Calculate Volume of Balloon to Lift 3670 kg Package

  • Thread starter Thread starter DDS
  • Start date Start date
  • Tags Tags
    Volume
Click For Summary

Homework Help Overview

The discussion revolves around calculating the volume of a helium-filled balloon required to lift a total mass of 4246 kg, which includes the weight of the balloon and a cargo load of 3670 kg. Participants are exploring the principles of buoyancy and the relationship between mass, volume, and density in the context of gases at standard temperature and pressure.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Archimedes' Principle and the need to account for the weight of the displaced air. There are attempts to calculate the necessary volume using various equations related to weight and density, while some participants express confusion about dimensional consistency and the need for specific formulas.

Discussion Status

Several participants have provided calculations and reasoning, but there is a lack of consensus on the correct approach. Some have pointed out the need to clarify the relationship between the weights of the balloon, cargo, and helium, while others have raised concerns about dimensional inconsistencies in their calculations. Guidance has been offered regarding the need to consider the weight of the helium and the density of air.

Contextual Notes

Participants are working under the constraints of using given masses and the density of helium, while also needing to determine the density of air at standard temperature and pressure, which is noted to be approximately 1.28 kg/m³. There is an ongoing exploration of how to integrate these values into a coherent solution.

DDS
Messages
171
Reaction score
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 balloon with a certain mass wants to lift an iditional mass therefore the combinded mass of the balloon 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??
 
Physics news on Phys.org
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 balloon with a certain mass wants to lift an iditional mass therefore the combinded mass of the balloon 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.
 
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??
 
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.
 
wow that just confused me because it seems like i have to find some many weights of so many things and I am 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
 
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 I am right
 
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 I am 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.
 
its 1.28 kg/m^3 but what do i do what that
 
would it be:

Vhe= Total mass/(Pair-PHe)
V=4246/1.1
v=4162.74
 
  • #10
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
4
Views
2K
Replies
6
Views
8K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 62 ·
3
Replies
62
Views
5K
  • · Replies 3 ·
Replies
3
Views
8K
Replies
2
Views
2K
Replies
18
Views
8K