What is the Efficiency of a Gas Machine in an A->B->C Change?

  • Thread starter Thread starter the_man
  • Start date Start date
  • Tags Tags
    Gas Temperature
Click For Summary
SUMMARY

The discussion focuses on calculating the heat transfer (Q) and the efficiency of a gas machine during an A->B->C thermodynamic process. The key equations utilized include W = pV and the ideal gas law PV = nRT. Participants highlight the need for the number of moles of gas, the gas constant R, and temperature values to derive Q effectively. The relationship between internal energy change (ΔU), heat transfer (ΔQ), and work done (ΔW) is also emphasized as crucial for understanding the machine's efficiency.

PREREQUISITES
  • Understanding of thermodynamic processes and the ideal gas law (PV = nRT)
  • Familiarity with the concepts of work (W) and internal energy (U)
  • Knowledge of heat transfer principles in thermodynamics
  • Basic proficiency in manipulating equations involving pressure, volume, and temperature
NEXT STEPS
  • Study the derivation and application of the first law of thermodynamics (ΔU = ΔQ + ΔW)
  • Learn about calculating heat transfer in various thermodynamic processes
  • Explore the significance of the gas constant (R) in thermodynamic equations
  • Investigate the efficiency metrics of gas machines and their practical applications
USEFUL FOR

Students in thermodynamics, engineers working with gas machines, and anyone interested in understanding heat transfer and efficiency in thermodynamic cycles.

the_man
Messages
14
Reaction score
0

Homework Statement



The task is to get Q, that gas has received through A-B-C change?
Second task is to get usefulness of the machine?

http://img40.imageshack.us/img40/8372/toplinaabc.jpg

Homework Equations



W=pV

The Attempt at a Solution



If I use W=pV i get 0 from A->B so i can't get usefulness. No idea.
 
Last edited by a moderator:
Physics news on Phys.org
To get the heat energy in or out of system one requires the value of the number of moles of gas and the gas constant R or the temperatures at some points on the graph.
 
So from A->B formula is PV=nRT then I get dT=((p2-p1)*v1)/nR?

from B->C formula is W=p2(v2*v1) and with ideal gas dT=(p2(v2-v1))/nR?

But how to get Q from that?
 
the_man said:
So from A->B formula is PV=nRT then I get dT=((p2-p1)*v1)/nR?

from B->C formula is W=p2(v2*v1) and with ideal gas dT=(p2(v2-v1))/nR?

But how to get Q from that?

from A->B formula is PV=nRT then I get dT=((p2-p1)*v1)/nR ...yes

from B->C formula is W=p2(v2*v1) ...replace * by -

dT=(p2(v2-v1))/nR ...yes

To get the heat transfer one can use \DeltaU = \DeltaQ + \DeltaW where U is the internal energy.
 
okay man, thanks!
 

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
7K
Replies
1
Views
2K
Replies
2
Views
2K
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K