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[Non-uniform Air Density] Pressure problem

  • #1

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?

Homework Equations



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?
 

Answers and Replies

  • #2
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3,979

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?



Homework Equations



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.
 
  • #3
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?
 
  • #4
19,666
3,979
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?
 
  • #5
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?
 
  • #6
19,666
3,979
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.
 
  • #7
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?
 
  • #8
19,666
3,979
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.
 

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