ΔT = Tgas - Treference, but on what basis do we choose the reference point and if it is any random point, how is it possible that a gas has two energy values, because if we change the reference value we would get difference heat exchange values.

 

[Chestermiller]

There is also one more thing. The final quantity we want to measure is Q (rate of energy) of the gas. So for that we need mass flow rate (which we calculate), Cp (specific heat capacity of gas, which we also calculate) and temperature of gas (which we again find out by iteration).

But what does ΔT represent? You mentioned ΔT = Tgas - Treference, but on what basis do we choose the reference point and if it is any random point, how is it possible that a gas has two energy values, because if we change the reference value we would get difference heat exchange values.

I don't know what is in your professor's mind as far as his definition of what constitutes the heat contained in the exhaust gases. What might make sense is to calculate the enthalpy of the exhaust gases relative to what it would be if we cooled them down to the ambient temperature. That may be what he's thinking.

Chet

 

[Jay_]

What might make sense is to calculate the enthalpy of the exhaust gases relative to what it would be if we cooled them down to the ambient temperature. That may be what he's thinking

So that makes it ΔT = Tgas - Tambient. Okay. I will show him all this in code tomorrow. Thanks Chet. I will get back to you for help. :-)

 

[Jay_]

The three constants - viscosity, heat capacity and thermal conductivity.

Now there are two places to go: pure substance to pure substance at given temperature and from there to the mixture at given temperature.

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

Lets do heat capacity first. I can use these links for the pure substances.

http://www.engineeringtoolbox.com/nitrogen-d_977.html >> for N2
http://www.engineeringtoolbox.com/carbon-dioxide-d_974.html >> for CO2
http://www.engineeringtoolbox.com/water-vapor-d_979.html >> H2O
http://www.engineeringtoolbox.com/carbon-monoxide-d_975.html >> CO

Since, I am using a 2001 Corolla, I think there would be some CO emissions? If I assume complete combustion, I wouldn't have any CO correct? Do you think I should then go with some literature data or calculate the value myself? My real question is can we assume complete combustion with a kind of old car?

Then I use the mole fraction mixing concept.

Handling viscosity next. BSL gives me the following: (Table 1.1-3)

N2 (@ 20 C) = 0.0175 mPa.s
H2O (@ 100 C) = 0.01211 mPa.s
CO2 (@ 20 C) = 0.0146 mPa.s

I know there is a Sutherland's relation to the equation I found in wikipedia, but its range is only 0 K to 555 K, the gas temperatures are going to be beyond that. So are you aware of any relation?

Also, I can't find thermal conductivity at different temperatures, so I thought I could just use the Prandtl number. Since you mentioned the Prandtl number is largely insensitive to temperature, its value is the same at any temperature. So I could find that, and then work backwards to find the thermal conductivity using

k = Cp*mu/Pr

After I get it for the individual ones at the given temperature, I would use the mixing rules. Would that be correct?

 

[Chestermiller]

Since, I am using a 2001 Corolla, I think there would be some CO emissions? If I assume complete combustion, I wouldn't have any CO correct? Do you think I should then go with some literature data or calculate the value myself? My real question is can we assume complete combustion with a kind of old car?

Considering that you are assuming an air:fuel ratio of 14.7, and the complete combustion relation gives a value of 15.1, I wouldn't worry too much about the CO. Also, the amount is going to be small, so how much of a difference could it make. Leaving it out is just equivalent to using the remaining gases as a surrogate for CO. This is not going to be the largest inaccuracy in your calculation. Also, whatever you're doing is certainly better than using properties for air.

Then I use the mole fraction mixing concept.

Handling viscosity next. BSL gives me the following: (Table 1.1-3)

N2 (@ 20 C) = 0.0175 mPa.s
H2O (@ 100 C) = 0.01211 mPa.s
CO2 (@ 20 C) = 0.0146 mPa.s

I know there is a Sutherland's relation to the equation I found in wikipedia, but its range is only 0 K to 555 K, the gas temperatures are going to be beyond that. So are you aware of any relation?

Also, I can't find thermal conductivity at different temperatures, so I thought I could just use the Prandtl number. Since you mentioned the Prandtl number is largely insensitive to temperature, its value is the same at any temperature. So I could find that, and then work backwards to find the thermal conductivity using

k = Cp*mu/Pr

You can find all the actual data you need in Chemical Engineers' Handbook, Perry, Chapter 3 for the pure gases. It gives viscosities and thermal conductivites on the pure gases up to even higher temeratures than you need. See the section of transport properties. It gives equations for the heat capacity of each gas as a function of temperature in an earlier section. All this is in tables (or, in the case of N2 and CO2 viscosity, alignment chart).

After I get it for the individual ones at the given temperature, I would use the mixing rules. Would that be correct?

Sure.

 

[Jay_]

Chemical Engineers' Handbook, Perry, Chapter 3

Okay. I got a pdf copy of the book. Now coming to mixing of viscosity of gases, the example 1.4-2 in BSL is not very clear. How is the last column of the table computed?

In any case, after that how do I go to the value at a given temperature?

 

[Chestermiller]

Okay. I got a pdf copy of the book. Now coming to mixing of viscosity of gases, the example 1.4-2 in BSL is not very clear. How is the last column of the table computed?

You already calculated ø_αβ in the previous column, so now you do a weighted sum over the mole fractions. (You hold α constant, and sum over the β's).

In any case, after that how do I go to the value at a given temperature?

The tables in Perry give the pure gas parameter values as a function of temperature. For any temperature you want, you either interpolate in the tables or use an equation fit to the temperature dependence of the parameters.

Chet

 

[Jay_]

The tables in Perry give the pure gas parameter values as a function of temperature.

The thing that is going to change in every iteration is the temperature. Mole fractions will remain constant. So we could just come up with a formula for the exhaust right? I didn't think the method would be so complicated. Because if we are going to do this calculation in each iteration, its going to be a lot of work.

Coming to the example and the last column. From the formula mentioned above the last column, this is the number I come up with the for the first element of the last column (which they have as 0.763) :

(0.133)*(1.000 + 0.730 +0.727) + (0.039)*(1.000 + 0.730 + 0.727) + (0.828)*(1.000 + 0.730 + 0.727) = 2.457

Nor is it this way that : (0.133)*(1.000 + 0.730 + 0.727) = 0.326781

How do they come up with 0.763?

 

[Chestermiller]

The thing that is going to change in every iteration is the temperature. Mole fractions will remain constant. So we could just come up with a formula for the exhaust right? I didn't think the method would be so complicated. Because if we are going to do this calculation in each iteration, its going to be a lot of work.

I don't see what's so complicated. Also, you have been concerned about the accuracy of the results. You can do some preliminary calculations to determine how far off you would be if you neglected the temperature iteration, and just assumed some nominal temperature for all cases.

Coming to the example and the last column. From the formula mentioned above the last column, this is the number I come up with the for the first element of the last column (which they have as 0.763) :

(0.133)*(1.000 + 0.730 +0.727) + (0.039)*(1.000 + 0.730 + 0.727) + (0.828)*(1.000 + 0.730 + 0.727) = 2.457

Nor is it this way that : (0.133)*(1.000 + 0.730 + 0.727) = 0.326781

How do they come up with 0.763?

(0.133)(1)+(0.039)(0.73)+(0.828)(0.727)

Chet

 

[Jay_]

I don't see what's so complicated.

Sorry, I am just getting lazy maybe lol

I see how the numbers come up. Thanks Chet. But in that case shouldn't the formula above the column be
∑(β = 1 to 3) x_β*ø_αβ.

The subscript on x should be β (because β is changing) and not α, right?

 

[Jay_]

Okay. I also checked chapter 9, example 9.3-3 on the thermal conductivity of gas mixes. Now, the last thing is the specific heat capacity of the mixture.

How do we get to that? Is it by weighted mole fractions again, or weighted masses?

 

[Chestermiller]

Sorry, I am just getting lazy maybe lol

I see how the numbers come up. Thanks Chet. But in that case shouldn't the formula above the column be
∑(β = 1 to 3) x_β*ø_αβ.

The subscript on x should be β (because β is changing) and not α, right?

In my book, the subscript on x is β.

Chet

 

[Chestermiller]

Okay. I also checked chapter 9, example 9.3-3 on the thermal conductivity of gas mixes. Now, the last thing is the specific heat capacity of the mixture.

How do we get to that? Is it by weighted mole fractions again, or weighted masses?

The specific heat of an ideal gas mixture is a straight weighted sum over mole fraction.

Chet

 

[Jay_]

Okay. I need to cite a source for each equation. Which source does this equation come from? :

$$\frac{T_{gas} - T_{inp}}{T_{otp} - T_{amb}} = \frac{h1}{h2}$$

I don't see that exact equation in BSL.

 

[Chestermiller]

This equation is derived from q = h1 (Tgas - Twall)=h2(Twall-Tair). It says that the heat flux from the gas to the wall is equal to the heat flux from the wall to the air. All the heat flow from the gas to the air passes through the wall and the resistances are in series .

Chet

 

[Jay_]

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

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?

 

[Chestermiller]

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

 

[Jay_]

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.

 

[Chestermiller]

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

 

[Jay_]

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?

 

[Chestermiller]

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

 

[Jay_]

The equation on slide 23 is for flow across a cylinder.

So couldn't we use this to find h_out?

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 => h_out = 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.

 

[Jay_]

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

 

[Chestermiller]

So couldn't we use this to find h_out?

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 => h_out = 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

 

[Chestermiller]

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

 

[Jay_]

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 T_otp and T_inp), and T_amb = 35 C, I get T_gas = 783.6 C Sound right?

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

And so (T_gas + T_wall)/2 = 409.3 => T_gas = 676.1 C

 

[Chestermiller]

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 T_otp and T_inp), and T_amb = 35 C, I get T_gas = 783.6 C Sound right?

Methodologically correct.

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

And so (T_gas + T_wall)/2 = 409.3 => T_gas = 676.1 C

Methodologically incorrect.

Chet

 

[Jay_]

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

 

[Chestermiller]

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.

 

[Jay_]

So what are we using for the air ultimately?

 

[Chestermiller]

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