Thermo Final

Isothermal Expansion

• Definition: Isothermal expansion refers to the process in which a gas expands at a constant temperature.
• Given Parameters:
    - Initial temperature ( 1) = Initial temperature ( 2) = 88 °C = 361 K
    - Initial pressure (p1) = 3 bar
    - Final pressure (p2) = 1 bar

Internal Energy and Enthalpy Changes

• Ideal Gas Assumption: The gas (dicadine, in this context) is treated as an ideal gas during the isothermal process.

• Key Equations:
    1. Change in Internal Energy, ΔU = 0
    2. Change in Enthalpy, ΔH = 0

• Reasoning: For an ideal gas, internal energy and enthalpy depend only on temperature. Since the temperature is constant, both changes are zero.

• First Law of Thermodynamics:
    - The first law states:
  $ext{ΔU} = Q + W$
    - Since ΔU = 0, it follows that:
  $Q = -W$

Work Done in Isothermal Expansion

• Expression for Work (W): Under the assumption of reversible processes, the work done is given by:
  $W = - ext{P}d ext{V}$

• Ideal Gas Law Substitution: The pressure (P) can be expressed as:
  $ext{P} = \frac{nRT}{V}$
    - Rearranging gives:
  $V = \frac{nRT}{P}$

• Integration Set Up: To find the work done, we convert the work done expression:
  $W = \frac{nRT}{P}$
    where dV becomes:
  $dV = -\frac{RT}{P^2}dP$

Integrating Work Done Expression

• Integration Steps:
    - Substitute the expression for dV into the work equation:
  
  $W = -\frac{nR}{P} imes \frac{T}{P^2}dP$
  
    - Resulting in:
  $W = nRT imes ext{ln}\frac{P2}{P1}$

Calculation of Work Done

• Substitution of Values:
    - Using R = 8.314 J/(mol∙K)
    - Temperature T = 361 K
    - Pressures p1 = 3 bar, p2 = 1 bar (convert to same units if needed)
    - Calculate Work (W):
    -
  $W = n imes 8.314 imes 361 imes ext{ln}\frac{1}{3}$
  
      - For n = 1 mole,
  $W = -3297.3 ext{ J/mole}$

• Result Summary:
    - Work Done (W) = -3297.3 Joules for 1 mole
    - Heat Transfer (Q),
  $Q = -W = +3297.3 Joules$

Key Takeaways

• Understanding Reversible Processes: Important to note that this calculation assumes the process is reversible, which simplifies the work expression.
• Real World Implications: Isothermal processes are significant in understanding heat engines and refrigeration cycles where temperature control is critical.

Adiabatic Process in Steam Turbine

• Definition: An adiabatic process allows for gas expansion or compression without heat transfer.
• Problem Context: Hardware is a steam-powered turbine; essential for finding real work, ideal work, and efficiency.

Key Assumptions and Concepts

• Assumptions: Steady-state condition, neglect KE and PE changes, and operates adiabatically.
  
• Energy Balance Equation:
  $W = m imes (H2 - H1)$
    where H1 and H2 are enthalpies at inlet and outlet conditions respectively.

Conditions Provided

• Input Conditions:
    - Inlet: 600 °C and 8 bar
    - Outlet: 400 °C and 1 bar
    - Enthalpy Values from Steam Tables:
    - H1 (600 °C, 8 bar) = 3700.1 kJ/kg
    - H2 (400 °C, 1 bar) = 3278.6 kJ/kg

Calculation of Real Work

• Calculation:
  $W_{ ext{real}} = H1 - H2 = 3700.1 - 3278.6 = -421.5 ext{ kJ/kg}$

Ideal Work Calculation

• Conditions for Ideal Work: For a reversible and adiabatic (isentropic) process, entropy must be constant.
    - Retrieve entropy values from the steam table:
      - S1 = S2 for ideal case.
      - Assuming ideal behavior allows calculation of conditions where this is true.

Process and Calculations

• Find corresponding temperature and enthalpy values from steam tables for pressure equal to 1 bar based on the equality of entropies.
• If calculations yield different enthalpy outputs or require interpolation or extrapolation from provided data, ensure reflective of physical parameters.

Final Assessment for Efficiency

• Efficiency Calculation:
     ext{Efficiency} = rac{W_{ ext{real}}}{W_{ ext{ideal}}} imes 100 ext{%}

Flash Calculation in Phase Equilibrium Problem

• Scenario Context: Equimolar mixture of methanol and butanol undergoing a flash calculation for separation.
• TXY Diagram: Understanding and interpreting this plot for equilibrium conditions.

Step-by-Step Process

1. Determine Liquid and Vapor Compositions:
      - Using intersection points on the TXY diagram (draw tie lines to identify compositions).
      - Example Liquid Composition:
   $X_{ ext{methanol}} = 0.3, ext{ thereby leading to } X_{ ext{butanol}} = 0.7$
      - Example Vapor Composition:
   $Y_{ ext{methanol}} = 0.78, ext{ thus } Y_{ ext{butanol}} = 0.22$

2. Molar Balances to Determine Flow Rates:
      - Establish molar balance equations, usable mix of species and overall balances.
      - Species Balance Example for Methanol:
       - $F imes Z = X_m imes L + Y_m imes V$
      - Overall balance ensuring F = L + V
      - Solve for vapor (V) and liquid (L) stream rates.

3. Heat Duty Calculation for Phase Equilibrium:
      - Using basic energy balance equation:
      - Calculate heat required for phase change and adjustments (depends on the molar enthalpy contributions from both liquid and vapor streams).
      - Be aware of signs of heat duty (positive for heating tasks).

4. Assumptions and Real-life Implications:
      - Ideal systems, model behavior, and applicability in large-scale separation operations.
   

Conclusion

• Final Review and Evaluation: Adherence to fundamental principles such as conservation laws, and reliance on validated empirical data from tables or charts when needed.
• Identification of Areas for Constant Review: Practicing calculations under varying assumptions or conditions can significantly enhance proficiency in thermodynamic analysis.

Summary and Final Insights

• Ensure understanding of each concept rather than rote memorization.
• Engage actively in peer discussions or seek clarification on difficult concepts.
• Regular revision can solidify understanding of thermodynamic processes for future applications in engineering and chemical systems analysis.