Second law thermodynamics FST 403

Second Law of Thermodynamics: Entropy

1. Introduction to the Second Law of Thermodynamics

• The second law of thermodynamics introduces the concept of entropy.
• Entropy is defined as a measure of disorder within a system:
  • The higher the disorder, the higher the entropy.
• Spontaneous Processes and Entropy:
  • Disorder occurs spontaneously, resulting in an increase in entropy.

2. States of Water and Entropy Change

• The entropy of a system increases as it transitions between states:
  • Ice (most ordered) → Liquid Water → Vapor (most disordered).
• As disorder increases from ice to vapor, the entropy of the system also increases.

3. Expression of the Second Law

• The second law may be mathematically expressed in terms of entropy:
  • 
    \Delta S_{tot} > 0
  • where ΔStot represents the total entropy change of the system and the surroundings.

3.1. Processes Classification

• Thermodynamically irreversible processes are spontaneous and are characterized by an increase in entropy.
• A reversible process is defined by:
  • 
    $\Delta S_{tot} = 0$
  • The system may change states during the process but returns to its initial state upon completion of the cycle.

4. Mathematical Formulation of Entropy Change

• The entropy change differential can be expressed as:
  • 
    $dS = \frac{dQ_{rev}}{T}$
  • where:
  • Qrev = Change in heat in a reversible process.
  • T = Absolute temperature at which the process occurs.

4.1. Closed System with Constant Pressure Process

• For a closed system undergoing a reversible and constant pressure process, we have:
  • 
    $dS = \frac{dQ<em>{rev}}{T} = \frac{dH</em>{rev}}{T}$
  • Rearranging this yields:
  • 
    $dH<em>{rev} = dQ</em>{rev} = TdS$

4.2. Measurable Change Between Two States

• For a measurable change between states 1 and 2, the equation can be represented as:
  • 
    $\Delta S = \int<em>{1}^{2} \frac{dQ</em>{rev}}{T}$

5. Entropy Change as a Function of Temperature

• The entropy change can be calculated as a function of temperature with the following equations:
  • 
    $\Delta S = C<em>p \ln \frac{T</em>2}{T_1}$
  • or
  • 
    $\Delta S = C<em>v \ln \frac{T</em>2}{T_1}$

6. Worked Example: Cooling Orange Juice

6.1. Problem Statement

• Cooling orange juice at a flow rate of 2000 kg·h⁻¹ from an initial temperature of 40°C using countercurrently chilled water entering at 15°C.
  • The temperature of approach at both ends is 10°C.
  • Objective: Find the entropy change of the juice, water, and total entropy change of the system.

6.2. Solution Steps

1. Define Outlet Temperatures:

   • Let the outlet temperatures of the hot fluid (orange juice) and cold fluid (chilled water) be defined as t2 and t4 respectively:
   • From the approach condition:
     • 
       40 - t4 = 10 ightarrow t4 = 30°C
     • 
       t2 - 15 = 10 ightarrow t2 = 25°C

2. Mass Flow Rate Calculation:

   • Designate the mass flow rate of chilled water as w kg·h⁻¹.

   - Using heat balance (no heat loss to surroundings):

   $2000 imes 3.73 (40 - 25) = w imes 4.2 (30 - 15)$

   - Solving for w gives:

   $w = 1776.19 ext{ kg·h}^{-1}$

3. Entropy Change Calculations:

   • The process is executed at constant pressure. The entropy change for each component is calculated as follows:
   • Entropy Change of Orange Juice:
     • 
       $\Delta S_{juice} = 2000 imes 3.73 \ln \frac{298}{313} = -366.36 ext{ kJ·h}^{-1}·K^{-1}$
   • Entropy Change of Chilled Water:
     • 
       $\Delta S_{water} = 1776.19 imes 4.2 \ln \frac{303}{288} = 378.76 ext{ kJ·h}^{-1}·K^{-1}$

4. Total Entropy Change Calculation:

   • 
     $Total \Delta S = -366.36 + 378.76 = 12.4 ext{ kJ·h}^{-1}·K^{-1}$

7. Conclusion

• The second law's relevance is emphasized, highlighting the tendency towards increasing entropy in natural processes, as illustrated in the cooling of orange juice using chilled water in thermal exchange systems.