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Introduction

• Doctor Ojo discussing thermodynamics, focusing on the first law and related principles.
• Problem presented involving an ICE engine with pressures and specific volume.

Problem Statement

• Initial conditions:
  • Pressure: $P_1 = 420 ext{ kN/m}^2$
  • Specific Volume: $V_1 = 0.14 ext{ m}^3/ ext{kg}$
• Final conditions:
  • Pressure: $P_2 = 35 ext{ kN/m}^2$
• Relationship: $PV^2 = C$

Calculation of Work Done

• Finding the work done by the fluid on the piston:

  • The equation $PV^2 = C$ relates initial and final states.
  • Initial state: $C = P_1 V_1^2$
  • Substitute values:
    • $C = 420 imes 10^3 imes (0.14)^2$
  • Simplification gives:
    • C
      ightarrow 8.59 ext{ N m/kg}

• Calculating final specific volume, $V_2$:

  • Formula for $V_2$:

  $V_2 = ext{sqrt}\frac{C}{P}$

  • Substitute $C$ and $P_2$:

  $V_2 = ext{sqrt}\frac{8.588}{35}$

  • Result:
  • V_2
    ightarrow 0.495 ext{ m}^3/ ext{kg}

• Work done (W):

  • Formula for work done:

  $W = ext{integral from } V_1 ext{ to } V_2 ext{ of } P ext{ dV}$

• Substitute $P$:

  $P = \frac{C}{V^2}$

• Work calculation:

  • $W = C imes ext{integral}\frac{1}{V^2}dV$

• Evaluate integral:

  • $W = C imes \bigg[-\frac{1}{V}\bigg]_{V_1}^{V_2}$

• Applying limits:

  • $W = 8.588 imes 1000 imes \bigg(-\frac{1}{0.143} + \frac{1}{0.495}\bigg)$

• Final answer for work done:

  • W
    ightarrow 42.71 ext{ kJ/kg}

First Law of Thermodynamics

• Definition: Related to the law of conservation of energy, stating:
  • Energy cannot be created or destroyed.
  • Energy can change forms.
• Equation: Change in internal energy () in a closed system can be expressed as:
  • $ext{ΔU} = Q - W$
  • Where:
  • $Q$: energy added to the system
  • $W$: work done by the system
• Analogy with Pressure Pot:
  • Energy in water due to intermolecular interactions.
  • Heating leads to increased pressure, allowing the release of pressure (e.g., whistle).

Work Done and Heat Flow

• Heat and Work Flow Behavior:
  • Work on the system from surroundings is negative.
  • Work done by the system on surroundings is positive.
  • Heat flow into the system is positive; heat flow out of the system is negative.
• Equilibrium State Concept:
  • An equilibrium state has an internal energy state variable.
  • Energy transition between equilibrium states involves heat transfer and work.

Process Terminologies

• Types of Processes:
  • Adiabatic Process: No heat transfer.
  • $Q = 0$, hence $ext{ΔU} = -W$ (positive change in internal energy).
  • Isothermal Process: Constant temperature.
  • $ext{ΔU} = 0$, therefore $Q = W$.
  • Isobaric Process: Constant pressure.
  • $W = P imes ext{ΔV}$.
  • Isochoric Process: Constant volume.
  • No work done $W = 0$, hence $ext{ΔU} = Q$.

Polytropic Process

• Definition: Involves various states characterized by heat capacities and indices.
• Example of Work Calculation:
  • Given: 500 J heat added to two moles of monatomic gas at 500 K.
  • Work done: 750 J
  • Use internal energy relationships for calculating final temperature.

Flow Processes

• Types of Flow:
  • Steady Flow: Parameters (velocity, pressure, density) are constant over time.
  • Unsteady Flow: Fluctuating parameters over time.
• Steady Flow Energy Equation (SFEE):
  • Total energy per unit mass:
    $PV = U + KE + PE$
  • Where KE: kinetic energy, PE: potential energy due to height.

Energy Change Equation

• Formulation for energy change:
  • $ext{ΔE} = P_2 V_2 + \bigg(U_2 + \frac{C_2^2}{2} + g z_2\bigg) - P_1 V_1 - \bigg(U_1 + \frac{C_1^2}{2} + g z_1\bigg)$
• Net Heat and Work Relation:
  • Energy change relates to heat supplied minus work done:
  • $Q - W = H_2 - H_1 + ext{(KE and PE contributions)}$

Mass Flow Rate

• Steady Flow Condition:
• Mass flow rate at inlet equals mass flow rate at outlet:
  • ext{ṁ} =
    ho_1 V_1 A_1 =
    ho_2 V_2 A_2 where:
  • 
    ho: Density,
  • $V$: Velocity,
  • $A$: Cross-sectional area

Pure Substances

• Definition: Substances with homogenous and invariable chemical composition.
• Characteristics:
  • Example substances: Nitrogen, Oxygen, Water (H2O), etc.
  • Phases of substances: solid (ice), liquid (water), gaseous (steam).
  • Properties change with phase transitions influenced by pressure and temperature.

Phase Change Terminologies

• Transitions:
  • Melting: solid to liquid
  • Evaporation: liquid to gas
  • Deposition: gas to solid
  • Sublimation: solid to gas
  • Triple Point: simultaneous presence of all phases.
• Steam Tables:
  • Capture intrinsic properties (pressure, volume, temperature).
• Boiling Point:
  • Water boils at 100°C at 1 atmosphere.

STEAM Table Properties

• Key Variables:
  • Saturated Pressure, Saturated Temperature, Specific Volume, Internal Energy, Enthalpy, Entropy, etc.
• Conversion of units (bar to Pascal):
  • $1 ext{ bar} = 10^5 ext{ N/m}^2$

Interpolation in Steam Tables

• Interpolation Concept:
• Used to derive unknown values between two known values:
  • $\frac{T_{ ext{unknown}} - T_{ ext{low}}}{T_{ ext{high}} - T_{ ext{low}}} = \frac{H_{ ext{unknown}} - H_{ ext{low}}}{H_{ ext{high}} - H_{ ext{low}}}$
• Example: Finding enthalpy at 31°C using known values at 30°C and 32°C.

Example Problem on Saturated System

• Case: Rigid tank with 10 kg of water at 90°C.
  • 8 kg in liquid state, 2 kg in vapor state.
• Steps to find:
  1. Determine tank pressure using steam table for 90°C; pressure ≈ 0.011 bar.
  2. Convert to SI units: $0.011 imes 10^5 ext{ N/m}^2$.
  3. Calculate total volume using mass and specific volumes of liquid and vapor.
  • Use dryness fraction for calculations:
  • $X = \frac{ ext{mass vapor}}{ ext{total mass}}$.

Conclusion

• These principles form the foundation of thermodynamics and its applications in systems analysis, understanding energy transfer, and process dynamics in engineering.
• The material serves to enhance comprehension of how various thermodynamic laws apply to real-world scenarios in fluid mechanics and thermodynamics.