CoCalc Public Filestmp / 2016-03-06-205753-plot.sagewsOpen in with one click!
Authors: Harald Schilly, ℏal Snyder, William A. Stein
%latex The following shows the pressure inside the chamber as a function of the piston displacement, assuming that the piston distends a membrane via a circular disk of diameter D1 and that the chamber has cylindrical dimensions D2 and L. I calculated the change in volume by rotating a solid about the x-axis. $$\mathrm{d}V = \int_0^\delta \! A(x) \, \mathrm{d}x$$ where $A(x)$ is the area of the extra volume created by the piston movement and delta is the piston displacement. Therefore $$V_2 = V_1 + \mathrm{d}V$$ and $$P2 = P1\frac{V1}{V2}$$ To stay within the range of the PNAS paper pressures, the min negative pressure is $$100-5.3329 = 94.6671 kPa$$
import numpy as np #constants D1 = .01 #Diameter of piston disk P1 = 100 #kpa D2 = .05 #meters Diameter of vacuum chamber L = .015 #meters length of cylinder CPM = 90 #cycles per minute (sucks per minute) ang_vel = CPM *2*pi()/60 #angular velocity of motor SSVacuum = 0 #steady state vacuum pressure kPa provided by secondary motor pistonDisplacement = .02 hose_V = 0 #placeholder for now cup_V = 0 #placeholder for now V1 = pi*D2*L + hose_V + cup_V def delta(t): """ Piston displacement as function of time. """ return pistonDisplacement/2*sin(ang_vel*t-pi()/2)+.01 #P2 equation assumes geometry constraints outlined above. P2(t) = P1*(V1/(V1+(D1*delta(t)/2 + (D2*delta(t)^2)/8 - (D1*delta(t)^2)/8))) plot(P2(t), (t, 0, 2),title='Pressure Curve as function of time', axes_labels=['$t$','Pressure (kPa)'])