Pascal's Principle — Question 10

Question 10

Equal pressure on both sides can still drive a rod piston

A horizontal cylinder contains a piston of full area $A=10\,\mathrm{cm^2}$ with a rod of area $a=2.0\,\mathrm{cm^2}$ extending through its right end. Both fluid chambers are connected to the same gauge pressure $p_g=500\,\mathrm{kPa}$. The exposed rod end is atmospheric. Neglect friction and inertia.

1. Find the net rightward hydraulic force on the piston-and-rod assembly. Explain why equal chamber pressure does not imply cancellation.

2. If it moves right by $x=0.050\,\mathrm{m}$ while both chambers remain supplied, find the net fluid volume drawn from the source and the pressure work delivered.

Question 10 - Solution

Equal pressure on both sides can still drive a rod piston

(a) Derivation and calculation

Using absolute pressure on every exposed face,$$F_x=p_{\mathrm{abs}}A-p_{\mathrm{abs}}(A-a)-p_{\mathrm{atm}}a.$$$$\boxed{F_x=(p_{\mathrm{abs}}-p_{\mathrm{atm}})a=p_ga}.$$$$\boxed{F_x=(500000)(2.0\times10^{-4})=100\,\mathrm N\ \text{rightward}}.$$

(b) Result and interpretation

$$\Delta V_{\mathrm{source}}=Ax-(A-a)x=ax.$$$$\boxed{\Delta V_{\mathrm{source}}=(2.0\times10^{-4})(.0500)=1.00\times10^{-5}\,\mathrm{m^3}}.$$$$\boxed{W=p_g\Delta V_{\mathrm{source}}=5.00\,\mathrm J=F_xx}.$$Equal chamber pressures act on unequal wetted areas; the rod area controls both net force and net source volume.

Physical check

If the rod area tends to zero, both the net force and the net volume demand vanish.