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ERT 206/4 Thermodynamics CHAPTER 2 The First Law and Other Basic Concepts Miss. Rahimah Bt. Othman Email: rahimah@unimap.edu.my COURSE OUTCOME 1 CO1) 1. Chapter 1: Introduction to Thermodynamics 2. Chapter 2: The First Law and Other Basic Concepts Define, discuss, apply and analyze internal energy, first law, energy balance-closed system, thermodynamic state and state function, equilibrium, the Phase Rule, reversible process, constant-V and constant-P processes, enthalpy and heat capacity. 3. Chapter 3: Volumetric properties of pure fluids 4. Chapter 4: Heat effects 5. Chapter 5: Second law of thermodynamics 6. Chapter 6: Thermodynamics properties of fluids System Closed system - Known as controlled mass system (mass constant). - Allow energy flow (in/out) and volume is not necessarily constant as a controlled. - Often undergo processes during the internal energy of the system changes. ∆(Energy of the system) + (∆Energy of surroundings) = 0 ∆(Energy of the system) = Q+W ∆(Energy of the system) =∆U+∆EK+ ∆EP Open system - Known as controlled volume system. - Allow mass and energy cross through the boundary of open system (= controlled surface). Total Energy E = U+ KE+ PE Microscopic Energy Internal energy, U - The internal energy of a system comes from the kinetic energy of the molecules motion and potential energy from the atomic bonds. - Eg: chemical energy, nuclear, sensible energy, latent energy. Makroscopic energy - Kinetic energy, KE - Potential energy, PE Thermodynamic state and State functions The phase rule (F = 2 – π +N) The reversible process Heat Capacity At constant Volume, Cv and Constant Pressure, CP CLOSED SYSTEM Equilibrium (Eg: heat, mechanical, phase, chemical) Entalphy (H = U +PV) Constant-V & Constant-P Processes dmCV (m) fs 0 dt (2.25) (m) fs m 3 m1 m 2 dmCV ( uA) fs 0 dt H (2.26) u 2 g z Q Ws 2 gc gc (2.32a) Energy Balance at Steady-state Flow Processes Mass Balance OPEN SYSTEM Energy Balance Mass Balance at Steady-state Flow Processes m u2 A2 u1 A1 u A V1 V1 V d (mU ) cv 1 2 [(U u zg ) m] fs Q work rate dt 2 d (mU )CV 1 H u 2 zg m Q W dt 2 fs (2.28) (2.27) Example 2.9 (HEAT CAPACITY) Air at 1 bar and 25oC (298.15 K) is compressed to 5 bar and 25oC (298.15 K) by two different 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 Q and W and ∆U and ∆H of the air for each path. Given ; Cv = 20.78 and CP = 29.10 Jmol-1K-1 Assume ; - For air, PV/T is a constant - At 25oC (298.15 K) and 1 bar the molar volume of air is 0.02479 m3mol-1. Solution In each case take the system as 1 mol of air contained in an imaginary piston/cylinder arrangement. Because the process considered are mechanically reversible, the piston is imagined to move in the cylinder without friction. The final volume is; V2 V1 P1 1 0.02479 0.004958 m3 P2 5 Example 2.9-cont’ (a) During the first step the air is cooled at the constant pressure of 1 bar until the final volume of 0.004958 m3 is reached. The temperature of the air at the end of this cooling step is; Whence; V2 0.004958 T ' T1 298.15 59.63 K V1 0.02479 Q H C P T (29.10)(59.63 298.15) 6941 J U H ( PV ) H PV 6941 (1 x 105 )(0.004958 0.02479) 4958 J During the second step the volume is held constant at V2 while the air is heated to its final state. By Eq. (2.19), U Q CV T (20.78)(298.15 59.63) 4958 J The complete process represents the sum of its steps. Hence, Q 6941 4958 1983 J And U 4958 4958 0 Example 2.9-cont’ Because the first law applies to the entire process, H Q W , and therefore, whence; 0 1983 W W 1983 J Equation (2.15), H U PV , also applies to the entire process. But T1 T2 , and therefore, P1V1 P2V2 . Hence, PV 0 and H U 0 (b) Two different steps are used in this case to reach the same final state of the air. In the first step the air is heated at a constant volume equal to its initial value until the final pressure of 5 bar is reached. The air temperature at the end of this step is; P2 5 T ' T1 298.15 1490.75K P1 1 For this step the volume is constant, and Q U CV T (20.78)(1490.75 298.15) 24788 J Example 2.9-cont’ In the second step the air is cooled at P = 5bar to its final state: Q H C P T (29.10)( 298.15 1490.75) 34703 J ΔU H PV H PV 34703 (5 x 105 )(0.004958 0.02479) 24788 J For the two steps combined, Q 24788 34703 9915 J ΔU 24788 24788 0 W and as before U Q 0 9915 9915 J H U 0 Conclusion: The property changes ∆U and ∆H calculated for the given change in state are the same for both paths. On the other hand the answers to parts (a) and (b) show that Q and W depend on the path. Example 2.11 (MASS ENERGY FOR OPEN SYSTEMS) Liquid n-hexane flows at a rate of ṁ = 0.75 kg s-1 in a pipe with inside diameter D = 5 cm. What are q, ṅ, and u? What would these quantities be for the same ṁ if D = 2 cm? Assume for liquid n-hexane that ρ = 659 kg m-3. Solution We have q m 1 whence M 1 n m and 0.75 kg s 1 3 1 q 0.00114 m s 3 659 kg m 0.75 kg s 10 n 1 g kg 1 1 8.703 mol s 86.177 g mol 1 3 Given ṁ, these quantities are independent of D. The velocity, however, depends on diameter through u = qA-1, where, for a circular cross-section, A = (π/4)D2. For D = 5 cm, 2 A whence 5 x 10 4 2 m 0.00196 m 2 0.00114 m 3s 1 1 u 0.582 m s 0.00196 m 2 Example 2.11-cont’ Similarly, for D = 2 cm, A 0.000314 m 2 and 0.00114 u 3.63 m s 1 0.000314 Example 2.14 (ENERGY BALANCE FOR OPEN SYSTEM) An insulated, electrically heated tank for hot tank for hot water contains 190 kg of liquid water at 60 oC (333.15 K) when a power outage occurs. If water is withdrawn from the tank at a steady state of ṁ = 0.2 kg s-1, how long will it take for the temperature of the water in the tank to drop from 60 to 35 oC (333.15 K to 308.15 K)? Assume: (a) Cold water enters the tank at 10 oC (283.15 K) (b) Heat losses from the tank is negligible. For liquid water let CV = CP = C, independent of T and P. Solution Here, Q W 0 . Additional assumption; (a) Perfect mixing of the contents of the tank. (b) m(flow rate in) = m(flow rate out) = mcv = constant. (c) Differences between inlet and outlet kinetic and potential energy can be neglected. Example 2.14-cont’ Equation (2.29) is therefore written as; dU dT C dt dt and H H1 C (T T1 ) Where unsubscripted quantities refer to the contents of the tank and H1 is the specific enthalpy of the water entering the tank. With CV = CP = C, dU m m ( H H1 ) 0 dt The energy balance then becomes, on arrangement, dt m dT m T T1 Integration from t = 0 (where T = T0) to arbitrary time t yields m T T1 t ln m T 0T1 Example 2.14-cont’ Substitution of numerical values into this equation gives, for the conditions of this problem, t 190 308.15 283.15 ln 658.5 s 0.2 333.15 283.15 Thus, it takes about 11 minutes for the water temperature in the tank to drop from 60 to 35 oC (333.15 K to 308.15 K). Example 2.16 (ENERGY BALANCE FOR OPEN SYSTEM) Air at 1 bar and 25oC (298.15 K) enters a compressor at low velocity, discharges at 3 bar, and enters a nozzle in which it expands to a final velocity of 600 ms-1 at the initial conditions of pressure and temperature. If the work of compression is 240 kJ per kilogram of air, how much heat must be removed during compression? Solution Because the air returns to its initial conditions of T and P, the overall process produces no change in enthalpy of the air. Moreover, the potentialenergy change of the air is presumed negligible. Neglecting also the initial kinetic energy of the air, we write Eq. (2.32a) as; 2 u Q 2 Ws 2 Example 2.16-cont’ The kinetic-energy term is evaluated as follows: 2 1 2 1 m m2 u2 600 180 000 2 2 2 s s m 2 kg 180 000 2 . 180 000 N m kg 1 180 kJ kg 1 s kg Then Q 180 240 60 kJ kg 1 Heat in the amount of 60 kJ must be removed per kilogram of air compressed. Example 2.17 (ENERGY BALANCE FOR STEADY-STATE FLOW PROCESSES OPEN SYSTEM) Water 93.5oC (366.65 K) is pumped from a storage tank at the rate of 3.15 x 10-3 m3 s-1. The motor for the pump supplies work at the rate of 1.5 kW. The water goes through a heat exchanger , giving up heat at the rate of 700 kW, and is delivered to a second storage tank at an elevation 15 m above the first tank. What is the temperature of the water delivered to the second tank? Solution This is a steady-state flow process for Eq. (2.3b) applies. The initial and final velocities of water in the storage tank are negligible, and the term ∆u2 / 2gc may be omitted. The remaining terms are expressed in units of kJ kg-1 through use of appropriate conversion factors. At 366.65 K the density of water is 958 kg m-3 thus the mass flow rate is; (3.15)(10-3)(958) = 3.018 kJ kg-1 from which we obtain Q = - 700/3.018 = - 231.9 kJ kg-1 Example 2.17-cont’ Since 1 kW is equivalent to 1 kJ s-1, the shaft work is; Ws = (1.5)(1) /3.018 = 0.497 kJ kg-1 If the local acceleration of gravity is taken as the standard value of 9.807 ms-2, the potential-energy term becomes; 9.807 159.807 0.147 kJ kg 1 g z 9.807 1000 gc Equation (2.32b) now yields ∆H: g H Q Ws z 231.9 0.497 0.147 231.55 kJ kg 1 gc The enthalpy of saturated water at 93.5 oC (366.65 K) is given in the steam tables as 391.6 kJ kg-1. Thus H H 2 H1 H 2 391.6 231.55 and H 2 391.6 231.55 160.05 kJ kg 1 Example 2.17-cont’ The temperature of water having this enthalpy is found from the steam tables to be; t 38.2 o C or T 311.35 K In this example Ws and (g/gc)∆z are small compared with Q, and for practical purposes they could be neglected. ERT 206/4 Thermodynamics CHAPTER 3 Volumetric Properties of Pure Fluids Miss. Rahimah Bt. Othman Email: rahimah@unimap.edu.my COURSE OUTCOME 1 CO1) 1. Chapter 1: Introduction to Thermodynamics 2. Chapter 2: The First Law and Other Basic Concepts 2. Chapter 3: Volumetric properties of pure fluids DESCRIBE and EXPLAIN PVT behavior of pure substances, Virial Equation of State, ideal gas, Virial EquationAPPLICATION, cubic equation of state, generalized correlations for gases and liquids. 4. Chapter 4: Heat effects 5. Chapter 5: Second law of thermodynamics 6. Chapter 6: Thermodynamics properties of fluids Objectives • Introduce the concept of a pure substance. • Discuss the physics of phase-change processes. • Illustrate the P-v, T-v, and P-T property diagrams and P-v-T surfaces of pure substances. • Demonstrate the procedures for determining thermodynamic properties of pure substances from tables of property data. • Describe the hypothetical substance “ideal gas” and the ideal-gas equation of state. • Apply the ideal-gas equation of state in the solution of typical problems. • Introduce the compressibility factor, which accounts for the deviation of real gases from ideal-gas behavior. • Present some of the best-known equations of state. Pure Fluids/ Substance A substance that has a fixed chemical composition throughout is called a Pure Substance. Pure Substance: - N2, O2, gaseous Air -A mixture of liquid and gaseous water is a pure substance, but a mixture of liquid and gaseous air is not. PHASES OF A PURE SUBSTANCE The molecules in a SOLID are kept at their positions by the large spring like inter-molecular forces. In a solid, the attractive and repulsive forces between the molecules tend to maintain them at relatively constant distances from each other. The arrangement of atoms in different phases: (a) molecules are at relatively fixed positions in a solid, (b) groups of molecules move about each other in the LIQUID phase, and (c) molecules move about at random in the GAS phase. Phase-Change Processes of Pure Substance Compressed liquid or a subcooled liquid: A liquid that is not about to vaporize. Saturated liquid: A liquid that is about to vaporize. Saturated vapor: A vapor that is about to condense. Saturated liquid-vapor mixture: the liquid and vapor phases coexist in equilibrium. Superheated vapor: A vapor that is not about to condense T-v diagram for the heating process of water at constant pressure. 2-1 Phase-Change Processes of Pure Substance Saturated temperature, Tsat: At a given pressure, the temperature at which a pure substance changes phase. Saturated pressure, Psat: At a given temperature, the pressure at which a pure substance changes phase. Latent heat: the amount of energy absorbed or released during a phase-change process. Latent heat of fusion: the amount of energy absorbed during melting. Latent heat of vaporization: the amount of energy absorbed during vaporization. PVT BEHAVIOR OF PURE SUBSTANCES/ FLUIDS Critical point – highest combination of pressure and temperature where the fluid exist in liq-vap equilibrium The 2-C line, also known as vaporization curve is where liquid-vapor is in equilibrium The 1-2 line, also known as sublimation curve is where solidvapor is in equilibrium Triple point, three phases exist in equilibrium (F=0) The 2-3 line, also known as fusion curve is where solidliquid is in equilibrium PV diagram Boundaries in PT diagram becomes region when illustrate with PV diagram Critical point becomes peak of the curve Triple point becomes horizontal line Compressed liquid region Saturated liquid line at boiling temperature Superheated vapor region Saturated vapor line at condensation temperature T >Tc, the line do not cross the boundary Isotherms in sub-cooled/ compressed lliquid region are steep, because liquid volumes change little with large changes in pressure SINGLE PHASE REGION At single phase regions in PV diagram, there is a relation connecting P,V and T. This relation known as PVT equation of state; f(P,V,T)=0 If V is considered as a function of T and P, then V=V(T,P) V V dV dT dP T P T T 1 V Volume exp ansivity V T P 1 V V T T isothermal compressib ility The combination will yield; dV dT dP V 3.4 Because the isotherms on the left side of PV diagram are very steep, both β and κ are small. Because of that, the liquid is known as incompressible fluid, where both constants are equal to zero However, this is just idealization, and in incompressible fluid, no equation of state exist, since V is independent of T and P If, we still want to calculate, for liquids, β is positive, and κ is positive as well. Integration of 3.4 yield ln V2 (T2 T1 ) ( P2 P1 ) V1 Try examples 3.1 3.5 Virial Equation of State The coefficients a(T), b(T), c(T), and so on, that are functions of temperature alone are called virial coefficients. Ideal-Gas Temperature; Universal Gas Constant Figure 3.4: Plot of PV vs. P for 4 gaseous at triple-point temperature of water. The limiting value of PV as P0 is the same for all of the gaseous. Limiting value (asterisk) PV* = a = f(T) This properties of gaseous is the basis for establishing an absolute temperature scale. The simplest procedure to define Kelvin scale: 1. (PV)* = a ≡ RT 2. (PV)*t = R x 273.16K(3.8) 3. PV * T / K PV * t 273.16 K * PV T / K 273.16 PV *t (3.9) •In the limit, P0, molecules separated by finite distance •Volumes becomes negligible compare with the total volume of the gas, and intermolecular forces approach zero • These condition define an IDEAL GAS state & Eq. 3.9 establishes the ideal-gas temperature scale • The proportionality constant R = universal gas constant Through the use of conversion factors, R may be expresses in various units. Commonly used values are given as above table. 2 Forms of the Virial Equation • Auxiliary thermodynamic property = • = Compressibility factor • • • • • • Z = 1 + B’P+C’P2 +D’P3 +… Z= 1 + B/V +C/V2 + D/V3 +… Virial expansion = Eq 3.11 & Eq 3.12 B’, C’, D’ …, B, C, D = virial coefficient B’ = 2nd virial coefficient C’ = 3rd virial coefficient…etc PV Z RT (3.10) (3.11) (3.12) Eq 3.11 & Eq 3.12 B B' (3.13a ) RT C B2 C' ( RT ) 2 (3.13b) D 3BC 2 B 3 D' ( RT )3 (3.13c) THE IDEAL GAS • Because the Eq 3.12 arise on account of molecular interactions, the virial coefficients B,C…etc = 0 were no such interaction to exist Z 1 or Z 1 or PV RT PV RT • Phase Rule = Internal energy of a real gas is a function of pressure as well as of temperature. THE IDEAL GAS • Equation of state: Any equation that relates the pressure, temperature, and specific volume of a substance. • The simplest and best-known equation of state for substances in the gas phase is the ideal-gas equation of state. This equation predicts the P-v-T behavior of a gas quite accurately within some properly selected region. Ideal gas equation of state R: gas constant M: molar mass (kg/kmol) Ru: universal gas constant U = U(T) (Ideal gas) Different substances have (3.15) different gas constants. Implied Property Relations for an Ideal Gas (f(T) only) • Heat Capacity for constant volume, Cv U dU (T ) Cv Cv(T ) dT T v (3.1) • Eq 2.11 applied to an Ideal Gas H U PV U (T ) RT H (T ) (3.17) Implied Property Relations for an Ideal Gas (f(T) only) Heat Capacity for constant pressure, Cp dH (T ) H CP CP dT T P (3.18) Useful relation between Cv & Cp CP dH dU R CV R dT dT (3.19) NOTE: This equation does not imply that Cp and Cv are themselves constant for an ideal gas, but only that they vary with temperature in such a way that their differences is equal to R For any ∆ of state of an ideal gas, Eq 3.16 and Eq.3.18 lead to: dU CV dT (3.20a ) U CV dT (3.20b) dH C P dT (3.21a ) H C P dT (3.21b) • T1 & T2 • a---b = Constant volume process • a---c & a---d ≠ constant volume The graph show that the ∆U= CV dT Equation for Process Calculation for Ideal Gas Working equation of dQ and dW depend on which pair of these variables is selected as independent With P=RT/V, dV V dP dQ C P dT RT P dQ CV dT RT dV V (3.22) dW RT (3.24) dW RdT RT (3.23) dP P (3.25) With V=RT/P and Cv given by Eq 3.19, dQ & dW written as Eq.3.24 & Eq.3.25 With T=PV/R, the work is simply dW=-PdV, and with Cv given by Eq.3.19, CV CP dQ VdP PdV R R (3.26) Isothermal Process • Q=-W V2 P2 Q W RT ln R ln V1 P1 (const T ) (3.27) Isobaric Process Q H CP dT (Const P) (3.28) Isochoric Process Q U CV dT (Const V ) (3.29) Adiabatic Process; Constant Heat Capacities dT R dV T CV V C P / CV V1 dT 1 RdV T2 T CV V T1 V2 similar to Eq.3.24 & Eq.3.26 lead to : T2 T1 P2 P 1 R / CP V1 P2 V P1 2 C P / CV Also, TV 1 const. (3.30a) PV const (3.30c) TP (1 ) / const. (3.30b) CP CV (3.31) IMPORTANT: Eq 3.30 are restricted in application to ideal gases with Constant Heat capacities undergoing mechanically reversible adiabatic expansion or compression For ideal gas, the WORK of any adiabatic closed-system process is given by: dW dU CV dT W U CV T (3.32) Because RT1 = P1V1 and RT2 = P2V2, W RT2 RT1 P2V2 P1V1 1 1 (3.33) Elimination of V2 from Eq 3.33 by Eq 3.30c, valid for mechanically reversible process, lead to: P1V1 P2 W 1 P1 1 / RT P 1 2 1 1 P1 1 / 1 Polytropic Process PV const (3.35a ) TV 1 const (3.35b) TP (1 ) / const RT1 P2 W 1 P1 (3.35c ) ( 1) / ( ) RT1 P2 Q ( 1)( 1) P1 1 ( 1) / (3.36) 1 (3.37) (3.34) Polytropic Process Isobaric process Isothermal process Adiabatic process Isochoric process APPLICATION OF VIRIAL EQ. All isotherms originate at Z=1 for P=0 PV BP Z 1 RT RT (3.38) Applied to vapors at sub critical temperature up to their saturation pressure When the virial equation is truncated to 3 terms, the appropriate form is: PV B C Z 1 2 RT V V (3.40) CUBIC EQ. OF STATE 1. The Van der Waals Equation of state RT a P 2 V b V (3.41) 2. A Generic Cubic Equation of State RT a(T ) P V b V b (V b) (3.42) 3. Determination of Eq-of State Parameters P 0 (3.43) V T ;cr 2P 2 0 (3.44) V T ;cr 4. Theorem of Corresponding State: Acentric Factor 1.0 log( Prsat )Tr 0.7 (3.48) NOTE: All fluid having the same value of , when compared at the same Tr And Pr have about the same value of Z, and all deviate from ideal gas behavior To about he same degree 5. Vapor& Vapor-like Root of the Generic Cubic Eq of state 6. Liquid & Liquid-like Root of the Generic Cubic Eq of state GENERALIZE CORRELATION FOR GASES GENERALIZE CORRELATION FOR GASES 1. Pitzer Correlation for the Compressibility Factor 2. Pitzer Correlation for the 2nd Virial Coefficient 3. Correlations for the 3rd Virial Coefficient 4. Condition of Approximate Vlidity of the Ideal-Gas Equation GENERALIZE CORRELATION FOR LIQUID Summary • Pure substance • Phases of a pure substance • Phase-change processes of pure substances – Compressed liquid, Saturated liquid, Saturated vapor, Superheated vapor – Saturation temperature and Saturation pressure • Property diagrams for phase change processes – The T-v diagram, The P-v diagram, The P-T diagram, The P-v-T surface • The ideal gas equation of state – Is water vapor an ideal gas? • Compressibility factor • Other equations of state Thank you Prepared by, MISS RAHIMAH OTHMAN