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Bouyancy force -- What mass can the balloon carry?

  • Thread starter Taylan
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Homework Statement:

a balloon is filled with helium. Volume of the balloon is 1000m^3 and the density of helium is 0.1785kg/m^3. The balloon will be released. the air pressure at the bottom is 1013hPa and the density of the air is 1.21kg/m^3.

a) what mass can the balloon can carry at the maximum?
b) to what height can the balloon reach, if its volume remains constant and the mass it is carrying is 500kg?

Homework Equations:

Fbouyancy= mg = density air * volume air *g

P0= p0 e ^ (-p0gh/p0)
In a I used FBouyancy - FBalloon -F mass = 0

rewrote mass as a product of density and volume to obtain;

m= v(densityair - densityHe)

m= 1031.5kg.


I am stuck in part b. I have this formula called Barometric formula which ı should use;

P = P0 * e((-P0*g*h)/ P0

So what I understood is that this formula would give me the pressure at a particular height but I am not sure about how to apply this on what is asked ie not sure about how to find how high the balloon can go up. can you give me any tips please?
 

Answers and Replies

  • #2
kuruman
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Is it only pressure that varies with height? The air also gets thinner as the ballon goes higher. What does "thinner" mean in terms of density?
 
  • #3
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Is it only pressure that varies with height? The air also gets thinner as the ballon goes higher. What does "thinner" mean in terms of density?
So it means density of air decreases too
 
  • #4
kuruman
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Right. At value of the air density will the balloon stop rising?
 
  • #5
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So it means density of air decreases too
I guess when density of air = density of helium
 
  • #6
kuruman
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That would be the case if the balloon carried no load.
 
  • #7
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I guess when density of air = density of helium
Hm so i have to use the 500kg somehow. Can you give me more tip please? Not sure how to include the 500 kg
 
  • #8
kuruman
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Assume that the buoyant force does not change from part (a) and draw a free body diagram of the balloon when it is hovering at some height with 500 kg hanging from it. Also assume that the buoyant force on the 500 kg mass is negligible.

On edit: Also you need to fix the barometric formula. You cannot have P0 in both numerator and denominator of the argument of the exponential.
 
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  • #9
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I am confused by your barometric formula because you seem to be using P for both density and pressure.
In a static air situation and treating air as compressible, we can use the ideal gas law to get

1) ##\rho=\frac{PM}{RT}##

where P is pressure
M is molar mass =28.97
R is the ideal gas constant = 8.31
T is temperature (can assume to be about ##298^{0}K##)

Because the weight of air must cancel out with the changing pressure, we can use the relationship
2) ##\frac{dP}{dh}=-\rho g=\frac{-PMg}{RT}##

Solving gives

3) ##P=P_{0}e^{\frac{-Mgh}{RT}}##

Using equation 1, we get the relationship

4) ##\rho _{0}=\frac{P_{0}M}{RT}##

Substituting back into equation 3 we get

5) ##P=P_{0}e^{\frac{-\rho _{0}gh}{P_{0}}}##

We can assume that ##\frac{P}{\rho }=\frac{P_{0}}{\rho _{0}}## so we get

6) ##\rho =\rho _{0}e^{\frac{-\rho _{0}gh}{P_{0}}}##

Which is probably the formula you were trying to get at.

You already have the weight of the balloon and 500 kg mass so now you just
need to set that equal to the buoyancy force which is ##\rho gV##
Plugging in equation 6 for density of air, you can solve for height.
 
  • #10
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thanks a lot!
 

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