Calculating Efficiency and Work in Otto Cycle Engine

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SUMMARY

The efficiency of an Otto cycle engine with a compression ratio of 6 and a heat input of 350 kJ is calculated to be 55.34%. The formula used for efficiency is Efficiency Otto = (1 - r^(1-k)), where r is the compression ratio and k is the specific heat ratio (Cp/Cv), which is 1.45 in this case. The work done by the engine is determined using the equation W = Efficiency * Q, resulting in 193.72 kJ. It is crucial to correctly apply the exponent notation in calculations for accurate results.

PREREQUISITES
  • Understanding of the Otto cycle and its principles
  • Familiarity with thermodynamic efficiency calculations
  • Knowledge of specific heat ratios (Cp/Cv)
  • Basic algebra for manipulating equations and exponents
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  • Study the derivation of the efficiency formula for the Otto cycle
  • Explore the implications of varying compression ratios on engine performance
  • Learn about the differences between Otto and Diesel cycle efficiencies
  • Investigate the impact of heat transfer methods on engine work output
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Homework Statement



An Engine works within Otto cycle where the compression ratio is 6, and 350 kJ of heat was added to it. Determine the efficiency of the cycle and the work done if Cp/Cv is 1.45

Homework Equations


Efficiency Otto = (1- r1-k)
Efficiency (eff.) =Work (W) /Heat Transferred (Q)

The Attempt at a Solution



Efficiency Otto = (1- r1-k) ( r is the compression ratio, k is Cp/Cv)

= 1-61-1.45

= 0.5534 = 55.34% --> The efficiency of the cycleW=eff. * Q = 0.5534 * 350

W=193.72 kJ --> The work done
 
Last edited by a moderator:
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Your result looks fine, but use the "x2" button for an exponent. The efficiency is 1-r(1-k) instead of 1-r1-k and 1-61-1.45 is not 55.34%.
 

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