Pascal's Principle — Question 6

Section 14-5 · Physics pages 2142–2143

Question 6 A small pump must deliver enough volume as well as pressure A hydraulic pump piston of area 1.0 cm2 makes delivery strokes of length 0.10 m. Ideal check valves refill it on each return stroke. A lift piston of area 80 cm2 must rise 0.050 m against a constant 1600 N load. Ignore height-head changes and losses. F1 F2 A1 A2 Pump and lift schematic; ideal check-valve operation is specified in the text. (a) Find the number of complete delivery strokes required and the force during a delivery stroke. (b) Find total input work and explain why check valves are essential to accumulating lift over repeated strokes.
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Question 6 - Solution A small pump must deliver enough volume as well as pressure F1 F2 A1 A2 (a) Derivation and calculation ∆Vstroke = Apxp = (10−4)(.100) = 10−5 m3. Vlift = ALxL = .00800(.0500) = 4.00 × 10−4 m3. N = Vlift/∆Vstroke = 40 . pg = 1600/.00800 = 200000 Pa, Fp = pgAp = 20.0 N . (b) Result and interpretation Win = NFpxp = 40(20.0)(.100) = 80.0 J . Wout = 1600(.0500) = 80.0 J. The outlet valve prevents backflow during refill, while the inlet admits replacement liquid. Without this sequence, a return stroke would tend to reverse the delivered volume. Physical check Forty full strokes supply both the required volume and the required work.

Question 6

A small pump must deliver enough volume as well as pressure

A hydraulic pump piston of area 1.0cm21.0\,\mathrm{cm^2} makes delivery strokes of length 0.10m0.10\,\mathrm{m}. Ideal check valves refill it on each return stroke. A lift piston of area 80cm280\,\mathrm{cm^2} must rise 0.050m0.050\,\mathrm{m} against a constant 1600N1600\,\mathrm{N} load. Ignore height-head changes and losses.

Diagram shown in the original SVG worksheet.

  1. Find the number of complete delivery strokes required and the force during a delivery stroke.

  2. Find total input work and explain why check valves are essential to accumulating lift over repeated strokes.

Question 6 - Solution

A small pump must deliver enough volume as well as pressure

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

ΔVstroke=Apxp=(10−4)(.100)=10−5m3.\Delta V_{\mathrm{stroke}}=A_px_p=(10^{-4})(.100)=10^{-5}\,\mathrm{m^3}.Vlift=ALxL=.00800(.0500)=4.00×10−4m3.V_{\mathrm{lift}}=A_Lx_L=.00800(.0500)=4.00\times10^{-4}\,\mathrm{m^3}.N=Vlift/ΔVstroke=40.\boxed{N=V_{\mathrm{lift}}/\Delta V_{\mathrm{stroke}}=40}.pg=1600/.00800=200000Pa,Fp=pgAp=20.0N.p_g=1600/.00800=200000\,\mathrm{Pa},\quad\boxed{F_p=p_gA_p=20.0\,\mathrm N}.

(b) Result and interpretation

Win=NFpxp=40(20.0)(.100)=80.0J.\boxed{W_{\mathrm{in}}=NF_px_p=40(20.0)(.100)=80.0\,\mathrm J}.Wout=1600(.0500)=80.0J.W_{\mathrm{out}}=1600(.0500)=80.0\,\mathrm J.The outlet valve prevents backflow during refill, while the inlet admits replacement liquid. Without this sequence, a return stroke would tend to reverse the delivered volume.

Physical check

Forty full strokes supply both the required volume and the required work.