When the rate of heat transfer remains constant during a process

Transcription

1 HERMODYNAMICS It has bee poited out that eergy ca either be created or destroyed; it ca oly chage forms. his priciple is the first law of thermodyamics or the coservatio of eergy priciple. Durig a iteractio betwee a system ad its surroudigs, the amout of eergy gaied by the system must be exactly equal to the amout of eergy lost by the surroudigs. Eergy ca cross the boudary of a closed system i the form of heat ad work. Heat rasfer Heat trasfer or just heat is defied as the form of eergy that is trasferred betwee two systems (or a system ad its surroudigs) by virtue of temperature differece. Hece, if two systems are the same temperature, there ca be o heat trasfer. A process durig which there is o heat trasfer is called a adiabatic process. A process ca be adiabatic if both the system ad its surroudigs are at the same temperature or if the system is well isulated. A adiabatic process is ot ecessarily a isothermal process! he amout of heat trasferred from state to state is demoted Q or just Q, it has uits kj. Q q m he heat trasfer rate is give as t Q Qdt t (kj/kg) (kj) Whe the rate of heat trasfer remais costat durig a process Q Q t (kj)

2 Modes of Heat rasfer here are three modes of heat trasfer: coductio, covectio ad radiatio. Coductio heat trasfer occurs i statioary solids, liquids ad gasses. It occurs due to radom molecular motio whe atoms or molecules of the system collide. he heat trasfer rate equatio is called the Fourier s law give below. Fourier s law: Q cod k A t d dx (W) Q cod kt A x (W) Where Δx is the costat thickess of the system, Δ is the temperature differece across the layer ad A is the area perpedicular to the heat flow. Fially, k t is called the thermal coductivity ad it is a measure of a materials ability to trasfer heat. Covectio heat trasfer occurs betwee a solid surface ad a movig fluid (liquid or gas). It occurs due to radom molecular motio ad bulk fluid motio. he first layer of fluid adjacet to the plate is statioary hece, heat trasfer is due to coductio. hereafter, it is due to hot fluid beig carried away ad replace by cooler fluid. he heat trasfer rate equatio is called the Newto s law of coolig give below. Newto s law of coolig: Q ha ) cov ( s f (W) Where h is the covectio heat trasfer coefficiet, A is the surface area through which heat trasfer takes place, s is the surface temperature ad f is the bulk fluid temperature. Radiatio is the eergy emitted by matter i the form of electromagetic waves. Ulike coductio ad covectio, it does ot require the presece of a iterveig medium. I fact it is fastest through vacuum. Solids, liquids ad gasses ca emit, absorb, reflect or trasmit radiatio.

3 he maximum rate of radiatio that ca be emitted by a surface at s is govered by the Stefa-Boltzma law as Q 4 rad A s (W) Where A is the area of the emittig surface ad σ = 5.67 x 0-8 is the Stefa- Boltzma costat. Oly a idealized blackbody ca emit this maximum amout of radiatio at s. Real objects emit less ad the rate of heat trasfer is expressed: Q 4 rad A s (W) Where ε is the emissivity ad measures the ability of a object to emit radiatio. It rages betwee 0 where ε = is a blackbody Absorbed radiatio also affects the eergy levels i a system it is give by Q abs Q ic (W) Where α is called absorptivity ad it measure the ability of a object to absorb radiatio. I the special case of a small object of emissivity ε, a surface area A ad absolute temperature s, is eclosed i a large surface at absolute temperature surr. he et rate of radiatio heat trasfer is give as: Q rad A 4 s 4 surr (W) Work As metioed, eergy ca cross a system boudary by heat trasfer ad work. herefore, if the eergy crossig the boudary of a closed is ot heat, it is work. he amout of work doe durig a process betwee state ad state is deoted by W or just W, it has uits kj. he work doe per uit mass is give as W w m (kj/kg) he work doe per uit time is called power. 3

4 Sig Covetio for Heat ad Work Heat trasfer ito a system is positive. Work doe by the system is positive. Heat trasfer out of a system is egative. Work doe to the system is egative. Electrical Work Whe calculatig rate of electrical work we use the expressio: W e VI (W) Where W e is the electrical power, V is the potetial differece ad I is the curret. I the case where V ad I varies with time, the electrical work doe durig a time iterval Δt is expressed: W e VIdt (kj) Whe V ad I are costat, expressio above is reduced to Mechaical Forms of Work W e We are familiar with the expressio VIt W Fs (kj) (kj) Where F is the force actig o a body ad s is the distace traveled be the body. If the force is ot costat tha W Fds (kj) here are two requiremets for work iteractio betwee a system ad its surroudigs to exist:. here must be a force actig o the boudary ad. he boudary must move. 4

5 Boudary Work Oe form of mechaical work frequetly ecoutered i practice is associated with the expasio or compressio of a gas i a pisto-cylider device. Durig this process, part of the boudary moves back ad forth. he expasio ad compressio work is called movig boudary work or just boudary work. he boudary work is aalysed for a quasi equilibrium process (or quasi static process), a process durig which the system remais i equilibrium at all times. A quasi equilibrium process is achieved whe the pisto moves at low velocities. Uder this coditio, the system work output is maximum ad the work iput is miimum. he differetial boudary work doe is W b Fds PAds PdV he total boudary work doe durig the etire process is the W b PdV Positive result idicates boudary work output (expasio). Negative result idicates boudary work iput (compressio). he area uder the process curve o a P- V diagram is equal, i magitude to the work doe durig a quasi-equilibrium expasio or compressio of a closed system. Area A da PdV W b 5

6 Polytropic Process Durig expasio ad compressio process of real gases, pressure ad volume are ofte related by PV = C, where ad C are costats. his is kid of process is called polytropic process. Hece Substitutig ito Sice W For a ideal gas where b P CV W b PdV, V V PV PV PdV CV dv C C PV PV PV mr he above equatio ca also be writte W b mr 6

7 Gravitatioal Work Gravitatioal work is defied as work doe by or agaist a gravitatioal force field. I a gravitatioal force field, F mg Where m is the mass of the body ad g is the acceleratio of gravity. he the work required to raise this body from level z to z is W g Fdz mg dz mg z z (kj) I.e. the chage i potetial eergy where z z is the vertical distace traveled. Acceleratio Work he work associated with the chage i velocity of a system is called acceleratio work. Hece W g Fds m dv dt F ma dv a dt dv F m dt Vdt mvdv mv V I.e. it is the chage i kietic eergy of the body. 7

8 Shaft Work A force F actig through a momet arm r geerates a torque τ Where is the umber of revolutios. Fr F r his force acts through a distace s where s r he shaft work is the W sh Fs r he shaft power is give by r W sh 8

9 Sprig work Work doe by a sprig is give as W sprig Fdx Where F kx Ad k is the sprig costat give i kn/m. Hece W sprig k x x (kj) x ad x are the iitial ad fial displacemet of the sprig respectively Cotiued 9

10 he First Law of hermodyamics he first law of thermodyamics as stated previously, is the coservatio of eergy priciple. I this sectio, we will use the first law of thermodyamics to relate heat, Q, work, W ad, total eergy, E. Recall that total eergy, E refers to the eergy a system possesses. Eergy Balace E U KE PE With regards to the first law of thermodyamics, the et chage (icrease ad decrease) i the total eergy of the system durig a process is equal to the differece betwee the total eergy eterig ad the total eergy leavig the system durig that process. Hece Where E i E out E system E i = total eergy eterig the system E out = total eergy leavig the system ENERGY BALANCE E system = chage i the total eergy of the system Eergy Chage of a System, Esystem : he eergy chage of a system is expressed as below Eergy chage = Eergy at fial state Eergy at iitial state Or E system E fial E iitial E E Or simply Where E U KE PE (Recall that E = U + KE + PE) 0

11 U m( u KE m( V PE mg( z u ) V ) z ) Most systems ecoutered are statioary systems i.e. the do ot chage i velocity or elevatio durig a process. Hece the chage i kietic ad potetial eergies are zero ( KE = PE = 0) Hece the eergy balace is very ofte reduced to E U. Mechaisms for Eergy rasfer, E i ad E out : Eergy ca be trasferred to a system i three forms: heat, work ad mass flow.. Heat trasfer, Q Heat trasfer to a system (heat gai) icreases the eergy of the system ad heat trasfer from the system (heat loss) decreases the eergy of the system.. Work, W Work trasfer to a system (i.e. work doe oe a system) icrease the eergy of the system ad work trasfer from a system (i.e. work doe by the system) decreases the eergy of the system. 3. Mass flow, m Whe mass eters a system, the eergy of the system icreases because mass carries eergy with it. Likewise, whe mass leaves the system it takes out some of the eergy of the system. Hece, the eergy balace ca be writte as E Qi Qout W i Wout Emass, i Emass out Esystem i Eout, he subscripts i ad out deote quatities that eter ad leave the system respectively. All six quatities o the right side of the equatio represet amouts, ad thus they are positive quatities. he directio of ay eergy trasfer is described by the subscript i ad out. hus we do ot eed to adopt a formal sig covetio.

12 O a per uit time basis the eergy balace is E i E out E system O a per uit mass basis the eergy balace is e i e out e system

13 For a closed system udergoig a cycle, the iitial ad fial states are idetical therefore E system E E - I terms of heat ad work iteractios W et, out Qet, i 0 Whe solvig a problem that ivolves a ukow heat or work iteractio, it is commo practice to assume Q W E Where we assume there is a heat iput ito the system ad work is doe by the system (work output). Whe a egative aswer is obtaied for Q ad W, it simply meas the assumed directio is wrog ad should be reversed. Questios. he radiator of a steam heatig system has a volume of 0 L ad is filled with superheated vapour at 300 kpa ad 50 C. At this momet both the ilet ad exit valves to the radiator are closed. Determie the amout of heat that will be trasferred to the room whe the steam pressure drops to 00 kpa. Also show the process o a P-v diagram with respect to saturatio lies.. A 0.5 m3 rigid tak cotais refrigerat-34a iitially at 00 kpa ad 40 percet quality. Heat is ow trasferred to the refrigerat util the pressure reaches 800 kpa. Determie (a) the mass of the refrigerat i the tak ad (b) the amout of heat trasferred. Also, show the process o a P-v diagram with respect to saturatio lies. 3

14 3. A well isulated rigid tak cotais 5 kg of a saturated liquid-vapour mixture of water at 00 kpa. Iitially, three-quarters of the mass is i the liquid phase. A electric resistor placed i the tak is coected to a 40-V source ad a curret of 8 A flows through the resistor whe the switch is tured o. Determie how log it will take to vaporize all the liquid i the tak. Also, show the process o a -v diagram with respect to saturatio lies. 4. A isulated tak is divided ito two parts by a partitio. Oe part of the tak cotais.5 kg of compressed liquid water at 60 C ad 600 kpa while the other part is evacuated. he partitio is ow removed, ad the water expads to fill the etire tak. Determie the fial temperature of the water ad the volume of the tak for a pressure of 0 kpa. 5. A pisto-cylider device cotais 5 kg of refrigerat-34a at 800 kpa ad 60 C. he refrigerat is ow cooled at costat pressure util it exists as a liquid at 0 C. Determie the amout of heat loss ad show the process o a -v diagram with respect to saturatio lies. 6. A pisto-cylider device cotais steam iitially at MPa, 350 C, ad.5 m 3. Steam is allowed to cool at costat pressure util it first starts codesig. Show the process o a -v diagram with respect to saturatio lies ad determie (a) the mass of the steam, (b) the fial temperature, ad (c) the amout of heat trasfer. 4

15 7. A pisto cylider device iitially cotais steam at 00 kpa. 00 C ad 0.5 m 3. At this state, a liear sprig is touchig the pisto but exerts o force o it. Heat is ow slowly trasferred to the steam, causig the pressure ad the volume to rise to 500 kpa ad 0.6m 3, respectively. Show the process o a P-v diagram with respect to saturatio lies ad determie (a) the fial temperature, (b) the work doe by the steam, ad (c) the total heat trasferred. 8. A pisto-cylider device iitially cotais 0.5 m 3 of saturated water vapour at 00 kpa. At this state, the pisto is restig of a set of stops, ad the mass of the pisto is such tat a pressure of 300 kpa is required to move it. Heat is ow slowly trasferred to the steam util the volume doubles. Show the process o a P-v diagram with respect to saturatio lies ad determie (a) the fial temperature, (b) the work doe durig this process, ad (c) the total heat trasfer. 9. wo rigid taks are coected by a valve. ak A cotais 0. m 3 of water at 400 kpa ad 80 percet quality. ak B cotais 0.5 m 3 of water at 00 kpa ad 50 C. he valve is ow opeed, ad the two taks evetually come to the same state. Determie the pressure ad the amout to heat trasfer whe the system reaches thermal equilibrium with the surroudig at 5 C. 5

16 Specific Heats It takes differet amouts of eergy to raise the temperature of idetical masses of differet substaces by oe degree. - We eed about 4.5 kj of eergy to raise the temperature of kg of iro to from 0 to 30 ºC - It takes about 4.8 kj of eergy to raise the temperature of kg liquid water from 0 to 30 ºC It is desirable to have a property that will eable us to compare the eergy storage capabilities of various substaces. his property is the specific heat. he specific heat is defied as the eergy required to raise the temperature of a uit mass of a substace by oe degree. I thermodyamics, we are iterested i the kids of specific heats: - Specific heat at costat volume (C v ) - Specific heat at costat pressure (C p ) he Costat Volume Heatig Of a Gas C v ca be viewed as the eergy required to raise the temperature of the uit mass of a substace by oe degree as the volume is maitaied costat. Let a mass of gas (m) be heated at costat volume such that it rises from to ad its pressure rises from P to P. he heat received my the gas is Q mcv Eq. 6

17 P P Applyig the first law eergy balace, assumig KE = PE = 0 ad o work is doe (Δν = 0) herefore: E i E out E system P V = V 3 Rewritig Eq. 3 produces herefore I differetial form C v C V C v v u Q W U KE PE Q U Eq. Hece, mc v U Eq. u u u du d v cos ta t 7

18 he Costat Pressure Heatig of a Gas As before, for a costat pressure process, heat received by the gas is give by Q mc p Eq. 4 he process o the P-v diagram is show below P P = P Work doe durig the process is give by the area uder the curve. herefore W P V V W P V Eq. 5 P V V V V Applyig the first law eergy balace, assumig KE = PE = 0 ad o work is doe (Δν = 0) herefore: E i E out Q W U Q U W mc mc mc C p p p p E system U U PV PV U P V U PV H H h h Eq. 6 Rewritig Eq. 6 produces 8

19 I differetial form C p C p h h h dh d p cos ta t Specific Heat Relatios for Ideal Gases A special relatioship betwee C p ad C v for ideal gases ca be obtaied by differetiatig the relatio h u Pv h u R dh du Rd We kow that dh C d ad du C d p v, therefore herefore, C d C d Rd p C p v C v R Aother importat ideal gas relatio is the specific heat ratio, γ which is give as γ =.4 for air at room temperature. C C p v 9

20 Iteral Eergy, Ethalpy, ad Specific Heat Relatios for Solids ad Liquids Specific Heat A substace whose specific volume is costat is called icompressible substace solids ad liquids. For icompressible substace (solids ad liquids) C p C v C Iteral Eergy Chages u C ( ) av Where C av is the average C value at average temperature. Ethalpy Chages By differetiatig h u Pv, we obtai dh du vdp Pdv Sice v is costat the dv = 0, yieldig Itegratig this equatio yields dh du vdp h u vp Cav vp For solids, the term v P is isigificat so h u C For liquids, there are two possible cases ) Costat pressure process (heater), ( P = 0) therefore, h u C ) Costat temperature process (heater), ( = 0) therefore, h vp av av 0

21 he Polytropic Process of a Gas A gas udergoig a polytropic process expad or cotracts accordig to PV C For a process betwee state ad state is the P V PV We have also established that the work doe durig a polytropic process is give by: Sice PV = mr W PV PV mr W he Combiatio of the Polytropic Law PV = C ad the Ideal Gas Equatio of a Perfect Gas he law PV C will eable calculatios to be made of the chages i pressure ad volume which occur durig a polytropic process. Combiig this iwht the characteristic equatio of a perfect gas will eable variatios i temperature to be foud. Cosider a polytropic process i which the state of a gas chages from P, V, to P, V,. By the polytropic law: PV PV Eq.

22 Hece: P P V V Eq. By the ideal gas equatio, PV Eq. 3 P V PV Eq. 4 P V

23 3 Hece, substitutig Eq ito Eq 4, V V V V Or, V V Eq. 5 Also, from Eq 4 P P V V Eq. 6 Substitutig Eq 6 ito Eq 4, P P Eq. 7 Combiig Eq. 5 ad Eq. 7 P P V V

24 Questios: ) kg of gas, occupyig 0.7 m 3, had a origial temperature of 5 ºC. If was the heated at costat volume util its temperature became 35 ºC. How much heat was trasferred to the gas ad what was its fial pressure? ake C v = 0.78 kj/kg K ad R = 0.87 kj/ kg K. ) A gas whose pressure, volume ad temperature are 75 kpa, 0.09 m 3 ad 85 ºC, respectively, has its state chaged at costat pressure util its temperature becomes 5 ºC. How much heat is trasferred form the gas ad the work doe o the gas durig the process? ake R = 0.87 kj/ kg K. ad C p =.005 kj/ kg K. 3) A gas whose origial pressure ad temperature were 300 kpa ad 5 ºC, respectively, is compressed accordig to the law PV.4 = C util its temperature becomes 80 ºC. Determie the pressure of the gas after it is compressed. 4) kg of gas at.4 MPa ad 80 ºC is expaded to four times the origial volume accordig to the law PV.3 = C. Determie a) the origial ad fial volume of the gas b) the fial pressure of the gas c) the fial temperature of the gas ake R = 0.87 kj/ kg K 5) 0.5 kg of air at pressure of 40 kpa occupies 0.5 m3 ad form this coditio it is compressed to.4 MPa accordig to the law PV.5 = C. Determie a) the chage of iteral eergy of the air b) the work doe o or by the air c) the heat received or rejected by the air ake C p =.005 kj/ kg K ad C v = 0.78 kj/kg K 4

25 he Adiabatic Process of a Gas Whe dealig with the geeral case of a polytropic expasio ad compressio, the process followed a law of the form PV = C. Now the adiabatic process is a particular case of a polytropic process i which o heat is allowed to eter of leave the system Cosider a adiabatic process i which the state of a gas chages from P, V, to P, V,. he, Also U mcv P V PV mr W Where gamma is kow as the adiabatic idex. From the polytropic law, V V Ad the characteristic equatio P V PV PV P P PV Applyig the first law eergy balace, Q W U But for a adiabatic process, Q = 0 Hece, mr W U mc v Hece, R Cv 5

26 R C v But R = C p - C v therefore, γ for air is.4 R C v Cv R C C C p v v Questio: A quatity of gas occupies a volume of 0.4 m 3 at a pressure of 00 kpa ad a temperature of 0 ºC. he gas is compressed isothermally to a pressure of 450 kpa ad the expaded adiabatically to its iitial volume. Determie, for this quatity of gas a) the heat trasfer durig the compressio b) the chage of iteral eergy durig the expasio c) the mass of the gas ake =.4 ad C p = kj/ kg K 6