Thermodynamics work required per unit mass problem

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SUMMARY

The discussion centers on calculating the work required per unit mass for water compressed isothermally from 100 kPa to 50 MPa at a constant temperature of 20°C. The user initially attempted to apply the ideal gas law but later discovered that the specific volumes indicate non-ideal behavior. The correct approach involves using the equation W = -P(V2-V1) for any process, including isothermal, given the specific volumes. The calculated work is approximately -2.68 kJ/kg, confirming the need for accurate methods when dealing with non-ideal gases.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically isothermal processes
  • Familiarity with the ideal gas law and its limitations
  • Knowledge of specific volume and its significance in thermodynamics
  • Basic calculus for integrating pressure-volume relationships
NEXT STEPS
  • Study the application of the equation W = -P(V2-V1) in various thermodynamic processes
  • Learn about the behavior of non-ideal gases and how to apply real gas equations
  • Explore curve fitting techniques for thermodynamic data analysis
  • Investigate the use of thermodynamic tables for various substances beyond ideal gases
USEFUL FOR

Students in thermodynamics courses, engineers working with fluid mechanics, and anyone involved in the analysis of gas behavior under varying pressure and temperature conditions.

nineeyes
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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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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.
 


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!
 

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