W (P∆V)out > W (P∆V) IN TEST PROCEDURE Prepare apparatus as per the upper illustration, DEPRESSURIZED SYSTEM (OFF), which disallows pressurization on both the Diamond Shaped Actuator (DSA) and the Conventional Piston (CP). Turn switch (SW1) on to PRESSURIZED SYSTEM mode, allowing both PRV’s to attain their pressure settings. (10 PSIG and 11 PSIG.) Observe the volume changes in TANKS 1 and 2. The minimum mechanical advantage of a DSA over a CP, reported by testing scientists is 15%. Please see the following scientists’ letters. The DSA, with 10 IN² walls and 10 PSIG pressure exerts a minimum force of 115 lb. The CP, with a 10.25 IN² boundary face and 11 PSIG pressure exerts a force of 112.75 lb. The DSA overpowers the CP (115 lb > 112.75 lb), forcing fluid from the CP. The volume change (∆V) in the DSA is 9.99 IN³; therefore, ∆V in TANK 1 is 9.99 IN³. The volume change in CP is 10.25 IN³; therefore, TANK 2 ∆V is 10.25 IN³ Using the fluidic formula Work = Pressure x Volume Change (W = P∆V) TANK 1’s work input potential……………..W = P∆V = 10.0 PSIG X 9.99 IN³ = 99.90 in-lb TANK 2’S work output potential……........W = P∆V = 11.0 PSIG X 10.25 IN³ = 1.12.75 in-lb CONCLUSION: Lesser work input can produce greater work output (Wout > Win ). 13.283 ON OFF HP AIR PRV SET 10 PSIG PRV SET 11 PSIG SW1 DEPRESSURIZED SYSTEM (OFF) CURRENT PRESSURE AT 0 PSIG TANK 1 TANK 2 2” 2” FLUID LEGEND SW1---PNEUMATIC SWITCH DSA---DIAMOND-SHAPED ACTUATOR CP ---CONVENTIONAL PISTON PRV---PRESSURE REDUCING VALVE HP ---HIGH PRESSURE ∆V ---VOLUME CHANGE IN³ ---CUBIC INCHES IN² ---SQUARE INCHES 10 IN² RETRACTED 10.25 IN² 2” DSA EACH WALL IS 10 IN² CP BOUNDARY FACE IS 10.25 IN² ON OFF PRV SET 10 PSIG CURRENT PRESSURE IS 10 PSIG PRV SET 11 PSIG CURRENT PRESSURE IS 11 PSIG HP AIR SW1 PRESSURIZED SYSTEM (ON) TANK 1 TANK 2 1.01” 3” 1” ∆V = 9.99 IN³ ∆V = 10.25 IN³ 1” DSA CP
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