# Super-pressure skin tension estimation

On this page we estimate the tension in the skin of a super-pressure balloon. We will make a number of identified assumptions along the way.

In order to estimate the surface tension we must first calculate the pressure differential between the inner and outer surfaces of the balloon.

The outer pressure is simply the local atmospheric pressure at the height the balloon is floating. The inner pressure is determined by the volume of the balloon, the mass of gas contained, and its temperature.

For the balloon to be floating at a constant altitude the balloon must be displacing the same weight of air as that of the balloon (including gas) plus payload. We can calculate the volume of air being displaced (and hence that of the balloon) by knowing the local air density.

The local atmospheric pressure, density and temperature can be determined from an atmospheric model.

If we assume a spherical balloon the skin tension can then be calculated with the standard sphere skin tension equation T = Pd . r / 4

Where Pd is the differential pressure (inside to out) and r is the sphere radius.

Here is an example:

3Kg balloon + 7Kg fabric envelope 1 Kg payload 13m^3 of Helium at launch - assume STP (Standard Temperature & Pressure) Floating at 25,000m/82,000ft

13m^3 of helium provides enough lift for the balloon + fabric envelope and payload and additional lift for a reasonable ascent rate.

Using the the 1962 NASA standard model at an altitude of 25,000m:

air pressure will be 2.48277kPa air density 0.03995 kg/m^3 temperature is -56.5C

Helium density is 0.179 kg/m^3 at STP and 4.003g/mol – so for 13m^3 we have 2.327kg of Helium = 581 moles

We have a total of 13.327kg to lift (3kg Latex balloon, 7kg fabric, 1kg payload, 2.327kg of helium)

So floating at 25,000m the balloon must displace 13.327kg of air – since the local air density is 0.03995 kg/m^3 the balloon will be displacing 333.6 m^3 of air.

Assume a sphere shape for the balloon balloon - volume V = (4/3) . Pi() . r^3 which gives an r of 4.3m – diameter 8.6m

We know the number of mols n of gas (581) and the volume V (333.6 m^3) temperature T (-56.5C = 216.5K assuming equalization and no thermal gain)

From the gas law PV = nRT we can calculate the internal pressure P = nRT/V

P (in kPa) = (581 . R . 216.5 / 333.6)

Where R is the universal gas constant 0.008314472 m^3.kPa/mol.K

Internal pressure Pi = 3.135kPa External Pressure Po = 2.48277kPa

Differential pressure Pd (Pi - Po) = 3.135kPa – 2.48277kPa = 0.6522kPa (about 0.08psi - very similar to observed latex balloon flights).

Surface Tension of a spherical container is given by T = (Pi – Po). r / 4

T = 0.6522kN/m^2 . 4.3m / 4 = 0.701115 kN/m

701.1N/m = 71.5kgf/m = 48lb/ft

This spreadsheet allows you to calculate the surface tension for a non-eleastic constraining envelope: constrainedfloaterv1.xls