Does a hydrogen baloon ascend faster than a helium baloon

[tonyjeffs]

Does a hydrogen baloon ascend faster than a helium baloon?
Ignoring the weight of the baloon fabric, and ignoring friction, but not ignoring the mass of the air, would a hydrogen baloon ascend twice as fast?

Would the same logic apply to lead and aluminium weights falling through water?

Thanks

Tony



ps. I'm very impressed with this forum. Some very high level discussions going on!

 

[Q_Goest]

Hi Tony,
As you probably know, helium is roughly twice as dense as hydrogen under standard conditions (STP). Assuming both balloons are equal in size and total weight (weight without gas), the "upward force" (F) is equal to the weight of the air displaced minus the weight of the balloon plus gas. That total upward force accelerates the balloon upward according to a=F/m where:
a = acceleration upward
F = force upward
m = mass of the balloon plus mass of helium or hydrogen

So the two balloons, one with helium, one with hydrogen, have different accelerations upwards, but the acceleration upward is not a ratio of the mass of the gas. It is a ratio of the mass given in the equation above, which is dependant on the balloon mass PLUS the mass of the gas.

Note also that wind resistance must be considered in the above equation, but we'll neglect that for the purposes of this argument.

 

[walkinginwater]

hi, Tony:
Q_Goest is right. But the sentence, " It is a ratio of the mass given in the equation above, which is dependant on the balloon mass PLUS the mass of the gas." , has some problem. Because the ratio of the speed of the balloon is as the same as the "a", the acceleration upward. So not only the mass of the balloon is involved, but also the upward force.

 

[FredGarvin]

No. There is nothing wrong with what Q posted. If you break down the balloon into a FBD and work the variables, you will see that the acceleration each balloon experiences is a function of the gas density, air density and the mass of the balloon and the gas. The upward force is a function of these as well.

This is going to take me a month of Sundays...

$$F_{net} = W_{air} - \left(W_{balloon} + W_{gas} \right)$$

$$F_{net} = \left(V_{balloon}\rho_{air} g \right) - W_{balloon} - \left(V_{balloon}\rho_{gas} g \right)$$

$$F_{net} = V_{balloon} g \left(\rho_{air} - \rho_{gas} \right) - W_{balloon}$$

Therefore:

$$A_{balloon} = \frac{V_{balloon} g \left(\rho_{air} - \rho_{gas} \right) - W_{balloon}}{m_{balloon}+m_{gas}}$$

$$A_{balloon} = \frac{V_{balloon} g \left[\frac{m_{air}}{V_{balloon}} - \frac{m_{gas}}{V_{balloon}}\right] - m_{balloon}g}{m_{balloon} + m_{gas}}$$

Finally...

$$A_{balloon} = \frac{g \left( m_{air}-m_{gas}-m_{balloon}\right)}{m_{balloon}+m_{gas}}$$

 

[rcgldr]

It's faster, but not twice as fast, because drag force is relative to v^2. If everything else is the same, and lift force (bouyancy) is twice as much, then velocity will be SQRT(2) = 1.41 times as fast.

Weather balloons typically use hydrogen:

http://en.wikipedia.org/wiki/Weather_balloon