Help figuring out what a balloon diameter should be with x amount of helium.

[MichaelB91]

I am planning to do a high altitude balloon launch and I was wondering if anyone on this forum could help me figure out a few things.
The payload+the balloon+the parachute= weigh around 2 pounds.
I am looking to get 3 pounds of free lift (each pound of free lift equals 300 feet per minute accent). I believe that one square foot of helium can lift 28grams.
The balloon that I am planning to use is the KCI 800 from this site http://kaymont.com/pages/sounding-balloons.cfm it has all the details about the balloon.
I am looking for the amount of helium that would be required to lift that weight at that rate, and I am wondering what the diameter of the balloon should be so I know that it is full.
Your help will be greatly appreciated,
Thank you.

 

[BruceW]

When you say you want 3 pounds of free lift, do you mean that you want 5 pounds of buoyant force (to overcome the 2 pounds of gravitational force)?
On the website, it gives a specific rate of ascent for the balloon, so I'm guessing they designed the balloon to go at a particular rate of ascent only.
The website says the volume at release is 1.76 cu. m and the rate of ascent is 320m.min
The buoyant force of the balloon is equal to the weight of the air displaced. So I would guess that the rate of ascent is proportional to the volume of the balloon.
BUT as more helium is pushed into the balloon, the density of the helium may increase, which would increase the mass of the balloon.

 

[MichaelB91]

Yes I do, I need to over the 2 pounds of gravitational force to get 3 pounds of buoyancy, which should give me 900 feet per minute accent.
Even though they designed the balloon to go at a certain rate its possible to change it all, by adding less helium it will rise slower but go higher.

 

[BruceW]

Assuming that my prediction that the volume of the balloon is proportional to the rate of ascent (which is not necessarily true, but its my first guess) then:
The website says 1.76 cubic metres gives an ascent of 320 metres/min.
And you want 900 feet/min ascent, which is 274 metres/min ascent.
Therefore, you should fill the balloon less, so that it is only 1.5 cubic metres, to get 900 feet/min ascent.
(But this all hinges on my guess that the volume is proportional to the rate of ascent).

 

[MichaelB91]

Would you happen to know what that should look like? In terms of the balloons diameter

 

[BruceW]

depends on the shape of the balloon. If the balloon is roughly spherical, then the volume is given by:
volume = \frac{4}{3} \pi r^3
(where r is the radius), and the radius is half the diameter, so in terms of the diameter (d) we get:
volume = \frac{1}{6} \pi d^3
But as I said in the last post, my assumptions for this problem are very simplistic. I think its likely that the true properties of the balloon are more complicated.