INTRODUCTION ON THERMODYNAMICS Basic Concepts Engr. Ann Lisette Geron Learning Objectives 1. To learn the concept of thermodynamics. 2. To calculate specific quantities utilized widely in thermodynamics. 3. To learn the variables that is being quantified and evaluated for most of the phenomena happening relative to thermodynamics. Topics • Thermodynamics and its History • Thermodynamic Properties and Quantities • Thermodynamic Systems • State and Path Functions • Zeroth Law THERMODYNAMICS AND ITS HISTORY Thermodynamics • from the Greek word “therme” (heat) and “dynamis” (power) • science of storage, transformation and transfer of energy 1st 2nd Law Law •Energy is conserved, thus, it is neither created nor destroyed •Expression of the conservation of energy principle •Energy has quality as well as quantity, and actual processes occur in the direction of decreasing quality of energy History of Thermodynamics Physics of Temperature/Heat Transfer Steam Engine Illustration of the Watt atmospheric engine for pumping water. The main pump is not shown. (Adapted from the engraving of Stuart, 1824, p 114.). Carnot Engine Thermodynamics History of Science Crash Course https://youtu.be/VpiLucwH-AQ Thermodynamics Microscopic and Macroscopic Approach 1. Macroscopic approach (classical thermodynamics) - The structure of the matter is not considered - A few variables are used to describe the state of the matter under consideration. - The values of these variables are measurable following the available techniques of experimental physics Ex.: A moving car, falling stone from a cliff, etc. 2. Microscopic Approach (statistical thermodynamics) - Knowledge of the molecular structure of matter under consideration is essential - A large number of variables are needed for a complete specification of the state of the matter Application of Thermodynamics automobile Refrigeration systems Industrial blowers Industrial heat exchangers Renewable energy Air-conditioning systems Industrial gas compressors Power plants THERMODYNAMIC PROPERTIES AND QUANTITIES Dimensions and Units Fundamental Dimensions • Primitives, for e.g, mass (m), Length (L), time (t), and Temperature (T) Derived Dimensions • Combination of fundamental dimensions, for e.g, Area (A), Volume (V), velocity (v) Dimensions and Units System of units → English Engineering Systems : used in United States → International System (le SysteΜme International in French), or the Metric SI system (mostly used) : meter-kilogram-second (mks), centimeter-gram-second (cgs) Dimensions and Units Dimensions and Units Some SI and English Units Force Weight W = weight m = mass g = gravitational acceleration The weight of a unit mass at sea level A body weighing 60 kgf on earth will weigh only 10 kgf on the moon Dimensions and Units Some SI and English Units Work = Force x Distance Heat/Energy A typical match yields about 1 Btu (or one kJ) of energy if completely burned Dimensions and Units Dimensional Homogeneity - All equations must be dimensionally homogeneous - To be dimensionally homogeneous, all the terms in an equation must have the same dimensions - All secondary units can be formed by combinations of primary units but can also be expressed more conveniently as unity conversion ratios for e.g, Force units can be expressed as: = Commonly used unity conversion ratios Measures of Amount or Size Three measures of amount or size in common use: Mass, m Total Volume, Vt Number of moles, n - Primitive without definition, may be divided by the molar mass M, to yield number of moles where n = no. of moles, mol M = Molar mass, g/mol m = mass - Representing the size of the system, is a defined quantity given as the product of three lengths. It may be divided by the mass or number of moles of the system to yield specific or molar volume * Specific or molar density is defined as the reciprocal of specific or molar volume: ρ = 1/V Force • Product of mass and acceleration • Derived from Newton’s Second law: F= ma, where m = mass and a = acceleration SI Unit: Newton, N = kg m/s2 F = ma , W = mg English: Pound-force, lbf = force which accelerates 1 lb mass 32.174 feet per second per second. *Newton’s law must here include a dimensional proportionality constant if it is to be reconciled with this definition. Thus, F= 1 ma ππ or F = 1 mg ππ Whence, 1 ππ‘ 1 lbf = x 1 lbm x 32.174 2 ππ π gc (gravitational constant) = 32.1740 πππ.ππ‘ πππ.π 2 Force / Weight Example 1.1 An astronaut weighs 730 N in Houston, Texas, where the local acceleration of gravity is g = 9.792 m/s2. What are the astronaut’s mass and weight on the moon, where g = 1.67 m/s2 ? Given: W (Texas) = 730 N , gTexas = 9.792 m/s2 ; gMoon = 1.67 m/s2 Required: Mass (m) and Weight (W) on moon Solution: W (Texas) = (mg)Texas 730 N = m (9.792 m/s2 ) mTexas = 74.55 kg = mmoon * N = kg m/s2 W(moon) = (mg)moon W(moon) = (74.55kg)(1.67 m/s2) W(moon) = 124.50 N m(texas) = m(moon) ππ‘ππ₯ππ ππ‘ππ₯ππ = ππ πππππ Temperature • It is the degree of hotness or coldness of body or control volume. • The property that indicates the direction of the flow of energy through a thermally conducting, rigid wall. • Measure of the average amount of energy of motion, or kinetic energy a system contains Temperature Conversions Temperature Steam Point: the boiling point of water Ice Point: the freezing point of water Absolute Zero: coldest temperature at which all molecular motion is zero and all the substance is perfectly ordered Temperature Phase Diagram of Water Absolute Zero Triple Point: The unique temperature and pressure at which the solid, liquid and gas phases of a substance are all in equilibrium. (273.16K and 0.01°C) Graph of Pressure Versus Temperature: Graph of pressure versus temperature for various gases kept at a constant volume. Note that all of the graphs extrapolate to zero pressure at the same temperature Phase Diagram of Water: In this typical phase diagram of water, the green lines mark the freezing point, and the blue line marks the boiling point, showing how they vary with pressure. The dotted line illustrates the anomalous behavior of water. Note that water changes states based on the pressure and temperature. Temperature International Temperature Scale of 1990 (ITS-90) - Used for calibration of scientific and industrial instruments - Based on assigned values of temperature for a number of reproducible phase-equilibrium states of pure substances (fixed points) and on standard instruments calibrated at these temperatures - The platinum-resistance thermometer is an example of a standard instrument; it is used for temperatures from 13.8 K (-259.35C) (the triple point of hydrogen) to 1234.93 K (961.78"C) (the freezing point of silver). Pressure • Normal force exerted by the fluid per unit area of the surface • πΉ π΄ = = π π2 πππ ππ2 = Pa (Pascal) [SI unit] = psi [English unit] Pressure The primary standard for pressure measurement is the dead-weight gauge in which a known force is balanced by a fluid pressure acting on a known area P= πΉ π΄ = ππ π΄ Where, m = mass of the piston, pan and Weights g = local acceleration of gravity A = cross-sectional area of the piston Pressure Since a vertical column of a given fluid under the influence of gravity exerts a pressure at its base in direct proportion to its height, pressure is also expressed as the equivalent height of a fluid column. Pressure can be expressed as: P= πΉ π΄ = ππ π΄πρπ = = ρgh π΄ π΄ Where, A = cross-sectional area of the column h = column height ρ = fluid density Manometer Pressure Absolute Pressure: The actual pressure at given position. It is measured relative to absolute vacuum (i.e., absolute zero pressure) Gage Pressure: The difference between the absolute pressure and the local atmospheric pressure; measured relative to the local atmospheric pressure and is thus, zero when the pressure is the same as atmospheric pressure Vacuum Pressure: Pressures below atmospheric pressure *Absolute pressures are used in Thermodynamic calculations Pressure The Manometer - Used to measure small and moderate pressure differences. A manometer contains one or more fluids such as mercury, water, alcohol or oil ; used to measure gauge pressure In a stacked-up fluid layers, the pressure change across a fluid layer of density ρ and height h is ρgh The basic Manometer Measuring the pressure drop across a flow section or a flow device by a differential manometer Pressure The Barometer and Atmospheric Pressure - Atmospheric pressure is measured by a device called a barometer; thus, the atmospheric pressure is often referred to as the barometric pressure. - Used to measure absolute pressure Pressure Pascal’s principle - Total pressure in a fluid is the sum of the pressures from different sources - A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. The area ratio A2/A1 is called the ideal mechanical advantage of the hydraulic lift A typical hydraulic system with two fluid-filled cylinders, capped with pistons and connected by a tube called a hydraulic line. A downward force F1 on the left piston creates a pressure that is transmitted undiminished to all parts of the enclosed fluid. This results in an upward force F2 on the right piston that is larger than F1 because the right piston has a larger area. Pressure Example 1.2 A dead-weight gauge with a piston diameter of 1 cm is used for the accurate measurement of pressure. If a mass of 6.14 kg (including piston and pan) brings it into balance, and if g = 9.82 m·s−2, what is the gauge pressure being measured? For a barometric pressure of 0.997 bar, what is the absolute pressure? Given: d= 1cm, m= 6.14kg, g = 9.82 m/s2 , Patm = 0.997 bar Required: Pgage, Pabs Solution: Pgage = ππ π΄ π = (6.14 ππ)(9.82 2 ) π π 1π 2 ( )(1 ππ π₯ ) 4 100ππ π = 7.677 x 105 π2 = 7.677 x 105 Pa or 767.70 kPa = Pgage *1 bar = 105 Pa [ Appendix A] Pabs = Pgage + Patm = 7.677 x 105 Pa + (0.997 x 105 Pa) Pabs = 8.674 x 105 Pa or 867.4 kPa Pressure Example 1.3 A gas is contained in a vertical, frictionless piston cylinder device. The piston has a mass of 4kg and cross-sectional area of 35 cm2. A compressed spring above the piston exerts a force of 60N on the piston. If the atmospheric pressure is 95kPa, determine the pressure inside the cylinder. g = 9.81 m/s2 Given: m, piston = 4kg ; Apiston = 35 cm2 ; Fspring = 60N , Patm = 95kPa Reqd: Pinside Solution: Pinside = Patm + Ppiston + Pspring π Pinside = 95kPa + 4 ππ (9.81 2 ) π 35 ππ2 Pinside = 123.35 kPa 2 1π 100 ππ + 60 π 35 ππ2 2 1π 100 ππ 1 πππ 1000 ππ THERMODYNAMICS SYSTEM System vs. Surroundings Everything external to the system is called the surroundings or environment. The system is separated from the surroundings by the system boundary. The boundary may be fixed or flexible. Classification of System Thermodynamic System X Mass Transfer √ Mass Transfer X Mass Transfer √ Energy Transfer √ Energy Transfer X Energy Transfer Properties of a System Behavior of a thermodynamic system - The behavior of the system depends upon the interaction of energy with or without mass transfer across the boundary. - There are 8 (eight) properties describing the behavior of a system: pressure (P), temperature(T), volume(V), entropy(S), internal energy(U), enthalpy(H), Gibbs function(G) and Helmholtz functions(H) Properties of a System Intensive properties physical property of a system that does not depend on the system size or the amount of material in the system Chemical potential, concentration, density, ductility, elasticity, electrical resistivity, hardness, magnetic field, malleability, pressure, specific energy, specific heat capacity, specific volume, temperature, viscosity Extensive properties system does depend on the system size or the amount of material in the system. Energy, Entropy, Gibbs Energy, Length, Mass, Particle Number, Momentum, Number of Moles, Volume, Magnetic Moment, Electrical Charge, Weight Thermodynamic Properties STATE AND PATH FUNCTIONS State and Path Functions State Function - Does not rely on the past history of the substance nor on the means by which it reaches a given state - Condition of a system identified by properties - P,V,T,U,H,S,A Path Function - Account for energy changes occur in the surrounding - Series of states through which a system passes during process - Q, W State and Path Functions State and Equilibrium A substance can be at various pressures & temperatures or in various states Thermal Equilibrium: temperature of system does not change when it is isolated from surroundings Mechanical Equilibrium: pressure of system does not change when it is isolated from surroundings Chemical Equilibrium: chemical composition does not change when it is isolated from surroundings Phase Equilibrium: mass of each phrase reaches an equilibrium level and stays there State and Path Functions Path and Processes Processes: Any change that a system undergoes from one equilibrium state to another The path of thermodynamic states that a system passes through as it goes from an initial state to a final state is known as the thermodynamic process. State and Path Functions Path and Processes These cannot be defined for a state (you cannot say a system has an amount of work at a specific set of conditions, only that it does a certain amount of work to get from one state to another, via a specified path). Two possible paths between states 1 and 2 Work done along the two paths Work • performed whenever a force acts through a distance (W) dW = Fdl Where, F = component of force acting along the line of the displacement dl SI Unit: N·m (Newton-meter) or J (Joule) Work (+) compression (when the piston moves into the cylinder to compress the fluid) Work (-) expansion (opposite direction / the fluid make the piston to move upward) Work An isobaric expansion of a gas requires heat transfer during the expansion to keep the pressure constant. Since pressure is constant, the work done is PΔV A path for compression of a gas from point 1, initial volume V1t at pressure P1, to point 2, volume V2t at pressure P2. This path relates the pressure at any point of the process to the volume. Work We consider three modes of work transfer across the boundary of a system, as shown in the following diagram: • Boundary Work - occurs when the volume V of a system changes. It is used for calculating piston displacement work in a closed system • Shaft Work - is the work done if the system turns the shaft of a motor or compressor. • Electrical Work – energy transferred to the system under the voltage potential, voltage is applied to a resistance in a system that results in current flow that in turn increases the internal energy of the system. Energy Energy may be contained (stored) in a system as Macroscopic form Energy related to motion or elevation of the 1 system e.g. KE = mv2 or PE = mgz Microscopic form Energy related to molecular structure called Internal energy(U) • Total Energy (E) = U + KE + PE = U + • Specific Total Energy, e = πΈ π 1 2 mv2 + mgz = u + ke + pe = u + • Specific Internal Energy, u = 2 π£2 2 + gz π π Energy transfers at the system boundary as heat and work Energy transfer at boundary Heat, Q Work, W Energy Internal Energy The portion of the internal energy of a system is associated with the: 1. Kinetic energies of the molecules (sensible energy) 2. Phase of a system (latent energy) 3. Atomic bonds in a molecule (chemical energy) 4. Strong bonds within the nucleus of the atom itself (nuclear energy) 5. Static energy (Stored in a system) 6. Dynamic energy: energy interactions at the system boundary (i.e. heat and work) Energy Kinetic Energy - The energy of a system possesses because of its velocity relative to the surroundings at rest shows that the work done on a body in accelerating it from an initial velocity u1 to a final velocity u2 is equal to the change in kinetic energy of the body. Conversely, if a moving body is decelerated by the action of a resisting force, the work done by the body is equal to its change in kinetic energy. Where m = mass in kg u = velocity in m/s dW = Fdl if F = ma dW = madl a = du/dt ππ’ ππ dW = m ( ππ‘ )dl = m (ππ‘)du u = dl/dt dW = m udu Energy Potential Energy - The energy of a system possesses because of the body force exerted on its mass by a gravitational or electromagnetic field with respect to reference surface work done on a body in raising it is equal to the change in the quantity mzg. Conversely, if a body is lowered against a resisting force equal to its weight, the work done by the body is equal to the change in the quantity mzg. Where m = mass in kg z = elevation in m g = gravitational acceleration in m/s2 Energy Energy Conservation Law of Conservation – states that energy is nor created or destroyed, thus the total energy of the universe is constant Energy Conservation Energy Example 1.4 An elevator with a mass of 2500 kg rests at a level 10 m above the base of an elevator shaft. It is raised to 100 m above the base of the shaft, where the cable holding it breaks. The elevator falls freely to the base of the shaft and strikes a strong spring. The spring is designed to bring the elevator to rest and, by means of a catch arrangement, to hold the elevator at the position of maximum spring compression. Assuming the entire process to be frictionless, and taking g = 9.8 m⋅s−2, calculate: (a) The potential energy of the elevator in its initial position relative to its base. (b) The work done in raising the elevator. (c) The potential energy of the elevator in its highest position. (d) The velocity and kinetic energy of the elevator just before it strikes the spring. (e) The potential energy of the compressed spring. (f) The energy of the system consisting of the elevator and spring (1) at the start of the process, (2) when the elevator reaches its maximum height, (3) just before the elevator strikes the spring, (4) after the elevator has come to rest. Energy Example 1.4 (a) The potential energy of the elevator in its initial position relative to its base. Given: m= 2500 kg ; z1 = 10m ; g = 9.8 m⋅s−2 Required: PE Solution: PE1 = mzg = (2500kg)(10m)(9.8 π π 2 ) PE1 = 245,000 J (b) The work done in raising the elevator. W = ΔPE = mg Δz = mg (z2-z1) = (2500kg)(9.8 W = 2,205,000 J π π 2 ) (100 m -10 m) Energy Example 1.4 (c) The potential energy of the elevator in its highest position. Given: m= 2500 kg ; z2 = 100m ; g = 9.8 m⋅s−2 Required: PE Solution: PE2 = mzg = (2500kg)(100m)(9.8 PE2 = 2,450,000 J π π 2 ) Energy Example 1.4 (d) The velocity and kinetic energy of the elevator just before it strikes the spring. ΔKE + ΔPE = 0 (KE3 - KE2 )+ (PE3 - PE2 )= 0 -> KE2 = PE3 = 0 KE3 = PE2 = 2,450,000 J 1 KE3 = ππ’2 2 2,450,000 J = = 1 2 2500ππ π’2 π2 u3 = 2(2450000 ππ 2 ) π 2500 ππ = 44.27 m/s = u3 Energy Example 1.4 (e) The potential energy of the compressed spring. ΔKE(elevator) + ΔPE(spring) = 0 (KE4 - KE3) + (PE4 – PE3) = 0 PE3 = 0 (initial potential energy of the spring before compression) KE4 = 0 (final kinetic energy of the spring after compression or elevator comes at rest) KE3 = PE4 = 2,450,000 J (final potential energy of the spring after compression) State 4 Energy Example 1.4 (f) The energy of the system consisting of the elevator and spring (1) at the start of the process, (2) when the elevator reaches its maximum height, (3) just before the elevator strikes the spring, (4) after the elevator has come to rest. PE2 f.1) PE1 = 245,000 J f.2) E = PE1 + ΔPE E = 245,000 J + 2,205,000 J E = 2,450,000 J f.3) KE3 = 2,450,000 J f.4) PE4 = 2,450,000 J ΔPE PE1 Base PE3 PE4 KE3 Energy Example 1.5 A team from Engineers Without Borders constructs a system to supply water to a mountainside village located 1800 m above sea level from a spring in the valley below at 1500 m above sea level. (a) When the pipe from the spring to the village is full of water, but no water is flowing, what is the pressure difference between the end of the pipe at the spring and the end of the pipe in the village? (b) What is the change in gravitational potential energy of a liter of water when it is pumped from the spring to the village? (c) What is the minimum amount of work required to pump a liter of water from the spring to the village? village 300m spring 1800m 1500m Sea level Energy Example 1.5 (a) When the pipe from the spring to the village is full of water, but no water is flowing, what is the pressure difference between the end of the pipe at the spring and the end of the pipe in the village? village 300m spring 1800m 1500m Sea level ΔP = ρgh = 1000kg/m3 (9.81 m/s2)(300m-0m) ΔP = 2,943,000 Pa or 2943kPa Energy Example 1.5 (b) What is the change in gravitational potential energy of a liter of water when it is pumped from the spring to the village? (c) What is the minimum amount of work required to pump a liter of water from the spring to the village? village 300m spring 1800m 1500m Sea level ρ = 1000 kg/m3 ρ = 1000 ππ x ( π3 ) (1L) = 1 kg π3 1000πΏ b) ΔPE = mg Δz = mg(z2-z1) = 1kg(9.81 m/s2)(300m-0m) ΔPE = 2943J c) W = ΔPE = 2943J Heat - Flows from a higher temperature to lower temperature - Leads to concept of temperature as a driving force for the transfer of energy as HEAT Units Calorie(cal) – quantity of heat which when transferred to one gram water raised its temperature to 1oC British Thermal Unit (BTU) – quantity of heat which when transferred to one pound mass raised its temperature to 1oF SI – Joule(N.m), J – mechanical work when a force of 1N acts through a distance of 1-meter - 1 ft.lbf = 1.3558179 J - 1 cal = 4.1840 J - 1 BTU = 1055.04 J Power(watt-W) – energy rate of 1 J per second - Watt(W) = J/s Heat - In thermodynamics, that part of the total energy flow across a system boundary that is caused by the temperature difference between the system and the surroundings. Q (+) – heat is transferred TO the system Q (-) – heat is transferred BY the system Q = UA ΔT • Where : Q – rate of heat transfer • U – empirical coefficient, over-all heat transfer coefficient • A – area for heat transfer • ΔT – change in temperature This equation is the mathematical expression of the first law for a steady-state, steady-flow process between one entrance and one exit if ΔKE, ΔPE, Ws = 0, thus, Q = ∫Cp ΔT = ΔH equations for calculating the heat change (Calorimetry) ZEROTH LAW OF THERMODYNAMICS Zeroth Law of Thermodynamics - When 2 bodies have equality of temperature with a 3rd body, then they have equality of temperature with each other TA TB TC If TA = TC & TB = TC Then, TA=TB Two bodies reaching thermal equilibrium after being brought into contact in an isolated enclosure Sources • Smith, J.M.; Van Ness, Hendrick; Abbott, Michael; and Swihart, Mark. (2017). • • • • Introduction to Chemical Engineering Thermodynamics. McGraw-Hill Education, 8th ed. LMD Gamer (April 21, 2017). History of Thermodynamics – Documentary 2017. https://youtu.be/up8LqQMLvK8 Module 1 Thermodynamics. https://nptel.ac.in/content/storage2/courses/112104113/lecture1/1_4.htm Thermodynamics Crash Course. https://www.youtube.com/watch?v=VpiLucwH-AQ&t=277s Cengel, Yunus A. Boles, Michael A. (2011). Thermodynamics: An Engineering Approach. McGraw Hill Inc.
