Isothermal, adiabatic, isovolumetric process

In summary, the air expands isobarically from 10 kg to 3 times its original volume, then contracts isothermally to the original temperature of 50 C. Molar mass of the air is M=29 g/mol, adiabatic quotient is k(kappa)=1,4. The thermal quotient is W/Q= (Qh-Qc)/Qh. The attempt at a solution was to find 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;
  • #1
Bassalisk
947
2

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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  • #2
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
 
  • #3
so the adiabatic part has no effect?
 
  • #4
Sorry but I am stuck with wrong result in the see of formulas, i simply cannot get the result...
 
  • #5
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
 
  • #6
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 [itex]\Delta Q = nC_p\Delta T = nC_p(T_2-T_1)[/itex]. 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 [itex]\Delta Q = nC_v\Delta T = nC_v(T_3-T_2)[/itex]. 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 [itex]\gamma[/itex])? 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 [itex]\Delta U = 0[/itex]?

AM
 
  • #7
ok, thank you, you have been very helpful, I got the final result and it matches.

Thank you
 

Related to Isothermal, adiabatic, isovolumetric process

What is an isothermal process?

An isothermal process is a thermodynamic process in which the temperature of a system remains constant throughout the process. This means that the system is in thermal equilibrium with its surroundings at all times.

What is an adiabatic process?

An adiabatic process is a thermodynamic process in which there is no transfer of heat between the system and its surroundings. This means that the system is insulated and does not exchange heat with its surroundings.

What is an isovolumetric process?

An isovolumetric process, also known as an isochoric process, is a thermodynamic process in which the volume of the system remains constant. This means that the system does not change in size or shape during the process.

What are the differences between isothermal, adiabatic, and isovolumetric processes?

The main difference between these processes is the change in temperature, heat transfer, and volume. In an isothermal process, the temperature remains constant, while in an adiabatic process, there is no heat transfer. In an isovolumetric process, the volume remains constant. These processes have different effects on the system, and they are used to study different aspects of thermodynamics.

How are these processes used in scientific research?

These processes are fundamental concepts in thermodynamics and are used to study the behavior of different systems. They are also used in engineering and industry to design and improve processes and machines that involve heat and energy transfer. In scientific research, these processes are used to understand and predict the behavior of various systems, such as gases, liquids, and solids, and to develop new technologies and materials.

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