Fundamentals of Thermodynamics SENG 205 PDF

Fundamentals of Thermodynamics SENG 205 School of Engineering Sciences University of Ghana Lecturer: Prof. Emmanuel Nyankson Email: [email protected] Teaching Assistant...

Fundamentals of Thermodynamics SENG 205 School of Engineering Sciences University of Ghana Lecturer: Prof. Emmanuel Nyankson Email: [email protected] Teaching Assistant: Credit Hours: Three (3) Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Chapter 3 (Robert DeHoff) Chapters 2, 6 (section 6.6) and 7 (section 7.2, 7.3) (Van Ness) Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Objectives 1. Introduce the laws of thermodynamics with emphasis on the first law. 2. Introduce the concept of internal energy 3. Introduce enthalpy 4. Introduce the concept of reversible processes/entropy generation/dissipation 5. Introduce state function and process variables 6. Introduce extensive and intensive properties Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Laws of Thermodynamics The laws of thermodynamics are highly condensed expressions of a broad body of experimental evidence. The three laws of thermodynamics are empirical: derived from experimental observations of how matter behaves. The laws of thermodynamics are: 1st law: there exists a property of the universe, called its energy, which cannot change no matter what processes occur. 2nd law: there exists a property of the universe, called its entropy, which always changes in the same direction no matter what processes occur. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Laws of Thermodynamics 1st law: there exists a property of the universe, called its energy, which cannot change no matter what processes occur. 2nd law: there exists a property of the universe, called its entropy, which always changes in the same direction no matter what processes occur. 3rd law: there exists a lower limit to the temperature that can be attained by matter, called the absolute zero of temperature (0 K , -273.15), and the entropy of all substances is the same at that temperature. Zeroth law: this law acknowledges that a temperature scale exists for all substances in nature and provides an absolute measure of their tendencies to exchange heat. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson The 1 Law of Thermodynamics st Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st 1st law: there exists a property of the universe, called its energy, which cannot change no matter what processes occur. Energy can be defined both physically and mathematically and can be measured with precision. Three broad categories of energy have been identified in scientific experience: 1. Kinetic energy 2. Potential energy 3. Internal energy The first law requires that: Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed Systems Except for a system that is isolated, changes that occur inside the system are always accompanied by changes in the condition of the matter in the vicinity of the system (surroundings). By the first law the total Surroundings energy of the universe cannot The sum of the changes that change for any process. occur in the system and the surroundings includes all of the changes in the universe System The only way that the internal associated with that process energy of a system can change is by transferring energy across its boundary. What are some of the ways that energy can be transferred across the boundary of a system? Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed System A mathematical statement of the first law is formulated from the statement: the change in internal energy of a system for a process must be equal to the sum of all energy transfers across the boundary of the system during the process Possible Energy Transfer Possible Energy Transfer Q-the quantity of heat Surroundings Q W-the mechanical work flows into the system done on the system by the during the process. force exerted by the System external pressure in the Possible Energy Transfer surroundings. W1-all other kinds of work W1 done on the system during W the process. The First Law of Thermodynamics ΔU = q + w Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st Relationship Between Heat and Work Benjamin Thompson, Count Rumford Cannon Boring Experiment The heat produced during the boring is proportional to the work performed during the boring Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st James Joules The First Law of Thermodynamics ΔU = q + w Same effect if Fundamentals water is placed in thermal of Thermodynamics demic Year Prof Nyankson contact 2022-2023 Aca with a source of heat  Laws of Thermodynamics Energy is added to the fluid as work, but is Joule’s Experiment transferred from the fluid as heat. What happens to this energy between its addition to and transfer from the fluid? It is contained in the fluid in another form called INTERNAL ENERGY. Kinetic energy of translation Kinetic energy of rotation Kinetic energy of internal vibration The First Law of Thermodynamics ΔU = q + w https://www.youtube.com/watch?v=ZOynwQ75pms Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed System The increase in the internal energy of the system during a process is the sum of the transfers across the boundary. Mathematical statement of the first law: For infinitesimal change in the condition of the system: Why The first law applies to any system taken through any Why process. Why Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed System WORKING EXAMPLE 1 When a system is taken from state a to state b along path acb, 100 J of heat flows into the system and the system does 40 J of work. a. How much heat flows into the system along path aeb if the work done by the system is 20 J? b. The system returns from b to a along path bda. If the work done on the system is 30 J, does the system absorb or liberate heat?Fundamentals How much? of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed System WORKING EXAMPLE 1 When a system is taken from state a to state b along path acb, 100 J of heat flows into the system and the system does 40 J of work. Data: Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed System Data: Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed System WORKING EXAMPLE 1 a. How much heat flows into the system along path aeb if the work done by the system is 20 J? Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed System WORKING EXAMPLE 1 a. How much heat flows into the system along path aeb if the work done by the system is 20 J? Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed System WORKING EXAMPLE 1 b. The system returns from b to a along path bda. If the work done on the system is 30 J, does the system absorb or liberate heat? How much? Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1st Law of Thermodynamics: Closed System WORKING EXAMPLE 1 b. The system returns from b to a along path bda. If the work done on the system is 30 J, does the system absorb or liberate heat? How much? Heat is liberated Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Equilibrium Equilibrium is a word denoting a static condition, the absence of change. In thermodynamics, it also means the absence of any tendency toward change on a macroscopic scale. A system is in equilibrium if it is in 1. Thermal equilibrium 2. Mechanical equilibrium 3. Chemical equilibrium Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Equilibrium Equilibrium is a word denoting a static condition, the absence of change. In thermodynamics, it also means the absence of any tendency toward change on a macroscopic scale. A system is in equilibrium if it is in 1. Thermal equilibrium 2. Mechanical equilibrium 3. Chemical equilibrium Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Equilibrium Equilibrium is a word denoting a static condition, the absence of change. In thermodynamics, it also means the absence of any tendency toward change on a macroscopic scale. A system is in equilibrium if it is in 1. Thermal equilibrium 2. Mechanical equilibrium 3. Chemical equilibrium Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Equilibrium Equilibrium is a word denoting a static condition, the absence of change. In thermodynamics, it also means the absence of any tendency toward change on a macroscopic scale. Phase Rule: Phase is a homogenous region of matter. N=number of components, F=degree of freedom, is the number of phases 2=non compositional variables (P,T). Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Equilibrium 𝐹 =2 − 𝜋 +𝑁 Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Working Example How many phase-rule variables must be specified to fix the thermodynamic state of each of the following systems? (a) Liquid water in equilibrium with its vapor. (b) Liquid water in equilibrium with a mixture of water vapor and nitrogen. (c) A three-phase system of a saturated aqueous salt solution at its boiling point with excess salt crystals present. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Working Example How many phase-rule variables must be specified to fix the thermodynamic state of each of the following systems? (a) Liquid water in equilibrium with its vapor. (b) Liquid water in equilibrium with a mixture of water vapor and nitrogen. (c) A three-phase system of a saturated aqueous salt solution at its boiling point with excess salt crystals present. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson Reversible Process The gas pressure is sufficient to balance the weight of the piston and all that it carries. In an equilibrium condition, the system has no tendency to change. Mass m is removed Oscillation Chaotic molecular motion Dissipation of work into internal energy Frictionless Piston (dissipation) Irreversible process. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Reversible Process Reversible Process: its direction can be reversed at any point by an infinitesimal change in external conditions 1. PROCESS 1: The pan and its contents are i ble lifted off the top of the piston. vers  The system experiences a sudden drop in e Ir r the force applied to it. Gas Gas  Process accompanied with dissipation, finite entropy production and permanent b changes in the universe. 2. PROCESS 2: The particles are removed a with a pair of tweezers one particle at a time Re and letting the system come to rest before the ve rsi next particle is removed bl Gas e Gas  No dissipation, no entropy production and no permanent changes. c Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Reversible Chemical Reaction Summary Remarks on Reversible Processes A reversible process: Is frictionless Is never more than differentially removed from equilibrium Traverses a succession of equilibrium states Can be reversed at any point by a differential change in external conditions When reversed, retracts its forward path, and restores the initial state of the system and surroundings. Minimum work input required or maximum work output attainable from a specified process Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Constant Volume and Constant Pressure Processes The energy balance for a homogenous closed system of n moles is: Hence This is the general first law equation for a mechanically reversible, closed system process. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Constant Volume and Constant Pressure Processes Constant Volume Process If the process occurs at constant volume, the work is zero. Thus integration yields Thus for a mechanically reversible, constant volume, closed system process, the heat transferred is equal to the internal energy of the system. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Constant Volume and Constant Pressure Processes Constant Pressure Process For a constant pressure change of state A new thermodynamic property needs to be defined H,U and V are molar volume or unit mass values. Hence Integration yields For a mechanically reversible, constant pressure, closed-system process, the heat transferred equals the enthalpy change of the system. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Constant Volume and Constant Pressure Processes Calculate ΔU and ΔH for 1 kg of water when it is vaporized at the constant temperature of 100°C and the constant pressure of 101.33 kPa. The specific volumes of liquid and vapor water at these conditions are 0.00104 and 1.673 m3·kg−1, respectively. For this change, heat in the amount of 2256.9 kJ is added to the water. Data T = 100°C P = 101.33 kPa Find VL = 0.00104 m3·kg−1 ΔU and ΔH VV = 1.673 m3·kg−1 Q = 2256.9 kJ Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Constant Volume and Constant Pressure Processes Data T = 100°C At constant pressure ΔH = Q = 2256.9 kJ P = 101.33 kPa VL = 0.00104 m3·kg−1 VV = 1.673 m3·kg−1 Q = 2256.9 kJ Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Constant Volume and Constant Pressure Processes Heat capacity: the heat capacity is defined as Two heat capacities can be defined Heat Capacity at Constant Volume Integration yields Hence, Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Constant Volume and Constant Pressure Processes Heat Capacity at Constant Pressure Reversible adiabatic Process Isothermal process Integration yields 𝐶 𝑃 >𝐶 𝑉 ,𝑊h𝑦 ? Hence, For an ideal gas Above room temperature, the temperature variation of Cp follows a relatively simple empirical relation: Fundamentals of Thermodynamics 2022-2023 Academic Year Prof Nyankson Constant Volume and Constant Pressure Processes Reversible adiabatic Process For an ideal gas Fundamentals of Thermodynamics 2022-2023 Academic Year Prof Nyankson The 1 Law of Thermodynamics st 𝛿𝑄=𝑑𝐻 =𝐶 𝑝 𝑑𝑇 𝐶 2 𝐶 𝑝 = 𝐴 + 𝐵𝑇 + + 𝐷 𝑇 𝑇2 𝑑𝑄=𝑑𝑈=𝐶𝑉 𝑑𝑇 Metal C / (J mol-1 K-1) selenium 24.9 gold 25.2 zinc 25.0 titanium 25.1 tungsten 25.0 Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson The 1 Law of Thermodynamics st 𝑐 2 𝐶 𝑝 = 𝑎+𝑏 𝑇 + + 𝑑 𝑇 𝑇2 If CuO is heated from 50 to 150 at constant pressure of 1 atm. Calculate the enthalpy change. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st 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: (a) Cooling at constant pressure followed by heating at constant volume. (b) Heating at constant volume followed by cooling at constant pressure. Calculate the heat and work requirements and ΔU and ΔH of the air for each path. The following heat capacities for air may be assumed independent of temperature: C V = 20.785 and C P = 29.100 J· mol −1 ·K −1 Assume also that air remains a gas for which PV/T is a constant, regardless of the changes it undergoes. At 298.15 K and 1 bar the molar volume of air is 0.02479 m 3· mol−1. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st 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: Data (a) Cooling at constant pressure followed by heating at constant volume Initial Final P1 = 1 bar 𝑷𝑽 Pf = 3 bar =𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 T1 = 298.15 K 𝑻 Tf = 298.15 K V1 = 0.02479 m3/mol Vf (b) Heating at constant volume followed by cooling at constant pressure Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st 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: Data (a) Cooling at constant pressure followed by heating at constant volume Initial Final 𝑷𝑽 P1 = 1 bar =𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 Pf = 3 bar T1 = 298.15 K 𝑻 Tf = 298.15 K V1 = 0.02479 m3/mol Vf Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st WORKING EXAMPLE 2 (a) Cooling at constant pressure followed by heating at constant volume. Initial V T P1 = 1 bar T1 = 298.15 K V1 = 0.02479 m3/mol initial initial V1 T1 final Intermediate P* = P1= 1 bar T* = V* = Vf V* final T* Final P*=P1 Pf Pf P P1 P Pf = 3 bar Tf = 298.15 K Vf =V* Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st WORKING EXAMPLE 2 (a) Cooling at constant pressure followed by heating at constant volume. V T Initial P1 = 1 bar T1 = 298.15 K initial initial V1 T1 V1 = 0.02479 m3/mol final Intermediate P* = P1= 1 bar T* = V* final T* V* = Vf P1 P1 Pf Pf P P Final Pf = 3 bar 𝑃𝑓 𝑉 𝑓 𝑃 ∗3 𝑉∗ ∗ 𝑉𝑓 1∗ 𝑉 ∗3 1 Tf = 298.15 K = ∗ = = ∗ 𝑇 𝑇=99.4 𝐾 𝑇𝑓 𝑇 298.15 𝑇∗ 298.15 ∗ Vf =V* Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st WORKING EXAMPLE 2 (a) Cooling at constant pressure followed by heating at constant volume. V T Initial P1 = 1 bar T1 = 298.15 K initial initial V1 T1 V1 = 0.02479 m3/mol final Intermediate P* = P1= 1 bar T* =99.4 K V* final T* V* = Vf P1 P1 Pf Pf P P Final Pf = 3 bar 𝑃 1𝑉 𝐼 𝑃∗ 1∗ ∗ 𝑉0.02479 1 ∗𝑉 ∗ Tf = 298.15 K = = ∗ 3 99.4𝑉 =0.00826 𝑚 =V f ∗ 𝑇1 𝑇 298.15 Vf =V* Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st WORKING EXAMPLE 2 (a) Cooling at constant pressure followed by heating at constant volume. V Initial P1 = 1 bar T1 = 298.15 K initial V1 V1 = 0.02479 m3/mol 𝑄1−𝑖 ,𝑊 1−𝑖 , ∆ 𝐻 1−𝑖 , ∆𝑈 1−𝑖 Intermediate P* = P1= 1 bar T* =99.4 K V* final V* = P1 Pf P Final Pf = 3 bar Tf = 298.15 K 𝑄𝑖 − 𝑓 ,𝑊 𝑖− 𝑓 ,∆ 𝐻 𝑖− 𝑓 ,∆ 𝑈𝑖 − 𝑓 Vf =V*= Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st WORKING EXAMPLE 2 (a) Cooling at constant pressure followed by heating at constant volume. V Initial P1 = 1 bar T1 = 298.15 K initial V1 V1 = 0.02479 m3/mol 𝑄1−𝑖 ,𝑊 1−𝑖 , ∆ 𝐻 1−𝑖 , ∆𝑈 1−𝑖 Intermediate P* = P1= 1 bar T* =99.4 K V* final V* = P1 Pf P Final Pf = 3 bar Tf = 298.15 K 𝑄𝑖 − 𝑓 ,𝑊 𝑖− 𝑓 ,∆ 𝐻 𝑖− 𝑓 ,∆ 𝑈𝑖 − 𝑓 Vf =V*= Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st WORKING EXAMPLE 2 (a) Cooling at constant pressure followed by heating at constant volume. V Initial P1 = 1 bar T1 = 298.15 K initial V1 V1 = 0.02479 m3/mol 𝑄1−𝑖 ,𝑊 1−𝑖 , ∆ 𝐻 1−𝑖 , ∆𝑈 1−𝑖 Intermediate P* = P1= 1 bar T* =99.4 K V* final V* = P1 Pf P Final Pf = 3 bar Tf = 298.15 K 𝑄𝑖 − 𝑓 ,𝑊 𝑖− 𝑓 ,∆ 𝐻 𝑖− 𝑓 ,∆ 𝑈𝑖 − 𝑓 Vf =V*= Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st WORKING EXAMPLE 2 (a) Cooling at constant pressure followed by heating at constant volume. V Initial P1 = 1 bar T1 = 298.15 K initial V1 V1 = 0.02479 m3/mol 𝑄1−𝑖 ,𝑊 1−𝑖 , ∆ 𝐻 1−𝑖 , ∆𝑈 1−𝑖 Intermediate P* = P1= 1 bar T* =99.4 K V* final V* = P1 Pf P Final Pf = 3 bar Tf = 298.15 K 𝑄𝑖 − 𝑓 ,𝑊 𝑖− 𝑓 ,∆ 𝐻 𝑖− 𝑓 ,∆ 𝑈𝑖 − 𝑓 Vf =V*= Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st WORKING EXAMPLE 2 (To be completed by students) (b) Heating at constant volume followed by cooling at constant pressure. T*,V* Initial V T P1 = 1 bar T1 = 298.15 K V1 = 0.02479 m3/mol initial T*,V* initial V1 T1 final Intermediate P* = Pf= 3 bar T* = V* =V1 = 0.02479 m3/mol final T* Final P1 Pf Pf P P1 P Pf = 3 bar Tf = 298.15 K Vf = Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st PRACTICE PROBLEM a. An ideal gas at 300 K has a volume of 15 L at a pressure of 15 atm. This system undergoes a reversible isothermal expansion to a pressure of 10 atm. Calculate 1. The final volume of the system R = 8.314 J/mol K 2. The work done by the system R = 0.08205 L atm/mol K Cv = 1.5 R 3. The heat entering or leaving the system 4. The change in the internal energy 5. The change in enthalpy b. Repeat the calculation if the process is taken through a reversible adiabatic expansion to a pressure of 10 atm Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st Solution (a) 1. The final volume of the system (22.5 L) 2. The work done by the system (-9244 J) 3. The heat entering or leaving the system (+9244 J) 4. The change in the internal energy ( 0 J) 5. The change in enthalpy (0 J) Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 1 Law of Thermodynamics st Solution (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) Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Mass and Energy Balances for Open Systems  Measure of Flow Open systems are characterize 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: Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Mass and Energy Balances for Open Systems Working Example 3 Liquid n-hexane flows at a rate of in a pipe with inside diameter What are ? What would these quantities be for the same if ? Assume for liquid n-hexane that The molar mass of n-hexane is 86.177 g/mol. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Mass and Energy Balances for Open Systems Working Example 3 Liquid n-hexane flows at a rate of in a pipe with inside diameter What are ? What would these quantities be for the same if ? Assume for liquid n-hexane that The molar mass of n-hexane is 86.177 g/mol. Data 5 cm M = 86.177 g/mol Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Mass and Energy Balances for Open Systems Working Example 3 Liquid n-hexane flows at a rate of in a pipe with inside diameter What are ? What would these quantities be for the same if ? Assume for liquid n-hexane that The molar mass of n-hexane is 86.177 g/mol. Data 2 cm M = 86.177 g/mol Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Mass and Energy Balances for Open Systems Practice Problem In a major human artery with an internal diameter of 5 mm, the flow of blood, averaged over the cardiac cycle, is 5 cm3·s−1. The artery bifurcates (splits) into two identical blood vessels that are each 3 mm in diameter. What are the average velocity and the mass flow rate upstream and downstream of the bifurcation? The density of blood is 1.06 g·cm−3. 3 mm 5 mm 3 mm Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  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 can be written as This equation is called the continuity equation. The control surface is extensible Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Mass Balance for Open Systems 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 When there is a single entrance and a single exit: Hence Also Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  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 with them energy in its internal, potential and kinetic forms. The rate of energy accumulation within the control volume is Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  General Energy Balance The work rate term includes All other work (): shaft work, work due to expansion or contraction of the control volume Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  General Energy Balance Considering the definition of enthalpy; H=U+PV Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  General Energy Balance Which is usually written as For the case where the kinetic and potential energies of the fluid are negligible Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Energy Balances for Steady State Processes Flow processes for which the accumulation term is zero is said to occur at steady state. For one entrance and one exit Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Energy Balances for Steady State Processes Dividing through by the mass flow rate gives For the case where the kinetic and potential energy terms are omitted because they are negligible when compared to others Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Energy Balances for Steady State Processes Working Example 4 Steam flows at a steady state through a converging insulated nozzle, 25 cm long and with inlet diameter of 5 cm. at the nozzle entrance (state 1), the temperature and pressure are 325 C and 700 kPa, and the velocity is 30 m/s. at the nozzle exit (state 2), the steam temperature and pressure are 240 C and 350 kPa. What is the velocity of the steam at the nozzle exit, and what is the exit diameter? Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Energy Balances for Steady State Processes Working Example 4 Steam flows at a steady state through a converging insulated nozzle, 25 cm long and with inlet diameter of 5 cm. at the nozzle entrance (state 1), the temperature and pressure are 325 C and 700 kPa, and the velocity is 30 m/s. at the nozzle exit (state 2), the steam temperature and pressure are 240 C and 350 kPa. What is the velocity of the steam at the nozzle exit, and what is the exit diameter? State 2 State 1 d2 = d1 = 5 cm P2 = 350 kPa P1 = 700 kPa T2 = 240 C T1 = 325 C u2 u1= 30 m/s L = 25 cm 2022-2023 Aca Fundamentals of Thermodynamics demic Year Prof Nyankson Energy Balances for Steady State Processes Working Example 4 State 2 State 1 d2 = d1 = 5 cm P2 = 350 kPa P1 = 700 kPa T2 = 240 C T1 = 325 C u2 u1= 30 m/s L = 25 cm Insulated so Q = 0 is negligible No shaft work, Ws = 0 ( ∆ 𝐻+ 1 2 2 ) 𝑢 + 𝑧𝑔 =𝑄 +𝑊 𝑠 Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Energy Balances for Steady State Processes Working Example 4 State 2 State 1 d2 = d1 = 5 cm P2 = 350 kPa P1 = 700 kPa T2 = 240 C T1 = 325 C u2 u1= 30 m/s L = 25 cm From the steam table, at P1 = 700kPa and T1 = 325 C , H1 = 3112.1 kJ/kg, V1 = 388.61 cm3/g, U1 = 2840.1 kJ/kg From the steam table, at P2 = 350kPa and T2 = 240 C , H2 = 2945.7 kJ/kg, V2 = 667.75 cm3/g, U2 = 2712.2 kJ/kg Read the enthalpy values from steam tables. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Energy Balances for Steady State Processes Working Example 4 State 2 State 1 d2 = d1 = 5 cm P2 = 350 kPa P1 = 700 kPa T2 = 240 C T1 = 325 C u2 u1= 30 m/s L = 25 cm Read the enthalpy values from steam tables. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Energy Balances for Steady State Processes Working Example 4 State 2 State 1 d2 = d1 = 5 cm P2 = 350 kPa P1 = 700 kPa T2 = 240 C T1 = 325 C u2 u1= 30 m/s L = 25 cm Read the enthalpy values from steam tables. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Energy Balances for Steady State Processes Working Example 4 State 2 State 1 d2 = d1 = 5 cm P2 = 350 kPa V1 = 388.61 cm3/g, U1 = P1 = 700 kPa T2 = 240 C 2840.1 kJ/kg T1 = 325 C u2 V2 = 667.75 cm3/g, U2 = u1= 30 m/s 2712.2 kJ/kg L = 25 cm Use the continuity equation (mass balance) Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Tables of Thermodynamic Properties In many instances, thermodynamic properties are tabulated. Check appendix F in Van Ness. Superheated steam originally at P1 and T1 expands through a nozzle to an exhaust pressure P2. Assuming the process is reversible and adiabatic, determine the down stream state of the steam and ΔH for the following condition P1=1000 kPa, t1 = 250 C and P2 = 200 kPa. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson 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. When the stream impinges on a blade attached to rotating shaft. ( 1 2 ˙ 𝑄+ ∆ 𝐻 + 𝑢 + 𝑧𝑔 𝑚= 2 ) ˙ 𝑊 ˙ 𝑠 Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd Processes observed in nature have a natural direction of change Motion picture ran backwards Videotape ran backwards  A wilted flower does not revive to full bloom and then curl into A pebble dropped into a pond a bud.  Time flows in one direction  The 2nd law of thermodynamics distills this aspect of experience and states it succinctly and quantitatively.  The 2nd law determines the direction and extent of a process that brings a system to equilibrium Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd Processes observed in nature have a natural direction of change Motion picture ran backwards Videotape ran backwards  A wilted flower does not revive to full bloom and then curl into A pebble dropped into a pond a bud.  Time flows in one direction  The 2nd law of thermodynamics distills this aspect of experience and states it succinctly and quantitatively.  The 2nd law determines the direction and extent of a process that brings a system to equilibrium Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd The initial state of a system can be; Natural or Its equilibrium state (state of rest) Spontaneous or No equilibrium state Irreversible Process Irreversible processes are accompanied by dissipation or A B degradation Mixing Once equilibrium state is reached, the capacity of the system for doing further work (to cause a change) is exhausted. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd There are two distinct spontaneous processes: Extent of degradation varies Conversion of heat to work from one process to another. Flow of heat down temperature gradient 1. Heat reservoir is at T2, Work is performed and the heat produced Q enters the heat reservoir. 𝑄 2. Heat reservoir at T2 is placed in thermal ∆ 𝑆= 𝑇 contact with another heat reservoir at T1. Q flows from T2 to T1. 3. Heat reservoir is at T1, Work is performed and the heat produced Q enters the heat reservoir. Quantitative measure of degree of irreversibility Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd Conduct the process in such a way that the degree of irreversibility is minimal. The degree of irreversibility is zero (no degradation/dissipation). Reversible Process. If a process is reversible, then the concept of spontaneity is not applicable. If spontaneity is removed, the system will be at equilibrium at all times. The process will pass through continuum of equilibrium states. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd Entropy is a state function, which 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. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd In every volume element of any system and surroundings that may be experiencing a change, at every instant in time, the entropy production is positive. The change in entropy of a system is not restricted to entropy transferred across its boundaries. Production of entropy-the first law is conservative law but the 2nd law is not. The transfer term may be positive or negative depending upon the nature of the process and the flows at the boundary of the system. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd The mathematical statement of the 2nd law: Consider the entropy change experienced by the surroundings to a system during a process. Treating the surroundings as the system: The entropy of the universe: This reduces to Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd The entropy of the universe: This reduces to Why? Thus, while the entropy of a system may increase or decrease during a process, the entropy of the universe, taken as system plus whatever surroundings are involved in producing the changes within the system, can only increase. For an infinitesimal change: Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  The 2 Law of Thermodynamics nd Reversible and Irreversible Process: 1. PROCESS 1: The pan and its contents are i ble lifted off the top of the piston. vers  The system experiences a sudden drop in e Ir r the force applied to it. Gas Gas  Process accompanied with dissipation, finite entropy production and permanent b changes in the universe. 2. PROCESS 2: The particles are removed a with a pair of tweezers one particle at a time Re and letting the system come to rest before the ve rsi next particle is removed bl Gas e Gas  No dissipation, no entropy production and no permanent changes. c Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson Relationship Between Entropy Transfer and Heat Absorbed For a system taken through a reversible process, the heat absorbed by the system during that step is: For any reversible cyclic path, the above expression is zero. is a state function which is defined to be the entropy of the system Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson Combined Statement of the 1 and 2 Laws st nd Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson Combined Statement of the 1 and 2 Laws st nd First Law: Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  3 Law of Thermodynamics rd A system consisting of one mole of carbon and one mole of pure silicon initially at absolute zero is heated from 0 to 1500 K and held at this temperature for Si to react with C and form SiC. The compound is then cooled to absolute zero. Si + C SiC Since entropy is a state function: II However, when individual entropy changes for steps I,11,and III are calculated from experimental T(K) Si + C I III SiC measurements it is found that: IV It can be deduced that 0 Si + C SiC Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  3 Law of Thermodynamics rd A system consisting of one mole of carbon and one mole of pure silicon initially at absolute zero is heated from 0 to 1500 K and held at this temperature for Si to react with C and form SiC. The compound is then cooled to absolute zero. Si + C SiC  The entropy change of the compound SiC is the II same for the pure Si and C at 0 K.  These empirical observations have established the third law of thermodynamics: T(K) Si + C I III SiC There exists a lower limit to the temperature that can be attained by matter, called the absolute zero of IV temperature, and the entropy of all substances is the same at that temperature. 0 Si + C SiC Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson  Example Alumina can be formed from aluminum and oxygen as described by the reaction below: Calculate the change in entropy for this reaction at 298 K. Fundamentals of Thermodynamics 2022-2023 Aca demic Year Prof Nyankson