Thermal efficiency of the engine in a PV diagram

In summary, a heat engine takes 2.0 moles of an ideal gas through the reversible cycle abca, where the path bc is an isothermal process. The temperature at c is 820 K, and the volumes at a and c are 0.010 m3 and 0.16 m3, respectively. The molar heat capacity at constant volume, Cv, is 37 J/mol·K. The thermal efficiency of the engine, calculated as the net work done on the system divided by the total heat input to the system, is close to 0.4396.
  • #1
YC_128
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A heat engine takes 2.0 moles of an ideal gas through the reversible cycle abca, on the pV diagram shown in the figure. The path bc is an isothermal process. The temperature at c is 820 K, and the volumes at a and c are 0.010 m3 and 0.16 m3, respectively. The molar heat capacity at constant volume, of the gas, is 37 J/mol·K. The thermal efficiency of the engine = (net workdone on the system) /(total heat input to the system) , is close to? (image: https://postimg.cc/QF6mj9db )1. Homework Statement
n=2 mol
Tb=Tc=820K (isothermal)
Va=0.010m3
Vc=0.16m3
Cv=37 J/mol·K

Homework Equations


Work done=nRT*ln(V2/V1)
Work done=P(V2-V1)
PV=nRT
e=(Work done)/(Qheat)

3. The Attempt at a Solution
I
first find the work done by Wbc-Wca

Wbc=nRT*ln(Vc/Va)=37785.94814J

Wca=P(0.16-0.01) (Where P is constant through c to a Pc=Pa PcVc=nRT Pa=Pc=85177.5Pa)

Wca=12776.624J

Wbca=Wbc-Wca=25009.32314J

PaVa=nRTa Ta=(nR)/(PaVa)=(2*8.31)/(85177.5*0.010)=51.25

Qheat=Qab=nCv(Tb-Ta)=2*37*(820-51.25) =56887.5J


e=Wbca/Qab=25009.32314/56887.5=0.439627741 which is wrong
 
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  • #2
What about the heat added in bc?
 
  • #3
It might be easier and more educational to organize your thoughts with problems of this sort by filling the blanks in a Table as shown below without using the first law of thermodynamics (figuring out what's left over.) Then you can check your work by verifying that the entries in any row satisfy the first law. After the Table is correctly completed, one can answer all questions related to the problem. For example the efficiency of the cycle is the sum of all the entries in column ##W## divided by the positive entries in column ##Q##. The starting point is figuring out the temperatures at the vertices and using them to fill in column ##\Delta U##.
$$\begin{array}{|c|c|c|c|}
\hline & \Delta U~~ & Q~~~~~ & W~~~~ \\
\hline a \rightarrow b & & & \\
\hline b \rightarrow c & & & \\
\hline c \rightarrow a & & & \\
\hline
\end{array}$$
 

1. What is thermal efficiency of an engine?

Thermal efficiency refers to the ratio of the work output of an engine to the heat input. It represents how much of the heat energy produced by the engine is converted into useful work.

2. How is thermal efficiency calculated?

Thermal efficiency is calculated by dividing the work output by the heat input and multiplying it by 100%. This can be represented by the equation: (Work output / Heat input) x 100%.

3. What does a PV diagram show?

A PV (pressure-volume) diagram is a graph that shows the relationship between the pressure and volume of a gas or fluid at different stages of a thermodynamic cycle. It is commonly used to visualize the efficiency of an engine.

4. How is thermal efficiency related to the shape of a PV diagram?

The thermal efficiency of an engine is affected by the shape of the PV diagram. It is determined by the area under the curve on the diagram, with a larger area representing a more efficient engine. A steeper slope of the curve also indicates a more efficient engine.

5. What factors can affect the thermal efficiency of an engine?

The thermal efficiency of an engine can be affected by several factors, including the design and size of the engine, the type of fuel used, the temperature and pressure of the engine, and the amount of friction and heat loss within the engine. Improvements in these factors can lead to an increase in thermal efficiency.

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