Thermodynamics closed system - first law

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SUMMARY

The discussion centers on calculating the total work done in a closed system involving 1 kg of water transitioning from a saturated mixture at 90 degrees Celsius to a superheated vapor at 250 degrees Celsius and 800 kPa. The first law of thermodynamics is applied, specifically the equation Q - W = ΔU. The initial state properties include H1 = 605.29 kJ/kg and U1 = 588.67 kJ/kg, while the final state properties are h2 = 2950.4 kJ/kg and u2 = 2715.9 kJ/kg. The correct formula for work done is W = m(h2 - h1), which incorporates enthalpy and internal energy.

PREREQUISITES
  • Understanding of the first law of thermodynamics for closed systems
  • Knowledge of thermodynamic properties of water, including enthalpy and internal energy
  • Familiarity with state changes in thermodynamics, specifically from saturated to superheated states
  • Ability to perform calculations involving thermodynamic equations
NEXT STEPS
  • Study the application of the first law of thermodynamics in closed systems
  • Learn about the properties of water at various temperatures and pressures
  • Explore the concept of enthalpy and its relationship with internal energy
  • Practice problems involving phase changes and work calculations in thermodynamic systems
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Students and professionals in mechanical engineering, thermodynamics enthusiasts, and anyone involved in energy systems analysis will benefit from this discussion.

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Homework Statement



1kg of water that is initially at 90 degrees celsius with a quality of 10% occupies a spring loaded piston cylinder device. The device is now heated until the pressure rises to 800kPa and the temperature is 250 degrees c
Determine the total work done during this process, in KJ


Homework Equations



The first law for closed systems
Q - W = ΔU
(h2-h1 - u2 - u1) = ΔU

The Attempt at a Solution



State 1

Saturated mixture (90 degrees)
x = 0.1
H1 = 605.29 KJ/kg
U1 = 588.67 KJ/kg

State 2 - super heated vapour

h2 = 2950.4 KJ/kg
u2 = 2715.9 KJ/kg

therefore,

ΔU = (2950.4 - 605.29) - (2715.9 - 588.67) = 217.88 KJ

which is incorrect, can you assist me with this question?
 
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You would actually just need to apply

W = m(h2-h1)

remember that enthalpy h includes internal energy u.
 

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