Increase of the size of an Helium balloon as it raises into the atmosphere

In this Post we shall determine the increase of size of a balloon as it raises up into the atmosphere. For this we shall set a practical scenario and start using some of the concepts and terminology that we have seen in previous Posts.

So let's consider a scenario where we want to send a balloon carrying a payload weighting 800g to an altitude of 30000 meters (30Km). From the Kaymont balloons manufacturer site lets choose the HAB-600 balloon. This is a 600 grams balloon recommended for individuals or groups launching a photographic mission for the first time. This balloon is recommended for payloads with a weight between 500 and 1000 grams.

Then lets use the CUSF balloon burst calculator and set the following (be sure that the selected gas is Helium):

• Payload mass (g): 800g
• Balloon mass (g): Kaymont - 600
• Target burst altitude (m): 30000

The result from the calculator shall be:

• Burst altitude: 30000m (as we have selected in the inputs for the calculator)
• Time to burst: 134m
• Launch volume: 1.81m3 (1811 L)
• Ascent rate: 3.72 m/s (223.2 m/min)
• Neck lift: 1258g

So now we know that by filling the 600-grams balloon with a volume of 1811 litres of Helium it is possible to send the balloon to an altitude of 30 Km carrying a payload weighting 800 grams. We discover also that this is possible with an ascent rate of 223.2 meters per minute - which is a bit far from the recommend ascent rate which is normally set to 320m/min - the consequence is that the flight will take longer, in this case, 134 minutes, which in my opinion is still reasonable. As we have seen in a previous post the neck lift is the difference between the gross lift (the total lift generated by the Helium gas) and the weight of the balloon (in this case 600 grams), meaning that is the lift available to the payload (if you imagine the balloon as the head, the nose will be in the beginning of the payload) and to generate the lift of the system. In this case the nose lift is 1258g, which means that if you subtract from that the value of the weight of the payload (800g) you get the amount of lift available to generate the lift of the system (458 grams).

To summarise we now have a 600-grams balloon at the Earth surface (let's consider at the sea level - 0 meters) filled in with 1811 litres of Helium. How can we then know what will be the size of the balloon at the surface and then at different altitudes?

Well for the size of the balloon let's apply the formula for the volume of a sphere (we are here considering the balloon to be a sphere):
$$Volume= V= \frac{4}{3}\pi r^{3}$$
Where:

• π= 3.14
• r= radius= ^diameter⁄₂

The units that we use for the radius will then have an impact on the resulting units for the volume. If we use meters for the radius, then we get the volume in m³. So, for the case above we have a volume of 1811 litres, which considering that 1 m³ corresponds to 1000 litres, gives a volume of 1.811 m³. In this way the sphere formula can be written in the following way:

$$1.811=\frac{4}{3}3.14r^{3}=4.19r^{3}\Leftrightarrow r^{3}=0.43\Leftrightarrow r=0.75m\Leftrightarrow diameter=d=r*2=1.5m$$
To summarise once again (but with more information), we have a 600-grams balloon at the Earth surface (let's consider at the sea level - 0 meters) filled in with 1811 litres of Helium and with a diameter of 1.5m.

Now let's determine the volume of this same balloon at a different altitude, for instance at 20km. As we have seen before, as the balloon rises, its volume increases and consequently the diameter of the balloon also increases. So let's use the ideal gas law to determine what will be the volume and diameter of the balloon at this altitude of 20km, but let's first deduce a general formula that allows to determine that. As we have seen before:

PV = nRT

Imagine that we have a balloon and that we have two different states, S₁ and S₂, associated to the balloon. In state 1 (S₁) the balloon is at the Earth's surface and in state 2 (S₂) the balloon will be at a different altitude, for instance, 20 km.
So now let's draw the ideal gas law for each one of the states S₁ and S₂:

For S₁:
P₁V₁ = n₁RT₁
^(P₁V₁)⁄_(n1RT1)= 1                        (eq. 1)


For S₂:
P₂V₂ = n₂RT₂
^(P₂V₂)⁄_(n2RT2)= 1                         (eq. 2)

As the first term of eq. 1 equals 1 and the first term of eq. 2 also equals 1, we can say that those terms are equal to each other:
^(P₁V₁)⁄_(n1RT1) = ^(P₂V₂)⁄_(n2RT2)

But as the amount of gas do not change between  S₁ and S₂, we can say that n =  n₁ = n₂, and consequently:
^(P₁V₁)⁄_(nRT1) = ^(P₂V₂)⁄_(nRT2) <=> ^nR(P₁V₁)⁄_(nRT1) = ^nR(P₂V₂)⁄_(nRT2) <=> ^(P₁V₁)⁄_(T1) = ^(P₂V₂)⁄_(T2) <=>
$$V_{2}=V_{1}\frac{P_{1}T_{2}}{P_{2}T_{1}}                       (eq. 3)$$
So the above formula is a general formula that allow us to determine the volume of the balloon in state 2 (S₂) given the balloon conditions in state 1 (S₁) and given the temperature and pressure of the balloon in state 2 (S₂). To make this more concrete let's apply to our example balloon. As we have seen before, in state 1 (S₁), we have the 600-grams balloon at sea level (0 meters) filled in with 1811 litres of Helium and with a diameter of 1.5m. Additionally let's consider that for the absolute pressure at sea level we have 101325 Pa and the temperature is 15ºC, as indicated by the International Standard Atmosphere (ISA):

State 1 (S₁)

• 600-grams balloon
• Altitude= 0 meters (sea-level)
• Pressure= P1= 101325 Pa
• Temperature= T1= 15 ºC= 288.15 k
• Volume= V1=  1811L=  1.811 m³
• Diameter= 1.5 m 

Then for state 2 (S₂) of the balloon, we would like to know the volume and consequently the diameter of the same balloon when at an altitude of 20000 meters. So  let's consider that for the absolute pressure at 20 km we have  5474.9 Pa  and the temperature is  -56.5 ºC, as indicated by the International Standard Atmosphere (ISA):

State 2 (S₂)

• 600-grams balloon
• Altitude=  20000 meters (20km)
• Pressure= P2= 5474.9 Pa
• Temperature= T2= -56.5 ºC= 216.65 k
• Volume= V2=  ?
• Diameter= ? 

Now it is just a question of using the eq.3 as follows:

$$V_{2}=V_{1}\frac{P_{1}T_{2}}{P_{2}T_{1}} <=> V_{2}=1811 \times \frac{101325 \times 216.65}{5474.9 \times 288.15} <=> V_{2}=1811 \times \frac{21952061.25}{1577592.435} <=>$$
$$<=> V_{2}=1811 \times 13.91 <=> V_{2}=25191.01L= 25.2m^{3}$$
And now that we have determined the volume of the balloon at state 2 (S₂), it is just a question of determining the new diameter of the balloon:
$$25.2=\frac{4}{3}3.14r^{3}=4.19r^{3}\Leftrightarrow r^{3}=6.01\Leftrightarrow r=1.82m\Leftrightarrow diameter=d=r*2=3.64m$$
Finally let's model a state 3 (S₃) of the balloon, where we would like to know the volume and consequently the diameter of the same balloon when at an altitude of 30000 meters. For the pressure and temperature at that altitude let's use again the ISA values but this time retrieved from the Engineering ToolBox site. So let's consider that for the absolute pressure at 30 km we have 0.01197 bar, which corresponds to 1197 Pa, and the temperature is 226.5k:

State 3 (S₃)

• 600-grams balloon
• Altitude=  30000 meters (30km)
• Pressure= P3= 1197 Pa
• Temperature= T3= 226.5 k
• Volume= V3=  ?
• Diameter= ? 

Now it is just a question of using the eq.3 as follows:

$$V_{3}=V_{1}\frac{P_{1}T_{3}}{P_{3}T_{1}} <=> V_{3}=1811 \times \frac{101325 \times 226.5}{1197 \times 288.15} <=> V_{3}=1811 \times \frac{22950112.5}{344915.55} <=>$$
$$<=> V_{3}=1811 \times 66.54 <=> V_{3}=120503.94L= 120.5m^{3}$$
And now that we have determined the volume of the balloon at state 3 (S₃), it is just a question of determining the new diameter of the balloon:
$$120.5=\frac{4}{3}3.14r^{3}=4.19r^{3}\Leftrightarrow r^{3}=28.76\Leftrightarrow r=3.06m\Leftrightarrow diameter=d=r*2=6.12m$$
The table bellow summarises the results for the three states, S₁, S₂, S₃:

	S1	S2	S3
Type	600-grams	600-grams	600-grams
Altitude (m)	0	20000	30000
Pressure (Pa)	101325	5474.9	1197
Temperature (K)	288.15	216.65	226.5
Volume (m3)	1.811	25.2	120.5
Diameter (m)	1.5	3.64	6.12

If you remember from the beginning of this post, we have used the CUSF balloon burst calculator and set the balloon parameters in such a way that the target burst altitude would be 30000m. Well, from the above calculations we have just determined that at 30000m the balloon will have a diameter of 6.12m. Consequently this means that the diameter of the balloon when bursting will be approximately 6.12m.
These results are very similar to the results presented in the Kaymont balloons manufacturer site regarding the weather forecasting (sounding) 600-grams balloon (with a 250g payload) - it is just a question of converting the units there in order to see that the resulting values are very similar to the ones we have just reached above. This means that we are now equipped to better understand the balloon specifications (and nomenclature) and the different possible configurations of the balloon and its consequences.