What is the Efficiency of a Cyclic Process?

Click For Summary
SUMMARY

The efficiency of a cyclic process is defined by the formula ##\eta=\frac{W}{Q}##, where W represents the work done and Q is the heat input. The work done can be calculated as the area enclosed within the loop on a pressure-volume graph, specifically as ##(P-P_o)V_o##. Heat input occurs during processes B->A and B->C, with equations for heat input being ##Q_1=nC_v\Delta T_1## and ##Q_2=nC_p\Delta T_2##. The relationship between molar specific heats at constant volume (C_v) and constant pressure (C_p) is crucial, particularly for monoatomic gases where C_v is ##\frac{3R}{2}## and C_p is ##\frac{5R}{2}##.

PREREQUISITES
  • Understanding of thermodynamic cycles and efficiency calculations
  • Familiarity with the concepts of molar specific heat (C_v and C_p)
  • Knowledge of the ideal gas law and its applications
  • Basic proficiency in calculus for area calculations in PV diagrams
NEXT STEPS
  • Research the derivation of the efficiency formula for cyclic processes
  • Study the relationship between specific heats and degrees of freedom in gases
  • Learn about the implications of different gas types (monoatomic vs diatomic) on thermodynamic calculations
  • Explore advanced thermodynamic concepts such as the Carnot cycle and its efficiency limits
USEFUL FOR

Students studying thermodynamics, engineers working with heat engines, and anyone interested in understanding the efficiency of cyclic processes in thermodynamic systems.

Saitama
Messages
4,244
Reaction score
93

Homework Statement


(see attachment, ignore the arrows made with the pen)


Homework Equations





The Attempt at a Solution


Efficiency of a cycle is defined as ##\eta=\frac{W}{Q}## where W is work done and Q is heat input.

W can be easily calculated by finding the area enclosed within the loop shown in the graph which is equal to ##(P-P_o)V_o##.

Heat input occurs only in the processes, B->A (##Q_1##) and B->C (##Q_2##).
##Q_1=nC_v\Delta T_1## where ##\Delta T_1=\frac{V_o}{nR}(P_o-P)##
##\Rightarrow Q_1=\frac{C_vV_o}{R}(P_o-P)##

##Q_2=nC_p\Delta T_2## and ##\Delta T_2=\frac{P_oV_o}{nR}##
##\Rightarrow Q_2=\frac{C_pP_oV_o}{R}##

\eta=\frac{W}{Q_1+Q_2}=\cfrac{(P_o-P)V_o}{\cfrac{C_vV_o}{R}(P_o-P)+\cfrac{C_pP_oV_o}{R}}
I don't have ##C_p## and ##C_v##, how am I supposed to solve this? :confused:

Any help is appreciated. Thanks!
 

Attachments

  • efficiency.jpg
    efficiency.jpg
    14.2 KB · Views: 1,354
Physics news on Phys.org
What is the relation between Cp and Cv, and how does Cv depend on the degrees of freedom of the molecules in the gas? Assume monoatomic gas first.

ehild
 
ehild said:
What is the relation between Cp and Cv, and how does Cv depend on the degrees of freedom of the molecules in the gas? Assume monoatomic gas first.

ehild

The molar specific heat at constant volume is fR/2 and at constant pressure is (f/2+1)R where f is the degrees of freedom. For a monoatomic gas, f=3, hence Cv=3R/2 and Cp=5/2. Substituting the values give ##P=2P_o/7## which is not correct. :confused:
 
f is not given, you get different pressures for different values of f. f=5 for diatomic gas and f=6 for three or more atomic molecules. For monoatomic gas, your solution should be correct.

ehild
 
ehild said:
f is not given, you get different pressures for different values of f. f=5 for diatomic gas and f=6 for three or more atomic molecules. For monoatomic gas, your solution should be correct.

ehild

I thought that it would be a bit difficult to put different values of f and find P. The answer in the key is A i.e ##0.3P_o##. I substituted it in the expression I ended up with and replaced ##C_v## and ##C_p## with fR/2 and (f/2+1)R respectively. Solving I got f=50/17. Is the question wrong? :confused:
 
2/7 is 0.3 with one significant digit. As you had to choose among different data, you should have figured out which was possible.

ehild
 
ehild said:
2/7 is 0.3 with one significant digit. The question is wrong in the sense that it did not indicate the type of the gas. ((mono-atomic).

ehild

Thank you ehild! :smile:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
865
Replies
35
Views
2K
Replies
1
Views
1K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K