# [Non-uniform Air Density] Pressure problem

• Gr33nMachine
In summary, the atmospheric pressure decreases as you go up, and the density becomes zero at some height. You need to use the differential form of the static equilibrium equation to determine the value of L.
Gr33nMachine

## Homework Statement

Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
What would be the height of the atmosphere if the air density decreased linearly to zero with height?

P = ρgh

## The Attempt at a Solution

Really not sure how to approach this. I've made density a function of height: ρ(h) = 1.3-1.3/h and I integrated this from 0 to h(max) so it now becomes 1.3 - 1.3ln(h), but where do I go from here?

Gr33nMachine said:

## Homework Statement

Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
What would be the height of the atmosphere if the air density decreased linearly to zero with height?

P = ρgh

## The Attempt at a Solution

Really not sure how to approach this. I've made density a function of height: ρ(h) = 1.3-1.3/h and I integrated this from 0 to h(max) so it now becomes 1.3 - 1.3ln(h), but where do I go from here?

You need to use the differential form of the static equilibrium equation. That is, what is the equation for dP/dz, where z is the distance upward? You also need to express the density as a linear function of z, with the density becoming equal to zero at some altitude L. The pressure at altitude L in this contrived example is also equal to zero. This should give you enough information to determine the value of L.

Chestermiller said:
You need to use the differential form of the static equilibrium equation. That is, what is the equation for dP/dz, where z is the distance upward? You also need to express the density as a linear function of z, with the density becoming equal to zero at some altitude L. The pressure at altitude L in this contrived example is also equal to zero. This should give you enough information to determine the value of L.

I don't understand how to express the density as a function of z...
y = mx + b
ρ(z) = -1.3/(z max) * z + 1.3 ? How does this integrate into anything useful?

Gr33nMachine said:
I don't understand how to express the density as a function of z...
y = mx + b
ρ(z) = -1.3/(z max) * z + 1.3 ? How does this integrate into anything useful?

This result is correct. You used z max instead of my L, but that's perfectly OK.

Now, the static equilibrium equation for variable density:

dP/dz = -ρg Substitute your density equation into this.

Boundary conditions: P = 1 atm (express this in N/M2) at z = 0 and P = 0 at z = z max

Can you see what the next step is?

Chestermiller said:
This result is correct. You used z max instead of my L, but that's perfectly OK.

Now, the static equilibrium equation for variable density:

dP/dz = -ρg Substitute your density equation into this.

Boundary conditions: P = 1 atm (express this in N/M2) at z = 0 and P = 0 at z = z max

Can you see what the next step is?

Well, dP/dz = ρg can be rearranged to dP = gρ(z)dz and integrate ρ(z) from 0 to h?
The result is P = g(-1.3/(2L)*h2+1.3h), and now I can solve for L = (1.3h2)/(2P/g-1.3h)

But what I can plug in for h?

Gr33nMachine said:
Well, dP/dz = ρg can be rearranged to dP = gρ(z)dz and integrate ρ(z) from 0 to h?
The result is P = g(-1.3/(2L)*h2+1.3h), and now I can solve for L = (1.3h2)/(2P/g-1.3h)

But what I can plug in for h?

You left out the minus sign in the hydrostatic equation. You also left out the boundary condition P = 1 atm at z = 0 in your integration. Also z and h are the same parameter. In the final equation, set P = 0 at z = L to find the value of L.

Chestermiller said:
You left out the minus sign in the hydrostatic equation. You also left out the boundary condition P = 1 atm at z = 0 in your integration. Also z and h are the same parameter. In the final equation, set P = 0 at z = L to find the value of L.

I'm really confused.
Why are there two boundary conditions for an single variable integration?
I know z and h are the same parameter.
If I set P=0 and z=h=L in the final equation, I get L = 1.3L2/-1.3L, which simplifies to -1.3L2=1.3L2 How does this help?

Gr33nMachine said:
I'm really confused.
Why are there two boundary conditions for an single variable integration?
I know z and h are the same parameter.
If I set P=0 and z=h=L in the final equation, I get L = 1.3L2/-1.3L, which simplifies to -1.3L2=1.3L2 How does this help?
In my previous reply, I also mentioned that you left out the constant of integration in your pressure equation. You need to make this correction also.

Another way of doing this problem is to recognize that, if the density varies linearly with z and is equal to zero at the "top of the atmosphere," then the average density is 1.3/2 = 0.65. Then you can use

Pground = ρgroundgL/2

You should confirm that this gives the same result as the integrated method.

## 1. What is non-uniform air density?

Non-uniform air density refers to the variation in density of air at different altitudes or locations. This can occur due to differences in temperature, humidity, and air pressure.

## 2. How does non-uniform air density affect pressure?

Non-uniform air density can affect pressure by creating differences in air pressure at different altitudes or locations. This can result in changes in wind patterns, weather conditions, and air flow.

## 3. What causes non-uniform air density?

Non-uniform air density can be caused by a variety of factors, including variations in temperature, humidity, and air pressure. These factors are influenced by factors such as altitude, geography, and weather patterns.

## 4. How is non-uniform air density measured?

Non-uniform air density can be measured using instruments such as weather balloons, anemometers, and barometers. These instruments measure factors such as temperature, humidity, and air pressure to determine air density.

## 5. What are the consequences of non-uniform air density?

Non-uniform air density can have various consequences, including changes in weather patterns, air flow, and wind patterns. It can also affect the performance of aircraft, as changes in air density can impact lift and drag forces.

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