Thermodynamics work required per unit mass problem

In summary, the speaker is having difficulty with a thermodynamics problem involving the calculation of work required per unit mass in an isothermal process. They have tried using tables and approximating with a curve fit, but it is best to use the ideal gas law for an isothermal process.
  • #1
nineeyes
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0
Having trouble with a thermodynamics problem

I'm having a bit of a problem with some of the homework in my thermodynamics class.
Question (Water at 20 C, 100 kPa is compressed isothermally to 50 MPa. Determine the work required per unit mass. )
using the tables I found:
State 1
[tex]T_1=20C[/tex]
[tex]P_1=.100MPa[/tex]
[tex]v_1=.001022[/tex] [tex]m^3/kg[/tex]
State 2
[tex]T_2=20C[/tex]
[tex]P_2=50MPa[/tex]
[tex]v_2=.0009804[/tex] [tex]m^3/kg[/tex]

However, according to the tables, both states are compressed/subcooled . The only method I found to solve for work in an isothermal process applied to ideal gases. I was thinking I needed to approximate this, I tried to plot as many points in between the states and do a curve fit to find function [tex]P(v)[/tex]. Then integrate [tex]Work = \int_{v_1}^{v_2}P(v) dv}[/tex]. If I can do it that way, what kind of line do I use? (2nd order polynomial, 3rd order polynomial, etc...)

Thanks for any help.
 
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  • #2
For this particular problem, it is best to use the ideal gas law for an isothermal process. The ideal gas law states that PV=nRT. Rearranging the equation yields W=nRTln(V2/V1). Plugging in the values from your states 1 and 2 yields W=nRTln(V2/V1)=nRTln(0.0009804/0.001022)=-2.68 kJ/kg.
 
  • #3


Hello there, it seems like you're having trouble with a thermodynamics problem. I understand that it can be challenging to solve problems involving different states and processes. Let me try to help you out.

Firstly, to solve for the work required per unit mass in an isothermal process, you can use the equation W = -nRT ln(V2/V1), where n is the number of moles, R is the gas constant, T is the temperature, and V1 and V2 are the initial and final volumes, respectively. This equation is applicable to ideal gases, so it might not work for the states you have given.

Since you have the specific volumes for both states, you can use the equation W = -P(V2-V1) to solve for the work required per unit mass. This equation is valid for any process, including isothermal processes, as long as the pressure remains constant.

As for your method of approximating the work by using a curve fit, it might not give an accurate result as the states are not ideal gases. You can try using a polynomial function, but it would be best to check with your professor or textbook for the appropriate method for solving problems involving non-ideal gases.

I hope this helps in solving your problem. Good luck!
 

1. What is the definition of work in thermodynamics?

In thermodynamics, work is defined as the transfer of energy from one system to another by means of a force acting through a distance.

2. How is work calculated in thermodynamics?

The work required per unit mass in thermodynamics is calculated by multiplying the force applied to the system by the displacement of the system in the direction of the force.

3. What is the unit of work in thermodynamics?

The unit of work in thermodynamics is Joules (J), which is equivalent to the unit of energy.

4. How is the work required per unit mass related to the first law of thermodynamics?

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat transferred to the system minus the work done by the system. Therefore, the work required per unit mass is a part of the energy balance in the first law of thermodynamics.

5. What factors affect the work required per unit mass in thermodynamics?

The work required per unit mass in thermodynamics is influenced by factors such as the force applied, the displacement of the system, and the properties of the system, such as temperature and pressure.

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