At cruise level around 10000 meters the cabin pressure is lower than surface pressure. The only reason for this, is to reduce stress on the structure of the plane, or fuselage, right?

My question is this:

Given the current state of technology, if manufacturers were to produce passenger aircraft, massive or robust enough to allow cruise flight with cabin pressure equal to surface pressure, then how much would the gross weight of the aircraft increase? How much would the range of the aircraft decrease or how much less cargo could it carry?

I suppose that the current planes could also fly at surface pressure, but the level of safety would decrease, due to increased forces on the structure. So, the question really is: Suppose we want to fly at ground pressure with the same levels of safety, how much would the gross weight increase, how much less cargo could a plane carry?

Many thanks.

  • $\begingroup$ Are you asking how much more the structure of the aircraft would have to weigh in order to contain the additional pressure? Or how much the additional air within the cabin would weigh? Or both? $\endgroup$ Sep 18 '16 at 5:38
  • $\begingroup$ My question is fairly broad, so the answer to your question is 'both'. The engineers have a reason, why pressure in the cabin at cruise flight is rather low. I want to know, how much tickets would be more expensive with a ground pressure cabin. The children often cry in the cabin. Probably also because of the noise of the engines. But also due to the pressure. $\endgroup$
    – user17025
    Sep 18 '16 at 5:45
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  • $\begingroup$ The pressure is kept lower not because the aircraft couldn't handle it, but because it would not be able to handle it for as long a lifetime. Metal fatigue and wear is dangerous as well as costs ineffective, so minimising it is a big engineering goal. $\endgroup$
    – Nij
    Sep 18 '16 at 7:56
  • $\begingroup$ Right, Nij. Agree. And my question is this: What are the negative side effects of building more robust airplanes that would survive as well with ground pressure in the cabin as our present planes with the 2000 Meters pressure. $\endgroup$
    – user17025
    Sep 18 '16 at 10:46

Not much. The current requirement for 8,000 ft equivalent pressure translates into 8.11 psi at 43,000 ft. Boeing studied the benefits of lower pressure and found that going below 6,000 ft equivalent pressure brought no additional benefits.

The full pressure differential between sea level and 43,000 ft is 12.3 psi or just 52% more than that at 8,000 ft. This would require a heavier fuselage, but would not be impossible to do at all. Statistical methods give a mass factor for the fuselage pressure differential of $(1 + ∆p)^{0.203}$ which yields 1.0933 for 8.11 psi (∆p = 0.552 atm) and 1.1314 for 12.3 psi (∆p = 0.837 atm). Increasing the pressure to its sea level value would increase the mass of a typical airliner fuselage by 3.5%.

If we use the volume of the cabin, we can also calculate how much heavier the air is: The A330-200 has a cabin volume of 372 m³, and with the density of air at 8,000 ft of 0.9633 kg/m³, this computes to 358.35 kg. Compared to the mass of the same volume at sea level (455.7 kg), the difference is roughly equivalent to the mass of a single passenger (not counting the amenities carried along for her/him, too). In other words: The added mass of air for sea level pressure will make no difference if one passenger more misses her/his flight that day.

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    $\begingroup$ @ymb1: It's not on the site but in the handbook, in the mass volume. I have not found an online source which lists this equation. $\endgroup$ Sep 18 '16 at 16:12

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