Isothermal, adiabatic, isovolumetric process

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SUMMARY

The discussion revolves around the thermodynamic processes involving air with a mass of 10 kg, initially at 15 bar and 50°C. The air undergoes isobaric expansion to three times its original volume, followed by isovolumetric cooling to 6 bar, an adiabatic expansion back to the original temperature, and finally isothermal contraction. Key equations utilized include Q=nCp*delta(T) for isobaric processes and Q=nCv*delta(T) for isochoric processes, with the adiabatic process having no heat flow (Q=0). The participants clarify that while the adiabatic process affects temperature, it does not contribute to heat flow calculations.

PREREQUISITES
  • Understanding of thermodynamic processes: isobaric, isochoric, adiabatic, and isothermal.
  • Familiarity with the ideal gas law (PV=nRT).
  • Knowledge of specific heat capacities (Cp and Cv) and the adiabatic quotient (kappa).
  • Ability to apply the first law of thermodynamics in cyclic processes.
NEXT STEPS
  • Study the derivation and application of the ideal gas law (PV=nRT) in various thermodynamic processes.
  • Learn about the relationships between Cp, Cv, and kappa in thermodynamics.
  • Explore the concept of work done in thermodynamic cycles, particularly in adiabatic processes.
  • Investigate the implications of heat flow in cyclic processes and how to calculate it effectively.
USEFUL FOR

Students and professionals in thermodynamics, mechanical engineers, and anyone involved in energy systems analysis will benefit from this discussion, particularly those working with gas processes and heat transfer calculations.

Bassalisk
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Homework Statement


Air with the mass of 10kg, with pressure 15 bar and temperature 50 C comes to a isobaric expansion to 3 times the original volume, then the air cools to the pressure of 6 bar(isovolumetricly). After that it comes to adiabatic expansion to original temperature (50 C i guess) and then isothermally contracts to the beginning state. Molar mass of the air is M=29 g/mol, adiabatic quotient is k(kappa)=1,4;

Find the thermal quotient. W/Q

Homework Equations


Q=nCp*delta(T);
Q=nCv*delta(T); (capital C=c*M
pV=nRT
cp/cv=k

The Attempt at a Solution



Q1=m/M*Cp(T2-T1) isobaric

T1=323 K

and from the pV/T=const, because we have always same amount of gas. We get the T=969 K and delta(T)=646 K;

After that same for isovolumetric condition we get that T3=387,6 K delta(T')=581,4 K, and the complicate route to go through adiabatic process and isothermal. I do not understand, where does work occur and where does heat come in and come out. I am stuck with that Q1 and i cannot break the mind to calculate it(its late) because i am not given cp, only kappa.

Can you assist me in this manner please?

Its very important.

Thanks
 
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Bassalisk said:
Q1=m/M*Cp(T2-T1) isobaric

T1=323 K

and from the pV/T=const, because we have always same amount of gas. We get the T=969 K and delta(T)=646 K;

After that same for isovolumetric condition we get that T3=387,6 K delta(T')=581,4 K, and the complicate route to go through adiabatic process and isothermal. I do not understand, where does work occur and where does heat come in and come out. I am stuck with that Q1 and i cannot break the mind to calculate it(its late) because i am not given cp, only kappa.
Work occurs when there is a change in volume. The work is the area inside the cycle graph (the area under the expansion parts of the cycle LESS the area under the compression parts). But it is complicated to find the work that way. It is easier to use heat flows:

W = Qh-Qc; so: W/Qh = (Qh - Qc)/Qh

Heat flow is into the system gas (air) only during isobaric expansion. Heat flow is out of the gas during the isochoric (isovolumetric) cooling and the isothermal compression. Calculate those heat flows and use the above expression to find W and W/Qh.

AM
 
so the adiabatic part has no effect?
 
Sorry but I am stuck with wrong result in the see of formulas, i simply cannot get the result...
 
Bassalisk said:
so the adiabatic part has no effect?
It certainly has an effect on the process. But since there is no heat flow during that part of the process you do not have to take that into account in calculating heat flow. If you were to measuring W directly you would have to calculate the area under that adiabatic curve. But if you are calculating W from W = Qh-Qc you can ignore the adiabatic part. It is not that the adiabatic part has no effect. The change in temperature that occurs during the adiabatic part affects the final temperature and, therefore, the heat flows in the other parts of the cycle.

AM
 
Bassalisk said:
Sorry but I am stuck with wrong result in the see of formulas, i simply cannot get the result...
Break it down into the four parts of the cycle:

Part 1: Isobaric expansion from V to 3V. Heat flow is \Delta Q = nC_p\Delta T = nC_p(T_2-T_1). Apply PV=nRT to find the change in T. Is Q positive or negative? If it is positive, it is heat flow INTO the system so it adds to Qh. If it is negative, it is heat flow it contributes to Qc.

Part 2: Isochoric compression from P=15 to P=6. Heat flow is \Delta Q = nC_v\Delta T = nC_v(T_3-T_2). Apply PV=nRT to find the change in T. Is Q positive or negative? If it is positive, it is heat flow INTO the system so it adds to Qh. If it is negative, it is heat flow it contributes to Qc.

Part 3: Adiabatic so Q=0. You have to calculate the volume and pressure at the beginning and end of this process in order to calculate the change in volume and heat flow in the next part. What is the relationship between T and V in this adiabatic process (hint: it involves \gamma)? From the change in V you can find change in P

Part 4: Isothermal compression from V = ? to V. Find the change in volume to find the heat flow. What is the heat flow if \Delta U = 0?

AM
 
ok, thank you, you have been very helpful, I got the final result and it matches.

Thank you
 

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