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ideas:super-pressure_balloon_skin_tension_calculations

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ideas:super-pressure_balloon_skin_tension_calculations [2018/06/14 21:45] rocketboyideas:super-pressure_balloon_skin_tension_calculations [2018/06/21 12:38] (current) rocketboy
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 The local atmospheric pressure, density and temperature can be determined from an atmospheric model. 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 skin tension equation T = Pd . r / 4+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. Where Pd is the differential pressure (inside to out) and r is the sphere radius.
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  3Kg balloon + 7Kg fabric envelope  3Kg balloon + 7Kg fabric envelope
  1 Kg payload  1 Kg payload
- 13cu m of Helium at launch - assume STP  (Standard Temperature & Pressure)+ 13m^3 of Helium at launch - assume STP  (Standard Temperature & Pressure)
  Floating at 25,000m/82,000ft   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:  Using the the 1962 NASA standard model at an altitude of 25,000m: 
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 We have a total of 13.327kg to lift (3kg Latex balloon, 7kg fabric, 1kg payload, 2.327kg of helium) 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 we must displace 13.327kg of air – since the local air density is 0.03995 kg/m^3 we will be displacing 333.6 m^3 of air.+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+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 which gives an r of 4.3m – diameter 8.6m
  
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 External Pressure Po =  2.48277kPa External Pressure Po =  2.48277kPa
  
-Differential pressure Pd (Pi - Po) = 3.135kPa – 2.48277kPa = 0.6522kPa (about 0.08psivery similar to observed latex balloon flights.+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 Surface Tension of a spherical container is given by T = (Pi – Po). r / 4
- 
-Surface tension units are force/length – so the units come out right (since kPa = kN/m^2) 
  
  T = 0.6522kN/m^2 . 4.3m / 4 = 0.701115 kN/m   T = 0.6522kN/m^2 . 4.3m / 4 = 0.701115 kN/m 
  
-701.1N/m = 71.5kgf/m = 157.6lbf/m = 48lb/ft   +701.1N/m = 71.5kgf/m = 48lb/ft  
- +
-Some of that tension will be borne by the latex balloon, some by the fabric. +
  
 +This spreadsheet allows you to calculate the surface tension for a non-eleastic constraining envelope:
 +{{:ideas:constrainedfloaterv1.xls|}}
  
ideas/super-pressure_balloon_skin_tension_calculations.1529012730.txt.gz · Last modified: 2018/06/14 21:45 by rocketboy

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