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Zorluk: OrtaPascal's Principle and Hydraulics

An industrial cardboard baler utilizes a closed hydraulic system to compress recyclable materials. The system features an input pump piston with a cross-sectional area of 3 in23 \text{ in}^2 and a large output driving piston with a cross-sectional area of 45 in245 \text{ in}^2. If the input piston is pushed downward for a single stroke distance of 15 inches15 \text{ inches}, how many inches will the output piston move? (Assume no fluid is lost and the fluid is incompressible.)

Cevap: 1 inches

Cevap

1
In a hydraulic system, the volume of fluid displaced must be conserved. The output piston has 15 times the cross-sectional area of the input piston (45÷3=1545 \div 3 = 15). Because it is 15 times larger, it will only travel 1/15th of the distance to displace the identical volume of fluid. Therefore, an input distance of 15 inches results in an output distance of exactly 1 inch.

Adım Adım Çözüm

1
Identify the relationship for fluid displacement in a closed hydraulic system.
A1×d1=A2×d2A_1 \times d_1 = A_2 \times d_2
Because liquids are incompressible, the volume of fluid pushed out by the input piston (A1×d1A_1 \times d_1) must equal the volume of fluid pushing up the output piston (A2×d2A_2 \times d_2).
2
Substitute the given values for the input piston (A1=3A_1 = 3, d1=15d_1 = 15) and output piston (A2=45A_2 = 45).
3×15=45×d23 \times 15 = 45 \times d_2
This sets up the equation with all known variables to solve for the unknown distance (d2d_2).
3
Calculate the total volume displaced by the input piston.
45=45×d245 = 45 \times d_2
Multiplying the input area by its travel distance gives a total displaced volume of 45 cubic inches.
4
Solve for the output distance (d2d_2) by dividing the total volume by the output piston's area.
d2=1d_2 = 1
Dividing 45 cubic inches by 45 square inches yields a travel distance of 1 inch.

Anahtar Kavram

Conservation of fluid volume and the distance-force tradeoff in hydraulics
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Konular

Pascal's Principle and Hydraulics

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