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Thermodynamic Properties are Measurements p,T,v, u ,h,s - measure directly -measure by change ∂s = ∂p ∂v T Tables Curve fits Property Data Tables Correlation's, Boyles Law pv=c@T=c Equations of State, pv=RT ∂T v Tables limited hand calculations Tables Calculation Modules NIST, EES, HYSYM interactive, callable P=1 atm vapor liquid Q Given T and P Super Heat Region if, p<p sat @T T>Tsat @p Compressed Liquid Region if, p>p sat @ T T<Tsat @ p STEAM PRESUE AND TEMPERATURE TABLES T Saturated Vapor Line Saturated Liquid Line hf hg uf ug sf sg vf vg Three Tables Temperature Table at spaced T’s Pressure Table at spaced P’s Superheat Table at spaced T and P 6 Properties Temperature Pressure Volume Internal Energy Enthalpy Entropy Solid-Liquid-Gas Phase Diagram Saturation liquid internal energy at 0 C 0. Table Base Saturated liquid enthalpy at 25 C 104.89 kJ/kg Saturated vapor entropy at 25 C 8.558 kJ/kg K Enthalpy at 20 C, 300 kPa assume saturated liquid enthalpy at 20 C 83.96 kJ/kg Temperature of saturated vapor with an internal energy of 2396.1 kJ/kg 15 C Enthalpy of vaporization at 10 C 2477.7 kJ/kg Table A-7 Metric T,p Table A-7E English T,p Table A-4 Metric, T Table A-5 Metric, p Table A-4E English, T Table A-5E English, p TEMPERATURE TABLE PRESSURE TABLE Table A-6, Metric, T,p Table A-6E, English, T,p SUPERHEAT TABLE R − 134a at 500 kPa and 6 C Tsaturation@ 500 kpa = 15.71C psaturation@ 6 C = 362.23 kpa water at 1400 kpa and 300 C Tsaturation@ 1400 kpa = 195.04 C psaturation@ 300 C = 8587.9 kPa p=constant T p=constant T 300 C 8587.9 kPa 1400 kpa 195.04 C superheated 15.71C 500 kPa 362.23 kpa 6C subcooled v v Two Phase Real Gas Properties V = Vf + V g mv = m f v f + m g v g x= mg Quality m v = (1 − x )v f + xv g v = v f + x × v fg v − vf x= v fg h = h f + x × h fg u = u f + x × u fg v = v f + x × v fg vf vg Quality, x = x= Find the properties of water at 90o F, v = 300 ft 3 /lb. v − vf vg − vf 300 − .0161 = .64 (64%) 467.7 − .0161 (v u = u f + x × (u f − u g ) u = u f + x × u fg u = 58.07 + .64 × 982.2 u = 686.7 BTU/lb h = h f + x × h fg Temperature Table g − vf ) (v f − v ) T 90o F u g =1040.2 u f =58.07 h f =58.07 s f =.11165 u h g =1107.7 h s g =2.0083 s h = 58.07 + .64 × 1042.7 h = 725.4 BTU/lb s = s f + x × s fg s = .11165 + .64 ×1.8966 s = 1.3254 BTU/lb o R v f = .0161ft 3 / lb v = 300. ft 3 /lb v g = 467.7ft 3 / lb Engineering Equation Solver - EES Fluid Property Information - 69 fluids available Thermophysical Functions - 25 properties calculated Equations Window h=enthalpy(steam, T=200.,P=200) h=enthalpy(steam,T=200.,X=1) u=intenergy(steam,T=200.,X=0.) p=pressure(steam,T=200.,X=0.) Thermophysical Functions entropy intenergy pressure quality density enthalpy isidealgas temperature volume superheated vapor saturated vapor saturated liquid saturation pressure Function Arguments H specific enthalpy P pressure S specific entropy T temperature U specific internal energy V specific volume X quality EES FLUIDS FUNCTIONS Table Saturation internal energy at 4 kPa EES Program Pressure Table u l = intenergy(steam, P = 4., X = 0) u l = 121.45kJ/kg u l = 121.3kJ/kg u g = 2415.2kJ/kg u g = 2414.29 u g = intenergy(steam, P = 4., X = 1.) Superheat Table Enthalpy and volume of water at 150 kPa and 30 C h l = h saturated liquid @ 30o C h l = 125.75 kJ/kg v l = v saturated liquid@30o C v l = .001004m 3 /kg h l = enthalpy(steam, T = 30., p = 500.) h l = 125.67kJ/kg v l = volume(steam, T = 30., p = 500.) v l = .001004m 3 /kg Temperature Table Internal energy u l = u saturated liquid @ 300 o C of water at 20 MPa u = 1332.0 kJ/kg l and 300 C u l = intenergy(steam, T = 300., p = 20000.) u l = 1333.446 kJ/kg Solve Windows Equations Windows Arrays Windows Plot Window 10 MPa saturated steam has an enthalpy of 2010 kJ/kg. What is its internal energy? h = h f + x h fg 2010 kJ/kg = 1407.56 + x 1317.1 x = .4574 u = u f + x u fg u = 1393.04 + .4574 × 1151.4 u = 1919.69 kJ/kg EES Program x = quality(steam,P = 10000.,h = 2010) x = .4576 u = intenergy(steam,P = 10000.,h = 2010) u = 1919.65 kJ/kg Steam at 20 C has an enthalpy of 1800 kJ/kg. What is the internal energy? h = h f + x h fg 1800 kJ/kg = 83.96 + x 2454.1 x = .7 u = u f + x u fg u = 83.95 + .7 × 2319.0 u = 1707.25 kJ/kg STEAM SUPERHEAT TABLE SUPERHEAT TABLE Enthalpy at 700 C and .10 Mpa 3928.2 kJ/kg Temperature at entropy of 8.8642 and .05 Mpa 400 C Enthalpy at .05 MPa and entropy of 10.6662 kJ/kg C 5147.7 kJ/kg Steam initially at a temperature of 1100 C and a pressure of .10 MPa undergoes a process during which its entropy remains constant to a pressure of .01 MPa. What is the enthalpy and temperature of the steam at the end of the process? T Entropy at 1100 C, .1 MPa 10.1659kJ/kg K Enthalpy at .01 MPa, entropy 10.1659 3705.4 kJ/kg Temperature at .01 MPa, entropy 10.1659 600 C .01 MPa s Linear Interpolation with 2 Variables h@(T = 450 C,p = 27 MPa) Steam Superheat Table Table Values p = 25 mPa T = 450 h = 2949.7 p = 27 MPa h = 2898.4 p = 30 MPa h = 2821.4 27 − 25 For the pressure table entry, = .4 30 − 25 The desired pressure, 27 kPa, is 40 % of the difference between table values. All the other properties at 27 kPa must be at the same difference. EES h = enthalpy(steam,T = 450.,p = 27000) h = 2901.7 kJ/kg .22% difference, table and interpolation SPREAD SHEET WORLD EXCELL ADD-IN 40 fluids 4 sets of units SPREADSHEET WORLD – THERMAL FLUIDS PROPERTIES A B C D E F G 1 2 T 200 3 P 15 4 5 CALL TTProps("AIR","EE_F", "P", $C$2,"T", $C$1) 6 7 P= v= T= u= s= h= 8 RETURN 15 16.29669 200 180.5582 1.085135 225.7937 9 psia ft^3/lbm deg F BTU/lbm BTU/lbm-R BTU/lbm H X= 1 nd I J STATE = ERROR = erheated va 0 A saturated mixture of 2 kg water and 3 kg vapor in contained in a piston cylinder device at 100 kpa. Heat is added and the piston, initially resting on stops, begins to move at a pressure of 200 kpa. Heating is stopped when the total volume in increased by 20%. Find: a) the initial and final temperatures. b) the mass of liquid water when the pressure reaches 200 kPa and the piston starts to move. c) the work done by the expansion. 3 kg vapor = .6 at100kPa, x = 5 kg total T = 99.61 v1 = vf + x × vfg 100 kpa 3 kg v1 = .001043+ .6× (1.694−.001043) v1 = 1.0168m3/kg u1 = uf + x × ufg Q 2 kg u1 = 417.40+ .6× 2088.2= 1670.62kJ/kg 3.88 v = constant,1 - 2 Q = ∆E + W p = constant,2 - 3 Point 2 at . 2 MPa, v1 = v 2 Q = ∆E + W Q = ∆E + p∆V V2 = 5 kg ×1.0161 = 5.08 m3 Interpolation from Table A − 6 u 2 = u@(v = 1.0161, P = .2 MPa ) W=0 Q = ∆E = ∆U Q = ∆H Q = m(u 2 − u1 ) u 2 = 2613.23 kJ/kg Q = m(h3 − h 2 ) h 2 = 2816.47 kJ/kg Q = 5 × (2988.65− 2816.47) Q = 4713.05 kJ Q2-3 = 860.9 kJ h 2 = h@(v = 1.0168, P = .2 MPa )) V3 = 1.2 × V2 = 6.096 alteratively, 3 6.096 m = 1.2192 5 kg v g = .8857 ⇒ superheated v3 = Q = 5 × (2613.23- 1670.62 ) W1-3 = Q1-3 − m × (u3 − u 2 ) 2 Point 3 at .2 MPa h 3 = h @ (v3 = 1.2192, P = .2 MPa ) 3 p h 3 = 2988.65 kJ/kg 2 3 1 2 W = ∫ pdV + ∫ pdV = 0 + p 2 (V3 − V2 ) ( 1 ) W = 0 + 200 kPa × 6.096 m3 − 5.08 m3 = 203.2 kJ v 3.88 Interpolation for h 3 Superheat Table A - 6 T 250 v 1.1989 1.2192 300 1.31623 enthalpy @ ( p = .2 MPa, v = 1.2192) h 2971.2 3092.1 1.2192 - 1.1989 = .173 1.31623- 1.1989 h 3 = 2988.65 kJ/kg ratio of v = Spread Sheet World module solution water 2 vapor 3 Properties at 1 given p1 and x1 x1 0.6 p1 100,000 P= 100000 Pascals 2 3 p 1 v= T= u= s= h= 1.016819 99.62524 1670.481 4.936567 1772.163 m^3/kg deg C kJ/kg kJ/kg-K kJ/kg X= 0.6 nd STATE = Mixed regio X= 1 nd STATE = erheated va V2=m x v2 P= v= T= u= s= h= 200000 1.016819 173.5216 2613.078 7.389259 2816.442 Pascals m^3/kg deg C kJ/kg kJ/kg-K kJ/kg 5.084094 V3= 1.2 * V2 v3=V3/m p3 6.100913 1.220183 200,000 X= 1 nd STATE = erheated va Properties at 2 given p2 and v2=v1 p2 200,000 Properties at 3 given p3 and v3 P= 200000 Pascals Q 2-3=m*(h3-h2) W 2-3 = Q 2-3-m*(u3-u2) Q 1-2=m*(u2-u1) 861.7029 203.363 4712.986 v= T= u= s= h= 1.220183 259.0808 2744.746 7.742318 2988.782 m^3/kg deg C kJ/kg kJ/kg-K kJ/kg v EES Solution 100 kpa 3 kg 2 kg Q 2 3 p 1 v 3 kg vapor and 1 kg liquid R-134a is contained in a rigid tank at 20 C. What is the volume of the tank? If the tank is heated until the pressure ? reaches .6 MPa? What is the quality, and enthalpy of the mixture of liquid and vapor? V1 = Vf + Vg V1 = m f × v f @ 20 C + m g v g @ 20 C page 842, TableA − 11 V1 = 1 kg × .0008157 + 3 kg × .0358 V1 = .1082 m 3 V1 .1082 m 3 v1 = = = .027 m 3 /kg m 4 kg After heating, T 20 C V = constant, m = constant ⇒ v = constant v 2 = v1 = v f @ .6 MPa + x × v fg @ .6 MPa page 843 , Table A - 12 .027 − .00008196 = .787(78.7%) .0341 − .0008196 h 2 =h f + x × h fg x2 = h 2 = 79.84 + .787 ×179.71 = 211.27 kJ/kg Q 2 .6 MPa 1 .5716 MPa v 3 kg of vapor and 2 kg of liquid R-134a is contained in a piston cylinder device. The volume of the vapor is .1074 cubic meters. What is the temperature and pressure? If the cylinder and its contents are heated until volume is .15 cubic meters what is the quality? v g1 = Vg1 m g1 .1074 m 3 = = .0358 m 3 /kg 3 kg T v g = .0358 m /kg @ 20 C, .5716 MPa page 842, TableA − 11 3 During the heating process the pressure is constant. 20 C at V2 = .15m 3 , V2 .15 = = .03 m 3 /kg m 5 v 2 = .03 m 3 /kg = v f @ 20 C + x × v f g @ 20 C 1 2 v2 = v v − vf .03 − .0008157 = = .834 (83.4%) x2 = 2 v fg .0358 − .0008157 x= property − (property)f (property)fg property = (property)f + x × (property)fg Q Thermodynamic Problem Solving Technique 1. Problem Statement 4. Property Diagram Carbon dioxide is contained in a cylinder with a piston. The carbon dioxide is compressed with heat removal from T1,p1 to T2,p2. The gas is then heated from T2, p2 to T3, p3 at constant volume and then expanded without heat transfer to the original state point. state points - processes - cycle T3,p3 p Q W T2,p2 2. Schematic W Q T1,p1 v 5. Property Determination 1 2 3 3. Select Thermodynamic System open - closed - control volume a closed thermodynamic system composed to the mass of carbon dioxide in the cylinder p T2,p2 T p v u h s v 6. Laws of Thermodynamics CO 2 Q=? W=? E=? material flows=? Ideal Gas Law AVOGADRO'S LAW One (1) mole of any gas = 22.4 liters. 6.023×1023 molecules /mole of gas at STP (1 atm and 0o C ) BOLYES LAW p1 × v1 = p 2 × v 2 IDEAL (PERFECT) GAS LAW pv = RT pV = mRT p - absolute pressure, psia,kPa T - absolute temperature, o R, o K R* R= molecular weight ft lbf o R R = 1545.15 lbmole kJ kPa m 3 * R = 8.314 or kmole o K kmole o K * CHARLES LAW v1 T1 = v 2 T2 p1 T1 = p 2 T2 mass = moles × Molecular Weight m = n × Molecular Weight pv = R * T pV = nR * T water Ideal Gas pv = RT constant specific heat What is the mass of 1.2 m3 of oxygen at 24o C and a gage pressure of 500 kPa. Atmospheric pressure is 97 kPa p = pgage + patmosphere = 500 kPa + 97 kPa = 597 kPa 8.314 kJ/kmole o K or kPa m3 /kmole o K = .259813 kPa m3 /kg R O2 = 32 pV 597 kPa ×1.2 m3 m= = = 9.28 kg RT .259813 kPa m3 /kg × ( 24o C + 273.16o K ) alsoTableA − 1 What is the volume mass of 1.2 lbm of air at 124o F and a gage pressure of 500 psia. Atmospheric pressure is 14.7 psia. p = pgage + patmosphere = 500 psia + 14.7 psia = 514 psia 1545.15 ft lbf / lbmole R ft lbf also Table A − 1E in BTU and psi units = 53.336 28.97 lbm o R o o o m R T 1.2 lbm × 53.336 ft lbf/ lbm R × (124 F + 459.69 R ) V= = 514 psia ×144 ft 2 /in 2 p R air = V = .5047ft 3 IDEAL GAS EQUATION FORMS - For Air P kPa kPa lb ft 2 lb ft 2 v = m m 3 m 3 = kmole = kg R kPa m3 8.314 kmole O K kPa m3 .287 kg O K T O O K K ft lbf lbmole O R ft lbf 53.35 lbm O R ft 3 = lbmole ft 3 = lbm psi ft 3 = lbm .3704 lb ft 2 ft 3 = lbm .06855 O 1545.15 psi lbf lbm O R BTU lbm O R kPa m3 R 8.314 / 28.96 = kmoleo K O O R R R O R ft lbf R = 1545.15 /28.96 lbmole O R ft lbf R = 1545.15 /144/28.96 O lbmole R ft lbf R 1545.15 /28.96/778 = O lbmole R The specific volume of air at 75o F and 14.7 psia R air = 1545.15 /28.96 = 53.35 ft lbf/lbm R ( RT 53.35 ft lbf/lbm R × 459.69o R + 75o F v= = p 14.7 lbf/in 2 × 144 in 2 /ft 2 ) v = 13.476 ft 3 /lb The specific volume of air at 24o C and 101.325 kPa kPa m 3 R = 8.314 / 28.96 O kg mole K ( RT .287 kPa m 3 /kg × 273.15o K + 24o F v= = p 101.325 kPa 2 v = .8417 m 3 /kg ) Air initially at a volume of 12 m 3 and a pressure of 225 kPa expands at a constant temperature to a volume of 23 m 3 . What is the final pressure? mass = constant, T = constant RT1 .286 × (273.15 + T1 ) m= = p1V1 224 kPa × 23 m 3 m= R T1 R T2 = p1V1 p 2 V2 p2 = R T2 p1 V1 V = p1 1 R T1 V2 V2 12 m 3 p 2 = 225 kPa 23 m 3 p 2 = 117.39 kPa m=const T=const p 12 m 3 v 2 3m 3 SPECIFIC HEATS FOR GASSES SPECIFIC HEAT ∂h Cp = ∂T p =const ∂u Cv = ∂T v =const ∆h = ∫ c p (T ) dT ∆u = ∫ c v (T ) dT For an ideal gas specific heats are assumed constant ∆h = c p ∫ dT ∆u = c v ∫ dT dh = c p dT du = c v dT h = u + pv By definition substituting pv = RT differentiating h = u + RT dh = du + RdT subsistuting for dh and du c p dT = c v dT + RdT cp = cv + R with same units and FOR IDEAL GAS ONLY k= cp cv cp − cv = R k −1 = R cv IDEAL GAS IMPROVEMENTS Enthalpy, h, internal energy, u, and entropy, s. are not absolute but Differences from a base. T h= ∫ c (T )dT p Table Base State T u= ∫ c (T )dT v Table Base State c p = c p (T ), c v = c v (T ) Tables A - 17, A - 17 to A - 22, A - 22E IDEAL GAS WITH VARIBLE SPECIFIC HEAT 2-104 Determine the enthalpy change, ∆h, of nitrogen in kJ/kg as it is heated from 600O K to 1000O K (726O C, 1340O F) using : a) empirical specific heat equation Table A − 2c , b) specific heat at the average temperature Table A - 2b, c) specific heat at room temperature Table A - 2a, d) Table A − 18E, e) EES. a) ∆h = ∫ cp (T )dT, cp = a + bT + cT 2 + dT 3 ( ∆h = 28.9(1000 − 600 ) + .5(.0001571) 1000 2 − 6001 + ( )( ) ( ) )( ) 1 1 .8081×10 −5 10003 − 6003 − 2.8738081× 10 −9 1000 4 − 600 4 = 12544 kJ/kmole 3 4 ∆h = 2544 kJ/kmole/28.013 = 447.8 kJ/kg b)Average specific heat over the temperature range, c p @800O K = 1.121 kJ/kgK ∆h = c p ∆T = 1.121 kJ/kg K (1000 − 600 ) = 448.5kJ/kg c)Room temperature specific heat, c p @800O K = 1.039448.5 kJ/kg ∆h = 1.039448 kJ/kg (1000 − 600 ) = 415.6 kJ/kg d) TableA − 18E ∆h = h@1000 O K − h@600O K = 30,129 kJ/kgmole − 17563 kJ/kgmole = 12566 kJ/kgmole ∆h = 12566kJ/kgmole/28.013kg/kgmole = 448.58 kJ/kg e)EES ∆h = enthalpy(nitrogen, T = 1000., p = 101.325) − enthalpy(nitrogen, T = 600., p = 101.325) ∆h = 449.46kJ/kg PRINCIPAL OF CORRERSPONDING STATES COMPRESSIBILITY FACTOR Z Z is about the same for all gasses at the same reduced temperature and the same reduced pressure where: P pv Z= RT pV Z= mRT PR = TR = p p critical T Tcritical (2 − 20) VAN DER WAALS EQUATION OF STATE - 1873 a (2 - 22) p + 2 (v − b ) = RT v a − intermolecular forces 2 v b − volume of gas molecules RTcritical a p= − 2 v critical − b v critical ∂p =0 dv Tcritical 2 27 R 2 Tcritical a= 64 p critical ∂p =0 ∂v T ∂ ∂p =0 ∂v ∂v T p critical point ∂ 2p 2 =0 dv Tcritical b= R Tcritical 8 p critical v THERMODYNAMIC PROPERTY MEASURMENT Thermodynamic properties are independent of path or process and are exact differentials. Heat and Work are not exact differentials but are dependent on process or path. u = f(s, v) ∂u ∂u du = ds + dv ∂s v ∂v s 6 thermodynamic properites 2 at a time, 60 equations ∂h = cp ∂T p ∂u = cv ∂T v ∂T = Joule Tompson Coefficient ∂p h ∂T = 0 for an ideal gas, a value for a real gas ∂p h used with the First Law ∂s ∂p = ∂ v ∂ T T v ∂T = JT Coefficient ∂p h h=constant Course Property Sources 1) Ideal Gas Law with constant specific heats 2) Tables Steam Refrigerant Air 3) EES CD NIST for home work convenience EQUATION OF STATE ERRORS nitrogen NIST Webbook Properties fttp://webbook.nist/gov/chemistry/fluid Temperature Table for Water in .1 degree increments from 40 to 40 degrees. Select Units Select Table Type Select fluid Set low and high temperature and temperature increment.