Quotes. Research Findings. The First Law of Thermodynamics. Introduction. Introduction. Thermodynamics Lecture Series

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1 8//005 Quotes Thermodynamcs Lecture Seres Frst Law of Thermodynamcs & Control Mass, Open Appled Scences Educaton Research Group (ASERG) Faculty of Appled Scences Unverst Teknolog MARA emal: "One who learns by fdg has sevenfold the skll of the one who learned by beg told. - Arthur Gutterman "The roots of educaton are btter, but the frut s sweet." -Arstotle Research Fdgs Research Fdgs Retenton % of Learng After 3 days perod Read only 0% Hear only 0% Say only 70% See only 30% See + hear 50% Say & do smultaneously - 90% The Frst Law of Thermodynamcs Goal: Identfyg sources of energy teractons and wrte energy balances thermodynamc processes CHAPTER 4 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Introducton Introducton Objectves:. State the conservaton of energy prcple.. rte an energy balance for a general system undergog any process. 3. rte the unt-mass bass and unt-tme bass (or rate-form bass) energy balance for a general system undergog any process. 4. Identfy the energes causg the system to change. Objectves: 5. Identfy the energy changes wth the system. 6. rte the energy balance terms of all the energes causg the change and the energy changes wth the system. 7. rte a unt-mass bass and unt-tme bass (or rate-form bass) energy balance terms of all the energes causg the change and the energy changes wth the system. Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005

2 8//005 Objectves: Introducton 8. State the condtons for statonary, closed system and rewrte the energy balance and the unt-mass bass energy balance for statonaryclosed systems. 9. Apply the energy conservaton prcple for a statonary, closed system undergog an adabatc process and dscuss ts physcal terpretaton. Introducton Objectves: 0.Apply the energy conservaton prcple for a statonary, closed system undergog an sochorc, sothermal, cyclc and sobarc process and dscuss ts physcal terpretaton..gve the meang for specfc heat and state ts sgnfcance determg ternal energy and enthalpy change for deal gases, lquds and solds..use the energy balance for problem solvg. SODA 5 C q Transfer -Heat Transfer P = 00 kpa T = 99.6 C Oven Nas Lemak Lem 0 C 00 C q Sat. Vapor 5 C H O: Sat. Lq. Q 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, q hat happens to the propertes of the system after the energy transfer? Example: A steam power cycle. Combuston Products Q Requred put Fuel Ar Steam Power Plant Pump The net work put s Steam Turbe Heat Exchanger Q Coolg ater net, = Mechancal to Generator Boundary for for Thermodynamc Analyss Analyss, k Desred put 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Mechancal work: Pston moves up Boundary work s done by system Transfer ork Done,,kJ Electrcal work s done on system No heat transfer T creases after some tme Copyrght The McGraw-Hll Companes, Inc. Permsson requred for reproducton or dsplay. FIGURE 4-46 Ppe or duct flow may volve more than one form of work at the same tme. H O: Sat. lqud H O: Super Vapor e = V e, = V t/00, kj Voltage, V 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005

3 8//005 Frst Law Transfer Frst Law Transfer s tal total energy s Can t change? How? hy? Movable boundary poston gone up E = U +KE +PE expands E = U +KE +PE or e = u +ke +pe, kj/kg Total energy E, E thermal equlbrum 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, A change has taken place. 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, E = U +KE +PE Intal Frst Law Transfer Movable boundary poston gone up, E A change has taken place Fal s fal energy s E =U +KE +PE expands 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Q How to relate changes to the cause,k q, or Q q, or, Q Q,k Frst Law Transfer Propertes wll change dcatg change of state E, P, T, V To E, P, T, V Heat as a cause (agent) of change 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Transfer Frst Law Transfer How to relate changes to the cause Propertes wll change dcatg change of state, k, ω, kj/kg, ω, kj/kg E, P, T, V To E, P, T, V,k How to relate changes to the cause Propertes wll change dcatg change of state Mass E, P, T, V m ϑ,k To E m k Mass mass, = ( ϑ ), E, P, T, V ( ϑ ),kj / kg ( ϑ ),kj / kg ork as a cause (agent) of change 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Mass transfer as a cause (agent) of change 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 3

4 8//005 How to relate changes to the cause Frst Law Transfer Mass Propertes wll change dcatg change of state E, P, T, V To Q Q E, P, T, V Dynamc Energes as causes (agents) of change Mass 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law-Conservaton of Prcple z =h z =h/ z =0 must be conserved any process. cannot be created nor destroyed. It can only change forms. Total before a process must equal total energy after process E=U+KE+PE = U+PE In any process, every bt of energy should be accounted for!! Known as Conservaton of Prcple 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law-Conservaton of Prcple z =h z =h/ z =0 must be conserved any process. cannot be created nor destroyed. It can only change forms. Total before a process must equal total energy after process E=U+KE+PE=U+0+PE E=U+KE+PE In any process, every bt of energy should be accounted for!! Known as Conservaton of Prcple 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 Frst Law-Conservaton of Prcple must be conserved any process. cannot be created nor destroyed. It can only change forms. Total before a process must equal total energy after process z =h z =h/ z =0 E=U+KE+PE E=U+KE+0 r ke υ f r υ pe z f z u T In any process, every bt of energy should be accounted for!! Known as Conservaton of Prcple 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 Frst Law Frst Law of Thermodynamcs Enterg - Leavg = Change of system s energy Enterg - Leavg = Change of system s energy Balance Balance Amount of energy causg change must be equal to amount of energy change of system 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, E E = E sys, kj or e e = e sys, kj/kg or E E = E sys,k 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 4

5 8//005 Frst Law of Thermodynamcs How to relate changes to the cause Mass Q Q Propertes wll change dcatg change of state E, P, T, V To E, P, T, V Dynamc Energes as causes (agents) of change Mass 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, FIGURE 4 7 The energy change of durg a process s equal to the net work and heat transfer between the system and ts surroundgs. Copyrght The McGraw-Hll Companes, Inc. Permsson requred for reproducton or dsplay. 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Interacton Energes The Agent E = Q + +E mass,,kj e = q + θ, kj/kg Frst Law - Interacton Energes The Agent E = Q + +E mass,,kj e = q + θ, kj/kg E = Q + + E mass, ; k E = Q + + E mass, ; k For Closed system: E mass, = 0, kj, θ = 0, kj/kg E mass, = 0, k 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, For Closed system: E mass, = 0, kj, θ = 0, kj/kg E mass, = 0, k 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law - s The Change Ith change wth the system, E sys = E -E s the sum of Frst Law Change The Change Ith E sys = U+ KE+ PE, kj e sys = u+ ke+ pe, kj/kg Internal energy change, U = U U E sys = U + KE + PE, k ketc energy change, KE = KE KE potental energy change, PE = PE PE 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, For Statonary system: KE= PE = 0, kj ke= pe=0, kj/kg KE = PE = 0 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 5

6 8//005 Frst Law General Frst Law General Enterg - Leavg = Change of system s energy General system Balance E E = E sys, kj or e e = e sys, kj/kg or E = E sys E,k 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Q + + E mass, Q -E mass, = U+ KE + PE, kj q + θ q ω θ = u+ ke + pe, kj/kg Q + + E mass, Q E mass, = U + KE+ PE, k 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Statonary Statonary system Q Q + + E mass, -E mass, q q ω + θ - θ = U+0+0 = u , kj/kg Q Q + + E mass, E mass, = U //005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Closed Closed system Q Q = U+ KE + PE, kj q q ω = u+ ke + pe, kj/kg Q = U + KE + PE, k Q 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Statonary & Closed Statonary Closed system Q Q = U q q ω = u , kj/kg Q Q = U Frst Law Hstorcal Perspectve Statonary Closed system-hstory (Q Q ) + ( - ) = (Q Q ) - ( - ) = Q net, net, = Q q ω = q net, ω net, = (q q ) (ω ω ) 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, //005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 6

7 8//005 Frst Law Adabatc Process Statonary Closed Adabatc: = U ω = u , kj/kg ω = ω b,compress and ω = ω b,expand kj/kg 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Adabatc Process Adabatc: ω ω = u , kj/kg Spontaneous Expanson: pstoncylder devce ω = 0, and ω = ω b,expand kj/kg ork s expanson work: 0 - ω = -ω b,expand = u< 0 u < u. Fal u s smaller than tal u, T drops 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Adabatc Process Adabatc: ω ω - 0 = u+0+0, kj/kg Compresson: pston-cylder devce ω = 0, and ω = ω b,compress kj/kg ork s compresson work: ω = ω b,compress = u> 0 u > u. Fal u s bgger than tal u; T creases 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Cyclc Process Cyclc: Q Q = E sys = 0 q q ω = e sys = 0, kj/kg q -q = ω ω or q net, = ω net, Expanson: q - 0 =ω = ω b,expand All heat absorbed s used to do expanson work 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, FIGURE 3- The net work done durg a cycle s the dfference between the work done by the system and the work done on the system. f ω b = Pdν f b = PdV 3-4 Copyrght The McGraw-Hll Companes, Inc. Permsson requred for reproducton or dsplay. ork done - Cyclc process Total work s area of A mus area of B. Total work s shaded area Output power b, =, k t Input power b, =, k t 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Q Q Frst Law Cyclc Process Cyclc: Q Q = E sys = 0 q q ω = e sys = 0, kj/kg q -q = ω ω Compresson: 0 - ω = -ω b,compress = q q = - q All compresson work s removed as heat 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 7

8 8//005 Frst Law Isochorc Process Isochorc : Q Q = U Rgd Tank q q ω = u, kj/kg Sce, ω - ω = ω others + 0 = ω Then, q -q = u 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law - Isobarc Process Isobarc : Q Q = U Pston-Cyl q q ω = u, kj/kg For expanson, ω - ω = ω - ω b,expand q q = ω b,expand + u, kj/kg 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Isobarc expanson q q q q Frst Law - Isobarc Process q q ω = u, kj/kg q q = ω b,expand + u, kj/kg fal = Pdν + u = P dν + u tal = P ν 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, P ν + u u Frst Law - Isobarc Process Isobarc expanson fal q q = Pdν + u = P dν + u q q tal = P ν 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, P ν + u u ( P ν u ) q q = P ν + u + q q = h h = h Frst Law - Closed system q q ω = u, kj/kg For pure substances, use property table and mathematcal manpulatons to determe u and u. Then u = u u. And h and h,. Then h = h h. Frst Law Specfc Heat Specfc heats to fd U U and H Specfc Heat Capacty C ν at constant volume, C p, at constant pressure Amount of heat necessary to crease temperature of a unt mass by K or degree Celcus 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, //005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 8

9 8//005 Frst Law Specfc Heat For Ideal Gases: : Estmate ternal energy change and enthalpy change Assume smooth change of C wth T, & approxmate to be lear over small T T (approx. a few hundred degrees) u = CV dt CV, avg ( T T ) = CV, avg T h kj kg = C dt C P avg ( T T ) = C P avg T K,, P 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, , Frst Law Specfc Heat For Ideal Gases : Estmate ternal energy change and enthalpy change Assume smooth change of C wth T, & approxmate to be lear over small T T (approx. a few hundred degrees) C ν, avg avg s determed usg terpolaton technque & use Table A-b A 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Specfc Heat For solds & Lquds: May consder as compressble or constant volume C v = C p = C, kj/kg K K, Hence the enthalpy change, u = C av (T - T ), kj/kg Enthalpy h = u + Pν, P So, dh =du= + νdp +Pdν,, kj/kg h h = u + ν P P + P ν, P, kj/kg h h = C av T T + ν P P + P ν, P, kj/kg 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Specfc Heat For solds & lquds: u = C av (T - T ), kj/kg C v = C p = C, kj/kg K K, For solds: Enthalpy, h h = C av T T + ν P P + 0,, kj/kg Sce, ν P P = 0, Then, h C av T, kj/kg 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Frst Law Specfc Heat Frst Law Example Prob 4-0 For Lquds: u = C av (T - T ), kj/kg Enthalpy, h h = C av T T + ν P P + 0,, kj/kg,,kj H O P = P Intal: V = 5 L phase sat. lq.. P = 50 kpa,, = 300 kj, Voltage, V t = 45 m x 60s/m t = 45 x 60s Current, = 8 A, Heaters where P P = 0, Pumps where T T = 0, h h = u C av T, kj/kg h h = ν P, kj/kg Or wrtten as h - h = ν(p - P ) 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Fal phase s sat. lq.-vapor mx at 50 kpa.. Qualty of steam s 0.5. Sce m g = m/, hence x = m g /m = m/m =0.5. 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 9

10 8//005,,kJ Frst Law Example Prob 4-0 H O P = P Intal: V = 5 L phase sat. lq.. P = 50 kpa,, = 300 kj, Voltage, V t = 45 m x 60s/m t = 45 x 60s Current, = 8 A, x = 0.5 ν = ν f@50 kpa, = m 3 /kg h = h f@50 kpa, = 467. kj/kg ν = [ν[ f + x v fg fg kpa kpa, = m 3 /kg h = [h[ f + xh fg kpa =580 kj/kg fg kpa 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, balance, Frst Law Example Prob 4-0 E - E = E sys = U , kj, + e, = U + Then where b e, = U U + PV PV, 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, , = PdV e, = U + PV ( U + PV ), e, = H H, = H, balance, Frst Law Example Prob 4-0 E - E = E sys Sce e, = H H, = H, V t Electrcal work done s = m h,, kj ( m h, ) Voltage source s V =, kj t balance, Frst Law Example Prob 4-0 E - E = E sys Voltage source s L *0 m / L 000 [( )( ) kj / kg 300 kj ] m /kg V = (8 A)(45 m x60 s / m) Note that the unt kj/s = Volts-Ampere or VA 30.9 kj V = = 30.9 As AV A = 30.9 V 8//005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, //005 Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, Copyrghts DR JJ, ASERG, FSG, UTM Shah Alam, 005 0