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Balloons- Rep

I've got a few questions which I need help with.

A man ties helium balloons to his chair, shooting upwards 4000 m when he cuts the chord. Combined mass of chair and man is 70kg. Calculate total volume of balloons required to lift man and his chair if weight of balloon fabric is negligible. Explain why viscous drag can be ignored.

Reply 1

definite_maybe

Any ideas?

I also have the density of air= 1.29
density of helium:0.18

Reply 2

definite_maybe

cheemaj187

does it say how many balloons he ties to his chair? or their mass?

No. The upthrust is said to be 13V where is the total volume of the helium baloons on cubic metres, and the first part of the question made me work out the weight of the helium in the balloons and the weight of the man and the chair, but the weight of the fabric of the baloon itself is negligble.

So stuck. + Rep for anybody who can help.

Reply 3

definite_maybe

cheemaj187

Ah ok, the first part of the question might have done the trick, as you have the weight of the balloons, you can work out the amount of energy they provide, as you know they rise by 4000m when they are released.

Using the formula for gravitational potential energy,

E=mgh

, you can see the amount of energy the balloons provide.

Then you have the weight of the man, which is 70 x g, where g is 9.81

As you have the density, where

d=m/v

, you can then work out the volume required by eliminating g from the weight of the helium.

That seems to make sense, but I could be wrong, do you have the answer?

I dont have the answer, but using your working out, is this what you would do?

E = 9.81 times 4000 * mass but how would I get the mass of the baloons with only the density of the helium.

Reply 4

definite_maybe

cheemaj187

is the man going to be lifted to 4000m as well?
or just lifted above the ground?

Well the second part to the question asks to calculate the total volume of balloons required just to lift the man and his chair from the ground.

Reply 5

Morbo

When the balloons reach 4km, they'll be in equilibirum. The upthrust at that altitude will equal their own weight.

At 4km
Weight Of Air Displaced (Upthrust) - Total Weight = 0.

\rho_{air}(4\mathrm{km}) V g = \rho_{bal} V g \Rightarrow \rho_{air}(4\mathrm{km}) = \rho_{bal}

Now, the density of air falls off as we ascend in altitude as

\rho_{air}(z) = \rho_{air}(0)\exp(-z/\zeta)

, where

\zeta = \frac{RT}{\mu g} \approx 8.4\mathrm{km}

.

So,

\rho_{air}(4\mathrm{km}) = \rho_{air}(0) \times 0.62 = 1.29 \mathrm{kg m}^{-3} \times 0.62 = 0.80 \mathrm{kg m}^{-3}

.

By our first equation, therefore, we find that the helium in the balloons has a density of

\rho_{bal} = 0.80 \mathrm{kg m}^{-3}

.

We can now calculate what happens at ground level, assuming that the balloons do not expand/contract i.e. the helium remains at the same density.

On the ground
Again:

Weight Of Air Displaced - Total Weight = 0 (the limit of lifting the man)

Ignoring the weight of air displaced by the man and chair (assuming they have been "weighed" as 70kg):

\rho_{air}(0) V g = (M+m)g

Where

M

is the mass of the man and chair and

m

is the mass of the helium in the balloons.

\\ \rho_{air}(0) V = M + \rho_{bal}V[br]\\ V = \frac{M}{\rho_{air}(0) - \rho_{bal}}[br]\\ = \frac{70\mathrm{kg}}{1.29 \mathrm{kg m}^{-3} - 0.80 \mathrm{kg m}^{-3}}[br]\\ = 143 \mathrm{m}^3

.

EDIT: This man's helium balloons must be at high pressure. If the helium in his balloons only had a density of 0.18 kg/m³, then they would go much higher than 4000m (assuming they don't burst!). In fact, using the formula from earlier, they would go up to where the air density is equal to this helium density, which is about 16km!