**Otto Cycle Analysis**An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 800 kJ/kg of heat is transferred to air during the constant-volume heat-addition process. Take into account the variation of specific heats with temperature. The gas constant of air is $ R = 0.287 \, \text{kJ/kg} \cdot \text{K} $.**Problem Statement:***Determine the thermal efficiency.* (You must provide an answer before moving on to the next part.)**Entered Solution:**The thermal efficiency is $ 451.78 \, \% $. **Note:** The entered value is incorrect as efficiencies greater than 100% are not physically possible. Please review the calculations based on the parameters provided.

Given:To find:Thermal efficiency.

Answered: An ideal Otto cycle has a compression…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

**Otto Cycle Analysis**

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 800 kJ/kg of heat is transferred to air during the constant-volume heat-addition process. Take into account the variation of specific heats with temperature. The gas constant of air is $ R = 0.287 \, \text{kJ/kg} \cdot \text{K} $.

**Problem Statement:**

*Determine the thermal efficiency.* (You must provide an answer before moving on to the next part.)

**Entered Solution:**

The thermal efficiency is $ 451.78 \, \% $.

**Note:** The entered value is incorrect as efficiencies greater than 100% are not physically possible. Please review the calculations based on the parameters provided.

Expert Solution

Step 1: Write the given data and what is to find:

Given:

$table row r equals 8 row cell T subscript 1 end cell equals cell 27 plus 273 equals 300 space K end cell row cell P subscript 1 end cell equals cell 95 space k P a end cell row cell q subscript i n end subscript end cell equals cell 800 space k J divided by k g end cell row R equals cell 0.287 space k J divided by k g minus K end cell row cell v subscript 3 end cell equals cell v subscript 2 end cell row cell v subscript 4 end cell equals cell v subscript 1 end cell end table$

To find:

Thermal efficiency.