Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-21
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Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-21: Discussion question: 333. The piston diameter is D=350 mm, the cylinder efficiency is η=0.9, and the air supply pressure for the actuator is Ps=500 KPa. Determine the output force of this piston actuator. (5 points are given only to those who provide the correct answer; additional points are awarded for a correct calculation process or detailed explanation.)
F=π/4×η ×D = 3.14/4×0.9×352×5 = 4300(×10N)
F=π/4×D×D×P×η = 3.14/4×35×35×5×0.9 = 4327.3 (Kg)
The piston diameter is D=350mm, the cylinder efficiency is η=0.9, and the air pressure supply for the actuator is Ps=500KPa. Determine the output force of this piston actuator. Solution: F = π/4 × η × D² × Ps = 3.14/4 × 0.9 × 352 × 5 = 4327.3 kgf (×10 N). Answer: The output force of this piston actuator is 4327.3 kgf (×10 N).
F=π/4×η×D2×Ps=3.14/4×0.9×352×5=4327.3kgf (×10N).
F=π/4×D×D×P×η = 3.14/4×35×35×5×0.9 = 4327.3 (Kg)
F=π/4×η×D2×Ps=3.14/4×0.9×352×5=43273N
The output force of this piston actuator is 4327.3 kgf (×10 N). Explanation: F=π/4×η×D²×Ps=3.14/4×0.9×352×5=4327.3kg.
F=π/4×D×D×P×η = 3.14/4×35×35×5×0.9 = 4327.3 (Kg)
F=π/4×D×D×P×η = 3.14/4×35×35×5×0.9 = 4327.3 (Kg)