Thermodynamic Properties w Property Table from direct measurement

Thermodynamic Properties w Property Table -from direct measurement w Equation of State -- any equations that relates P, v, and T of a substance

Ideal -Gas Equation of State w Any relation among the pressure, temperature, and specific volume of a substance is called an equation of state. The simplest and best-known equation of state is the ideal-gas equation of state, given as where R is the gas constant. Caution should be exercised in using this relation since an ideal gas is a fictitious substance. Real gases exhibit ideal-gas behavior at relatively low pressures and high temperatures.

Universal Gas Constant Universal gas constant is given on Ru = 8. 31434 k. J/kmol-K = 8. 31434 k. Pa-m 3/kmol-k = 0. 0831434 bar-m 3/kmol-K = 82. 05 L-atm/kmol-K = 1. 9858 Btu/lbmol-R = 1545. 35 ft-lbf/lbmol-R = 10. 73 psia-ft 3/lbmol-R

Example Determine the particular gas constant for air and hydrogen.

Ideal Gas “Law” is a simple Equation of State

Percent error for applying ideal gas equation of state to steam

Question …. . . Under what conditions is it appropriate to apply the ideal gas equation of state?

Ideal Gas Law Good approximation for P-v-T behaviors of real gases at low densities (low pressure and high temperature). w Air, nitrogen, oxygen, hydrogen, helium, argon, neon, carbon dioxide, …. ( < 1% error). w

Compressibility Factor w The deviation from ideal-gas behavior can be properly accounted for by using the compressibility factor Z, defined as Z represents the volume ratio or compressibility.

Ideal Gas Real Gases Z=1 Z > 1 or Z<1

Real Gases w Pv = ZRT or w Pv = ZRu. T, where v is volume per unit mole. Z is known as the compressibility factor. w Real gases, Z < 1 or Z > 1. w

Compressibility factor What is it really doing? w It accounts mainly for two things w • Molecular structure • Intermolecular attractive forces

Principle of corresponding states w. The compressibility factor Z is approximately the same for all gases at the same reduced temperature and reduced pressure. Z = Z(PR, TR) for all gases

Reduced Pressure and Temperature where: PR and TR are reduced values. Pcr and Tcr are critical properties.

Compressibility factor for ten substances (applicable for all gases Table A-3)

Where do you find critical-point properties? Table A-7 Mol (kg-Mol) R (J/kg. K) Tcrit (K) Pcrit (MPa) Ar 28, 97 287, 0 (---) O 2 32, 00 259, 8 154, 8 5, 08 H 2 2, 016 4124, 2 33, 3 1, 30 H 2 O 18, 016 461, 5 647, 1 22, 09 CO 2 44, 01 188, 9 304, 2 7, 39

Reduced Properties This works great if you are given a gas, a P and a T and asked to find the v. w However, if you are given P and v and asked to find T (or T and v and asked to find P), trouble lies ahead. w Use pseudo-reduced specific volume. w

Pseudo-Reduced Specific Volume w When either P or T is unknown, Z can be determined from the compressibility chart with the help of the pseudoreduced specific volume, defined as not vcr !

Ideal-Gas Approximation w The compressibility chart shows the conditions for which Z = 1 and the gas behaves as an ideal gas: w (a) PR < 0. 1 and TR > 1

Exercise 3 -21 Steam at 600 o. C & 1 MPa. Evaluate the specific volume using the steam table and ideal gas law T o. C Tsat = 180 o. C therefore is superheated steam 600 374 (Table A-3) Volume = 0, 4011 m 3/kg 1 MPa Tcr 180 v

Solution - page 1 Part (b) ideal gas law Rvapor = 461 J/kg. K v = RT/P , v = 461 x 873/106 = 0. 403 m 3/kg Steam is clearly an ideal gas at this state. Error is less than 0. 5% Check: PR = 1/22. 09 = 0. 05 & TR = 873/647 = 1. 35

TEAMPLAY Find the compressibility factor to determine the error in treating oxygen gas at 160 K and 3 MPa as an ideal gas.

Other Thermodynamic Properties: Isobaric (c. pressure) Coefficient v P T

Other Thermodynamic Properties: Isothermal (c. temp) Coefficient v T P

Other Thermodynamic Properties: We can think of the volume as being a function of pressure and temperature, v = v(P, T). Hence infinitesimal differences in volume are expressed as infinitesimal differences in P and T, using k and b coefficients If k and b are constant, we can integrate for v:

Other Thermodynamic Properties: Internal Energy, Enthalpy and Entrop

Other Thermodynamic Properties: Specific Heat at Const. Volume u v T

Other Thermodynamic Properties: Specific Heat at Const. Pressure h P T

Other Thermodynamic Properties: Ratio of Specific Heat

Other Thermodynamic Properties: Temperature v s 1 T u

Ideal Gases: u = u(T) 0 Therefore,

We can start with du and integrate to get the change in u: Note that Cv does change with temperature and cannot be automatically pulled from the integral.

Let’s look at enthalpy for an ideal gas: w h = u + Pv where Pv can be replaced by RT because Pv = RT. w Therefore, h = u + RT => since u is only a function of T, R is a constant, then h is also only a function of T w so h = h(T)

Similarly, for a change in enthalpy for ideal gases:

Summary: Ideal Gases w For ideal gases u, h, Cv, and Cp are functions of temperature alone. w For ideal gases, Cv and Cp are written in terms of ordinary differentials as

For an ideal gas, wh = u + Pv = u + RT Cp = Cv + R

Ratio of specific heats is given the symbol, g

Other relations with the ratio of specific heats which can be easily developed:

For monatomic gases, Argon, Helium, and Neon

For all other gases, w Cp is a function of temperature and it may be calculated from equations such as those in Table A-5(c) in the Appendice w Cv may be calculated from Cp=Cv+R. w Next figure shows the temperature behavior …. specific heats go up with temperature.

Specific Heats for Some Gases w Cp = Cp(T) a function of temperature

Three Ways to Calculate Δu and Δh w Δu = u 2 - u 1 (table) w Δh = h 2 - h 1 (table) w Δu = w Δh = w Δu = Cv, av ΔT w Δh = Cp, av ΔT

Isothermal Process w Ideal gas: PV = m. RT

For ideal gas, PV = m. RT We substitute into the integral Collecting terms and integrating yields:

Polytropic Process w PVn = C

Ideal Gas Adiabatic Process and Reversible Work w w 1. 2. 3. 4. 5. What is the path for process with expand or contract without heat flux? How P, v and T behavior when Q = 0? To develop an expression to the adiabatic process is necessary employ: Reversible work mode: d. W = Pd. V Adiabatic hypothesis: d. Q =0 Ideal Gas Law: Pv=RT Specific Heat Relationships First Law Thermodynamics: d. Q-d. W=d. U

eal Gas Adiabatic Process and Reversible Wo First Law: Using P = MRT/V Integrating from (1) to (2)

Ideal Gas Adiabatic Process and Reversible Work (cont) Using the gas law : Pv=RT, other relationship amid T, V and P are developed accordingly:

Ideal Gas Adiabatic Process and Reversible Work (cont) An expression for work is developed using PVg = constant. i and f represent the initial and final states

Ideal Gas Adiabatic Process and Reversible Work (cont) The path representation are lines where Pvg = constant. w For most of the gases, g 1. 4 w The adiabatic lines are always at the righ of the isothermal lines. w The former is Pv = constant (the exponent is unity) w P f f Q =0 t. ns co T= i v

Polytropic Process A frequently encountered process for gases is the polytropic process: PVn = constant Since this expression relates P & V, we can calculate the work for this path.

Polytropic Process Constant pressure w Constant volume w Isothermal & ideal gas w Adiabatic & ideal gas w Exponent n 0 1 k=Cp/Cv

Boundary work for a gas which obeys the polytropic equation

We can simplify it further n The constant c = P 1 V 1 = P 2 V 2 n

Polytropic Process

Exercise 3 -11 A bucket containing 2 liters of water at 100 o. C is heated by an electric resistance. a) Identify the energy interactions if the system boundary is i) the water, ii) the electric coil b) If heat is supplied at 1 KW, then how much time is needed to boil off all the water to steam? (latent heat of vap at 1 atm is 2258 k. J/kg) c) If the water is at 25 o. C, how long will take to boil off all the water (Cp = 4. 18 J/kgºC)

Solution - page 1 Part a) If the water is the system then Q > 0 and W = 0. There is a temperature difference between the electric coil and the water. If the electric coil is the system then Q < 0 and W < 0. It converts 100% of the electric work to heat!

Solution - page 2 Part b) The mass of water is of 2 kg. It comes from 2 liters times the specific volume of 0. 001 m 3/kg To boil off all the water is necessary to supply all the vaporization energy: Evap = 2285*2=4570 KJ The power is the energy rate. The time necessary to supply 4570 KJ at 1 KJ/sec is therefore: Time = 4570/1 = 4570 seconds or 1. 27 hour

Solution page 3 Part c) The heat to boil off all the water initially at 25 o. C is the sum of : (1) the sensible heat to increase the temperature from 25 o. C to 100 o. C, (2) the vap. heat of part (b) The sensible heat to increase from 25 o. C to 100 o. C is determined using the specific heat (CP = 4. 18 k. J/kgo. C) Heat = 4. 18*2*(100 -25)=627 KJ The time necessary to supply (4570+627)KJ at 1 KJ/sec is : Time = 5197/1 = 5197 seconds or 1. 44 hour

Exercise 3 -12 A bucket containing 2 liters of R-12 is left outside in the atmosphere (0. 1 MPa) a) What is the R-12 temperature assuming it is in the saturated state. b) the surrounding transfer heat at the rate of 1 KW to the liquid. How long will take for all R 12 vaporize?

Solution - page 1 Part a) From table A-2, at the saturation pressure of 0. 1 MPa one finds: • Tsaturation = - 30 o. C • Vliq = 0. 000672 m 3/kg • vvap = 0. 159375 m 3/kg • hlv = 165 KJ/kg (vaporization heat)

Solution - page 2 Part b) The mass of R-12 is m = Volume/v. L, m=0. 002/0. 000672 = 2. 98 kg The vaporization energy: Evap = vap energy * mass = 165*2. 98 = 492 KJ Time = Heat/Power = 492 sec or 8. 2 min

Exercise 3 -17 A rigid tank contains saturated steam (x = 1) at 0. 1 MPa. Heat is added to the steam to increase the pressure to 0. 3 MPa. What is the final temperature?

Solution - page 1 P Tcr = 374 o. C superheated 0. 3 MPa 0. 1 MPa 133. 55 o. C 99. 63 o. C v Process at constant volume, search at the superheated steam table for a temperature corresponding to the 0. 3 MPa and v = 1. 690 m 3/kg -> nearly 820 o. C.

Exercise 3 -30 a) b) c) d) Air is compressed reversibly and adiabatically from a pressure of 0. 1 MPa and a temperature of 20 o. C to a pressure of 1. 0 MPa. Find the air temperature after the compression What is the density ratio (after to before compression) How much work is done in compressing 2 kg of air? How much power is required to compress 2 kg per second of air?

Solution - page 1 w In a reversible and adiabatic process P, T and v follows: P f f Q =0 t. ns co T= i v

Solution - page 2 Part a) The temperature after compression is Part b) The density ratio is

Solution - page 3 Part c) The reversible work: Part d) The power is:

Exercícios – Capítulo 3 Propriedades das Substâncias Puras Exercícios Propostos: 3. 6 / 3. 9 / 3. 12 / 3. 16 / 3. 21 / 3. 22 / 3. 26 / 3. 30 / 3. 32 / 3. 34 Team Play: 3. 1 / 3. 2 / 3. 4

Ex. 3. 1) Utilizando a Tabela A-1. 1 ou A-1. 2, determine se os estados da água são de líquido comprimido, líquido-vapor, vapor superaquecido ou se estão nas linhas de líquido saturado ou vapor saturado. a) Vapor superaquecido w a) P=1, 0 MPa; T=207 ºC b) Líquido comprimido w b) P=1, 0 MPa; T=107, 5 ºC w c) P=1, 0 MPa; T=179, 91 ºC; x=0, 0 c) Líquido saturado w d) P=1, 0 MPa; T=179, 91 ºC; x=0, 45 d) Líquido-Vapor e) Líquido comprimido w e) T=340 ºC; P=21, 0 MPa f) Vapor superaquecido w f) T=340 ºC; P=2, 1 MPa w g) T=340 ºC; P=14, 586 MPa; x=1, 0 g) Vapor saturado w h) T=500 ºC; P=25 MPa h) Vapor superaquecido - Fluido i) Líquido comprimido - Fluido w i) P=50 MPa; T=25 ºC

w Ex. 3. 2) Encontre o volume específico dos estados “b”, “d” e “h” do exercício anterior. b) P=1, 0 MPa; T=107, 5 ºC v vl=0, 001050 (utilizar referência T=107, 5ºC) d) P=1, 0 MPa; T=179, 91 ºC; x=0, 45 v=0, 08812 [v=(1 -x)vl+x(vv)] h) T=500 ºC; P=25 MPa v=0, 011123 m 3/kg (Tabela A-1. 3 Vapor Superaquecido)

w Ex. 3. 4) Amônia a P=150 k. Pa, T=0 ºC se encontra na região de vapor superaquecido e tem um volume específico e entalpia de 0, 8697 m 3/kg e 1469, 8 k. J/kg, respectivamente. Determine sua energia interna específica neste estado. u =1339, 345 k. J/kg (u = h - P v)