Hydraulic Press — Efficiency Expression For a hydraulic press with load W on the ram (area A) and effort P on the plunger (area a), what is the expression for efficiency?

Introduction:
A hydraulic press multiplies force via Pascal’s principle. Efficiency compares the actual mechanical advantage to the theoretical (ideal) mechanical advantage and accounts for losses such as seal friction and minor leakage.

Given Data / Assumptions:

• Ram area A, plunger area a.
• Load W applied by the ram; effort P applied to the plunger.
• Quasi-static operation; small losses represented in efficiency.

Concept / Approach:
Ideal mechanical advantage IMA = A / a (from equal pressure in connected cylinders). Actual mechanical advantage AMA = W / P. Efficiency eta = AMA / IMA = (W / P) / (A / a) = (W * a) / (P * A).

Step-by-Step Solution:
Compute AMA: W / P.Compute IMA: A / a.Form efficiency: eta = (W / P) / (A / a) = (W * a) / (P * A).

Verification / Alternative check:
For an ideal, lossless press, W / P = A / a, so eta = 1 as predicted by the formula.

Why Other Options Are Wrong:
(W * A)/(P * a) and (P * A)/(W * a): reciprocals or rearrangements that do not equal AMA/IMA.(P * a)/(W * A): would give eta < 1 even in ideal conditions.

Common Pitfalls:
Swapping ram/plunger areas; remember IMA = A / a.

Final Answer:
eta = (W * a) / (P * A)