Thermodynamic Proerties are Measurements,T,, u,h,s - measure directly -measure by change s Tables T T Proerty Data Cure its Tables Correlation's, Boyles Law Tables c@tc limited hand calculations Equations o State, RT Tables Calculation Modules NIST, EES, HYSYM interactie, callable
P atm aor liquid Q
Gien T and P Suer Heat Region i, < sat sat @T T>T @ Comressed Liquid Region i, > @ T sat T<T @ sat
STEAM PRESUE AND TEMPERATURE TABLES
T Saturated Liquid Line Saturated Vaor Line h u s h u s g g g g
Three Tables Temerature Table at saced T s Pressure Table at saced P s Suerheat Table at saced T and P 6 Proerties Temerature Pressure Volume Internal Energy Enthaly Entroy
Solid-Liquid-Gas Phase Diagram
Saturation liquid internal energy at 0 C 0. Table Base Saturated liquid enthaly at 5 C 04.89 kj/kg Saturated aor entroy at 5 C 8.558 kj/kg K Enthaly at 0 C, 00 kpa assume saturated liquid enthaly at 0 C 8.96 kj/kg Temerature o saturated aor with an internal energy o 96. kj/kg 5 C Enthaly o aorization at 0 C 477.7 kj/kg
Table A-7 Metric T, Table A-7E English T, Table A-4 Metric, T Table A-5 Metric, Table A-4E English, T Table A-5E English, TEMPERATURE TABLE PRESSURE TABLE Table A-6, Metric, T, Table A-6E, English, T, SUPERHEAT TABLE
water T saturation saturation at 400ka and 00 C @400 ka 95.04C @ 00C 8587.9kPa R 4a at 500 kpa and 6 C Tsaturation@ 500 ka 5.7C @ 6 C 6.ka saturation T 00C 8587.9kPa 95.04 400ka C constant T suerheated 5.7C 500 kpa 6.ka 6 C subcooled constant
Two Phase Real Gas Proerties ( ) g g g g g g g x x x x Quality m m x m m m V V V + + + + g g g g x u x u u h x h h + + +
Quality, x g 00.06 x.64 (64%) 467.7.06 u u + x (u u ) g u u + x u g u 58.07 +.64 98. u 686.7 BTU/lb h h + x h g h 58.07 +.64 04.7 h 75.4 BTU/lb T 90 o Find the roerties o F Temerature Table u58.07 h58.07 ( ) s.65 water at 90 ( g ) o u h s F, 00 t g g g /lb. u 040. h 07.7 s.008 s s + x s s.65 +.64.8966 s g o.54 BTU/lb R.06t / lb 00. t /lb g 467.7t / lb
Engineering Equation Soler - EES Fluid Proerty Inormation - 69 luids aailable Thermohysical Functions - 5 roerties calculated Equations Window henthaly(steam, T00.,P00) suerheated aor henthaly(steam,t00.,x) saturated aor uintenergy(steam,t00.,x0.) saturated liquid ressure(steam,t00.,x0.) saturation ressure Thermohysical Functions entroy intenergy ressure quality density enthaly isidealgas temerature olume Function Arguments H seciic enthaly P ressure S seciic entroy T temerature U seciic internal energy V seciic olume X quality
EES FLUIDS FUNCTIONS
Table EES Program Saturation internal energy at 4 kpa u l u Pressure Table g.45kj/kg 45.kJ/kg u u u u l l g g intenergy(steam, P 4.,X 0).kJ/kg intenergy(steam, P 44.9 4., X.) Enthaly and olume o water at 50 kpa and 0 C Internal energy u o water at 0 MPa u and 00 C l l Suerheat Table h h saturated liquid @ 0 h l l l l 5.75 kj/kg saturated liquid@0.00004m /kg Temerature Table u saturated liquid @ 00.0 kj/kg o o o C C C u u l l h h l l l l enthaly(steam,t 5.67kJ/kg olume(steam,t.00004m /kg intenergy(steam,t.446 kj/kg 0., 0., 00., 500.) 500.) 0000.)
Sole Windows Equations
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0 MPa saturated steam has an enthaly o 00 kj/kg. What is its internal energy? h h + x h g 00 kj/kg 407.56 + x 7. x.4574 u u + x u g u 9.04 +.4574 5.4 u 99.69 kj/kg EES Program x quality(steam,p 0000.,h 00) x.4576 u intenergy(steam,p 0000.,h 00) u 99.65 kj/kg
Steam at 0 C has an enthaly o 800 kj/kg. What is the internal energy? h h + x h 800 kj/kg 8.96 + x 454. x.7 u u + x u g g u 8.95 +.7 9.0 u 707.5 kj/kg
STEAM SUPERHEAT TABLE
SUPERHEAT TABLE Enthaly at 700 C and.0 Ma 98. kj/kg Temerature at entroy o 8.864 and.05 Ma 400 C Enthaly at.05 MPa and entroy o 0.666 kj/kg C 547.7 kj/kg
Steam initially at a temerature o 00 C and a ressure o.0 MPa undergoes a rocess during which its entroy remains constant to a ressure o.0 MPa. What is the enthaly and temerature o the steam at the end o the rocess? T Entroy at 00 C,. MPa 0.659kJ/kg K Enthaly at.0 MPa, entroy 0.659 705.4 kj/kg Temerature at.0 MPa, entroy 0.659 600 C.0 MPa s
Linear Interolation with Variables h@(t 450 C, 7 MPa) Steam Suerheat Table Table Values 5 mpa 7 MPa 0 MPa T 450 h 949.7 h 898.4 h 8.4 7 5 For the ressure table entry,.4 0 5 The desired ressure, 7 kpa, is 40 % o the dierence between table alues. All the other roerties at 7 kpa must be at the same dierence. EES h enthaly(steam,t 450., 7000) h 90.7 kj/kg.% dierence, table and interolation
SPREAD SHEET WORLD EXCELL ADD-IN 40 luids 4 sets o units
SPREADSHEET WORLD THERMAL FLUIDS PROPERTIES 4 5 6 7 8 9 CALL A B C D E F G H I J T 00 P 5 TTPros("AIR","EE_F", "P", $C$,"T", $C$) P T u s h X STATE ERROR RETURN 5 6.9669 00 80.558.0855 5.797 erheated a 0 sia t^/lbm deg F BTU/lbm BTU/lbm-R BTU/lbm nd
A saturated mixture o kg water and kg aor in contained in a iston cylinder deice at 00 ka. Heat is added and the iston, initially resting on stos, begins to moe at a ressure o 00 ka. Heating is stoed when the total olume in increased by 0%. Find: a) the initial and inal temeratures. b) the mass o liquid water when the ressure reaches 00 kpa and the iston starts to moe. c) the work done by the exansion. at 00kPa, T 99.6 x kg aor.6 5kg total u u.0004+.6 (.694.0004).068m /kg u + x + x u g g 47.40+.6 088. 670.6kJ/kg Q 00 ka kg kg.88
Point V Interolation romtablea 6 u u h h V h h g W 5 kg.06 5.08m u@ h@. V h @ at. MPa, (.06,P. MPa) 6.kJ/kg (.068,P. MPa) 86.47kJ/kg.8857 suerheated Point at. MPa (.9,P. MPa) 988.65 kj/kg dv + 6.096 W 0 + 00 kpa 6.096 m 5 kg.9 dv 0 + ) ( V V ) ( 6.096 m 5.08m ) 0.kJ constant,- Q E+ W Q E+ V Q H Q m h Q W - - ( h ) alteratiely, Q - Q 5 (988.65 86.47) 860.9 kj m ( u u ) constant,- Q E + W W 0 Q E U Q m u ( u ) Q 5 (6.-670.6) Q 47.05 kj.88
Interolation or h Suerheat TableA -6 enthaly @ (. MPa,.9) T 50 00.989.9.6 h 97. 09..9-.989 ratio o.6-.989 h 988.65 kj/kg.7
Sread Sheet World module solution water aor Proerties at gien and x x 0.6 00,000 P T u s h X STATE 00000.0689 99.654 670.48 4.96567 77.6 0.6 Mixed regio Pascals m^/kg deg C kj/kg kj/kg-k kj/kg nd Proerties at gien and 00,000 P T u s h X STATE 00000.0689 7.56 6.078 7.8959 86.44 erheated a Pascals m^/kg deg C kj/kg kj/kg-k kj/kg nd Vm x 5.084094 V. * V 6.009 V/m.08 00,000 Proerties at gien and P T u s h X STATE 00000.08 59.0808 744.746 7.748 988.78 erheated a Pascals m^/kg deg C kj/kg kj/kg-k kj/kg nd Q -m*(h-h) 86.709 W - Q --m*(u-u) 0.6 Q -m*(u-u) 47.986
EES Solution Q 00 ka kg kg
kg aor and kg liquid R-4a is contained in a rigid tank at 0 C. What is the olume o the tank? I the tank is heated until the ressure? reaches.6 MPa? What is the quality, and enthaly o the mixture o liquid and aor? V V V V Ater heating, V constant, m constant constant x h h V m kg.000857 + kg.058.08 m V m + V @ 0 C + m.08 m 4 kg @.6 MPa + x.07.0000896.04.000896 h + x h g g.07 m /kg.787(78.7%) 79.84 +.787 79.7.7 kj/kg g g @ 0 C age 84, TableA g @.6 MPa age 84, Table T 0 C A - Q.6 MPa.576 MPa
kg o aor and kg o liquid R-4a is contained in a iston cylinder deice. The olume o the aor is.074 cubic meters. What is the temerature and ressure? I the cylinder and its contents are heated until olume is.5 cubic meters what is the quality? x g g at V V m g g.058 m /kg @ 0 C,.5m.0 m /kg g.074 m kg,.5 5.0 m /kg.058 m /kg.576 MPa During the heating rocess the ressure V m @ 0 C + x g @ 0 C is constant..0.000857.84 (8.4%).058.000857 age 84, TableA 0 C T x roerty (roerty) (roerty) g roerty (roerty) + x (roerty) g Q
Thermodynamic Problem Soling Technique. Problem Statement Carbon dioxide is contained in a cylinder with a iston. The carbon dioxide is comressed with heat remoal rom T, to T,. The gas is then heated rom T, to T, at constant olume and then exanded without heat transer to the original state oint.. Schematic. Select Thermodynamic System oen - closed - control olume a closed thermodynamic system comosed to the mass o carbon dioxide in the cylinder CO 4. Proerty Diagram state oints - rocesses - cycle Q T, T, W Q W T, 5. Proerty Determination T, T u h s 6. Laws o Thermodynamics Q? W? E? material lows?
AVOGADRO' S LAW Ideal Gas Law IDEAL (PERFECT) GAS LAW One() moleo any gas.4 liters. 6.0 0 o STP (atm and 0 C) BOLYES LAW molecules/moleo gas at CHARLES LAW T T RT V mrt - absolute ressure, sia,kpa o o T - absolute temerature, R, K * R R molecular weight * R 545.5 o t lb R lbmole kj kmole K * R 8.4 or o kpa m o kmole K T T mass moles Molecular Weight m n Molecular Weight V * R T * nr T
water Ideal Gas RT constant seciic heat
o What is the mass o. m o oxygen at 4 C and a gage ressure o 500 kpa. Atmosheric ressure is 97 kpa + 500 kpa + 97 kpa 597 kpa gage atmoshere o o 8.4 kj/kmole K or kpa m /kmole K R O.598 kpa m /kg alsotablea V 597 kpa. m m 9.8 kg o o RT.598 kpa m /kg 4 C + 7.6 K ( ) o What is the olume mass o. lbm o air at 4 F and a gage ressure o 500 sia. Atmosheric ressure is 4. 7 sia. + 500 sia + 4.7 sia 54 sia air gage atmoshere 545.5 t lb / lbmole R t lb R 5.6 also Table A E in BTU and si units m R T V V.5047t o 8.97 lbm R ( 9R o ) o o. lbm 5.6 t lb/ lbm R 4 F + 459.6 54 sia 44 t /in
IDEAL GAS EQUATION FORMS - For Air P m R T kpa m O kpa m kmole 8.4 K O kmole K kpa m O kpa m kpa m kg.87 K R 8.4 / 8.96 O o kg K kmole K lb t lb O t lbmole 545.5 R O t lbmole R lb t lb O t lb t lbm 5.5 R R 545.5 /8.96 O O t lbm R lbmole R si lb O t lb si t lbm.704 R O R 545.5 /44/8.96 O lbm R lbmole R lb BTU O t lb t lbm.06855 R R 545.5 /8.96/ O 778 O t lbm R lbmole R
The seciic olume o R.476 t The seciic olume o kpa m R 8.4 kg mole air 545.5 /8.96 5.5 t lb/lbm R RT RT /lb.847 m /kg O air at 75 5.5 t lb/lbm R 4.7 lb/in 44 in air at 4 / 8.96 K.87 kpa m /kg 0.5 kpa o F and4.7 sia o o ( 459.69 R + 75 F) o /t C and0.5 kpa o o ( 7.5 K + 4 F)
Air initially at a olume o m and a ressure o 5 kpa exands at a constant temerature to a olume o m. What is the inal ressure? mass constant, RT m V R T m V R T V m 5 kpa m 7.9 kpa R T V R T V ( 7.5 + T ) T constant.86 4 kpa m V V m mconst Tconst m
SPECIFIC HEATS FOR GASSES
h SPECIFIC HEAT C T const T h c ( T) dt u c ( T) h c dh c dt dt u c subsistuting or dh and du c with same units and k FOR IDEAL GAS ONLY du c dt c dt dt c c c c c const For an ideal gas seciic heats are assumed constant By deinition h u + substituting RT h u + RT dierentiating dh du + RdT C u dt + R c dt + RdT R k R c
IDEAL GAS IMPROVEMENTS Enthaly, h, internal energy, u, and entroy, s. are not absolute but Dierences rom a base. h u c T c c Table Base State T c Table Base State ( T) ( T) dt dt ( T),c c ( T) Tables A -7, A -7 to A -, A - E
IDEAL GAS WITH VARIBLE SPECIFIC HEAT
b) a) -04 600 Table A - a, h + O h e)ees K to000 seciic heat at the aerage temerature Table A - b, d) ( ) c T dt, c a + bt + ct + dt 8.9( 000 600) +.5(.00057)( 000 600 ) 5 9 4 4 (.808 0 )( 000 600 ) (.87808 0 )( 000 600 ) b)aerage seciic heat oer the temerature range, c)room temerature seciic heat, d) TableA 8E h h@000 Determine the enthaly change, h, o nitrogen in kj/kg as it is heated rom O K (76 C,40 Table A 8E, O O K h@600 O O F) using : e) 4 c EES. @800 a) h c O emirical seciic heat equation Table A c, h 544 kj/kmole/8.0 @800 T.kJ/kg K K.09448.5 kj/kg h.09448 kj/kg K. kj/kgk ( 000 600) ( 000 600) 448.5kJ/kg 45.6 kj/kg K 0,9 kj/kgmole 756 kj/kgmole 566 kj/kgmole h 566kJ/kgmole/8.0kg/kgmole 448.58 kj/kg h enthaly(nitrogen, T 000., 0.5) enthaly(nitrogen, T 600., 0.5) c c) O seciic heat at room temerature 544 kj/kmole 447.8 kj/kg h 449.46kJ/kg
PRINCIPAL OF CORRERSPONDING STATES COMPRESSIBILITY FACTOR Z Z is about the same or all gasses at the same reduced temerature and the same reduced ressure where: P Z Z RT V mrt P T R R T critical T critical ( 0)
VAN DER WAALS EQUATION OF STATE - 87 a + ( b) RT ( - ) a intermolecular orces b olume o gas molecules d a RT critical T critical critical 7 R T 64 b 0 critical critical a critical d b T critical R T 8 critical critical 0 T 0 0 critical oint T
THERMODYNAMIC PROPERTY MEASURMENT Thermodynamic roerties are indeendent o ath or rocess and are exact dierentials. Heat and Work are not exact dierentials but are deendent on rocess or ath. gas real alue or a gas,a 0 or an ideal T Joule Tomson Coeicient T c T u c T h 60 equations time, at a 6 thermodynamic roerites d u ds s u du ) (s, u h h s + T T s used with the First Law
T JT Coeicient h hconstant
Course Proerty Sources ) Ideal Gas Law with constant seciic heats ) Tables Steam Rerigerant Air ) EES CD NIST or home work conenience
EQUATION OF STATE ERRORS nitrogen
NIST Webbook Proerties tt://webbook.nist/go/chemistry/luid Temerature Table or Water in. degree increments rom 40 to 40 degrees. Select Units Select Table Tye
Select luid
Set low and high temerature and temerature increment.