How Does Adding Heat Affect Pressure in a Sealed Steam Container?

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SUMMARY

The discussion focuses on the thermodynamic analysis of a sealed steam container subjected to heat addition, specifically determining the heat added, work performed, change in internal energy, and final pressure. Initial conditions include a temperature of 200°C, a mass of 10 kg, and a volume of 1 m³, with corresponding pressures of 1554.9 KPa and 16529 KPa at final conditions of 350°C. The work done is zero due to the closed system, and the change in internal energy is calculated to be 1759 KJ. The final pressure can be determined using the ideal gas equation.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the first law of thermodynamics.
  • Familiarity with steam tables and properties of saturated vapor systems.
  • Knowledge of the ideal gas law and its application in closed systems.
  • Ability to perform calculations involving internal energy changes in thermodynamic systems.
NEXT STEPS
  • Study the first law of thermodynamics in detail, focusing on closed systems.
  • Learn how to utilize steam tables for various thermodynamic properties.
  • Explore the ideal gas law and its implications for pressure and temperature changes.
  • Investigate the relationship between heat addition and pressure changes in sealed containers.
USEFUL FOR

This discussion is beneficial for students and professionals in mechanical engineering, particularly those studying thermodynamics, as well as anyone involved in the design and analysis of steam systems.

TheBelgiumWaff
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Homework Statement



Given: A sealed, rigid container (cannot change shape or size) containing steam at the shown conditions has heat added until it reaches the final shown state.

Determine: a) heat added, b) work performed, c) change in internal energy of the system, and d) final system pressure

Homework Equations



Q = W + Δu

Also given: T1= 200 deg C, mass = 10kg, volume = 1m^3 (for initial conditions)
T2 = 350 C (for final conditions)

The pressures I found in the thermo tables for the corresponding temperatures are:
P1 = 1554.9 KPa
P2 = 16529 KPa

The Attempt at a Solution



What I know is it's a saturated vapor system since it stated steam in the problem statement and closed system so no work is done.

From that I was able to answer b) W = 0 and c) Q = 0 + m(u1-u2) --> 10Kg(2594.2-2418.3) KJ/kg = 1759 KJ of internal energy. The u's I got from thermo tables.

The two I'm having a hard time with is a) the heat added to the system and d) final system pressure.
 
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First of all, the "conditions" are not mentioned. But we can proceed nevertheless.

The OP answered (b) and (c). The answer of (a) is but the answer of (c). The answer to (d) can be obtained using the ideal gas equation.
 

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