Thermodynamics: Change in specific enthelpy

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SUMMARY

The change in specific enthalpy for saturated water vapor compressing from 100 bar to 150 bar at constant temperature can be calculated using the enthalpy values from steam tables. The initial approach using the equation delta h = delta e + p * delta v was incorrect due to miscalculating delta e and delta v. The correct method involves directly finding the enthalpy values at both pressures and taking their difference, resulting in a change of specific enthalpy of 190049 kJ/kg.

PREREQUISITES
  • Understanding of thermodynamic properties of water vapor
  • Familiarity with steam tables for enthalpy values
  • Knowledge of the concept of specific enthalpy
  • Basic principles of thermodynamics, particularly the first law
NEXT STEPS
  • Research how to read and interpret steam tables for various substances
  • Study the first law of thermodynamics and its applications in real-world scenarios
  • Learn about the properties of saturated and superheated steam
  • Explore the implications of pressure changes on enthalpy in thermodynamic systems
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Students and professionals in thermodynamics, mechanical engineers, and anyone involved in the study of phase changes and energy transfer in fluids.

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Homework Statement



What is the change in specific enthalpy if saturated water vapor at 100 bar compresses to a pressure of 150 bar? The temperature is constant.

Homework Equations



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The Attempt at a Solution



I used this equation to solve the problem: delta h = delta e + p * delta v
I found the value of delta e and delta v from steam tables. The answer I'm getting is 190049 kj/kg which is incorrect. Any help would be greatly appreciated, TIA!
 
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I got it. Had to lookup values of enthalpy at 100 bar and 150 bar then take the difference of them.
 

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