Temperature of fluid flowing through pipe

[Jay_]

I don't have the data on the mpg (mileage) rating of the Toyota Corolla for various speeds. I guess if I did, I could just plug the numbers in and there would be a gallons per minute for that.

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But if I approximate with some averaged numbers, here goes...

It says for city, its mileage is 28 mpg (the car I will be using is a Toyota Corolla, the Honda Accord is out of city). Let's say we drove at an average of 35 mph.

That equals 0.58333 miles per minute. So my gallons per minute is 0.5833/28 = 0.020832 gallons per minute of fuel. The density of gasoline according to Google is 6.073 lbs/US gal.

So the mass of fuel coming in is 0.126514 lbs/minute.

Since the air-to-fuel ratio is taken in terms of their masses if fuel is 1, air is 14.7. That means the air coming in is 1.85975 lbs/minute.

This sums it up to 1.986264 lbs/minute or 0.01501582 kg/second or 15.016 grams/second. The same that goes in, comes out. So that is the exhaust mass flow rate. Correct?

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Let's rewind a little bit

In post #21 you mentioned that equations (2) and (3) are the same thing, but isn't (2) convection heat and (3) the conductive heat through the wall? I imagine that they are both different.

Of course, it is (2) that I want. The real issue now is to get to the gas temperature, just knowing the temperature at the outside surface of the pipe.

You mentioned its reasonable to assume that the temperature of the inside wall is the same as the outside wall. But what would the temperature of the gas inside be? Ask me to go through the relevant posts if you've already taught me. My mind is all over the place, I may have forgot.

In post #36, I typed an equation based on a statement you made in post #35 is there relevance between this and the Nusselt number and Reynolds number we had equations for in post #47

Starting from the matter of post #36. (I type it again here) :

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

That gives me,

T_{gas} - T_{inp} = \frac{(T_{otp} - T_{amb})*h1}{h2}

T_{gas} = \frac{(T_{otp} - T_{amb})*h1}{h2}+T_{inp}

Now, since you mentioned the Nusselt number is the dimensionless heat transfer coefficient,. Is there a way I can find h1 and h2, which are the heat transfer coefficients for the inside and outside of the pipe respectively using that?

You mentioned using the literature. But if I know h1, h2 and I can approximate the temperature of the pipe inside and outside to be roughly the same, then I can find the temperature of the gas using the last equation, because everything to the right side of the equation will be known to me (if I know h1 and h2 as well). Right?

 

[Chestermiller]

I don't have the data on the mpg (mileage) rating of the Toyota Corolla for various speeds. I guess if I did, I could just plug the numbers in and there would be a gallons per minute for that.

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

But if I approximate with some averaged numbers, here goes...

It says for city, its mileage is 28 mpg (the car I will be using is a Toyota Corolla, the Honda Accord is out of city). Let's say we drove at an average of 35 mph.

That equals 0.58333 miles per minute. So my gallons per minute is 0.5833/28 = 0.020832 gallons per minute of fuel. The density of gasoline according to Google is 6.073 lbs/US gal.

So the mass of fuel coming in is 0.126514 lbs/minute.

Since the air-to-fuel ratio is taken in terms of their masses if fuel is 1, air is 14.7. That means the air coming in is 1.85975 lbs/minute.

This sums it up to 1.986264 lbs/minute or 0.01501582 kg/second or 15.016 grams/second. The same that goes in, comes out. So that is the exhaust mass flow rate. Correct?

Nice job. Let's worry about the effect of speed on the mpg's later. For right now, this is adequate.

The next step is to calculate a value of the Reynolds number. To do this, you need an estimate of the tailpipe diameter and the gas viscosity at the bulk gas temperature. Do you have any approximate idea of what the bulk gas temperature in the tailpipe will be? (If not, we will have to find it by trail and error.) I think you mentioned an approximate bulk gas temperature of about 650 C. We can correct this later. For now, we are going to be approximating the viscosity of the gas by treating it as air (since most of the exhaust gas is nitrogen). We can sharpen our pencils on this later also. Right now, I want you to go through the steps of getting the Reynolds number, and then the Nussult Number, and then the heat transfer coefficient. So, what is the viscosity of air at 650 C. Plug this viscosity and the tube diameter into your equation for the Reynolds number (in terms of the mass flow rate).

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Let's rewind a little bit

In post #21 you mentioned that equations (2) and (3) are the same thing, but isn't (2) convection heat and (3) the conductive heat through the wall? I imagine that they are both different.

Let's come back to this later.

Of course, it is (2) that I want. The real issue now is to get to the gas temperature, just knowing the temperature at the outside surface of the pipe.

You mentioned its reasonable to assume that the temperature of the inside wall is the same as the outside wall. But what would the temperature of the gas inside be? Ask me to go through the relevant posts if you've already taught me. My mind is all over the place, I may have forgot.

In practice, the temperature of the gas at the inside wall is going to be essentially equal to the temperature of the gas at the outside wall. This is because the resistance to heat transfer through the wall is going to be much much lower than from the wall to the bulk flow and from the wall to the ambient air.

Here is a bit of a primer.
The bulk temperature of the gas in the tube is going to be higher than the tube wall temperature. There is a thin layer of gas immediately adjacent to the wall within which all this temperature change takes place. This thin boundary layer of gas will be on the order of less than 1/10 the radius of the tube. Heat is conducted through this thin boundary layer from the bulk gas flow to the wall. Throughout most of the tube cross section, the bulk fluid temperature is nearly constant. The same type of situation prevails outside the tube, between the tube wall and the ambient air.

In post #36, I typed an equation based on a statement you made in post #35 is there relevance between this and the Nusselt number and Reynolds number we had equations for in post #47

Sure. We're trying to determine h1 and h2, but first we need to determine Nu1 and Nu2.

Starting from the matter of post #36. (I type it again here) :

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

That gives me,

T_{gas} - T_{inp} = \frac{(T_{otp} - T_{amb})*h1}{h2}

T_{gas} = \frac{(T_{otp} - T_{amb})*h1}{h2}+T_{inp}

Now, since you mentioned the Nusselt number is the dimensionless heat transfer coefficient,. Is there a way I can find h1 and h2, which are the heat transfer coefficients for the inside and outside of the pipe respectively using that?

Yes. That's what we're trying to do.

You mentioned using the literature. But if I know h1, h2 and I can approximate the temperature of the pipe inside and outside to be roughly the same, then I can find the temperature of the gas using the last equation, because everything to the right side of the equation will be known to me (if I know h1 and h2 as well). Right?

Yes. But you need to use the literature correlations (as we are doing) to get h1 and h2.

Chet

 

[Jay_]

Right now, I want you to go through the steps of getting the Reynolds number, and then the Nussult Number, and then the heat transfer coefficient. So, what is the viscosity of air at 650 C. Plug this viscosity and the tube diameter into your equation for the Reynolds number (in terms of the mass flow rate).

Okay.I am plugging the value into this equation to find Re :

Re=\frac{4W}{μπD}

The estimated diameter of the pipe (from my head) is 7 cm (at that point) so 0.07 m.

The viscosity of air at 650 C is : 4.06633E-5 (from this link, http://www.lmnoeng.com/Flow/GasViscosity.php , I selected the gas as Standard Air)
So,

D = 0.07 m
W = 0.01501582 kg/sec
π = 3.14159...
μ = 4.06633E-5 S.I.

Everything is in S.I.

I get Re = 6716.74 Correct? (Please verify). The value seems to be in the unstudied region.

2100 < Re < 8000. What do we do now? Do we still proceed to use the same formulas?

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For the Prandtl number, we need :

Cp = Specific Heat Capacity
η= dynamic viscosity
λ = thermal conductivity

Do gases follow proportions on these as far as computing the constants is concerned? For instance if I had 70% Nitrogen and 30% Oxygen, then would they properties be like

Cp_gas = 0.7*(Cp_Nitrogen) + 0.3*(Cp_Oxygen)? If this is the case I can get these values easily.

Also tell me after I get the Nusselt number is it that the ratio of the Nusselt numbers are equal to the ratio of the heat transfer coefficients?

 

[Chestermiller]

Okay.I am plugging the value into this equation to find Re :

Re=\frac{4W}{μπD}

The estimated diameter of the pipe (from my head) is 7 cm (at that point) so 0.07 m.

The viscosity of air at 650 C is : 4.06633E-5 (from this link, http://www.lmnoeng.com/Flow/GasViscosity.php , I selected the gas as Standard Air)
So,

D = 0.07 m
W = 0.01501582 kg/sec
π = 3.14159...
μ = 4.06633E-5 S.I.

Everything is in S.I.

I get Re = 6716.74 Correct? (Please verify). The value seems to be in the unstudied region.

2100 < Re < 8000. What do we do now? Do we still proceed to use the same formulas?

From Fig. 14.3-2, when the Reynolds number is 6717, the ordinate on the figure is 0.004. So,

Nu=0.004RePr^{1/3}(\frac{μ_b}{μ_0})^{0.14}

-----------------------------------------------------
For the Prandtl number, we need :

Cp = Specific Heat Capacity
η= dynamic viscosity
λ = thermal conductivity

Do gases follow proportions on these as far as computing the constants is concerned? For instance if I had 70% Nitrogen and 30% Oxygen, then would they properties be like

Cp_gas = 0.7*(Cp_Nitrogen) + 0.3*(Cp_Oxygen)? If this is the case I can get these values easily.

If you Google the Prantdl number of air, you find that at 650C, the Prantdl number will be about 0.68. It is very insensitive to temperature.

Let's assume that the viscosity correction factor is very close to unity. What do you get for the Nussult number? What do you get for the heat transfer coefficient?

Also tell me after I get the Nusselt number is it that the ratio of the Nusselt numbers are equal to the ratio of the heat transfer coefficients?

No. You need to do the heat transfer coefficients separately (unless the thermal conductivites of the gases on both sides of the wall are the same).

Chet

 

[Jay_]

Hey Chet,

I understand now the ordinate is the coefficient in front of that equation. Is there a more detailed graph (I can look closely into). I want to get a more accurate value.

Also, how can we just assume things. Like temperature being 650 C, or the viscosity correction factor being close to 1?

Given the information of Re = 6717 and Pr = 0.68 and you say mu_b / mu_0 = 1?

In that case, we get Nu = 23.6268 but I don't see how we can just estimate the temperature, when that is what we are trying to calculate.

 

[Chestermiller]

Hey Chet,

I understand now the ordinate is the coefficient in front of that equation. Is there a more detailed graph (I can look closely into). I want to get a more accurate value.

BSL notes that the data that was used to produce the graph as accurate to plus or minus 10%. So trying to get it more accurate than this is fruitless. Anyway, the other uncertainties you are working with are somewhat larger than this: mpg vs speed, range of velocities, tube diameter, etc.

Also, how can we just assume things. Like temperature being 650 C, or the viscosity correction factor being close to 1?

As I said before, this is just our first cut, only to get our feet wet. We are going to be "sharpening our pencils" when we repeat the calculations more seriously. So please be patient. We are going to be using your measurements of the tube wall temperature to estimate the gas temperature. We are going to be using the real gas composition, rather than modeling it as air. We are going to be using a range of speeds, rather than just one speed to check that. We are going to be taking into account the difference between the bulk viscosity of the gas, and the gas viscosity at the wall. But, for now, as a experienced modeler, I can tell you it is of great value to see a number right away, even if it is not a very accurate number. This is just the beginning.

Given the information of Re = 6717 and Pr = 0.68 and you say mu_b / mu_0 = 1?

In that case, we get Nu = 23.6268 but I don't see how we can just estimate the temperature, when that is what we are trying to calculate.

It is going to be an iterative process. We are going to be solving iteratively for the gas temperature from your equation.

So, if Nu = 24, what do you get for the heat transfer coefficient on the inside tube side? How does that compare with the typical range of values that they give for convective heat transfer in BSL's table?

Understand that, all this is just to determine h1. We are also going to have to use BSL to estimate h2.

Chet

 

[Jay_]

Which equation am I supposed to use to find 'h'?

Nu = hL/k it says for air the value of k = 0.024 (S.I.), L = diameter of pipe = 0.07 and Nu = 24.

So that gives me h = 8.2286 (S.I.)

In BSL book's table 14.1-1, this corresponds to "Gases". Explain the iterative process for me Chet. I am a little behind on this model creating, and I don't mean to be disrespectful and rush you into answers. I understand this is a learning process I need to take. But I would like to know where we are going (step wise). Thank you sir.

 

[Chestermiller]

Which equation am I supposed to use to find 'h'?

Nu = hL/k it says for air the value of k = 0.024 (S.I.), L = diameter of pipe = 0.07 and Nu = 24.

So that gives me h = 8.2286 (S.I.)

The value of the thermal conductivity should be for 650 C, not room temperature. Please try again.

In BSL book's table 14.1-1, this corresponds to "Gases".

BSL give the range from 10-100 W/m^2K, so when you correct the k value, we will be in that range.

Explain the iterative process for me Chet. I am a little behind on this model creating, and I don't mean to be disrespectful and rush you into answers. I understand this is a learning process I need to take. But I would like to know where we are going (step wise). Thank you sir.

You have a measured value for the wall temperature. You guess the bulk gas temperature in the tube, and calculate the heat transfer coefficient on that basis. Then from your equation involving the two h's, you calculate a new value for the gas temperature. Then you redo the calculation with this value of the gas temperature. You continue redoing the calculation until the calculated gas temperature stops changing.

Chet

 

[Jay_]

you calculate a new value for the gas temperature. Then you redo the calculation with this value of the gas temperature. You continue redoing the calculation until the calculated gas temperature stops changing.

Okay. I have to make this into a MATLAB code/model. A model is just a bunch of equations right? I am good with MATLAB, but I need the "model", if I understand it correctly its the equations. Okay to find the thermal conductivity at 650 C, I am using this link :

http://bouteloup.pierre.free.fr/lica/phythe/don/air/air_k_plot.pdf

650 C = 923 K, using this value in the equation pasted :

k (@ 923K) = 0.0641756431 ≈ 0.0642

So using Nu = 24, L = 0.07, k = 0.0642, I get

h = 22.01143

This comes in the range of forced convection for gases. I calculate the temperature again using this?

We are going to do this iterative process for every data point of temperature? I am going to be having like 1800 data points, given that I am sampling every second in a half an hour drive. Does doing the iterative process for each data point sound reasonable?

 

[Jay_]

Hey Chet,

I was studying the three constants from this video. Is the "L" of the Nusselt number the same as the diameter, it seems to be the thickness of the boundary layer. Please verify this. In another pdf, it seems to be the same as the diameter. In wikipedia under "laminar flow" it says diameter again.

But in this video he seems to say "L" is a small layer. Go to 5:40. Have we done it correctly?

 

[Chestermiller]

Hey Chet,

I was studying the three constants from this video. Is the "L" of the Nusselt number the same as the diameter, it seems to be the thickness of the boundary layer. Please verify this. In another pdf, it seems to be the same as the diameter. In wikipedia under "laminar flow" it says diameter again.

But in this video he seems to say "L" is a small layer. Go to 5:40. Have we done it correctly?

Yes. We have done it correctly.

The length scale that you use in the Nusselt number and/or the Reynolds number depends on the context. Sometimes you use a variable distance z to represent the length scale if the heat transfer coefficient is varying with position, and sometimes you even can use a boundary layer thickness. In any event, since these results are all either correlations of experimental data (in the case of turbulent flow) or analytic solutions to the fluid mechanics and energy equations (in the case of laminar flow), as long as things are done in a consistent manner, there is no problem.

Chet

 

[Chestermiller]

Okay. I have to make this into a MATLAB code/model. A model is just a bunch of equations right? I am good with MATLAB, but I need the "model", if I understand it correctly its the equations.

Yes. There's the math model that consists of the set of equations you are solving, and there's the computer model, which represents the computer code you are using to solve the equations.

Okay to find the thermal conductivity at 650 C, I am using this link :

http://bouteloup.pierre.free.fr/lica/phythe/don/air/air_k_plot.pdf

650 C = 923 K, using this value in the equation pasted :

k (@ 923K) = 0.0641756431 ≈ 0.0642

So using Nu = 24, L = 0.07, k = 0.0642, I get

h = 22.01143

This comes in the range of forced convection for gases. I calculate the temperature again using this?

No yet. To use your equation, we need to determine the heat transfer coefficient h2 on the outside of the tailpipe.

See this link: http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html. You will notice that they give an equation for the heat transfer coefficient in air flow past an object (in our case, the tailpipe): h = 10.45 - v + 10 v^1/2. It would be very convenient if we could use this equation for h2, since it already has the air physical properties inherently built into the correlation. However, as you can see, there is no length scale in the correlation, so I am uncomfortable with it, and don't know how accurate it would be for your situation. If it were me working on this, I would test this equation out against the correlations in BSL for heat transfer in flow past a sphere and over a cylinder to see how it compares with these confirmed correlations. I would like to know if there is any range of diameters for either case in which this equation gives accurate predictions. However, that's just me.

If you are comfortable using this equation, then you can use it to get h2. But, don't expect it to be as accurate as the results for h1, considering the above caveats. My feeling is that the heat transfer coefficient h2 is going to depend on some length scale, and there is going to be a difference between cross flow and axial flow. In your case, the air flow is axial along the outside of the tailpipe. I also think that h2 may vary with the placement of your measuring device. For example it may vary with how far back the device is from the exit of the muffler.

Is there any way of calibrating your system, by doing some tests where you measure both the temperature on the tailpipe (your main experimental method) and also in the exhaust gas leaving the tailpipe (temporary set up)? This would help calibrate the value of h2.

If you are comfortable using the toolbox equation to get h2, then we are close to the end of our journey. You just substitute the speed of the car for v in the equation.

We are going to do this iterative process for every data point of temperature? I am going to be having like 1800 data points, given that I am sampling every second in a half an hour drive. Does doing the iterative process for each data point sound reasonable?

Are you required to provide the results on streaming data, or are you allowed to record the data during the test and analyze them afterwards? If it is the latter, then no problem. Even if it's the former, there still shouldn't be a problem, considering how fast computers are now-a-days.

Chet

 

[Jay_]

If you are comfortable using the toolbox equation to get h2, then we are close to the end of our journey. You just substitute the speed of the car for v in the equation.

I am tempted to complete this problem quickly. But I took a look at that function in a grapher. It attains a local maxima at v = 25 m/s and becomes zero before v = 120 m/s, past this it becomes negative. The graph shown in the website also seems to be for values before 20 m/s, so I guess its been modeled for that range (v < 20 m/s). Since I will be driving the car at speeds of up to 80 mph (35.76 m/s), maybe its not a good idea. If there are other models on the heat transfer coefficient of air, you could tell me about those. But its okay if its not there. Let's proceed to do it the same way. I think only the temperature would change instead of the 650 C, it would be 35 C. Right?

Are you required to provide the results on streaming data, or are you allowed to record the data during the test and analyze them afterwards? If it is the latter, then no problem.

Its the latter.

 

[Jay_]

Would you be able to find any papers that could give an equation? I search and got one on thermal conductivity. But we don't want that.

 

[Chestermiller]

I am tempted to complete this problem quickly. But I took a look at that function in a grapher. It attains a local maxima at v = 25 m/s and becomes zero before v = 120 m/s, past this it becomes negative. The graph shown in the website also seems to be for values before 20 m/s, so I guess its been modeled for that range (v < 20 m/s). Since I will be driving the car at speeds of up to 80 mph (35.76 m/s), maybe its not a good idea. If there are other models on the heat transfer coefficient of air, you could tell me about those. But its okay if its not there. Let's proceed to do it the same way. I think only the temperature would change instead of the 650 C, it would be 35 C. Right?

We can't use the same correlation for outside the tube that we did for inside the tube, because the gas flow outside is different. Let's look at 2 things:

1. Let's try completing the problem using the equation in toolbox. I just want you to get some experience. We already assumed a car speed of 35 mph for determining the gas flow inside the tube. Let's now assume a 35 mph air flow rate outside the pipe, since this is the relative velocity of the air with respect to the tube. Use the toolbox equation to get h2, and then use your equation to get the exhaust gas temperature for some assumed measured value of the wall temperature (say 350 C?). See what you come up with. Or, if the exhaust gas temperature actually was 650 C, what would that imply for the value of the measured wall temperature?

It might be possible that the ratio of h1 to h2 may not vary much with the car speed. If this were the case, then we may be able to extrapolate to higher car speeds. We can try some other car speeds to see what we get.

2. As far as the heat transfer coefficient h2 outside the tube is concerned, we an get an upper bound to this heat transfer coefficient by assuming that the air flow is perpendicular to the tailpipe, rather than parallel (as in our real world case). We can get this upper bound by using the correlation in BSL for flow over a cylinder. We can then compare this upper bound with the values that we get from the toolbox equation (for various values of the relative air velocity). Maybe we can deduce an equivalent diameter for the tube that will give the same value as for the toolbox equation. This also might enable us to extrapolate to higher car speeds.

What I'm saying here is that we have to "play with" the correlations and numbers until we arrive at something that is comfortable for us. This is called "doing scouting calculations."

Also, see Eqns. 11-13 of this reference: https://www.google.com/search?q=hea...ome..69i57.18410j0j1&sourceid=chrome&ie=UTF-8

Chet

 

[Jay_]

Hey Chet,


I am using these equations :

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

T_{gas} = \frac{(T_{otp} - T_{amb})*h1}{h2}+T_{inp}

We calculated h1 (inside) = 22.01143

Now using the equation in engineering toolbox, and the velocity as 35 mph = 15.6464 m/s.

Using this I get h2 = 34.3591

Putting this in equation above I get :


T_{gas} = \frac{(T_{otp} - T_{amb})*22}{34}+T_{inp}

So for example if the ambient temperature is 35 C, the pipe outside is 150 C. And we assume the same temperature inside, we have

Temperature of gas = 224 C.

Sound reasonable?

 

[Chestermiller]

Hey Chet,


I am using these equations :

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

T_{gas} = \frac{(T_{otp} - T_{amb})*h1}{h2}+T_{inp}

We calculated h1 (inside) = 22.01143

Now using the equation in engineering toolbox, and the velocity as 35 mph = 15.6464 m/s.

Using this I get h2 = 34.3591

Putting this in equation above I get :


T_{gas} = \frac{(T_{otp} - T_{amb})*22}{34}+T_{inp}

So for example if the ambient temperature is 35 C, the pipe outside is 150 C. And we assume the same temperature inside, we have

Temperature of gas = 224 C.

Sound reasonable?

No. The starting equation should be:
\frac{T_{gas} - T_{inp}}{T_{otp} - T_{amb}} = \frac{h2}{h1}
So the final equation should be:
T_{gas} = \frac{(T_{otp} - T_{amb})*34}{22}+T_{inp}

Also, why don't you try a higher temperature for the wall temperature than 150 C, like 250C? I would like to show you how the iteration works for correcting the gas temperature.

I feel like we're getting closer to where we want to be now. Are you starting to feel more comfortable? But, we still have more work to do.

Have you had a chance to try out that correlation for flow over a cylinder to assess the upper bound for h2?

Chet

 

[Jay_]

Have you had a chance to try out that correlation for flow over a cylinder to assess the upper bound for h2?

I haven't tried that yet. I will though.

Are you starting to feel more comfortable? But, we still have more work to do.

Yea I do. But to get the equation right, its :

T_{gas} = \frac{(T_{otp} - T_{amb})*34}{22}+T_{inp} and assuming outer wall temperature for 250 C, and the inner wall temperature being the same, and the ambient 35 C, we get

T_gas = 582.27 C

Now, I assume we start from post # 53

STEPS FOR GAS TEMPERATURE ITERATION

1) Find μ at given temperature (now it is 582 C) (http://www.lmnoeng.com/Flow/GasViscosity.php)
2) Use Re = 4W/(μπD)
3) Find the corresponding ordinate value in the graph of BSL from Re (M say)
4) Find the corresponding Prandtl number (http://www.mhtl.uwaterloo.ca/old/onlinetools/airprop/airprop.html)
5) Use the relation below to find the Nusselt number (I have a question on this *):

Nu=MRePr^{1/3}(\frac{μ_b}{μ_0})^{0.14}

6) Find the thermal conductivity (http://bouteloup.pierre.free.fr/lica/phythe/don/air/air_k_plot.pdf)

7) Find the heat transfer coefficient using

h = \frac{kNu}{L}

8) Find the heat transfer coefficient of standard ambient temperature air.

9) Find gas temperature using
T_{gas} = \frac{(T_{otp} - T_{amb})*h_{out}}{h_{in}}+T_{inp}


Repeat till different in the results of previous iterations are below the acceptable error.

* The equation we use in the step 5 seems to be slightly different in different sources. The difference being that some ignore the viscosity factor, and the number M and the power on the Reynolds number are different. What's the actual equation?

Another question, we are using air for finding the values of the constants in steps 1, 4 and 6. But the gas inside is a mix of N₂, H₂O and CO₂ and small fractions of CO, NO_X and HC. Is the value going to be roughly the same? Is it safe to make that assumption?

 

[Chestermiller]

I haven't tried that yet. I will though.



Yea I do. But to get the equation right, its :

T_{gas} = \frac{(T_{otp} - T_{amb})*34}{22}+T_{inp} and assuming outer wall temperature for 250 C, and the inner wall temperature being the same, and the ambient 35 C, we get

T_gas = 582.27 C

Now, I assume we start from post # 53

STEPS FOR GAS TEMPERATURE ITERATION

1) Find μ at given temperature (now it is 582 C) (http://www.lmnoeng.com/Flow/GasViscosity.php)
2) Use Re = 4W/(μπD)
3) Find the corresponding ordinate value in the graph of BSL from Re (M say)
4) Find the corresponding Prandtl number (http://www.mhtl.uwaterloo.ca/old/onlinetools/airprop/airprop.html)
5) Use the relation below to find the Nusselt number (I have a question on this *):

Nu=MRePr^{1/3}(\frac{μ_b}{μ_0})^{0.14}

6) Find the thermal conductivity (http://bouteloup.pierre.free.fr/lica/phythe/don/air/air_k_plot.pdf)

7) Find the heat transfer coefficient using

h = \frac{kNu}{L}

8) Find the heat transfer coefficient of standard ambient temperature air.

9) Find gas temperature using
T_{gas} = \frac{(T_{otp} - T_{amb})*h_{out}}{h_{in}}+T_{inp}


Repeat till different in the results of previous iterations are below the acceptable error.

Excellent, excellent, excellent. You're rapidly becoming a heat transfer man.

* The equation we use in the step 5 seems to be slightly different in different sources. The difference being that some ignore the viscosity factor, and the number M and the power on the Reynolds number are different. What's the actual equation?

This is all for flow in a pipe, correct? (Of course, other geometries have different relationships.) Give me some examples so I can comment better. Don't forget that, for flow in a pipe, this is just a correlation of experimental data. If the viscosity factor is ignored, then one is implicitly assuming that the temperature difference between the wall and the bulk flow is not large.

Another question, we are using air for finding the values of the constants in steps 1, 4 and 6. But the gas inside is a mix of N₂, H₂O and CO₂ and small fractions of CO, NO_X and HC. Is the value going to be roughly the same? Is it safe to make that assumption?

Not necessarily. We would like to check this out. The physical properties of mixtures of gases can be estimated using the techniques discussed in the first chapter of each section in BSL. (Momentum transport and Heat Transport). To do this, you need to know the mole fractions of the various gases in the exhaust. We're going to assume complete combustion, and no side reactions. So we need to do some stoichiometry. Let's start simple by assuming that the fuel is pure octane. Write the balanced chemical reaction equation for combustion of octane to yield water and CO2. If you have 1 mole of octane, how many moles of O2 are consumed, and how many moles of H20 and CO2 are produced? How many moles of N2 end up in the exhaust gas? What are the mole fractions of N2, H20, and CO2 in the exhaust gas. To check to see how good an approximation using octane is, calculate the air to fuel mass ratio and compare it with 14.7 (our assumed ratio). If it's close to this value, using the calculated mole fractions will be adequate.

Chet

 

[Jay_]

Give me some examples so I can comment better.

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.

The physical properties of mixtures of gases can be estimated using the techniques discussed in the first chapter of each section in BSL. (Momentum transport and Heat Transport).

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

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?

We're going to assume complete combustion, and no side reactions. So we need to do some stoichiometry. Let's start simple by assuming that the fuel is pure octane. Write the balanced chemical reaction equation for combustion of octane to yield water and CO2. If you have 1 mole of octane, how many moles of O2 are consumed, and how many moles of H20 and CO2 are produced?

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

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

 

[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