We regard a sector of a cylinder of height b and radius R with an angle θ. We can derive the differential equation for the change in pressure dP in a small volume of the section dV at a distance r from the cylinder center when we change the distance to r + dr.
| (1) | |
| (2) |
The increase in force and pressure in the direction of r is caused by the mass of the air in the volume element dV due to the centrifugal acceleration aC:
| (3) |
So the pressure differential is then:
| (4) |
The density ρ(r) can be expressed from pressure P(r) using the ideal gas law:
| (5) |
So we can insert this into equation (4):
| (6) |
Now bringing the pressure term to the left hand side gives the follong DE:
| (7) |
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We have to integrate both sides: ∫ 1/P dP = ln(P):
| (8) |
We assume a constant temperature T and rotation rate ω. Solving the integrals yields:
| (9) |
Applying the limits:
| (10) | |
| (11) |
To get rid of the logarithms we raise both sides to the power of e:
| (12) |
Bringing Po to the right gand side we get:
| (13) |
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