Balloon Jet Propulsion.High Altitude Calculation.

Everyone knows how if you let go of a balloon allowing the gas to escape quickly, the balloon shoots across the room. This is an attempt to quantify this balloon jet propulsion for the particular case of balloons venting up in the stratosphere. Could the balloon, approaching its bursting point on the back of Archimedes's Principle be saved and driven higher and if so by how much?

Although the calculations are used to obtain numbers for helium, He, they would apply equally for any other gas.

THE MAIN ASSUMPTIONS.

( a ) The balloon is filled with gas at 1 atmos. and 298*K, ( S.T.P. ) and the gas retains this state when it is vented. Of course, even in a well insulated envelope, the gas would be cooler in the frigid high atmosphere. This could be corrected for.( See below )

( b ) That the gas vents from the balloon axially and downwards along the vertical venting pipe with a velocity equal to the atoms mean random molecular speed ( c ) characterised by their temperature and pressure. This is optimistic, to put it mildly, but it gives an upper bound to what may be possible.

( c ) The solid material of the balloon has zero buoyancy. This will be the case when all the gas has vented but initially there will of course be buoyancy as the venting starts with the balloon at or near equilibrium, neither going up or down. This assumption underestimates the lift as does the assumption made that the loss of He does not reduce the balloons initial mass.



MASS OF HELIUM ( m ).

For a spherical balloon of radius 1m, the volume is 4.2m(3). The density of He at S.T.P. is 0.175 kg/m(3).
m=0.74kg.


MEAN RANDOM MOLECULAR SPEED ( c ).

For He at S.T.P of density r and pressure P,
P=(r*c(2))/3. With r=0.175kg/m(3) and P=10(5)Pa, c=1.3*10(3)=1300m/s.

IMPULSE AND CHANGE OF MOMENTUM.

Ft=m*(final velocity )-m*(initial velocity ). F=force in N, t=time in secs.
( initial velocity )=0 as inside the balloon the atoms of He are moving randomly in all directions.
Therefore, Ft=mc N-sec.
With the venting pipe pointing vertically downwards F becomes the upward force on the balloon and t the venting time.
If we set p=mc, then F=mc/t=p/t N.

NET UPWARD FORCE ON BALLOON.
If the total mass of balloon and load, assumed constant even though He is being lost, is M, and as stated above the initial buoyancy is ignored, the net upward force is
F-Mg=p/t-Mg N

UPWARDS ACCELERATION OF BALLOON.
From Newton II,
p/t-Mg=Ma where a is the upwards acceleration.
Therefore, a=(p/t-Mg)/M m/sec(2)

MAXIMUM DISTANCE BALLOON IS DRIVEN VERTICALLY.

s=ut+(at(2)/2.
Assuming balloon is not moving up fast before venting, u=0.

s=(a*t*t)/2.=(p/(tM)-g)*t*t/2= pt/(2M)-gt(2)/2.


To find t for maximum s, a spot of calculus.
(ds)/(dt)=p/(2M)-gt when (ds)/(dt)=0 we get the maximum value of s and the corresponding value of t.
So, at (ds)/(dt)=o , p/(2M)-gt=0 and t=p/(2Mg), the optimum t.

Finally, substituting this value of t into s=(a*t*t)/2, we get

the maximum extra height gained as

s=(p/M)*(p/M)/(8g)

EXAMPLES.

For our first example of a Im radius balloon carrying 0.74 kgs of He, p=1000kg-(m/s), M=2kg, venting time=25secs and

s=3125m or about 9400 ft.

Upwards force due to jet effect F=p/t=1000/25=40 N.


With a balloon of 1.5m radius carrying 2.4kgs of He and with a total mass now M=5kg, venting time= 31 secs.and
p=3100kg-(m/s), F=3100/31=100 N and
s=4800m or another 14400 ft, WoW!. A slightly bigger balloon and we get all the way to the moon!
No, but seriously any comments feedback very welcome.

Sorry about indices notation t*t means t squared as does t(2).

TEMPERATURE CORRECTION.

If C is random speed at 300*K and c is that at 250*K,then,

c=C*Square-root(250/300)=0.91*C. This gives c=1200 m/sec at 250*K instead of 1300 m/sec at 300*K.
Small potatoes!