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• Fundamentals of Thermodynamics is the subject of SENG 205, taught at the School of Engineering Sciences, University of Ghana.
• The lecturer is Professor Emmanuel Nyankson, reachable at [email protected].
• The course carries a credit of 3 hours.
• Chapter 3 of Robert DeHoff, Chapters 2, 6 (section 6.6) and 7 (section 7.2, 7.3) of Van Ness, are required readings for the course.

Objectives

• Introduce thermodynamic laws, especially the first law.
• Introduce internal energy.
• Introduce Enthalpy.
• Introduce reversible processes, entropy generation, and dissipation.
• Introduce state function and process variables.
• Introduce extensive and intensive properties.

Laws of Thermodynamics

• The laws of thermodynamics are condensed expressions of broad experimental evidence.
• The three laws of thermodynamics are empirical, derived from observations of matter.
• 1st Law: Energy is a property of the universe that cannot change, regardless of processes.
• 2nd Law: Entropy is a property of the universe that always changes in the same direction.
• 3rd Law: There is a lower temperature limit, absolute zero (0 K, -273.15), where all substances have the same entropy.
• Zeroth Law: A temperature scale exists for all substances, providing an absolute measure of their heat exchange tendencies.

1st Law of Thermodynamics

• Energy can be defined and measured precisely, both physically and mathematically.
• Kinetic, potential, and internal energy are three categories of energy.
• The first law requires that for a system that is isolated, changes inside the system are always be changes in surrounding matter.
• The sum of changes in a system and its surroundings includes all universe changes associated with a process.
• The total energy of the universe remains constant in any process.
• The only way a system's internal energy changes is by transferring energy across its boundary.

1st Law of Thermodynamics for Closed Systems

• Change in internal energy of a system equals the sum of all energy transfers across its boundary during the process.
• Possible energy transfers include heat (Q) flowing into system and work (W) done on system.
• The First Law of Thermodynamics is ΔU = q + w, where,
  • q is the quantity of heat flows into the system
  • w is mechanical work done on the system

Heat and Work

• Heat produced during boring is proportional to work performed during the same boring.
• When energy is added to fluid as work, it is transferred from the fluid as heat
• Energy is stored in a fluid as internal energy.
• Internal energy is comprised of kinetic energy of translation, kinetic energy of rotation, and kinetic energy of internal vibration

First Law Application

• The increase in the internal energy of the system during a process is the sum of energy transfers across the boundary.
• The first law applies to any system taken through any process,
• To calculate heat flow into the system along path "aeb" when work done by system is 20 J.
• To determine if a system absorbes or liberates heat and how much when the system returns from "b to a" along path "bda" doing 30 J of work.
• If the system returns from b to a along path bda, it liberates heat.

Equilibrium

• Defined as a static condition with no change.
• In thermodynamics, it means no tendency for change on a macroscopic scale.
• A system is in equilibrium if it is in thermal, mechanical, and chemical equilibrium.

Phase Rule

• Phase - a homogenous region of matter.
• N is the number of components.
• F is the degree of freedom.
• π is the number of phases.
• 2 represents non-compositional variables (P, T).
• The formula for equilibrium is F = 2 - π + N.

Reversible Process

• Involves a gas pressure sufficient to balance the piston's weight.
• In an equilibrium condition, the system has no tendency to change.
• Removing mass may cause oscillation, chaotic molecular motion, and dissipation work into internal energy.
• They involve frictionless pistons with dissipation and irreversible processes.

Reversible Chemical Reaction

• A reversible process is frictionless
• It is never more than differentially removed from equilibrium.
• It traverses a succession of equilibrium states and can be reversed.
• When reversed, it restores the system's initial state and surroundings.
• Reversible reactions need minimum work input or maximum work output

Constant Volume and Constant Pressure Process

• The energy balance for a homogenous closed system of n moles is required,
• For a mechanically reversible, constant volume, closed system process, the heat transferred equals the system's internal energy.
• For constant pressure change of state, a new thermodynamic property needs to be defined.
• H, U, and are V molar volume or unit mass values.
• For a mechanically reversible, constant pressure, closed-system process, the heat transferred equals the enthalpy change of the system.

Calculating AU and AH

• To calculate AU and AH for 1 kg of water vaporized at a constant temperature of 100°C and a constant pressure of 101.33 kPa.
• Specific volumes of liquid and vapor water must be known under those conditions.
• Under this process, heat in the amount of 2256.9 kJ is added to the water.
  • T = 100°C
  • P = 101.33 kPa
  • V₁ = 0.00104 m³·kg-1
  • Vv = 1.673 m³·kg-1
  • Q = 2256.9 kJ
• At constant pressure ∆H = Q = 2256.9 kJ

Heat Capacity

• The heat capacity is defined as two heat capacities that can be defined.
  • Heat Capacity at Constant Volume
  • Heat Capacity at Constant Pressure
  • Heat Capacity > Constant Volume (Cp>Cv, Why?)

First Law Equations

• Key Equations:
  • 6Q=dH=CdT
  • C₁=A+BT+___+DT2
  • dQ=dU=CvdT

Working Example 2

• Air at 1 bar and 298.15 K is compressed to 3 bar and 298.15 K by two different closed-system mechanically reversible processes.
• The two processes are:
  • Cooling at constant pressure followed by heating at constant volume.
  • Heating at constant volume followed by cooling at constant pressure.
• When solving, calculate the heat and work requirements and AU and AH of the air for each path. The following heat capacities for air may independent of temperature with the following values:
  • Cᵥ = 20.785
  • Cₚ = 29.100 J·mol −1 ·K −1
• When solving, assume air remains a gas for which PV/T is a constant, regardless of the changes it undergoes.
• Also, the molar volume of air is is 0.02479 m³-mol-1 at 298.15 K and 1 bar.

Practice Problem

• An ideal gas at 300 K has a volume of 15 L at a pressure of 15 atm.
• The gas undergoes a reversible isothermal expansion to a pressure of 10 atm, calculate the following:
  • Final volume of the system
  • Work done by the system
  • Heat entering or leaving the system
  • The change in the internal energy
  • The change in enthalpy
• Values to use
  • R = 8.314 J/mol K
  • R = 0.08205 L atm/mol K
  • Cᵥ = 1.5 R
• The same can be performed through a reversible adiabatic expansion to a pressure of 10 atm.

Solutions

• For the isothermal expansion version (a.):
  1. The final volume of the system is (22.5 L)
  2. The work done by the system is (-9244 J)
  3. The heat entering or leaving the system is (+9244 J)
  4. The change in the internal energy is (0 J)
  5. The change in enthalpy is (0 J)
• For the adiabatic expansion version (b.):
  1. The final volume of the system (19.13 L)
  2. The work done by the system (-5130 J)
  3. The heat entering or leaving the system (0 J)
  4. The change in the internal energy (-5130J)
  5. The change in enthalpy (-8529.9 J)

Mass and Energy Balances for Open Systems

• Open systems are characterized by flowing streams, for which there are four common measures of flow:
  • Mass flowrate.
  • Molar flow rate.
  • Volumetric flow rate.
  • Velocity.
• The measures of flow are interrelated.
• Liquid n-hexane flows at a rate of in a pipe with inside diameter.
• Molar mass of n-hexane is 86.177 g/mol.

Practice Problem

• A major human artery has a internal diameter of 5 mm and a flow of blood averaged over the cardiac cycle, is 5 cm³.s-¹.
• The artery then splits into two identical blood vessels that are each 3 mm in diameter.
• Determine the following:
  • Average velocity
  • The mass flow rate upstream
  • The mass flow rate downstream of the bifurcation
• The density of blood is 1.06 g.cm-3.

Mass Balance For Open Systems

• The rate of change of mass within the control volume equals the net rate of flow of mass into the control volume.
• This equation is called the continuity equation.
• The control surface is extensible.
• For a steady state process
• The condition within the control volume does not change with time.
• The accumulation terms is zero
• The continuity equation reduces to zero if, there is a single entrance and a single exit

General Energy Balance

• The rate of change of energy within a control volume equals the net rate of energy transfer into the control volume.
• Streams flowing into and out of the control volume have associated energy in internal, potential and kinetic forms.
• The rate of energy accumulation within the control volume is
• All other work () includes shaft work and work due to expansion or contraction of the control volume.
• Considering the definition of enthalpy requires H=U+PV

Energy Balance

• Flow processes when the accumulation term is zero is a steady state.

Superheated Steam Parameters

• Superheated steam originally at P1 and T1 expands through a nozzle to an exhaust pressure P2. Given these paramters:
  • P1=1000 kPa
  • t1 = 250 C
  • P2 = 200 kPa
• Assuming the process is reversible and adiabatic, need the downstream of the steam and ΔH.

Turbines (Expanders)

• The expansion of a gas through a nozzle produce a high-velocity stream in a process that converts internal energy into kinetic energy.
• The kinetic energy is in turn converted to shaft work, and the stream impinges on a rotating shaft.
• The heat flow equals Q, the shaft work equals Wₛ and changes equal
• Steam turbines follow three formula:

The Second Law of Thermodynamics

• Processes observed in nature have a natural direction of change.
• A wilted flower cannot revive to full bloom and then curl into a bud.
• The 2nd law determines the direction and extent of a process that brings a system to equilibrium.
• The second law means time flows in one direction
• After equilibrium, the capacity of system to perform further work is exhausted.
• Entropy: a state function, when summed up for both system and surroundings for any process, always changes in the same direction.
• In every volume element of any system and surroundings that may be experiencing change, at every instant in time, the entropy production:
  • Is positive.
  • Is not restricted to entropy transferred across its boundaries.
• The amount of degradation can vary from processes.
• Measure the degrees of irreversibility

Reversible Processes

• Conduct the process in such a way that the degree of irreversibility is minimal,
• The degree of irreversibility is zero (no degradation/dissipation)
• If a process is reversible, then concept of spontaneity doesn't apply
• If spontaneity is removed, the system will be at equilibrium at all times
• the process will pass through continuum of equilibrium states

Third Law of Thermodynamics

• Consider a system of one mole of carbon and one mole of pure silicon initially at absolute zero.
• It is heated from zero to 1500 kelvin and held, allowing the materials to form silicon carbide.
• Since entropy is a state function calculations determine that the entropy change of any process must be path indepedent.
• The entropy change of the silicon carbide is the same for the pure silicon and the carbon at zero kelvin.
• Empirical observations lead to:
  • The third law that says there exists a lower limit to the temperature that can be attained by all matter
  • The entropy of all substances at the temperature that is at that limit must be at the same value.