Indicator diagram factor — relation between actual and theoretical mean effective pressures Given pa = actual mean effective pressure, pm = theoretical mean effective pressure, and K = diagram factor, which relation is correct?

Difficulty: Easy

Correct Answer: pa = pm × K

Explanation:


Introduction / Context:
Real engine indicator diagrams deviate from the ideal theoretical diagrams due to throttling, wire-drawing, heat losses, and valve timing. The diagram factor K captures the ratio of the actual diagram area to the theoretical diagram area, and thus relates actual to theoretical mean effective pressure (MEP).


Given Data / Assumptions:

  • pm is computed from a theoretical cycle for the same boundary conditions.
  • pa is obtained from the measured indicator diagram.
  • Diagram factor K = actual area / theoretical area.


Concept / Approach:
Since MEP is proportional to the diagram area for a given cylinder, K scales the theoretical MEP to match reality. Therefore, actual MEP equals theoretical MEP multiplied by the diagram factor.



Step-by-Step Solution:
By definition: K = Area_actual / Area_theoretical.MEP ∝ diagram area (for fixed stroke and bore).Thus, pa = K * pm.Select the matching expression: pa = pm × K.


Verification / Alternative check:
Worked examples compute pm from an idealized cycle and then multiply by K (often 0.7–0.9) to obtain pa before calculating indicated power.



Why Other Options Are Wrong:

  • pa = pm/K inverts the effect, implying pa > pm for K < 1, which is unphysical.
  • Sum or ratio forms lack basis in the definition of K.


Common Pitfalls:
Treating K as an additive correction rather than a multiplicative factor derived from area ratios.



Final Answer:
pa = pm × K

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