3.6 KiB
Buckingham Pi Theorem
Buckingham π theorem states that an equation involving n number of physical variables which are expressible in terms of k independent fundamental physical quantities can be expressed in terms of p = n - k dimensionless parameters.
from pint import UnitRegistry ureg = UnitRegistry() Q = ureg.Quantity
To start with a very simple case, consider that you want to find a dimensionless quantity involving the magnitudes V, T and L with dimensions [length]/[time], [time] and [length] respectively.
>>> from pint import pi_theorem >>> pi_theorem({'V': '[length]/[time]', 'T': '[time]', 'L': '[length]'}) [{'V': 1.0, 'T': 1.0, 'L': -1.0}]
The result indicates that a dimensionless quantity can be obtained by multiplying V by T and the inverse of L.
Which can be pretty printed using the Pint formatter:
>>> from pint import formatter >>> result = pi_theorem({'V': '[length]/[time]', 'T': '[time]', 'L': '[length]'}) >>> print(formatter(result[0].items())) T * V / L
You can also apply the Buckingham π theorem associated to a Registry. In this case, you can use derived dimensions such as speed:
>>> from pint import UnitRegistry >>> ureg = UnitRegistry() >>> ureg.pi_theorem({'V': '[speed]', 'T': '[time]', 'L': '[length]'}) [{'V': 1.0, 'T': 1.0, 'L': -1.0}]
or unit names:
>>> ureg.pi_theorem({'V': 'meter/second', 'T': 'second', 'L': 'meter'}) [{'V': 1.0, 'T': 1.0, 'L': -1.0}]
or quantities:
>>> Q = ureg.Quantity >>> ureg.pi_theorem({'V': Q(1, 'meter/second'), ... 'T': Q(1, 'second'), ... 'L': Q(1, 'meter')}) [{'V': 1.0, 'T': 1.0, 'L': -1.0}]
Application to the pendulum
There are 3 fundamental physical units in this equation: time, mass, and length, and 4 dimensional variables, T (oscillation period), M (mass), L (the length of the string), and g (earth gravity). Thus we need only 4 − 3 = 1 dimensionless parameter.
>>> ureg.pi_theorem({'T': '[time]', ... 'M': '[mass]', ... 'L': '[length]', ... 'g': '[acceleration]'}) [{'T': 2.0, 'g': 1.0, 'L': -1.0}]
which means that the dimensionless quantity is:
$$\Pi = \frac{g T^2}{L}$$
and therefore:
$$T = constant \sqrt{\frac{L}{g}}$$
(In case you wonder, the constant is equal to 2 π, but this is outside the scope of this help)
Pressure loss in a pipe
What is the pressure loss p in a pipe with length L and diameter D for a fluid with density d, and viscosity m travelling with speed v? As pressure, mass, volume, viscosity and speed are defined as derived dimensions in the registry, we only need to explicitly write the density dimensions.
>>> ureg.pi_theorem({'p': '[pressure]', ... 'L': '[length]', ... 'D': '[length]', ... 'd': '[mass]/[volume]', ... 'm': '[viscosity]', ... 'v': '[speed]' ... }) # doctest: +SKIP [{'p': 1.0, 'm': -2.0, 'd': 1.0, 'L': 2.0}, {'v': 1.0, 'm': -1.0, 'd': 1.0, 'L': 1.0}, {'L': -1.0, 'D': 1.0}]
The second dimensionless quantity is the Reynolds Number