Pressure Difference due to Altitude

[millsjj]

Hi All,

Want to bring this back to basics for a minute. I work for a gas company and someone came in with a simple question.

A sealed pipe of natural gas rises 100m. the pressure at the bottom is 3kPa. What is the pressure at the top?

I believe the pressure is less due to Bernoulli's equation:
P1 - P2 = w(Z2 - Z1).

Some of my collegues have disputed this reasoning that as gas is lighter than air the pressure would be greater at the top. My opinion is that as the pipe is sealed then that is of no consequence.

Can someone please answer this so that we can put this one to rest!

Regards

 

[BigInch]

That's technically not Bernoulli, just a head lift due to pressure. Bernoulli actually describes the total energy = Static Reference + Pressure head + velocity head.

In a sealed tube, air gives no buoyant force to the gas contained within, but even that is of no consequence, as pressure is always greater at the bottom of a column of liquid or gas due to the weight of the fluid mass above. Applying a buoyant force by surrounding it with air would simply mean that the total weight of the gas column would be less when weighed in air, so the gas' pressure at the bottom would also be less when it is weighed in air, but that would still be higher than the pressure measured at the top of the gas column.

BTW, with compressible fluids, the w(Z2-Z1) is not correct, because as the fluid compresses, the w term, weight, in each Z step you take increases as you go downwards, or decreases as you go upwards, which is why the atmospheric pressure at Denver is (usually) less than the atmospheric pressure in Houston. Summing weight of a gas column in the going up direction, each step up adds just a little bit less weight than the last step up you took.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[millsjj]

Thanks for the response BigInch. I am aware that natural gas is not a non-compresssible fluid but for simplicity of this exercise we chose to treat it as such.

That equation is not the full version of Bernoulli but that is what it can be worked down to when the fluid is static. The velocity head goes to zero leaving only pressure, specific weight and static reference.

I have used that equation to work out the pressure at the top but my collegues (engineers and gas fitters of 30 years of experience) are all divided as to the actual answer. Can anyone supply me with a documented answer so that we can rectify this problem.

Thanks

 

[BigInch]

Yes, but then why use Bernoulli, that's energy/unit mass. You only need to look at weight and pressure.

If I sign and seal my answer above, then I have to charge you $250. Check the NASA website for K12 content. You will find standard atmosphere pressures at various altitudes. That's air with no buoyancy. Subtract buoyancy... any constant buoyant force you like from each elevation and you will see that pressure is still greater at the bottom than at the top, or just do the math yourself.



Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[katmar]

You and your dissenting colleagues are both correct. In the vertical column of gas the pressure must decrease as you go higher up the column for exactly the reasons that you and BigInch have pointed out. But remember that you are talking absolute pressure when you are talking of pressure inside a sealed tube.

What your colleagues have noticed over their 30 years of experience is that the gauge pressure in a low pressure gas supply network increases as you go higher up a building because air is heavier than low pressure natural gas and so the absolute pressure of the air decreases by a larger amount than does the gas - making the difference (or gauge pressure) higher as you go up.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[millsjj]

Mate,

If I could do all of this out and explain it properly I wouldnt be asking here. All im looking for is a simple answer to a simple question. Yes you have stated that the pressure will be greater at the bottom, as I already believed, but I just want it spelt out so that it can universally be understood.

 

[ione]

The force exerted by the weight of a column of gas from Newton's second law and gravitational acceleration is:
F = m*g
Where m is the mass of gas
The pressure P resulting from the column of gas on cross sectional area A
P = m*g/A
The mass of gas is given by the product of its density rho and its volume For a cylindrical column, the volume V of gas is the product of the cross-sectional area A and the height of the column h.
P = m*g/A = rho*V*g/A = rho*h*A*g/A = rho*g*h

I think you don't need any more reference than basic physic laws quoted above.
Do not forget the comment from BigInch on gas density variation due to compressibility effect.

 

[katmar]

Well mate, if this isn't simple enough for you then let me put some numbers to it and maybe you can achieve universal understanding.

The 3 kPa you measure at the bottom is a gauge pressure. Let us assume the air pressure at the base of the column is 101,3 kPa absolute, making the gas pressure in the base of the column 104,3 kPa abs. If we assume that the gas is pure methane and the temperature is 20 C then the gas density is 0.688 kg/m[sup]3[/sup]. Using a slight variation on your formula the change in pressure up the column is
[Δ]P = density x gravity x change in height = 0,688 x 9,81 x 100
= 675 Pascal or about 0,7 kPa
This makes the gas pressure at the top of the column 104,3 - 0,7 = 103,6 kPa absolute.

This very low change in pressure makes the assumption of constant density justifiable.

Now we do the same calculation for the air. The density of air at 20 C and 101,3 kPa abs is 1,2 kg/m[sup]3[/sup]
[Δ]P[sub]air[/sub] = density x gravity x change in height = 1,2 x 9,81 x 100
= 1177 Pascal or about 1,2 kPa
This makes the air pressure at 100 m high 101,3 - 1,2 = 100,1 kPa abs.

The gauge pressure of the gas is the difference between the absolute pressure inside the column and the air outside, i.e. 103,6 - 100,1 = 3,5 kPag

In summary, the gas pressure at the base of the column is 104,3 kPa abs and at the top it is 103,6 kPa so this shows that the absolute pressure is highest at the base of the column.

On the other hand, the gas pressure at the base of the column is 3 kPag and at the top it is 3,5 kPag so this shows that the gauge pressure is highest at the top of the column.

Simple enough?

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[BigInch]

"absolute pressure of the air decreases by a larger amount than does the gas making the difference (or gauge pressure) higher as you go up. "

Good point Katmar. If you continue and carry the cylinder into outer space you will have an abs pressure difference of 14.73 psia, with no chance of buoyant forces from any other gas affecting the answer.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[ione]

I must confess I’ve found myself lost with Katmar’s example.

Assuming density being constant, I thought the pressure component which varies with height was hust the hydrostatic component (the one related to weight of the fluid). So if you read 3 kPa gauge at the bottom of the column at an elevation of 100 m you should read 2.3 kPa gauge (3 – 0.7), so gauge pressure is decreased. Now if you wanted to turn to absolute pressure you should have to use the atmospheric pressures at the corresponding elevation. At the bottom of the column you’ll get 104.3 kPa absolute ( 101.3 + 3) and at the top 102.4 kPa absolute (100.1 + 2.3).

 

[BigInch]

I didn't care a 100 m problem and didn't do the math. The math was much easier taking it to outer space. Corresponding abs pressure at that altitude is 0 psia... and the pressure at the bottom or at the top of the column are equal due to zero gravity.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[katmar]

ione, it depends where you are referring the "atmospheric" side of the gauge pressure to. The way I have done it in my example was to compare the absolute pressure of the gas inside the pipe to the air directly outside the pipe i.e. at the same height as the point where the gas absolute pressure is measured.

You are comparing it to the atmospheric pressure at the base of the column.

BigInch has quoted the extreme case, and it does illustrate it well. If you have a closed vessel with 101,3 kPa absolute inside it then at normal sea level the gauge pressure would be 0.0 kPa. However, if you take that same closed vessel into outer space then its gauge pressure becomes 101,3 kPa (or very close to that). If you only go part of the way into space, say where the atmospheric pressure is only 50 kPa abs, then the pressure in the same vessel is now 51,3 kPag. If you only go 100 m towards space then the difference is the small amount I calculated earlier.

This is actually a very relevant and important problem in the gas reticulation business because household appliances are designed to work with only a few inches of water column pressure. This is of course a gauge pressure because we are looking at the flow of gas from the burner into the local atmosphere. In a tall building this effect can quickly add an inch or two to the gauge pressure and this can impact on the burner control.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[katmar]

ione, I think I misunderstood your problem with my example. As you go up the column it is the absolute pressure that decreases and not the gauge pressure because the force of gravity on the molecules inside the sealed tube is not affected by the atmosphere outside. Even if there was a total vacuum around the column, 100 m up it would be the absolute pressure that has decreased by 0,7 kPa.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[zdas04]

Think of a gas well. The gas gradient is

P(bot)=P(top)*exp(0.01875*SG*h/T(avg)/Z(avg))

The exponential term accounts for the density change from bottom to top (all terms are in absolute units). And no, there is no part of this that involves the Bernoulli equation at all (dang I wish people would stop blaming Danny-boy for every fluid problem on the planet). And yes, the pressure at the bottom of the column will be greater than the pressure at the top. Gravity does only go one way after all.

In the sealed tube example (just like in a gas well), the pressure at the top is the atmospheric pressure at the elevation of the top of the column. For 50m that is going to be a small thing, but then the difference between P(top) and P(bot) in the equation over 50m is a really small thing too.

If "experienced engineers" are confused about how gravity works then I'm really worried about our profession.

David