Hot air balloon releasing sand

In summary, the balloon with a 7000 kg total mass and a 1000 kg sand bag payload leaks 50 kgs of sand per second.
  • #1
albega
75
0

Homework Statement


A hot air balloon with a total mass of 7000 kg carries a gondola (or basket) which includes 1000 kg of sand bags fixed to the outside. While hovering at a fixed altitude, sand starts leaking out of thebags at a rate of 50 kgs-1. Determine the velocity of the balloon at the moment when all the sand has leaked out. You may assume that the acceleration due to gravity and the upward force from the hot air remains constant. You may also neglect air resistance.

Homework Equations


Going from first principles,
Δp=p(t+dt)-p(t)
F=dp/dt

The Attempt at a Solution


Let the balloon be moving upwards at speed v at time t with mass m. Then at time t+dt it moves upwards with speed v+dv, mass m-dm, whilst the 'sand packet' has mass dM and moves upwards still at speed v (because it is not propelled out, only dropped).

Then dp(t)=mv and dp(t+dt)=(m-dm)(v+dv)+vdM
dp=mdv-vdm+vdM to first order
dp/dt=mdv/dt-vdm/dt+vdM/dt
Then dM/dt=-dm/dt so dp/dt=mdv/dt-2vdm/dt. I believe this step is wrong. I can't see a flaw in my reasoning though.

Then the next issue is deciding what dp/dt is in terms of the forces. I have an expression for dP/dt for the whole system so I guess I need the force that acts on the whole system. But I'm not sure what to do... Thanks for any help :biggrin:
 
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  • #2
It is not a rocket - the leaking sand won't give any velocity change. The (broken) force balance between hot air and the sand is the important point here.
 
  • #3
mfb said:
It is not a rocket - the leaking sand won't give any velocity change. The (broken) force balance between hot air and the sand is the important point here.

Hmm I'm lost.

I have let the velocity of the balloon be v+dv at time t+dt because it must be changing... I don't really know where to go.
 
  • #4
albega said:
Let the balloon be moving upwards at speed v at time t with mass m. Then at time t+dt it moves upwards with speed v+dv, mass m-dm, whilst the 'sand packet' has mass dM and moves upwards still at speed v (because it is not propelled out, only dropped).

Then dp(t)=mv and dp(t+dt)=(m-dm)(v+dv)+vdM
I think you mean
p(t)=mv and p(t+dt)=(m-dm)(v+dv)+vdM
dp=mdv-vdm+vdM to first order
dp/dt=mdv/dt-vdm/dt+vdM/dt
Then dM/dt=-dm/dt
No, you did -dm and +dM, so in fact dm and dM are the same.
Then the next issue is deciding what dp/dt is in terms of the forces. I have an expression for dP/dt for the whole system so I guess I need the force that acts on the whole system.
No, your equation did not include sand that had previously leaked out.
 
  • #5
haruspex said:
I think you mean
p(t)=mv and p(t+dt)=(m-dm)(v+dv)+vdM

No, you did -dm and +dM, so in fact dm and dM are the same.

No, your equation did not include sand that had previously leaked out.

Yes I did mean that!

Ah I realize now my system isn't the whole mass is it, it's the mass of the balloon plus the differential sand mass released at any time. Thus this system experiences a force F-mg with F the thrust which is constant. I'm still a little bit unsure about the fact that this thrust force F acts on the m-dm rather than the m, whilst my system is the whole m

Thanks for your help - I think If I can get things to be consistent with the rocket equation derivation that will be the best way of understanding everything.
 
Last edited:
  • #6
albega said:
I'm still a little bit unsure about the fact that this thrust force F acts on the m-dm rather than the m, whilst my system is the whole m
In the equation F-mg = a(m-dm), the dm is insignificant.
 
  • #7
haruspex said:
In the equation F-mg = a(m-dm), the dm is insignificant.

I have an answer of
v=(F/a)ln[m0/(m0-aT)]-gT
where m0 is the initial mass, a the rate of decrease of the sand on board, T the time for the sand supply to be exhausted, F the thrust force which is constant.

F=7000g from the initial condition, a=50kgs-1 T=20s and m0=7000kg. This gives v=15.5ms-1. Not sure if this is correct.
 
  • #8
albega said:
I have an answer of
v=(F/a)ln[m0/(m0-aT)]-gT
where m0 is the initial mass, a the rate of decrease of the sand on board, T the time for the sand supply to be exhausted, F the thrust force which is constant.

F=7000g from the initial condition, a=50kgs-1 T=20s and m0=7000kg. This gives v=15.5ms-1. Not sure if this is correct.

I concur with all that.
 
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1. How does a hot air balloon release sand?

Hot air balloons release sand through a trap door located at the bottom of the basket. This allows the pilot to control the altitude of the balloon by adjusting the weight of the basket.

2. Why is sand used in hot air balloons?

Sand is used in hot air balloons as ballast to help control the altitude of the balloon. As the sand is released, the balloon becomes lighter and rises, and as more sand is added, the balloon becomes heavier and descends.

3. How much sand is typically used in a hot air balloon?

The amount of sand used in a hot air balloon varies depending on the size of the balloon and the weight of the passengers. On average, a hot air balloon can carry up to 100 pounds of sand.

4. Can hot air balloons release sand while in flight?

Yes, hot air balloons can release sand while in flight. This is a common technique used by pilots to maintain control of the balloon's altitude and to safely land the balloon.

5. Is releasing sand the only way to control the altitude of a hot air balloon?

No, releasing sand is not the only way to control the altitude of a hot air balloon. Pilots can also adjust the temperature of the air inside the balloon to increase or decrease the buoyancy, allowing them to ascend or descend without releasing any sand.

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