Find Area Enclosed by Thermodynamic Cycle

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SUMMARY

The area enclosed by the thermodynamic cycle ABCD is calculated using the equations P=0.8+8.6e^(-0.3V) for line AB and P=1.2+9.6e^-(0.1V) for line CD. The method involved calculating the area of the rectangular shape formed by points C and B, then subtracting the integral of the equation for AB from the limits 2.076 to 0.159. The final computed area is 67.7, which is confirmed to be correct when rounded to three significant figures.

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  • Understanding of integral calculus
  • Familiarity with thermodynamic cycles
  • Knowledge of exponential functions
  • Ability to perform numerical integration
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  • Explore thermodynamic properties and their graphical representations
  • Learn about the applications of the area under curves in thermodynamics
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Students and professionals in engineering, particularly those focused on thermodynamics, as well as anyone involved in solving problems related to area calculations in physics.

jackscholar
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Homework Statement


The question is pretty much explained in the picture.
The area ABCD is enclosed, i need to find that area. The equation for AB is P=0.8+8.6e^(-0.3V) and the equation for CD is P=1.2+9.6e^-(0.1V)
What I did was find the area enclosed by the rectangular shape (C-B)*9. From there I subtracted the integral of the equation AB with the upper and lower limits 2.076 and 0.159 respectively. I then surmised that if I subtracted the integral of AB from (25 (upper) to 2.076 (lower)) from CD then it would give me the rest of the Area. I ended up getting 67.7 doing this. Is this correct?
 

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jackscholar said:

Homework Statement


The question is pretty much explained in the picture.
The area ABCD is enclosed, i need to find that area. The equation for AB is P=0.8+8.6e^(-0.3V) and the equation for CD is P=1.2+9.6e^-(0.1V)
What I did was find the area enclosed by the rectangular shape (C-B)*9. From there I subtracted the integral of the equation AB with the upper and lower limits 2.076 and 0.159 respectively. I then surmised that if I subtracted the integral of AB from (25 (upper) to 2.076 (lower)) from CD then it would give me the rest of the Area. I ended up getting 67.7 doing this. Is this correct?

That answer is OK, rounded to 3 significant figures.
 

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