Re: What is the formula for amount of helium to lift wieght?

Question:
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What is the formula for amount of helium to lift weight?

References:
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The following material was taken from the book "Physics for Scientists
& Engineers" by Raymond A. Serway third edition pages 399 to 402.  
However, this material should be in most any college level physics text 
book and some high school level texts.

Answer:
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Being a graduate student, I hear this type of question a lot and I would 
like to take an opportunity to caution you against using formulas this 
way.  Just finding a formula in a book and using it is not usually a wise 
thing to do.  In order to use formulas correctly, you must know how they 
are derived and you must know the assumptions which go into them so you 
don't use them incorrectly.

In order to solve this problem let me give a little background.  Air and 
helium are considered fluids, mainly because they conform to any 
container you put them in.  This is not the technical definition of a 
fluid, but it will do for our purposes.  When we speak of fluids we often 
use the idea of "pressure" to describe their behavior.  Pressure is just 
the amount of force exerted on an area, or P = Force/Area.  This will be 
very useful in our treatment of the problem.

The pressure in a fluid varies with how deep we are in the fluid.  More 
than likely you have been swimming, and know that the deeper you swim the 
more pressure you have on your ears.  Also, if you are driving around and 
go down a long hill your ears will pop because of the pressure difference 
(depending on how long the hill is).  These are all common experiences, 
but lets put an equation to them.  Basically the pressure difference in a 
fluid can be represented as follows:


                   ----------------- P1 = pressure at higher level
                         /\
                         |
                         | height difference = h
                         |
                         \/
                   ----------------- P2 = pressure at lower level


If the fluid has a density of (Df), and the gravitational constant is 
g = 9.81 meters/second^2, then:

                  P2 = P1 + (Df)gh

What this equation says is that the pressure at the lower level is the 
pressure at the higher level plus the weight of fluid pushing down.

The equation is often written as: (P2 - P1) = (Df)gh

Buoyancy is the tendency of objects to be lighter in a fluid than they 
would normally be.  Again, calling on the example of the swimming pool.  
You more than likely have noticed that you "feel" lighter when you are 
swimming than when you are walking around the pool.  This is due to the 
bouyant force.

If I have a submerged object, then the part of the object which is deeper 
in the fluid is experiencing a greater pressure.  Lets take for instance 
a cube with all three sides being of length "h":


                      /----------/  /\
                     /          /|   |
                    /----------/ |   | height of cube = h
                    |         |  |   |
                    |         |  /  \/
                    |         | /
                    |---------|/


So, the pressure at the bottom of the cube will be greater than the 
pressure at the top of the cube.  (Pbottom - Ptop) = (Df)gh  Where Df is 
the density of the fluid the cube is submerged in (just like our 
expression above).  This creates a net force pushing up on the cube by 
the following expressions:

                    (Pbottom - Ptop)   = (Df)gh
                    (Fbottom - Ftop)/A = (Df)gh

Remember that P = F/A, where A is the surface area of one side of the 
cube.  The bouyant force is then the difference in the forces on the cube 
(Fbottom - Ftop) = Fbouyant:

Multiplying both sides by A gives:

                    Fbouyant = (Df)ghA

However, we know that the volume of the cube is Vcube = hA.  This gives:
 
                    Fbouyant = (Df)gV  (V is the volume of the cube)

The buoyant force is the force pushing upward on our cube.  Now we have 
to balance that by the weight of the cube pushing down.  Let the density 
of the cube be (Dc).  So the total mass of the cube is (Dc)V, which makes 
the weight of the cube (Dc)gV.

                   Fbuoyant = (Df)gV
                   Wcube    = (Dc)gV  (where V = volume of cube)

So, summing the forces on the cube gives:

                 (Fbuoyant - Wcube) = (Df - Dc)gV

We subtract the forces because they act in opposite directions.  But this 
is the equation of total force on our little cube.

                   Ftotal = (Df - Dc)gV

What this equation says is:

   (1) if the density of the fluid (Df) is greater than the density of
       the material (Dc) the object will float because Ftotal will
       be positive.

   (2) if the density of the fluid (Df) is equal to the density of the
       material (Dc) the object will suspend in the fluid and the total
       force Ftotal will be zero.

   (3) if the density of the fluid (Df) is less than the density of the
       material (Dc) the object will sink and the total force Ftotal will
       be negative.

This all fits with our everyday experience of wood floating and pieces of 
metal sinking.  However, we want to lift a balloon so we need more 
equations.

In order to lift an object using the buoyant force, the buoyant force 
must exceed the weight of the object.  This means:

                 Ftotal > Weight of object

If our balloon has a mass of M, it has a weight of Mg where g = 9.81 
meters/second^2 and our equation becomes:

                       (Df - Dc)gV > Mg  (V is now the volume of helium)

Solving the inequality for V gives us:

                       V > M/(Df - Dc)

Where:    V  = volume of helium required
          M  = mass of balloon
          Df = density of air (fluid our balloon is immersed in)
          Dc = density of helium (our bouyant gas)

This is the proper equation for finding how much helium to use to fly a 
baloon of a given mass.