# I Velocity of gases in a flue / stack

1. Jul 18, 2016

### Jehannum

I know the flow rate of gas (products of combustion + air) entering a natural draught (no fans) flue.

If the flue is removing all the gas:

velocity of flue gas = flow rate (corrected for thermal expansion) / flue csa​

As the flue gas moves up the flue what happens to its velocity? In particular, I need to know the exit velocity at the flue terminal. Obviously, this will depend upon the flue height, but what is the relationship between height and velocity?

I know the gas will lose heat as it moves up in contact with the flue wall, slowing it down. Are there any other factors involved? Does the constant upthrust from the pressure difference between the top and bottom of the flue mean the gas continues to accelerate throughout its ascent?

2. Jul 18, 2016

### Staff: Mentor

If you neglect the cooling effect, you can apply the compressible form of the Bernoulli equation to get an answer.

3. Jul 18, 2016

### Andrew Mason

The flow rate in terms of mass or moles per unit time has to be the same at the top of the chimney as at the bottom, assuming that the chimney is operating normally. To work out the density of the gas at the top, you would have to take into account the pressure and temperature gradient within the chimney. That will enable you to determine the speed of the gas at the top. The volume per unit time at the top would be the volume that contains the same mass or moles per unit time of the gas entering the chimney (using the ideal gas equation, V = nRT/P).

It is not a case of higher pressure at the bottom and lower pressure at the top. A properly working chimney will have lower than ambient pressure at the bottom in order to draw gases in. The upper part, however, will have higher than ambient pressure in order to expel the gases against ambient pressure. The gas rises in the chimney due to buoyancy. Velocity of the gas slows as it rises. The slower moving upper gas is compressed by the incoming rising gas below it.

AM

Last edited: Jul 18, 2016
4. Jul 19, 2016

### Jehannum

Thank you. This is very helpful.

A further couple of questions:

1. What slows the flue gases down as they move up the flue? Is it purely temperature loss? In a perfectly heat-insulating flue would the velocity still decrease?

2. How do flue sizing charts work? A typical table from a gas standard gives a correct flue diameter for a 4% CO2 flue gas concentration. The table requires only the appliance heat input (or gas rate) and the height of the flue to be known. There is no indication of the heat loss of the flue, or the flue material. Has that been cancelled out somehow?

5. Jul 19, 2016

### Andrew Mason

Think of the chimney just starting to draw. There is stationary air at ambient pressure in the chimney. Hot air is introduced in the bottom. It begins to rise due to buoyancy. As it rises it creates a low pressure below it that draws more hot gases in. The rising gas pushes on the air above it, compressing it. As the chimney stack stabilises, a column of moving air exists within the chimney with negative (lower than ambient) pressure in the bottom half and positive pressure in the upper half. The kinetic energy of the gas flow at the bottom has to be used to a) gain gravitational potential energy as it rises and b) do compression work on and lift the air above it.

There should be no significant loss of heat within a properly constructed chimney. That is why they are insulated or made of brick.

AM

6. Jul 19, 2016

### Staff: Mentor

The steady state compressible flow of the gas through the chimney can actually be quantified.

For the inviscid adiabatic reversible flow of an ideal gas through the chimney, the enthalpy (per unit mass) variation must satisfy 3 separate equations:

1. The differential form of the open system steady state flow version of the first law of thermodynamics:$$dh+vdv+gdz=0\tag{1}$$where v is the velocity and z is the elevation above the flue.
2. The enthalpy property relationship for an ideal gas: $$dh=c_pdT\tag{2}$$
3. The constraint that the process is adiabatic and reversible (constant entropy):$$dh=Tds+\frac{dP}{\rho}dP=\frac{dP}{\rho}\tag{3}$$where $\rho$ is the gas density.
If we combine Eqns. 2 and 3 with the ideal gas law, we obtain:
$$P=C\rho^\gamma\tag{4a}$$
or equivalently,
$$\rho=kP^{1/\gamma}\tag{4b}$$with $\rho=\frac{PM}{RT}$.
Eqn. 1 can be integrated immediately to yield:
$$\Delta h+\Delta \left(\frac{v^2}{2}\right)+g\Delta z = 0\tag{5}$$
where $$v=\frac{w}{\rho A}=\frac{w}{kP^{1/\gamma}A}\tag{6}$$with w representing the mass rate of flow.
If we substitute Eqn. 4b into Eqn. 3 and integrate, we obtain:
$$\Delta h=\frac{\gamma}{k(\gamma - 1)}\Delta P^{\frac{\gamma - 1}{\gamma}}\tag{7}$$
The constant k can be evaluated from the conditions at the flue:
$$\rho_0=\frac{P_0M}{RT_0}=kP_0^{1/\gamma}\tag{8a}$$or
$$k=\frac{MP_0^{\frac{\gamma -1}{\gamma}}}{RT_0}\tag{8b}$$Combining Eqns. 7 and 8b yields:
$$\Delta h=\frac{RT_0\gamma}{M(\gamma - 1)}\Delta \left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}=\frac{c_0^2}{(\gamma - 1)}\Delta \left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}\tag{9}$$where $c_0=\sqrt{\frac{\gamma RT_0}{M}}$ is the sonic velocity at the base of the chimney.
Similarly, from Eqns. 6 and 8b, $$v=\frac{w}{\rho A}=\frac{w}{\rho_0A(P/P_0)^{1/\gamma}}=\frac{v_0}{(P/P_0)^{1/\gamma}}$$So, $$v^2=\frac{v_0^2}{(P/P_0)^{2/\gamma}}\tag{10}$$
If we substitute Eqns. 9 and 10 into Eqn. 5, and reduce the resulting relationship to dimensionless form, we obtain:
$$\frac{1}{(\gamma - 1)}\left[\lambda^{(\gamma-1)/\gamma}-1\right]+\frac{M_0^2}{2}\left[\frac{1}{\lambda ^{2/\gamma}}-1\right]+Z=0\tag{11}$$
where $\lambda = P_{T}/P_0$ is the ratio of the pressure at the top of the chimney to the pressure at the bottom, $M_0=\frac{v}{c_0}$ is the Mach number at the bottom of the chimney (and also the dimensionless mass flow rate at the bottom $\frac{w}{\rho_0 c_0 A}$), and $Z=\frac{gz}{c_0^2}$ is the dimensionless height of the chimney.

We don't know the ratio $\lambda$ of the top pressure to the bottom pressure within the column, but we can get a handle on it by finding the value that makes good on mass flow rate at the bottom (i.e., the measured Mach number at the bottom) and the height of the column, and then we can compare the value with the corresponding pressure ratio outside the chimney so that we can assess the draft. The "game plan" would be to solve for $M_0^2$ as a function of $\lambda$ and Z, and then prepare a master plot of $M_0^2$ vs $\lambda$ with Z as a parameter. The starting point in preparing the master plot would be to solve Eqn. 10 for $M_0^2$ as follows:

$$M_0^2=\frac{2Z}{\left[1-\frac{1}{\lambda ^{2/\gamma}}\right]}+\frac{2\left[\lambda^{(\gamma-1)/\gamma}-1\right]}{(\gamma-1)\left[1-\frac{1}{\lambda ^{2/\gamma}}\right]}\tag{12}$$

Last edited: Jul 21, 2016
7. Jul 21, 2016

### Staff: Mentor

@Jehannum and @Andrew Mason : I have added some development to post #6 that you may be interested in.

Chet

8. Jul 28, 2016

### Jehannum

As a non-scientist, this has been taking me some time to go through. Doing so has been educational.

It seems equation 10 (in non-squared form) could be the most useful for me.

I get a little lost after that. Could you explain the concept of 'dimensionless height', and why using dimensionless quantities is useful?

9. Jul 28, 2016

### Staff: Mentor

In physical problems, expressing the equations in dimensionless form boils down the problem to its bare essence and reduces the number of variables you need to consider to an absolute minimum. In the original formulation, the height z, the gravitiational acceleration g, the pressure at the bottom, the pressure at the top, the velocity, and the temperature were separate parameters that would need to be considered independently. By expressing the equations in dimensionless form, we see that there are really only 3 key parameters involved: the Mach number, the pressure ratio, and the dimensionless height of the chimney. This dimensionless version of the formulation contains all the information that the original formulation contained, but requires only establishing the relationship between the 3 key parameters. Once this is done, you can use it to find the answer for any combination of the original parameters. Dimensional analysis is a very powerful technique.

10. Jul 28, 2016

### cjl

I suspect that the actual compression is quite small, and any density change is largely driven by temperature gradient, not pressure.

11. Jul 28, 2016

### Staff: Mentor

The analysis suggests that, if the chimney is essentially adiabatic and the expansion of the parcels of gas traveling up the chimney is nearly reversible (i.e., gradual, pressing against adjacent gas parcels), the pressure and temperature of the gas will change in tandem according to the usual adiabatic reversible expansion relationships.

12. Jul 28, 2016

### Andrew Mason

There is not supposed to be much heat lost through the chimney walls of a properly made chimney. The pressure in the bottom half of the chimney has to be lower than ambient pressure in order to draw in the hot flue gases. The pressure at the top has to be greater than ambient pressure in order to expel the gases at the top of the stack. So there has to be a pressure difference. There does not have to be a temperature difference. In fact, a temperature decrease within the stack, makes it more difficult to maintain the flow.

Just commenting on Chet's last post, I think this is correct. The compression should be adiabatic so the gases actually should be a bit hotter at the top than at the bottom.

AM

Last edited: Jul 28, 2016
13. Jul 28, 2016

### cjl

You're pretty much assuming your conclusion here. Of course the gas behaves adiabatically if the situation is adiabatic. I suspect it is not though.

14. Jul 28, 2016

### cjl

There may not be much heat lost, but the heat capacity of gases is relatively low, so there wouldn't need to be much heat lost for it to be a significant effect. Also, I would be willing to bet quite a bit that the pressure at the bottom of the chimney is higher than it is at the top, since the gas wouldn't flow up the chimney otherwise. For fluid to flow against an adverse pressure gradient, you need significant momentum effects, and I doubt that the flow speed is anywhere near high enough here for that to be at all significant.

Keep in mind, the flow is being driven by buoyancy. The external pressure at the top of the chimney is lower than the pressure at the bottom, and the gravitational pressure gradient in the chimney is smaller than the one outside due to the lower density of the gas. For a 10m chimney though, the external pressure difference will only be on the order of 10-15 pascals though, which is why I strongly suspect that compression or compressibility effects are completely negligible, except those related to thermal expansion.

15. Jul 28, 2016

### Staff: Mentor

This all sounds somewhat reasonable. How would you modify the model in post #6 to better fit your concept of what is happening? Add chimney heat loss to the model?

Chet

16. Jul 28, 2016

### Andrew Mason

If it was higher at the bottom than at the top the chimney would not draw air. A chimney has to have negative pressure in the lower part and positive pressure in the upper part. Otherwise it will not draw. It seems counter intuitive but if you make a hole in the side of the chimney in the lower half, it will suck air through the hole, into and up the chimney.

AM

17. Jul 28, 2016

### cjl

Not necessarily. The pressure at the top of the chimney needs to be higher than the ambient pressure at the top. The pressure at the bottom needs to be lower than the ambient pressure at the bottom. Both of these can be true AND have the pressure at the top of the chimney lower than the pressure at the bottom (and have your statement about a hole remain true).

The reason this works is because ambient pressure at the top of the chimney is lower than ambient pressure at the bottom, due to gravitational effects. For a 10m tall chimney, as I stated above, I'd expect the ambient pressure to be about 10-15 pascals lower at the top of the chimney than at the bottom. If the pressure at the bottom is 2-3 pascals below ambient, and there's a 5 pa gradient from the bottom to the top of the chimney (with the top having lower pressure), the pressure at the top of the chimney will still be a couple pascals above ambient, allowing the chimney to flow.

18. Jul 28, 2016

### cjl

Yeah, though figuring out a reasonable rate for the heat loss would be rather difficult I suspect. You'd probably have to just go measure some chimneys to get some sense of what a reasonable value was.

19. Jul 28, 2016

### Staff: Mentor

I'm not so sure about this. The upward velocity of the gases at the bottom of the chimney might correspond to a lower pressure inside the chimney than outside (at the bottom), but the pressure inside the chimney might still decrease with elevation. I tried come calculations in which the pressure was assumed to increase with elevation, but I was not able to satisfy my final equation in post #6 unless the gas velocity at the bottom of the chimney were above sonic. I'm interested in hearing what cjl says about this.

20. Jul 28, 2016

### Staff: Mentor

That could be played with in the model by adding some Newton cooling.

21. Jul 29, 2016

### Andrew Mason

If there is a breach of the chimney in the upper part you will have flue gases coming out. But not in the lower part. I am not sure how that could occur if the pressure in the chimney decreases with height.

AM

22. Jul 29, 2016

### Staff: Mentor

See cjl's post #17.

23. Jul 29, 2016

### cjl

Keep in mind that external pressure also decreases in height. As long as the pressure gradient in the chimney is lower than the gradient outside the chimney, this still occurs.

24. Jul 30, 2016

### Andrew Mason

But the pressure in the chimney is equal to the ambient pressure at about the mid-point of the stack. This is the neutral pressure plane. That is, if you make a hole in the chimney at that point, the gases will not flow into or out of the chimney. Above it, flue gases will flow out and below it outside air will flow in. So the pressure gradient inside for the bottom half AND the top half has to be greater than the pressure gradient outside. I agree that, theoretically, there could still be a negative gradient over-all inside the chimney. It would be interesting to see actual measurements of pressure in a working chimney

AM