Thermodynamics Lecture Handout PDF

Basic Concepts of Thermodynamics Reading Problems 2-1 → 2-10 2-44, 2-59, 2-78, 2-98 Thermal Sciences Thermodynamics Heat Transfer Conservatio...

Basic Concepts of Thermodynamics Reading Problems 2-1 → 2-10 2-44, 2-59, 2-78, 2-98 Thermal Sciences Thermodynamics Heat Transfer Conservation of mass Conduction Conservation of energy Convection cs Second law of thermodynamics Radiation mi Properties Conjugate He na at dy Tr mo an Thermal er sfe Th Systems r Engineering Fluids Mechanics Fluid Mechanics Fluid statics Conservation of momentum Mechanical energy equation Modeling Thermodynamics: the study of energy, energy transformations and its relation to matter. The anal- ysis of thermal systems is achieved through the application of the governing conservation equations, namely Conservation of Mass, Conservation of Energy (1st law of thermodynam- ics), the 2nd law of thermodynamics and the property relations. Heat Transfer: the study of energy in transit including the relationship between energy, matter, space and time. The three principal modes of heat transfer examined are conduction, con- vection and radiation, where all three modes are affected by the thermophysical properties, geometrical constraints and the temperatures associated with the heat sources and sinks used to drive heat transfer. 1 Fluid Mechanics: the study of fluids at rest or in motion. While this course will not deal exten- sively with fluid mechanics we will be influenced by the governing equations for fluid flow, namely Conservation of Momentum and Conservation of Mass. Thermodynamics Microscopic: tracking the movement of matter and energy on a particle by particle basis Macroscopic: use the conservation equations (energy and mass) to track movement of matter and energy on an average over a fixed domain (referred to as classical thermodynamics) Energy the total energy of the system per unit mass is denoted as e and is given as   E kJ e= m kg if we neglect the contributions of magnetic, electric, nuclear energy, we can write the total energy as mV 2 E = U + KE + P E = U + + mgz 2 Dimensions and Units SI: International System SI is the preferred because it is logical (base 10) and needs no correction factors unit convention: Parameter Units Symbol length, L meters m mass, m kilograms kg time, t seconds s temperature, T kelvin K velocity, V meter per second, ≡ L/t m/s acceleration, a meter per second squared ≡ L/t2 m/s2 force, F newton, ≡ m · L/t2 N energy, E joule ≡ m · L2 /t2 J 2 Thermodynamic Systems Isolated Boundary Surroundings rk Wo at Surroundings He System - everything that interacts with the system System Boundary System (real or imaginary - may be as simple fixed or deformable) as a melting ice cube - or as complex as a nuclear power plant SYSTEM: Closed System: composed of a control (or fixed) mass where heat and work can cross the boundary but no mass crosses the boundary. Open System: composed of a control volume (or region in space) where heat, work, and mass can cross the boundary or the control surface weights by-pass flow g fan piston gas system engine core boundary Closed System Open System WORK & HEAT TRANSFER: work and heat transfer are NOT properties → they are the forms that energy takes to cross the system boundary 3 Thermodynamic Properties of Systems Basic Definitions Thermodynamic Property: Any observable or measurable characteristic of a system. Any math- ematical combination of the measurable characteristics of a system Intensive Properties: Properties which are independent of the size (or mass) of the system they are not additive ⇒ XA+B = XA + XB examples include: pressure, temperature, and density Extensive Properties: Properties which are dependent of the size (or mass) of the system they are additive ⇒ XA+B = XA + XB examples include: volume, energy, entropy and surface area Specific Properties: Extensive properties expressed per unit mass to make them intensive prop- erties extensive property specific property (intensive) −→ mass Measurable Properties P, V, T, and m are important because they are measurable quantities. Many other thermo- dynamic quantities can only be calculated and used in calculations when they are related to P, V, T, and m – Pressure (P ) and Temperature (T ) are easily measured intensive properties. Note: They are not always independent of one another. – Volume (V ) and mass (m) are easily measured extensive properties Pressure F orce N P ressure = ; → ≡ Pa Area m2 – in fluids, this is pressure (normal component of force per unit area) – in solids, this is stress 4  Pressure gauge pressure Patm absolute vacuum pressure pressure ABSOLUTE ATMOSPHERIC absolute PRESSURE vacuum pressure Temperature temperature is a pointer for the direction of energy transfer as heat TA > TB TA < TB Q Q TB TB TA TA 0th Law of Thermodynamics: if system C is in thermal equilibrium with system A, and also with system B, then TA = TB = TC State and Equilibrium State Postulate how long does the list of intensive properties have to be in order to describe the intensive state of the system? 5  same substance A B mA = 10 kg mB = 0.1 kg T = 500 K P = 0.1 MPa 3 v = 0.5 m /kg u = 3.0 kJ/kg... System A and B have the same intensive state, but totally different extensive states. State Postulate (for a simple compressible system): The state of a simple compressible system is completely specified by 2 independent and intensive properties. note: a simple compressible system experiences negligible electrical, magnetic, gravita- tional, motion, and surface tension effects, and only P dV work is done in a single phase system, T, v, and P are independent and intensive (in a multiphase system however, T and P are not independent) if the system is not simple, for each additional effect, one extra property has to be known to fix the state. (i.e. if gravitational effects are important, the elevation must be specified and two independent and intensive properties) it is important to be able to: – find two appropriate properties to fix the state – find other properties when the state is fixed (we will discuss this later) Thermodynamic Processes the process is any change from one equilibrium state to another. (If the end state = initial state, then the process is a cycle) 6  the process path is a series intermediate states through which a system passes during the process (we very seldom care what the process path is) processes are categorized based on how the properties behave: – isobaric (P = constant) – isothermal (T = constant) – isochoric or isometric (V = constant) – isentropic (s = constant) – isenthalpic (h = constant) – adiabatic (no heat transfer) Stored Energy how is energy stored in matter? Stored Energy = E = KE + P E + U Kinetic Energy: Energy due to motion of the matter with respect to an external reference frame (KE = mV 2 /2) Potential Energy: Energy due to the position of the matter in a force field (gravitational, magnetic, electric). It also includes energy stored due to elastic forces and surface tension (P E = mgz) Internal Energy = microscopic forms of energy, U ) – forms of the energy in the matter due to its internal structure (independent of external reference frames) Transit Energy Heat transit form of energy that occurs when there is ΔT (a temperature gradient) notation - Q (kJ ), q (kJ/kg), Q̇ (kW ), q̇ (kW/kg) Work transit form of energy that occur due to all other driving forces notation - W (kJ ), w (kJ/kg), Ẇ (kW ), ẇ (kW/kg) 7 Properties of Pure Substances Reading Problems 3-1 → 3-7 3-49, 3-52, 3-57, 3-70, 3-75, 3-106, 3-9 → 3-11 3-121, 3-123 Pure Substances a Pure Substance is the most common material model used in thermodynamics. – it has a fixed chemical composition throughout (chemically uniform) – a homogeneous mixture of various chemical elements or compounds can also be con- sidered as a pure substance (uniform chemical composition) – a pure substance is not necessarily physically uniform (different phases) Phases of Pure Substances a pure substance may exist in different phases, where a phase is considered to be a physically uniform 3 principal phases: Solids: – strong molecular bonds – molecules form a fixed (but vibrating) structure (lattice) Liquids: – molecules are no longer in a fixed position relative to one another – molecules float about each other Gases: – there is no molecular order – intermolecular forces ≈ 0 Behavior of Pure Substances (Phase Change Processes) Critical Point: liquid and vapor phases are not distinguishable Triple point: liquid, solid, and vapor phases can exist together 1  P substances that substances that expand on freezing contract on freezing melting condensation freezing LIQUID critical point mel ting lting n vaporization tio za ori me vap SOLID triple point sublimation VAPOR on ati sublimation b lim su T P0 P0 P0 P0 P0 G G L L S L S dQ dQ dQ dQ dQ T critical point sa tur line ate iquid d fusion va P0 l po ated r li line ne satur G L L+G S+L S triple point line S+G v 2 T − v Diagram for a Simple Compressible Substance consider an experiment in which a substance starts as a solid and is heated up at constant pressure until it all becomes as gas depending on the prevailing pressure, the matter will pass through various phase transforma- tions. At P0 : 1. solid 2. mixed phase of liquid and solid 3. sub-cooled or compressed liquid 4. wet vapor (saturated liquid-vapor mixture) 5. superheated vapor The Vapor Dome general shape of a P − v diagram for a pure substance is similar to that of a T − v diagram P critical point gas (vapor) subcooled liquid two-phase region Tcr T (saturated liquid & saturated vapor) L+G Psat(T) T saturated saturated vfg vapor line liquid line vf vg v express values of properties for conditions on the vapor dome as: specific volume: vf , vg and vf g = vg − vf internal energy: uf , ug and uf g = ug − uf specific enthalpy: hf , hg and hf g = hg − hf specific entropy: sf , sg and sf g = sg − sf 3  in the two phase region, pressure and temperature cannot be specified independently Psat = P (Tsat) ⇔ Tsat = T (Psat) this only holds true under the vapor dome in the two-phase region. The Two-Phase Region mg mg Quality ⇒ x= = m mg + mf total mass of mixture ⇒ m = mg + mf x = mg/m G mass fraction of the gas ⇒ x = mg /m L mass fraction of the liquid ⇒ 1 − x = mf /m 1-x = mf/m Properties of Saturated Mixtures all the calculations done in the vapor dome can be performed using Tables. – in Table A-4, the properties are listed under Temperature – in Table A-5, the properties are listed under Pressure use if given mixture: 1) is saturated, 2) has a quality, 3) liquid & vapor are present V Vf + Vg mf vf + mg vg v = = = = (1 − x) vf + x vg m m m = vf + x(vg − vf ) = vf + x vf g v − vf x = vf g 4 Properties of Superheated Vapor superheated vapor is a single phase (vapor phase only). – T and P are independent of each other in the single-phase region – properties are typically calculated as a function of T and P T L P = 100 kPa o T = 200 C P G L+G Tsat(P) state: 200 oC and 100 kPa P v Properties of Sub-cooled Liquid T P G L L+G P = 100 kPa Tsat(P) o T = 20 C P o state: 20 C and 100 kPa v 5  can be treated as incompressible (independent of P ) v = v(T, P ) ⇒ v ≈ v(T ) = vf (T ) u = u(T, P ) ⇒ u ≈ u(T ) = uf (T ) s = s(T, P ) ⇒ s ≈ s(T ) = sf (T ) while property values for the sub-cooled liquids are not available as a function of temperature and pressure, they are available as a function of temperature only. for enthalpy: h(T, P ) = u(T, P ) + v(T, P )P ≈ uf (T ) + vf (T )P = uf (T ) + vf (T )Psat(T ) +vf (T )[P − Psat(T )]    hf (T ) = hf (T ) + vf (T )[P − Psat(T )] Equation of State for Gaseous Pure Substances Thermodynamic relations are generally given in three different forms: Tables: water (Tables A-4 → A-8), R134a (Tables A-11 → A-13) Graphs: water (Figures A-9 & A-10), R134a (Figure A-14) Equations: air (Tables A-2 & 3-4) since most gases are highly superheated at atmospheric conditions. This means we have to use the superheat tables – but this can be very inconvenient a better alternative for gases: use the Equation of State which is a functional relationship between P, v, and T (3 measurable properties) Ideal Gases gases that adhere to a pressure, temperature, volume relationship P v = RT or P V = mRT referred to as ideal gases 6  – where R is the gas constant for the specified gas of interest (R = Ru/M̃ ) Ru = Universal gas constant, ≡ 8.314 kJ/(kmol · K) M̃ = molecular wieght (or molar mass) of the gas (see Table A-1)) When is the ideal gas assumption viable? – for a low density gas where: ∗ the gas particles take up negligible volume ∗ the intermolecular potential energy between particles is small ∗ particles act independent of one another – Under what conditions can it be used? ∗ low density ∗ high temperatures - considerably in excess of the saturation region ∗ at very low pressures Real Gases experience shows that real gases obey the following equation closely: P v = ZRT (T and P are in absolute terms) – this equation is used to find the third quantity when two others are known – Z is the compressibility factor – Note: Rair = 0.287 kJ/kg · K, RH2 = 4.124 kJ/kg · K what is the compressibility factor (Z)? – Z charts are available for different gases, but there are generalized Z charts that can be used for all gases – if we “reduce” the properties with respect to the values at the critical point, i.e. P reduced pressure = Pr = Pc = critical pressure Pc T reduced temperature = Tr = Tc = critical temperature Tc 7 Reference Values for u, h, s values of enthalpy, h and entropy, s listed in the tables are with respect to a datum where we arbitrarily assign the zero value. For instance: Tables A-4, A-5, A-6 & A-7: saturated and superheated water - the reference for both hf and sf is taken as 0 ◦ C. This is shown as follows: uf (@T = 0 ◦ C) = 0 kJ/kg hf (@T = 0 ◦ C) = 0 kJ/kg sf (@T = 0 ◦ C) = 0 kJ/kg · K Tables A-11, A-12 & A-13: saturated and superheated R134a - the reference for both hf and sf is taken as −40 ◦ C. This is shown as follows: hf (@T = −40 ◦ C) = 0 kJ/kg hf (@T = −40 ◦ C) = 0 kJ/kg sf (@T = −40 ◦ C) = 0 kJ/kg · K Others: sometimes tables will use 0 K as the reference for all tables. While this standard- izes the reference, it tends to lead to larger values of enthalpy and entropy. Calculation of the Stored Energy for most of our 1st law analyses, we need to calculate ΔE ΔE = ΔU + ΔKE + ΔP E for stationary systems, ΔKE = ΔP E = 0 1 in general: ΔKE = m(V22 − V12 ) and ΔP E = mg(z2 − z1 ) 2 how do we calculate ΔU ? 1. one can often find u1 and u2 in the thermodynamic tables (like those examined for the states of water). 2. we can also explicitly relate ΔU to ΔT (as a mathematical expression) by using the thermodynamic properties Cp and Cv. 8 Specific Heats: Ideal Gases for any gas whose equation of state is exactly P v = RT the specific internal energy depends only on temperature u = u(T ) the specific enthalpy is given by h = u + Pv where h(T ) = u(T ) + RT Note: Since u = u(T ), and R is a constant, enthalpy is only a function of temperature. for an ideal gas du Cv = ⇒ Cv = Cv (T ) only dT dh Cp = ⇒ Cp = Cp (T ) only dT From the equation for enthalpy, RT = h(T ) − u(T ) If we differentiate with respect to T dh du R= − dT dT R = Cp − Cv 9  the calculation of Δu and Δh for an ideal gas is given as  2 Δu = u2 − u1 = Cv (t) dT (kJ/kg) 1  2 Δh = h2 − h1 = Cp (t) dT (kJ/kg) 1 to carry out the above integrations, we need to know Cv (T ) and Cp (T ). These are available from a variety of sources Table A-2a: for various materials at a fixed temperature of T = 300 K Table A-2b: various gases over a range of temperatures 250 K ≤ T ≤ 1000 K Table A-2c: various common gases in the form of a third order polynomial Specific Heats: Solids and Liquids solids and liquids are incompressible. (i.e., ρ = constant). for solids and liquids it can be shown (mathematically) that for incompressible materials: Cp = Cv = C and dQ = mCdT Table A-3 gives C for incompressible materials as with ideal gases, for incompressible materials C = C(T )  2 Δu = C(T ) dT ≈ Cavg ΔT 1 Δh = Δu + Δ(P v) = Δu + vΔP ≈ Cavg ΔT + vΔP – for solids: Δh = Δu = Cavg ΔT (vΔP is negligible) – for liquids: If P = const, Δh = Cavg ΔT If T = const, Δh = vΔP (we already used this to do more accurate calculations in the subcooled region). 10 First Law of Thermodynamics Reading Problems 4-1 → 4-6 4-19, 4-24, 4-42, 4-61, 4-65 5-1 → 5-5 5-37, 5-67, 5-84, 5-104, 5-120, 5-141, 5-213 Control Mass (Closed System) A thermodynamic analysis of a system can be performed on a fixed amount of matter known as a control mass or over a region of space, known as a control volume. Conservation of Mass Conservation of Mass, which states that mass cannot be created or destroyed, is implicitly satisfied by the definition of a control mass. Conservation of Energy The first law of thermodynamics states: Energy cannot be created or destroyed it can only change forms. energy transformation is accomplished through energy transfer as work and/or heat. Work and heat are the forms that energy can take in order to be transferred across the system boundary. the first law leads to the principle of Conservation of Energy where we can stipulate the energy content of an isolated system is constant. energy entering − energy leaving = change of energy within the system 1 Sign Convention There are many potential sign conventions that can be used. Cengel Approach Heat Transfer: heat transfer to a system is positive and heat transfer from a system is negative. Work Transfer: work done by a system is positive and work done on a system is negative. Culham Approach Anything directed into the system is positive, anything directed out of the system is negative. 2 Example: A Gas Compressor Performing a 1st law energy balance: ⎧ ⎫ ⎧ ⎫ ⎪ Initial ⎪   ⎪ ⎨ ⎬ + Energy gain W1−2 ⎨ F inal ⎪ ⎬ Energy = Energy ⎪ ⎩ ⎭ − ⎪ Energy loss Q1−2 ⎪ ⎩ ⎪ ⎭ E1 E2 A first law balance for a control mass can also be written in differential form as follows: dE = δQ − δW Note: d or Δ for a change in property and δ for a path function Forms of Energy Transfer Work Versus Heat Work is macroscopically organized energy transfer. Heat is microscopically disorganized energy transfer. 3 Heat Energy Notation: – Q (kJ ) amount of heat transfer – Q̇ (kW ) rate of heat transfer (power) – q (kJ/kg) - heat transfer per unit mass – q̇ (kW/kg) - power per unit mass modes of heat transfer: – conduction: diffusion of heat in a stationary medium (Chapters 8 & 9) – convection: it is common to include convective heat transfer in traditional heat transfer analysis. However, it is considered mass transfer in thermodynamics. (Chapters 10 & 11) – radiation: heat transfer by photons or electromagnetic waves (Chapter 12) Work Energy Notation: – W (kJ ) amount of work transfer – Ẇ (kW ) power – w (kJ/kg) - work per unit mass – ẇ (kW/kg) - power per unit mass work transfer mechanisms in general, are a force acting over a distance Mechanical Work force (which generally varies) times displacement  2 W12 = F ds 1 4 Moving Boundary Work compression in a piston/cylinder, where A is the piston cross sectional area (frictionless) the area under the process curve on a P − V 2 diagram is proportional to P dV 1 the work is: – +ve for compression – −ve for expansion sometimes called P dV work or compression /expansion work  2  2  2 W12 = − F ds = − P · A ds = − P dV 1 1 1 Polytropic Process: where P V n = C examples of polytropic processes include: Isobaric process: if n = 0 then P = C and we have a constant pressure process Isothermal process: if n = 1 then from the ideal gas equation P V = RT and P V is only a function of temperature Isometric process: if n → ∞ then P 1/nV = C 1/n and we have a constant volume process Isentropic process: if n = k = Cp /Cv then we have an isentropic process 5 Gravitational Work Work is defined as force through a distance  2 W12 = F ds 1 Since in the case of lifting an object, force and displacement are in the same direction, the work will be positive and by definition positive work is into the system.  2  2  2 W12 = F ds = mg ds = mg dz 1 1 1 integrating from state 1 to state 2 gives W12 = mg(z2 − z1 ) the potential energy of the system increases with an addition of gravitational work, ΔP E = W = mg(z2 − z1 ) Acceleration Work if the system is accelerating, the work associated with the change of the velocity can be calculated as follows:  2  2  2  2 dV ds W12 = F ds = ma ds = m ds = m dV 1 1 1 dt 1 dt  V and we can then write  2   V22 V12 W12 = mV dV = m − 1 2 2 if we drop an object with the assistance of gravity, the first law balance gives ΔP E + ΔKE = 0. Potential energy decreases and kinetic energy increases. 6 Charge Transfer Work (Electrical Work) current, I is the rate of charge transfer d q+ coulomb I≡ , = Ampere dt sec where q + = −N e with N being the number of electrons and e the charge of the electron. the electrical work done is given as δWe = (1 − 2 ) dq + =  dq + J oule where  is the electrical potential difference with units volt = Coulomb the electrical work done per unit time is power δWe Ẇe = P ower = = I (W ) dt  2 We =  I dt 1 =  IΔt 7 Control Volume (Open System) The major difference between a Control Mass and and Control Volume is that mass crosses the system boundary of a control volume. CONSERVATION OF MASS: Unlike a control mass approach, the control volume approach does not implicitly satisfy conservation of mass, therefore we must make sure that mass is neither created nor destroyed in our process. m OUT m cv m IN cv ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎪ ⎨ ⎪ rate of increase ⎬ ⎪ ⎪ ⎨ net rate of ⎬ ⎪ ⎨ net rate of ⎪ ⎬ ⎪ of mass within ⎪ = ⎪ mass f low ⎪ − ⎪ mass f low ⎪ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ the CV IN OU T d (mCV ) = ṁIN − ṁOUT dt where:  mCV = ρ dV V ṁIN = (ρ V A)IN ṁOUT = (ρ V A)OUT with V = average velocity 8 CONSERVATION OF ENERGY: The 1st law states: ECV (t) + δQ + δWshaf t + (ΔEIN − ΔEOUT )+ (δWIN − δWOUT ) = ECV (t + Δt) (1) where: ΔEIN = eIN ΔmIN ΔEOUT = eOUT ΔmOUT δW = flow work E V2 e = = u +  + gz m 2  internal  potential kinetic 9 What is flow work? This is the work required to pass the flow across the system boundaries. When mass enters/leaves a control volume, some work is done on/by the control volume. volume    ΔmIN = ρIN AIN VIN Δt δWIN = F · distance = PIN AIN · VIN Δt       F Δs PIN ΔmIN = ρIN with 1 v= ρ δWIN = (P v Δm)IN → flow work (2) Similarly δWOUT = (P v Δm)OUT (3) 10 Substituting Eqs. 2 and 3 into Eq. 1 gives the 1st law for a control volume ECV (t + Δt) − ECV (t) = δQ + δWshaf t + ΔmIN (e + P v)IN − ΔmOUT (e + P v)OUT (4) Equation 4 can also be written as a rate equation → divide through by Δt and take the limit as Δt → 0 d ECV = Q̇ + Ẇshaf t + [ṁ(e + P v)]IN − [ṁ(e + P v)]OUT dt where: ṁ = ρ v ∗ A Note that: (v ∗ )2 e + Pv = u + Pv +    + gz 2 = h(enthalpy) + KE + P E By using enthalpy instead of the internal energy to represent the energy of a flowing fluid, one does not need to be concerned with the flow work. 11 Some Practical Assumptions for Control Volumes Steady State Process: The properties of the material inside the control volume do not change with time. For example P2 > P1 P1 P2 Diffuser: P changes inside the control volume, but the pressure at each point does not change with time. V2 < V1 Steady Flow Process: The properties of the material crossing the control surface do not change with time. For example T = T(y) Inlet Pipe: T at the inlet may be different at differ- ent locations, but temperature at each boundary T ≠ T(t) point does not change with time. y x The steadiness refers to variation with respect to time if the process is not steady, it is unsteady or transient often steady flow implies both steady flow and steady state Uniform State Process: The properties of the material inside the control volume are uniform and may change with time. For example o 50 C t=0 o 50 C 50 oC o 50 C Heating Copper: Cu conducts heat well, so that it Q heats evenly. o 60 oC t=10 60 C o 60 oC 60 C 12 Uniform Flow Process: The properties of the material crossing the control surface are spatially uniform and may change with time. For example P ≠ P(y) Inlet Pipe: P at the inlet is uniform across y. y x Uniformity is a concept related to the spatial distribution. If the flow field in a process is not uniform, it is distributed. Steady-State, Steady-Flow Process Idealizations: the control volume does not move relative to the reference frame the state of the mass at each point with the control volume does not change with time the flows in and out of the control volume are steady, i.e. there is no mass accumulation within the control volume the rates at which work and heat cross the control volume boundary remain constant 13 Control Volume Analysis for Electrical Devices We recall from the 1st law for a control volume dEcv = Q̇ − Ẇ + ṁin(e + P v)in − ṁout(e + P v)out dt We can consider the following analogy Let N = number of charged particles q+ = +ve charge on each particle N q+ = total charge mq = mass of each charged particle N mq = total mass of the “charge gas” Ṅ mq = flow rate of the “charge gas”  = electrical potential ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ V2 (N q + ) ⎟ ṁ(e + P v) = (Ṅ mq ) ⎜ ⎜u + Pv + + gz + ⎟ ⎟ ⎜ 2   (N m q ) ⎟ ⎜    ⎟ ⎜ ⎟ ⎝ negligible electrical potential⎠ per unit mass   (N q + ) ṁ(e + P v) = (Ṅ mq (N mq ) 14  = Ṅ q +  = I where I = current ≡ Ṅ q + = −Ṅ e The first law takes the form dEcv = Q̇ − Ẇ + (Ṅ q + )in − (Ṅ q + )out dt 15 Entropy and the Second Law of Thermodynamics Reading Problems 6-1, 6-2, 6-7, 6-8, 6-11 6-81, 6-87, 7-1 → 7-10, 7-12, 7-13 7-39, 7-46, 7-61, 7-89, 7-111, 7-122, 7-124, 7-130, 7-155, 7-158 Why do we need another law in thermodynamics? Answer: While the 1st law allowed us to determine the quantity of energy transfer in a process it does not provide any information about the direction of energy transfer nor the quality of the energy transferred in the process. In addition, we can not determine from the 1st law alone whether the process is possible or not. The second law will provide answers to these unanswered questions. A process will not occur unless it satisfies both the first and the second laws of thermody- namics. 1. Direction: Consider an isolated system where Q = W = 0 From the 1st law we know ⎧ ⎪ ⎨ same both ways (U + KE)1 = U2 1→2     ⎪ ⎩ E1 E2 2→1 possible gas gas impossible 1: cold system 2: warm system propeller & propeller & gas rotating gas stationary The physical interpretation of this is: 1  State 1: Most of the energy is in a highly organized form residing in the macroscopic KE of the propeller and the rotating gas. State 2: All of the energy is now in a disorganized form residing in the microscopic E, i.e. U of the propeller and the gas. The process 1 → 2 has resulted in a higher state of molecular chaos. EN T ROP Y is the thermodynamic property that describes the degree of molecular disorder in matter. Hence, S2 > S1. Entropy can be considered a quantitative index that describes the quality of energy. The process 2 → 1 is impossible because it would require disorganized KE to pro- duce macroscopically organized KE. That is S2 < S1 which is impossible for an isolated system. A thermodynamic process is only possible if it satisfies both the 1st and 2nd laws simultaneously. 2. Quality of Energy: A heat engine produces reversible work as it transfers heat from a high temperature reservoir at TH to a low temperature reservoir at TL. If we fix the low temperature reservoir at TL = 300 K, we can determine the relationship between the efficiency of the heat engine, QL TL η =1− =1− QH TH as the temperature of the high temperature reservoir changes. In effect we are determin- ing the quality of the energy transferred at high temperature versus that transferred at low temperature. TH QH reversible heat engine Wrev QL TL 2  TL (K) TH (K) η 300 1000 0.700 300 800 0.625 300 600 0.500 300 400 0.250 Since the purpose of the heat engine is to convert heat energy to work energy, we can clearly see that more of the high temperature thermal energy can be converted to work. Therefore the higher the temperature, the higher the quality of the energy. Second Law of Thermodynamics The second law of thermodynamics states: The entropy of an isolated system can never decrease. When an isolated system reaches equilibrium, its entropy attains the maximum value possible under the constraints of the system Definition PS = Sgen = S2 − S1 ≥ 0 → 2nd law the 2nd law dictates why processes occur in a specific direction i.e., Sgen cannot be −ve The second law states, for an isolated system: (ΔS)system + (ΔS)surr. ≥ 0 where Δ ≡ f inal − initial Gibb’s Equation From a 1st law energy balance when KE and PE are neglected Energy Input = Energy Output + Increase in Energy Storage 3  δQ = δW + dU  (1)  amount dif f erential We know that the differential form of entropy is δQ dS = (2) δW = P dV (3) T Combining Eqs. 1, 2 and 3 dU P dV du P dv dS = + ⇒ ds = + T T  T  T  per unit mass 2nd Law Analysis for a Closed System (Control Mass) MER TER TTER dW dQ CM We can first perform a 1st law energy balance on the system shown above. dU = δQ + δW (1) For a simple compressible system δW = −P dV (2) From Gibb’s equation we know TT ER dS = dU + P dV (3) 4 Combining (1), (2) and (3) we get TT ER dS = δQ Therefore δQ (dS)CM = + dP    T   S ≡ storage  TER ≡ production ≡ entropy f low Integrating gives Q1−2 (S2 − S1 )CM = + Sgen TT ER    ≥0 where Q1−2 - the entropy associated with heat transfer across a TT ER finite temperature difference, i.e. T > 0 5 2nd Law Analysis for Open Systems (Control Volume) A S= -dQ 1-2 A S= -s Am1-2 A TTER TER A FR A A mA dQ 1-2 1-2 S=0 MER dW1-2 SCV CV B mB dQ 1-2 1-2 B TER FR B B S= sBm1-2 B S= dQ1-2 B TTER isolated Sgen ≥ 0 δSgen = (ΔS)sys + (ΔS)sur   δQA 1−2 δQB 1−2 δSgen = ΔS  CV + −sA mA 1−2 + sB m B 1−2 − + TTAER TTBER system    surroundings or as a rate equation   ⎛ ⎞ ⎛ ⎞ dS Q̇ Q̇ Ṡgen = + ⎝sṁ + ⎠ − ⎝sṁ + ⎠ dt CV TT ER TT ER OUT IN This can be thought of as generation = accumulation + OU T − IN 6 Reversible Process Example: Slow adiabatic compression of a gas A process 1 → 2 is said to be reversible if the reverse process 2 → 1 restores the system to its original state without leaving any change in either the system or its surroundings. → idealization where S2 = S1 ⇒ Sgen = 0 T2 > T1 ⇒ increased microscopic disorder V2 < V1 ⇒ reduced uncertainty about the whereabouts of molecules    + Adiabatic Reversible  P rocess ⇒ Isentropic   P rocess  Sgen =0 Q=0 S1 =S2 7 Calculation of Δs in Processes T 1 process path dA = T ds 2 s ds the area under the curve on a T − s diagram is the heat transfer for internally reversible processes  2 qint,rev = T ds and qint,rev,isothermal = T Δs 1 Tabulated Calculation of Δs for Pure Substances Calculation of the Properties of Wet Vapor: Use Tables A-4 and A-5 to find sf , sg and/or sf g for the following s = (1 − x)sf + xsg s = sf + xsf g Calculation of the Properties of Superheated Vapor: Given two properties or the state, such as temperature and pressure, use Table A-6. Calculation of the Properties of a Compressed Liquid: Use Table A-7. In the absence of compressed liquid data for a property sT,P ≈ sf @T 8 Calculation of Δs for Incompressible Materials - for an incompressible substance, dv = 0, and Cp = Cv = C du dT ds = =C T T  2 dT s2 − s1 = C(T ) 1 T T2 Δs = Cavg ln where Cavg = [C(T1 ) + C(T2 )]/2 T1 Calculation of Δs for Ideal Gases For an ideal gas with constant Cp and Cv Ideal Gas Equation ⇒ P v = RT du = Cv dT ⇒ u2 − u1 = Cv (T2 − T1 ) dh = Cp dT ⇒ h2 − h1 = Cp (T2 − T1 ) There are 3 forms of a change in entropy as a function of T & v, T & P , and P & v. T2 v2 s2 − s1 = Cv ln + R ln T1 v1 T2 P2 = Cp ln − R ln T1 P1 v2 P2 = Cp ln + Cv ln v1 P1 By setting the above equations to zero (isentropic, i.e. Δs = 0) we can obtain the isentropic equations  (k−1)/k  (k−1) T2 P2 v1 = = T1 P1 v2 where k = Cp /Cv which can be found tabulated in Table A-2 for various gases. 9 The Carnot Cycle an ideal theoretical cycle that is the most efficient conceivable based on a fully reversible heat engine - it does not include any of the irreversibilities asso- ciated with friction, viscous flow, etc. in practice the thermal efficiency of real world heat engines are about half that of the ideal, Carnot cycle T P3 = P2 W QH TH Q P4 = P1 in out TL QL s1 s4 s Process State Points Description Pump 1→2 isentropic compression from TL → TH to return vapor to a liquid state Heat Supply 2→3 heat is supplied at constant temperature and pressure Work Output 3→4 the vapor expands isentropically from the high pressure and temperature to the low pressure Condenser 4→1 the vapor which is wet at 4 has to be cooled to state point 1 The cycle is totally reversible. The reversed Carnot cycle is called the Carnot refrigeration cycle. 10 Conduction Heat Transfer Reading Problems 17-1 → 17-6 17-35, 17-57, 17-68, 17-81, 17-88, 17-110 18-1 → 18-2 18-14, 18-20, 18-34, 18-52, 18-80, 18-104 Fourier Law of Heat Conduction x=L insulated Qx+Dx x+Dx x x x=0 g Qx A The general 1-D conduction equation is given as   ∂ ∂T ∂T k + ġ = ρC ∂x ∂x     internal  ∂t longitudinal heat thermal generation inertia conduction where the heat flow rate, Q̇x , in the axial direction is given by Fourier’s law of heat conduction. ∂T Q̇x = −kA ∂x Thermal Resistance Networks Resistances in Series The heat flow through a solid material of conductivity, k is kA Tin − Tout L Q̇ = (Tin − Tout) = where Rcond = L Rcond kA 1 The total heat flow across the system can be written as T∞1 − T∞2  4 Q̇ = where Rtotal = Ri Rtotal i=1 This is analogous to current flow through electrical circuits where, I = ΔV /R Resistances in Parallel L k1 R1 Q1 k2 R2 Q2 T1 T2 In general, for parallel networks we can use a parallel resistor network as follows: where 1 1 1 1 = + + + ··· Rtotal R1 R2 R3 2  R1 T1 R2 T2 T1 Rtotal T2 = R3 and T1 − T2 Q̇ = Rtotal Cylindrical Systems k r2 r1 T1 r Qr A=2prL T2 L Steady, 1D heat flow from T1 to T2 in a cylindrical systems occurs in a radial direction where the lines of constant temperature (isotherms) are concentric circles, as shown by the dotted line in the figure above and T = T (r). Q̇r Q̇r r2 T2 − T1 = − (ln r2 − ln r1 ) = − ln 2πkL 2πkL r1 Therefore we can write   T2 − T1 ln(r2 /r1 ) Q̇r =   where R = ln(r2 /r1 ) 2πkL 2πkL 3 Critical Thickness of Insulation Consider a steady, 1-D problem where an insulation cladding is added to the outside of a tube with constant surface temperature Ti. What happens to the heat transfer as insulation is added, i.e. we increase the thickness of the insulation? The resistor network can be written as a series combination of the resistance of the insulation, R1 and the convective resistance, R2 ln(ro /ri) 1 Rtotal = R1 + R2 = + 2πkL h2πro L Note: as the thickness of the insulation is increased the outer radius, ro increases. Could there be a situation in which adding insulation increases the overall heat transfer? To find the critical radius, rc , where adding more insulation begins to decrease heat transfer, set dRtotal =0 dro dRtotal 1 1 = − =0 dro 2πkro L h2πro2 L k rc = h 4 Heat Generation in a Solid Heat can be generated within a solid as a result of resistance heating in wires, chemical reactions, nuclear reactions, etc. A volumetric heat generation terms will be defined as follows: Ėg ġ = (W/m3 ) V for heat generation in wires, we will define ġ as I 2 Re ġ = πro2 L Slab System 5       T1 + T2 T1 − T2 x q̇L2 x 2 T = − + 1− 2 2 L 2k L Cylindrical System ⎛  2 ⎞ ġr02 r T = Ts + ⎝1 − ⎠ 4k r0 where dT BC1 : = 0 @r = 0 dr BC2 : T = Ts @r = r0 Heat Transfer from Finned Surfaces The temperature difference between the fin and the surroundings (temperature excess) is usually expressed as θ = T (x) − T∞ 6 which allows the 1-D fin equation to be written as d2 θ − m2 θ = 0 dx2 where the fin parameter m is  1/2 hP m= kAc and the boundary conditions are θ = θb @ x = 0 θ → 0 as x → ∞ The solution to the differential equation for θ is θ(x) = C1 sinh(mx) + C2 cosh(mx) substituting the boundary conditions to find the constants of integration cosh[m(L − x)] θ = θb cosh(mL) The heat transfer flowing through the base of the fin can be determined as   dT Q̇b = Ac −k dx @x=0 = θb (kAc hP )1/2 tanh(mL) Fin Efficiency and Effectiveness The dimensionless parameter that compares the actual heat transfer from the fin to the ideal heat transfer from the fin is the fin efficiency actual heat transfer rate Q̇b η= = maximum heat transfer rate when hP Lθb the entire fin is at Tb 7 If the fin has a constant cross section then tanh(mL) η= mL An alternative figure of merit is the fin effectiveness given as total fin heat transfer Q̇b f in = = the heat transfer that would have hAc θb occurred through the base area in the absence of the fin Transient Heat Conduction Performing a 1st law energy balance on a plane wall gives TH − Ts Ts − T∞ Ėin − Ėout ⇒ Q̇cond = = Q̇conv = L/(k · A) 1/(h · A) where TH − Ts L/(k · A) internal resistance to H.T. = = Ts − T∞ 1/(h · A) external resistance to H.T. hL = = Bi ≡ Biot number k Rint 500, 000 3. Combined Laminar and Turbulent Boundary Layer Flow, Isothermal (UWT) average, turbulent, UWT, hL L N uL = = (0.037 Re0.8 L − 871) P r 1/3 ⇒ 0.6 < P r < 60, ReL > 500, 000 k 4. Laminar Boundary Layer Flow, Isoflux (UWF) N ux = 0.453 Re1/2 x P r 1/3 ⇒ local, laminar, UWF, P r ≥ 0.6 5. Turbulent Boundary Layer Flow, Isoflux (UWF) N ux = 0.0308 Re4/5 x P r 1/3 ⇒ local, turbulent, UWF, P r ≥ 0.6 6 Flow Over Cylinders and Spheres 1. Boundary Layer Flow Over Circular Cylinders, Isothermal (UWT) The Churchill-Berstein (1977) correlation for the average Nusselt number for long (L/D > 100) cylinders is ⎡  5/8 ⎤4/5 average, UWT, Re < 107 ReD + f (P r) ReD ⎣1 + ∗ 1/2 ⎦ N uD = SD ⇒ 0 ≤ P r ≤ ∞, Re · P r > 0.2 282000 ∗ where SD = 0.3 is the diffusive term associated with ReD → 0 and the Prandtl number function is 0.62 P r 1/3 f (P r) = [1 + (0.4/P r)2/3 ]1/4 All fluid properties are evaluated at Tf = (Tw + T∞ )/2. 2. Boundary Layer Flow Over Non-Circular Cylinders, Isothermal (UWT) The empirical formulations of Zhukauskas and Jakob are commonly used, where hD N uD ≈ = C Rem D Pr 1/3 ⇒ see Table 19-2 for conditions k 3. Boundary Layer Flow Over a Sphere, Isothermal (UWT) For flow over an isothermal sphere of diameter D average, UWT,  !  1/4 0.7 ≤ P r ≤ 380 ∗ 1/2 2/3 μ∞ N uD = SD + 0.4 ReD + 0.06 ReD Pr 0.4 ⇒ 3.5 < ReD < 80, 000 μw ∗ where the diffusive term at ReD → 0 is SD =2 and the dynamic viscosity of the fluid in the bulk flow, μ∞ is based on T∞ and the dynamic viscosity of the fluid at the surface, μw , is based on Tw. All other properties are based on T∞. 7 Internal Flow Lets consider fluid flow in a duct bounded by a wall that is at a different temperature than the fluid. For simplicity we will examine a round tube of diameter D as shown below The velocity profile across the tube changes from U = 0 at the wall to a maximum value along the center line. The average velocity, obtained by integrating this velocity profile, is called the mean velocity and is given as 1  ṁ Um = u dA = Ac Ac ρmAc where the area of the tube is given as Ac = πD 2 /4 and the fluid density, ρm is evaluated at Tm. The Reynolds number is given as Um D ReD = ν For flow in a tube: ReD < 2300 laminar flow 2300 < ReD < 4000 transition to turbulent flow ReD > 4000 turbulent flow 8 Hydrodynamic (Velocity) Boundary Layer the hydrodynamic boundary layer thickness can be approximated as  −1/2 Um x 5x δ(x) ≈ 5x = √ ν Rex 9 Thermal Boundary Layer the thermal entry length can be approximated as Lt ≈ 0.05ReD P rD (laminar flow) for turbulent flow Lh ≈ Lt ≈ 10D 1. Laminar Flow in Circular Tubes, Isothermal (UWT) and Isoflux (UWF) For laminar flow where ReD ≤ 2300 N uD = 3.66 ⇒ fully developed, laminar, UWT, L > Lt & Lh N uD = 4.36 ⇒ fully developed, laminar, UWF, L > Lt & Lh developing laminar flow, UWT,  1/3  0.4 P r > 0.5 ReD P rD μb N uD = 1.86 ⇒ L < Lh or L < Lt L μw For non-circular tubes the hydraulic diameter, Dh = 4Ac /P can be used in conjunction with 10 Table 10-4 to determine the Reynolds number and in turn the Nusselt number. In all cases the fluid properties are evaluated at the mean fluid temperature given as 1 Tmean = (Tm,in + Tm,out) 2 except for μw which is evaluated at the wall temperature, Tw. 2. Turbulent Flow in Circular Tubes, Isothermal (UWT) and Isoflux (UWF) For turbulent flow where ReD ≥ 2300 the Dittus-Bouler equation (Eq. 19-79) can be used turbulent flow, UWT or UWF, 0.7 ≤ P r ≤ 160 ReD > 2, 300 n = 0.4 heating N uD = 0.023 Re0.8 D Pr n ⇒ n = 0.3 cooling For non-circular tubes, again we can use the hydraulic diameter, Dh = 4Ac /P to determine both the Reynolds and the Nusselt numbers. In all cases the fluid properties are evaluated at the mean fluid temperature given as 1 Tmean = (Tm,in + Tm,out) 2 11 Natural Convection What Drives Natural Convection? a lighter fluid will flow upward and a cooler fluid will flow downward as the fluid sweeps the wall, heat transfer will occur in a similar manner to boundary layer flow however in this case the bulk fluid is stationary as opposed to moving at a constant velocity in the case of forced convection We do not have a Reynolds number but we have an analogous dimensionless group called the Grashof number buouancy force gβ(Tw − T∞ )L3 Gr = = viscous force ν2 where g = gravitational acceleration, m/s2 β = volumetric expansion coefficient, β ≡ 1/T (T is ambient temp. in K) 12  Tw = wall temperature, K T∞ = ambient temperature, K L = characteristic length, m ν = kinematic viscosity, m2 /s The volumetric expansion coefficient, β, is used to express the variation of density of the fluid with respect to temperature and is given as   1 ∂ρ β=− ρ ∂T P Natural Convection Heat Transfer Correlations The general form of the Nusselt number for natural convection is as follows: N u = f (Gr, P r) ≡ CRamP r n where Ra = Gr · P r C depends on geometry, orientation, type of flow, boundary conditions and choice of char- acteristic length. m depends on type of flow (laminar or turbulent) n depends on the type of fluid and type of flow 1. Laminar Flow Over a Vertical Plate, Isothermal (UWT) The general form of the Nusselt number is given as ⎛ ⎞1/4 ⎛ ⎞1/4 hL ⎜ gβ(Tw − T∞ )L3 ⎟ ⎜ ν ⎟ ⎜ ⎟ ⎜ ⎟ 1/4 N uL = =C ⎜ ⎟ ⎜ ⎟ = C GrL P r 1/4 kf ⎝ 2 ν ⎠ ⎝ α ⎠       Ra1/4 ≡Gr ≡P r where gβ(Tw − T∞ )L3 RaL = GrL P r = αν 13 2. Laminar Flow Over a Long Horizontal Circular Cylinder, Isothermal (UWT) The general boundary layer correlation is ⎛ ⎞n ⎛ ⎞n hD ⎜ gβ(Tw − T∞ )D 3 ⎟ ⎜ ν ⎟ ⎜ ⎟ ⎜ ⎟ n N uD = =C⎜ ⎟ ⎜ ⎟ = C GrD P rn kf ⎝ ν 2 ⎠ ⎝ α ⎠        Ran ≡Gr ≡P r D where gβ(Tw − T∞ )D 3 RaD = GrD P r = αν All fluid properties are evaluated at the film temperature, Tf = (Tw + T∞ )/2. 14 Natural Convection From Plate Fin Heat Sinks Plate fin heat sinks are often used in natural convection to increase the heat transfer surface area and in turn reduce the boundary layer resistance 1 R ↓= hA ↑ For a given baseplate area, W × L, two factors must be considered in the selection of the number of fins more fins results in added surface area and reduced boundary layer resistance, 1 R ↓= hA ↑ more fins results in a decrease fin spacing, S and in turn a decrease in the heat transfer coefficient 1 R ↑= h↓A A basic optimization of the fin spacing can be obtained as follows: Q̇ = hA(Tw − T∞ ) 15 where the fins are assumed to be isothermal and the surface area is 2nHL, with the area of the fin edges ignored. For isothermal fins with t < S   L Sopt = 2.714 Ra1/4 with gβ(Tw − T∞ L3 ) Ra = Pr ν2 The corresponding value of the heat transfer coefficient is h = 1.31k/Sopt All fluid properties are evaluated at the film temperature. 16 Radiation Heat Transfer Reading Problems 21-1 → 21-6 21-21, 21-24, 21-41, 21-61, 21-69 22-1 → 21-5 22-11, 22-17, 22-26, 22-36, 22-71, 22-72 Introduction It should be readily apparent that radiation heat transfer calculations required several additional considerations in addition to those of conduction and convection, i.e. optical aspects: the manner in which an emitting body “sees” its neighbors surface conditions Blackbody Radiation A blackbody is an ideal radiator that absorbs all incident radiation regardless of wavelength and direction 1  at a given temperature and wavelength, no surface can emit more energy than a blackbody emitted radiation is a function of wavelength and temperature but is independent of direction, i.e. a black body is a diffuse emitter (independent of direction) Definitions 1. Blackbody emissive power: the radiation emitted by a blackbody per unit time and per unit surface area Eb = σ T 4 [W/m2 ] ⇐ Stefan-Boltzmann law where σ = Stefan-Boltzmann constant = 5.67×10−8 W/(m2 ·K 4 ) and the temperature T is given in K. 2. Spectral blackbody emissive power: the amount of radiation energy emitted by a black- body per unit surface area and per unit wavelength about the wavelength λ. The following relationship between emissive power, temperature and wavelength is known as Plank’s dis- tribution law C1 Eb,λ = [W/(m2 · μm)] λ5 [exp(C2 /λT ) − 1] where C0 = 2.998 × 108 [m/s] (vacuum conditions) C1 = 2πhC02 = 3.743 × 108 [W · μm4 /m2 ] C2 = hC0 /K = 1.439 × 104 [μ · K] K = Boltzmann constant ≡ 1.3805 × 10−23 [J/K] h = Plank s constant ≡ 6.63 × 10−34 [J · s] Eb,λ = energy of radiation in the wavelength band dλ per unit area and time If we integrate the spectral emissive power between dλ = λ2 − λ1 we will obtain the blackbody emissive power given as Eb (T ) = σT 4. 2 3. Blackbody radiation function: the fraction of radiation emitted from a blackbody at tem- perature, T in the wavelength band λ = 0 → λ  λ  λ C1 Eb,λ (T ) dλ dλ 0 λ5 [exp(C2 /λT ) − 1] f0→λ =  0∞ = Eb,λ (T ) dλ σT 4 0 let t = λT and dt = T dλ, then  λ C1 T 5 (1/T )dt 0 t5 [exp(C2 /t) − 1] f0→λ = σT 4 C1  λT dt = σ 0 t5 [exp(C2 /t) − 1] = f (λT ) f0→λ is tabulated as a function λT in Table 21.2 We can easily find the fraction of radiation emitted by a blackbody at temperature T over a discrete wavelength band as fλ1 →λ2 = f (λ2 T ) − f (λ1 T ) fλ→∞ = 1 − f0→λ 3 Radiation Properties of Real Surfaces The thermal radiation emitted by a real surface is a function of surface temperature, T , wavelength, λ, direction and surface properties. Eλ = f (T, λ, direction, surface properties) ⇒ spectral emissive power while for a blackbody, the radiation was only a function of temperature and wavelength Eb,λ = f (T, λ) → diffuse emitter ⇒ independent of direction Definitions 1. Emissivity: defined as the ratio of radiation emitted by a surface to the radiation emitted by a blackbody at the same surface temperature. radiation emitted by surface at temperature T (T ) = radiation emitted by a black surface at T  ∞  ∞ Eλ (T ) dλ λ (T )Ebλ (T ) dλ E(T ) =  0∞ = 0 = Ebλ (T ) dλ Eb (T ) σT 4 0 where  changes rather quickly with surface temperature. 2. Diffuse surface: properties are independent of direction. 4 3. Gray surface: properties are independent of wavelength. 4. Irradiation, G: the radiation energy incident on a surface per unit area and per unit time An energy balance based on incident radiation gives G = ρG + αG + τ G where G = incident radiation or irradiation, W/m2 ρG = reflected radiation, W/m2 αG = absorbed radiation, W/m2 τ G = transmitted radiation, W/m2 with the associated surface properties being ⎫ ρ = reflectivity ⎪ ⎬ α = absorptivity ⎪ ⇒ function of λ & T of the incident radiation G ⎭ τ = transmissivity  = emissivity ⇒ function of λ & T of the emitting surface If we normalize with respect to the total irradiation α+ρ+τ =1 In general  = α. However, for a diffuse-gray surface (properties are independent of wave- length and direction) =α diffuse-gray surface 5  These unsubscripted values of α, ρ and τ represent the average properties, i.e. due to incident radiation energy from all directions over a hemispherical space and including all wavelengths. We can just as easily define these properties for a specific wavelength, such that Gλ = ρλ Gλ + αλ Gλ + τλ Gλ where ρλ = spectral reflectivity = f (λ, T ) αλ = spectral absorptivity = f (λ, T ) τλ = spectral transmissivity = f (λ, T ) and ρλ + αλ + τλ = 1 λ depends strongly on the temperature of the emitting surface but not at all on the irradiation field Gλ. 5. Radiosity, J : the total radiation energy leaving a surface per unit area and per unit time. For a surface that is gray and opaque, i.e.  = α and α + ρ = 1, the radiosity is given as J = radiation emitted by the surface + radiation reflected by the surface =  Eb + ρG = σT 4 + ρG Since ρ = 0 for a blackbody, the radiosity of a blackbody is J = σT 4 Diffuse-Gray Surfaces,  = α Conditions for  = α in many radiation problems, the calculations are greatly simplified when  = α, which defines a gray surface. Here, we seek the necessary conditions for  = α. This is a 3-step procedure. 6  the first step involves adopting Kirchhoff’s law which is stated here without proof: (λ, T, φ, θ) = α(λ, T, φ, θ) This equation is always applicable because both (λ, T, φ, θ) and α(λ, T, φ, θ) are inher- ent surface properties the second step involves finding the requirements for (λ, T ) = α(λ, T ). This is done using the definitions hemispherical radiation, as given below:  2π  π/2 ⎫ ⎪ ⎪ λ,θ cos θ sin θdθdφ ⎪ ⎪ ⎪ ⎪ (λ, T ) = 0  0 ⎪ ⎪ 2π  π/2 ⎪ ⎪ ⎪ ⎪ cos θ sin θdθdφ ⎪ ⎪ 0 0 ⎬ ⎪ ⇒ ??? (λ, T ) = α(λ, T )  2π  π/2 ⎪ ⎪ ⎪ ⎪ αλ,θ Iλ cos θ sin θdθdφ ⎪ ⎪ ⎪ ⎪ α(λ, T ) = 0  0 ⎪ 2π  π/2 ⎪ ⎪ ⎪ Iλ cos θ sin θdθdφ ⎪ ⎭ 0 0 From this, we can see that for (λ, T ) = α(λ, T ), we must have either a diffuse surface or a diffuse irradiation. As indicated before, in most engineering calculations, we use direction averaged properties which amounts to the assumption of a diffuse surface. having adopted (λ, T, φ, θ) = α(λ, T, φ, θ) and (λ, T ) = α(λ, T ), the third step involves finding the requirements for  = α; that is:  ∞  ∞ λ Eλ,b dλ αλ Gλ dλ (T ) = 0 ∞ =? = 0 ∞ = α(T ) Eλ,b dλ Gλ dλ 0 0 From this, we can see that for (T ) = α(T ), we must have either Gλ = Eλ,b which means that the irradiation originated from a blackbody or (λ, T ) = constant and α(λ, T ) = constant. To see the last point, remember that if (λ, T ) = constant then (T ) = (λ, T ) and similarly α(T ) = α(λ, T ). But at step two, we have already established that (λ, T ) = α(λ, T ), hence it follows that (T ) = α(T ). by definition, a gray surface is one for which (λ, T ) and α(λ, T ) are independent of λ over the dominant spectral regions of Gλ and Eλ. 7 View Factor (Shape Factor, Configuration Factor) Definition: The view factor, Fi→j is defined as the fraction of radiation leaving surface i which is intercepted by surface j. Hence Q̇i→j radiation reaching j Fi→j = = A i Ji radiation leaving i AiFi→j = Aj Fj→i This is called the reciprocity relation. consider an enclosure with N surfaces  N Fi→j = 1 ; i = 1, 2,... , N j=1 This is called the summation rule. Note that Fi→i = 0 for a concave surface. For a plane or convex surface Fi→i = 0. Hottel Crossed String Method Can be applied to 2D problems where surfaces are any shape, flat, concave or convex. Note for a 2D surface the area, A is given as a length times a unit width. (total crossed) − (total uncrossed) A1 F12 = A2 F12 = 2 8 A1 and A2 do not have to be parallel 1 A1 F12 = A2 F21 = [(ac + bd) − (bc + ad)] 2       crossed uncrossed Radiation Exchange Between Diffuse-Gray Surfaces Forming an Enclosure an energy balance on the i th s