Finding Minimum Work to Compress Water

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SUMMARY

The discussion focuses on calculating the minimum work required to isothermally compress 1 kg of water from an initial pressure of 1 bar and temperature of 120°C to a volume that is one-third of the original. The water is identified as superheated vapor, and properties such as specific volume, internal energy, enthalpy, and entropy are derived from superheated tables. The energy balance equation is applied, with the participant considering neglecting kinetic and potential energy, and exploring the use of TdS equations for heat transfer calculations.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the energy balance equation.
  • Familiarity with superheated steam tables and property calculations.
  • Knowledge of the TdS equation and its application in thermodynamic processes.
  • Basic concepts of closed systems and phase changes in water.
NEXT STEPS
  • Study the application of the TdS equation in thermodynamic processes.
  • Learn how to utilize superheated steam tables for property determination.
  • Research energy balance applications in closed systems for various thermodynamic processes.
  • Explore the implications of neglecting kinetic and potential energy in energy balance calculations.
USEFUL FOR

Students and professionals in thermodynamics, particularly those working on energy balance calculations and phase change analysis in water systems.

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Homework Statement


Find the minimum work to compress 1 kg of water isothermally from p1=1 bar, T1=120C to a volume that is 1/3 the original volume.

Homework Equations


Energy balance Q-W=U+KE+PE

The Attempt at a Solution


So first I found the phase of the water (p<psat for that temp so superheated vapor)
I found the properties (v,u,h,s) for that temperature and pressure in the superheated table.
I can use the specific volume at 1 to find the specific volume at 2 because mass stays constant (closed system) and it's just 1/3 v1. I used that specific volume to determine the state at 2 (in between vf and vg so it's in the saturated liquid vapor phase). I found the quality using tables so I can calculate u, h, s.

I'm mostly confused about the energy balance: I'm pretty sure I can neglect KE and PE. Is there a Q value though? When I went over this problem with the professor he hinted at using TdS equations (Gibbs?) for finding the heat. The Tds equation that I know is Tds=du+Pdv (which is similar to the energy balance since Pdv is the work and du is the the change in internal energy). Do I do W=du+Tds?

Or maybe it is okay to just assume no heat transfer and do W=U and then do m(u2-u1). I'm mostly confused about when you can assume things when doing energy balance.
 
Last edited:
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You know that the temperature is constant right, and for minimum work, Q = ∫TdS, so Q=TΔS. Just get the Δs from your tables and calculate Q.
 

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