Volume of Ballon

DDS

171
0
A 576 kg weather balloon is designed to lift a 3670 kg package. What volume should the balloon have after being inflated with helium at standard temperature and pressure in order that the total load can be lifted?

Am i missing something or is there more to this question there , there seems.

The ballon with a certain mass wants to lift an iditional mass therefore the combinded mass of the baloon is 4246 Kg

The dnesity of Helium at STP is 0.18 kg/m^3

and the relationship describing density is :

D=M/V rearrange for volume and and get Volume= mass/ density

V=4246 kg/ 0.18kg/m^3
V=23588.89 m^3 which is wrong...what am i missing here??
 

OlderDan

Science Advisor
Homework Helper
3,023
1
DDS said:
A 576 kg weather balloon is designed to lift a 3670 kg package. What volume should the balloon have after being inflated with helium at standard temperature and pressure in order that the total load can be lifted?

Am i missing something or is there more to this question there , there seems.

The ballon with a certain mass wants to lift an iditional mass therefore the combinded mass of the baloon is 4246 Kg

The dnesity of Helium at STP is 0.18 kg/m^3

and the relationship describing density is :

D=M/V rearrange for volume and and get Volume= mass/ density

V=4246 kg/ 0.18kg/m^3
V=23588.89 m^3 which is wrong...what am i missing here??
Your are missing Archimedes' Principle. The buoyant force on the balloon is equal to the weight of air that is displaced by the balloon.
 

DDS

171
0
so the weight of the air displaced is equal to the weight of the balloon correct?

so

w=mg
w=5650.56 N

then i add this to the weight of my cargo load and divid over the density of helium??

which gives me

wc=mg
wc=(3670)(9.81)
wc=36002.7

36002.7 + 5650.56 = 41653.26

v=m/d
v=41653.26/0.18
v=231407 m^3??
 

OlderDan

Science Advisor
Homework Helper
3,023
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DDS said:
so the weight of the air displaced is equal to the weight of the balloon correct?

so

w=mg
w=5650.56 N

then i add this to the weight of my cargo load and divid over the density of helium??

which gives me

wc=mg
wc=(3670)(9.81)
wc=36002.7

36002.7 + 5650.56 = 41653.26

v=m/d
v=41653.26/0.18
v=231407 m^3??
41653.26 Newtons is the combined weight of the balloon and cargo. That is all you have so far. It is not a mass. You need to work on being consistent about dimensions, so that you don't do things like use a weight in a place where you need mass.

Use the weight you have calculated to continue the problem. You are going to fill a balloon with helium to displace the amount of air needed to lift the balloon. When you do this, don't forget that the helium in the balloon has weight of its own that must be lifted by the buoyant force. Once you correctly figure out the weight of the displaced air, you need to find the mass of the air before you calculate its volume.
 

DDS

171
0
wow that just confused me because it seems like i have to find some many weights of so many things and im only given two masses and a density of helium...is there any way to be more specifc with that i have to do... i learn and grasp things better via formula description
 

DDS

171
0
This is what i have so far:

Wc=36002.7
Wb=5650.56
Wt=41653.26

Dhe=0.180

Fb= D*V*G
fb=(1000)(0.180)(9.81)
Fb=Whe=1765.8

now i believe that the weight of air is the sum of all my weight thus

Wa=43419.06


thus the mass of air is 43419.06/9.81=m
m=4426

v=m/d
4426/0.180
v=24588.89

please tell me im right
 

OlderDan

Science Advisor
Homework Helper
3,023
1
DDS said:
This is what i have so far:

Wc=36002.7
Wb=5650.56
Wt=41653.26

Dhe=0.180

Fb= D*V*G
fb=(1000)(0.180)(9.81)
Fb=Whe=1765.8

now i believe that the weight of air is the sum of all my weight thus

Wa=43419.06


thus the mass of air is 43419.06/9.81=m
m=4426

v=m/d
4426/0.180
v=24588.89

please tell me im right
More dimensional inconsistencies

0.18 is a density, not a volume. You don't know the volume of the helium until the problem is solved.

Downward forces:
Weight of balloon [Newtons] Mb*g
Weight of cargo [Newtons] Mc*g
Weight of helium [Newtons] Mh*g = Dh*V*g

Upward forces:
Weight of displaced air [Newtons] Ma*g = Da*V*g

V is the same for the air and the helium

Upward forces must be the same magnitude as the downward forces. You will need to find the density of air at standard temperature and pressure. It depends on humidity, but you can assume dry air.
 

DDS

171
0
its 1.28 kg/m^3 but what do i do what that
 

DDS

171
0
would it be:

Vhe= Total mass/(Pair-PHe)
V=4246/1.1
v=4162.74
 

OlderDan

Science Advisor
Homework Helper
3,023
1
DDS said:
its 1.28 kg/m^3

would it be:

Vhe= Total mass/(Pair-PHe)
V=4246/1.1
v=4162.74
My last post gives you everything you need to write an equation
Weight of balloon [Newtons] Mb*g
Weight of cargo [Newtons] Mc*g
Weight of helium [Newtons] Mh*g = Dh*V*g

Upward forces:
Weight of displaced air [Newtons] Ma*g = Da*V*g

V is the same for the air and the helium
Da*V*g = Dh*V*g + Mb*g + Mc*g

You now know every quantity in the equation except for V. Solve the equation for V. Notice that g is common to every term.
 

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