Thermodynamic Cycle: Calculating Heat Transfer and Work

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SUMMARY

The discussion focuses on calculating heat transfer and work in a thermodynamic cycle involving three processes. The initial conditions are defined with pressure p1 at 1 bar and volume V1 at 1.6 m3, transitioning to a final volume of V2 at 0.2 m3. The work done during the compression process (1-2) is calculated as 333 kJ, and the pressure at state 2 (p2) is determined to be 8 bar. The heat transfer for process 2-3 can be calculated using the equation ΔQ = ΔU + W, where ΔU is the change in internal energy.

PREREQUISITES
  • Understanding of thermodynamic cycles and processes
  • Familiarity with the ideal gas law and its applications
  • Knowledge of internal energy and its relation to heat transfer
  • Proficiency in calculus for integrating work done (W = ∫p dV)
NEXT STEPS
  • Study the derivation of the ideal gas law and its implications in thermodynamics
  • Learn about the First Law of Thermodynamics and its applications in heat transfer calculations
  • Explore the concept of enthalpy and its relevance in constant pressure processes
  • Investigate advanced thermodynamic cycles, such as the Carnot cycle and its efficiency
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Students and professionals in mechanical engineering, thermodynamics researchers, and anyone involved in energy systems analysis will benefit from this discussion.

kdinser
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Sorry about the double post, I had technical difficulties that I was working on and the title of my orginal post got screwed up.

I'm having problems getting started on this one.

A gas undergoes a thermodynamic cycle consisting of 3 processes

process 1-2 compression with pressure(p)*volume(V) = constant, from
[tex]p_{1} = 1 bar[/tex]
[tex]V_{1} = 1.6m^3[/tex]
to
[tex]p_{2} = ?[/tex]
[tex]V_{1} = .2m^3[/tex]

[tex]U_{2}-U_{1}=0[/tex]

process 2-3
Constant pressure to [tex]V_{3}=V_{1}[/tex]

process 3-1
Constant Volume, [tex]U_{1}-U_{3} = -3549kJ[/tex]

There are no significant changes in kinetic or potential energy.
Determine the heat transfer and work for process 2-3 in kJ.

I don't have any problems finding[tex]p_{2}[/tex] or the work needed to compress the gas, but I'm not really sure where to go from there.

[tex]p_2=\frac{p_1V_1}{V_2}[/tex]

[tex]W=\int p dV[/tex]

When I work these out, I end up with 333kJ for W and 8 bar for p2.

If someone could give me a quick push in the right direction, that would be great.
 
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kdinser said:
A gas undergoes a thermodynamic cycle consisting of 3 processes

process 1-2 compression with pressure(p)*volume(V) = constant, from
[tex]p_{1} = 1 bar[/tex]
[tex]V_{1} = 1.6m^3[/tex]
to
[tex]p_{2} = ?[/tex]
[tex]V_{1} = .2m^3[/tex]

[tex]U_{2}-U_{1}=0[/tex]

process 2-3
Constant pressure to [tex]V_{3}=V_{1}[/tex]

process 3-1
Constant Volume, [tex]U_{1}-U_{3} = -3549kJ[/tex]

There are no significant changes in kinetic or potential energy.
Determine the heat transfer and work for process 2-3 in kJ.
.
The work done between 2-3 is just [itex]P_2\Delta V = P_2(V_3-V_2) = P_2(7*V_2)[/itex]

The change in internal energy is [itex]U_3-U_2 = U_3-U_1[/itex], since [itex]U_2=U_1[/itex]

Use [itex]\Delta Q = \Delta U + W[/itex] to find the energy (heat) flow into the system.

AM
 
Last edited:

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