Temperature of fluid flowing through pipe

[Jay_]

Hi Chet,

Okay. So, we have 4 temperatures :

1. temperature of gas(T_gas), 2. temperature of inside of pipe(T_inp), 3. temperature of outside of pipe(T_otp), 4. ambient temperature(T_amb)

h1 (heat transfer coefficient of inside of pipe), h2 (heat transfer coefficient of outside of pipe).

Now you have mentioned:

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

That gives me,

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

I am guessing there is a relation between T_gas and T_inp based on the specific heat of the inside of the pipe or something. If that's true, then we could have an equation for just T_gas on one side and all known quantities on the other.

 

[Chestermiller]

Hi Chet,

Okay. So, we have 4 temperatures :

1. temperature of gas(T_gas), 2. temperature of inside of pipe(T_inp), 3. temperature of outside of pipe(T_otp), 4. ambient temperature(T_amb)

h1 (heat transfer coefficient of inside of pipe), h2 (heat transfer coefficient of outside of pipe).

Now you have mentioned:

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

That gives me,

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

I am guessing there is a relation between T_gas and T_inp based on the specific heat of the inside of the pipe or something. If that's true, then we could have an equation for just T_gas on one side and all known quantities on the other.

No. The way to do this is to look up in the literature how to estimate the heat transfer coefficient from the bulk gas to the inside pipe wall, and the heat transfer coefficient from the outside wall to the ambient air. (It might be a reasonable approximation to neglect the heat transfer resistance of the wall, and to assume that the inside wall temperature and the outside wall temperature are equal).

Chet

 

[Jay_]

Would you be able to guide me on what sort of papers (like key words) I should search on? Also, how do you think I can get the heat transfer coefficients of the inside and outside of the pipe, would it also be available from the literature?

Also is there a source/reference for the equation I typed in post #36 of this thread. Because I need to cite that in the report I am typing.

Thank you for helping me Chet.

 

[Chestermiller]

Would you be able to guide me on what sort of papers (like key words) I should search on? Also, how do you think I can get the heat transfer coefficients of the inside and outside of the pipe, would it also be available from the literature?

Also is there a source/reference for the equation I typed in post #36 of this thread. Because I need to cite that in the report I am typing.

Thank you for helping me Chet.

What you are looking for should all be in Chapter 14 of Bird, Stewart, and Lightfoot, but I'm sure you don't want me to spoon feed it to you. You need to digest the material in the chapter. Figure 14.3-2 should have what you need to get the heat transfer coefficient inside the tailpipe. I leave it up to you to figure out what info to use to get the heat transfer coefficient outside the tailpipe.

Chet

 

[Jay_]

Hi Chet,

From what I understand from the book, the two heat transfer coefficients they call h_loc and h_in and they say (page 424) that most experimentalists use h_in which is easier to measure.

Which one is for the outside of the pipe and which is for the inside of the pipe? The comment below the graph (Fig. 14.3-2) gives h_in = (3/2)*h_loc which is useful.

I guess using this the other equation you mentioned, and which I put in post #36 can be simplified? Would that be correct?

 

[Jay_]

Hi Chet,

To clear things out on what is going on in my mind, let me put it in equations:

Okay since,

$$h_{in} = \frac{3}{2}*h_{loc}$$ , and if it is like h_in is the constant for the inside of the pipe and h_loc is for the outside, we have :


$$\frac{T_{gas} - T_{inp}}{T_{otp} - T_{amb}} = \frac{3}{2}$$

That gives me :

$$T_{gas} - T_{inp} = \frac{3(T_{otp} - T_{amb})}{2}$$

and hence :
$$T_{gas} = \frac{3(T_{otp} - T_{amb})}{2} \ +\ T_{inp}$$

Here T_otp is the temperature of the outside of the pipe at the point where the exhaust starts from the engine - this is measured by my temperature sensor.

T_inp is the temperature of the inside of the pipe at the same point, which I hope to get from literature data based on the T_otp.

T_amb is the ambient temperature of the air that's easily available.

T_gas is the temperature of the gas which we can find by plugging in all the values.

Would all this be correct?

 

[Chestermiller]

Hi Chet,

To clear things out on what is going on in my mind, let me put it in equations:

Okay since,

$$h_{in} = \frac{3}{2}*h_{loc}$$ , and if it is like h_in is the constant for the inside of the pipe and h_loc is for the outside, we have :


$$\frac{T_{gas} - T_{inp}}{T_{otp} - T_{amb}} = \frac{3}{2}$$

That gives me :

$$T_{gas} - T_{inp} = \frac{3(T_{otp} - T_{amb})}{2}$$

and hence :
$$T_{gas} = \frac{3(T_{otp} - T_{amb})}{2} \ +\ T_{inp}$$

Here T_otp is the temperature of the outside of the pipe at the point where the exhaust starts from the engine - this is measured by my temperature sensor.

T_inp is the temperature of the inside of the pipe at the same point, which I hope to get from literature data based on the T_otp.

T_amb is the ambient temperature of the air that's easily available.

T_gas is the temperature of the gas which we can find by plugging in all the values.

Would all this be correct?

Actually, none of it is correct. Every last bit of it is wrong. Please see my private message to you.

Chet

 

[Jay_]

Hi Chet,

I just read the chapter. Well, h_ln is the heat transfer coefficient based on the log mean temperature, and he mentions its used in most calculations. Since the h values depend on the geometry as well as the fluid properties, I would have to get that from Toyota's literature, if they have it that is.

But the Fig. 14.3-2 in Bird's book is still hard to understand. You mentioned that the gas flow in my case is turbulent, so Re > 8000. Would that be correct? So in that case I need to be looking at the right side of the graph.

On that side the Re seems to be increase for decreasing value heat transfer coefficient, because the line has a negative slope (and is not perfectly straight though). Does this go to give me a relation between the heat transfer coefficient and the flow rate, because he mentions flow rate is the Reynold's number.

I don't see the concept (equation) you mentioned in post #35, even though the term (T_b2 - T_b1)/(T₀ - T_b)_ln looks similar, but I realize it only mentions the temperature of the inside wall, and the bulk gas temperatures.

Thanks.

 

[Chestermiller]

Hi Chet,

I just read the chapter. Well, h_ln is the heat transfer coefficient based on the log mean temperature, and he mentions its used in most calculations. Since the h values depend on the geometry as well as the fluid properties, I would have to get that from Toyota's literature, if they have it that is.

It's good that your area of expertise is circuits and electronics, since that can help you greatly with your understanding of heat transfer. I assume that you are familiar with the concept of resistors in series. This is what you are dealing with in your problem. There is a resistance to heat flow from the bulk gas flow to the tube wall, and there is a second resistance to heat flow from the tube wall to the ambient air.

In the analogy between heat transfer and electronics, the temperature is analogous to voltage and the heat flux (W/m^2) is analogous to current density. The heat transfer coefficient gives the ratio of the heat flux to the temperature difference. So, it's like the reciprocal of resistance.

Do you remember studying heat conduction in freshman physics. For heat conduction through a wall, the heat flux is equal to the temperature difference times the thermal conductivity divided by the thickness of the wall. In convective heat transfer, the thickness of the wall is replaced by the thickness of the "thermal boundary layer." The heat transfer coefficient is equal to the thermal conductivity of the gas divided by the thermal boundary layer thickness (which is typically very thin).

You need to determine the two series heat transfer resistances (inside and outside the pipe wall) so that you can determine the temperature at the wall, knowing the temperature of the exhaust gas and the air temperature. This is where the correlations in BSL chapter 14 come in.

But the Fig. 14.3-2 in Bird's book is still hard to understand. You mentioned that the gas flow in my case is turbulent, so Re > 8000. Would that be correct?

Yes.

So in that case I need to be looking at the right side of the graph.

Yes.

On that side the Re seems to be increase for decreasing value heat transfer coefficient, because the line has a negative slope (and is not perfectly straight though). Does this go to give me a relation between the heat transfer coefficient and the flow rate, because he mentions flow rate is the Reynold's number.

You have to look at the total relationship. Actually, the heat transfer coefficient increases with the Reynolds number (as the flow rate increases, the thermal boundary layer gets thinner). Look at the full definition of the ordinate on the graph.

To use this correlation, you are going to have to estimate the physical properties of the gas: thermal conductivity, heat capacity, viscosity, density. You know the composition of the exhaust gas from the chemical reaction mass balance on the combustion. BSL gives relationships for estimating the thermal properties of a gas or gas mixture. You are going to be dealing primarily with a mixture of carbon dioxide, water vapor, and nitrogen. On the air side of the tailpipe, you are going to be dealing with air (a mixture of oxygen and nitrogen).

Start working on trying to estimate the heat transfer coefficient on the gas side of the wall.

Chet

 

[Jay_]

You are going to be dealing primarily with a mixture of carbon dioxide, water
vapor, and nitrogen. On the air side of the tailpipe, you are going to be dealing with
air (a mixture of oxygen and nitrogen).

Right. But I am still confused on how to establish the heat transfer coefficient on both walls of the pipe because I don't know what the ordinate of the graph represents.

Look at the full definition of the ordinate on the graph. To use this correlation, you are going to have to estimate the physical properties of the gas: thermal conductivity, heat capacity, viscosity, density. You know the composition of the exhaust gas from the chemical reaction mass balance on the combustion. BSL gives relationships for estimating the thermal properties of a gas or gas mixture.

Okay so what I am thinking :

1. Get the reaction that happens when the exhaust gas is formed
2. Calculate the mass composition of each constituent
3. Use their properties to find that of the mixture

But what does the ordinate represent exactly? From the first equation on the ordinate, I think "D" (diamter) is related to the pipe, and k, Re, Pr, mu_b, mu_0, h_ln are related to the gas. After calculating this what do I have? What does it represent?

----

Now quoting post #35 :

gallons per minute = (miles per minute)/(miles per gallon)

Here wouldn't we have to account for the air mixing with the fuel?

The data they have on the car's miles per gallon takes into consideration only the burning gallons of the fuel right?

But what comes out the tailpipe in gallons per minute is also the air-reacted carbon dioxide.

 

[Chestermiller]

Right. But I am still confused on how to establish the heat transfer coefficient on both walls of the pipe because I don't know what the ordinate of the graph represents.

Maybe it would be easier for you to work with Eqn. 14.3-16 which is equivalent to the behavior of the graph at high Re. The Nussult number is the dimensionless heat transfer coefficient. The Reynolds number is given by:

$$Re=\frac{ρvD}{μ}$$
For flow inside the tailpipe, this can be combined with the equation for the mass flow, which is given by
$$W=ρv\frac{πD^2}{4}$$
What do you get when you combine these two equations to eliminate ρv?

Okay so what I am thinking :

1. Get the reaction that happens when the exhaust gas is formed
2. Calculate the mass composition of each constituent
3. Use their properties to find that of the mixture

Basically, yes. All you need to know is the air/fuel mass flow ratio. From the internet, I found that this is about 14.7

But what does the ordinate represent exactly? From the first equation on the ordinate, I think "D" (diamter) is related to the pipe, and k, Re, Pr, mu_b, mu_0, h_ln are related to the gas. After calculating this what do I have? What does it represent?

See my comment above about using Eqn. 14.3-16. To get your feet wet, try doing the calculation assuming the gas has a viscosity equal to that of air. What would this give you for the heat transfer coefficient?

----

Now quoting post #35 :


Here wouldn't we have to account for the air mixing with the fuel?

The data they have on the car's miles per gallon takes into consideration only the burning gallons of the fuel right?

But what comes out the tailpipe in gallons per minute is also the air-reacted carbon dioxide.

Once you know the air/fuel ratio (~14.7), you have everything you need to estimate the composition of the exit gas (using the reaction stoichiometry).

To get the heat transfer coefficient for the boundary layer outside the pipe, you use the information in chapter 14 for Submerged Objects. The exhaust pipe is submerged in the air flow moving past the exhaust pipe as the car is moving. The relative velocity of the air flow is the car speed, and it is flowing parallel to the axis of the exhaust pipe (mainly).

Chet

 

[Jay_]

Hey thanks Chet. The equation I would get for the flow is :

$W=\frac{ReμπD}{4}$

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

Or are we looking for the Nusselt number?

$$Nu_{ln}=0.026(\frac{4W}{μπD})^{0.8}Pr^{1/3}(\frac{μ_b}{μ_0})^{0.14}$$

That's what I get for equation 14.3-16, eliminating ρv.

 

[Chestermiller]

Hey thanks Chet. The equation I would get for the flow is :

$W=\frac{ReμπD}{4}$

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

Or are we looking for the Nusselt number?

$$Nu_{ln}=0.026(\frac{4W}{μπD})^{0.8}Pr^{1/3}(\frac{μ_b}{μ_0})^{0.14}$$

That's what I get for equation 14.3-16, eliminating ρv.

Good. Yes, we're focusing on getting the Nussult number. This will lead directly to the heat transfer coefficient.

The first thing I would like you to do is to calculate a typical Reynolds number for the exhaust gas flow through the tail pipe. To do this, you need to get an estimate of the mass flow rate through the tailpipe. Temporarily, we can use the viscosity of air or nitrogen at the approximate temperature of the exhaust gas. We can refine the calculation later. We just want to get our feet wet and find out what "ballpark" we are playing in.

Do you have any idea how to estimate the mass flow rate of exhaust gas? If so, let's see how you do it.

Chet

 

[Jay_]

The mass flow rate of the exhaust is roughly the same as the intake of the engine. This can be calculated from the AFR (air-fuel ratio) and the concept:

gallons per minute = (miles per minute)/(miles per gallon)

Here miles per minute is measured (vehicle speed), and miles per gallon (mileage) of the car is something we will know (hopefully), so we can end up with gallons of fuel burned per minute. Using the AFR we find how much air is also used, and the exhaust flow rate has to be the combination of the two I guess. Correct?

 

[Chestermiller]

The mass flow rate of the exhaust is roughly the same as the intake of the engine. This can be calculated from the AFR (air-fuel ratio) and the concept:

gallons per minute = (miles per minute)/(miles per gallon)

Here miles per minute is measured (vehicle speed), and miles per gallon (mileage) of the car is something we will know (hopefully), so we can end up with gallons of fuel burned per minute. Using the AFR we find how much air is also used, and the exhaust flow rate has to be the combination of the two I guess. Correct?

So, let's see some numbers.

Chet

 

[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.

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

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?

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

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).

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

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?

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