UNIT-I BASIC CONCEPTS AND FIRST LAW

Transcription

1 hermodyamics UNI-I BASI ONEPS AND FIRS LAW hermodyamics is a brach of sciece that deals with the relatioshi betwee the heat ad mechaical eergy. hermodyamics Laws ad Alicatios hermodyamics laws are formulated based o the exeriece ad the results of exerimets. hese laws gover the riciles of eergy trasfer. hese laws are used i the field of eergy trasfer, like, steam ad uclear ower lats, gas turbies, iteral combustio egies, air coditioig lats, refrigeratio, jet roulsio, comressors, chemical rocess lats, etc. Macroscoic ad Microscoic cocets I macroscoic aroach, the behaviour of the matter is studied cosiderig a certai quatity of matter (cosistig of may molecules), without the evets occurrig i the molecular level. Actio of may molecules ca be erceived by huma seses. I microscoic aroach, the matter is comosed of myriads of the molecules ad the behaviour of the matter is described by summig u behaviour of the idividual molecules. ocet of otiuum I macroscoic aalysis we cosider a small amout (small volume) of matter. his small volume is very large comared to molecular dimesios. Eve a very small volume of a matter cotais large umber of molecules. osider the mass δm i a volume δ. he average mass desity is defied as δm/δ. δm/δ Domai of molecular effect δm/δ Domai of cotiuum δm/δ ρ δ, δm If δ 0, δm/δ If δm 0, δm/δ 0 he smallest volume which may be regarded as cotiuous is δ. he desity ρ of the system at a oit is defied as δ δ

2 hermodyamic System, Surroudigs ad Boudary A thermodyamic system is a secified quatity of matter or a regio i a sace uo which the attetio is focused o to aalyse the roblem. System Boudary Surroudigs Everythig outside the system is called surroudigs. he system ad surroudigs is searated by a boudary. he boudary may be real or imagiary, fixed or movig. here are three tyes of systems: Oe system. Oe system. losed system 3. Isolated system System Mass out Mass trasfer Boudary Eergy trasfer he oe system is oe i which both matter (mass) ad eergy (work or heat) cross the boudary of the system. Examles: Steam turbie, Gas turbie, Rotary comressors, Heat exchagers, etrifugal um, Water turbie, etc. Work rasfer W d losed system he closed system is a system of fixed mass. I closed system, there is o mass trasfer, but eergy trasfer takes lace. System No mass trasfer Boudary Eergy trasfer Examles: I..egie cylider with the both valves closed, Recirocatig air comressor with both the valves closed.

3 Isolated system Work trasfer W d I isolated system, there is o mass or eergy trasfer across the boudary of the system. System No mass trasfer Boudary No eergy trasfer Uiverse he system ad surroudigs together is called Uiverse. hermodyamic Equilibrium he system uder Mechaical, hermal ad hemical equilibrium is said to be uder hermodyamic equilibrium. A system is said to be i mechaical equilibrium whe there is o ubalaced force withi the system. (Uiformity of force). A system is said to be i chemical equilibrium whe there is o chemical reactio betwee differet arts of the system. (Absece of chemical reactio). A system is said to be i thermal equilibrium whe there is o temerature chage betwee differet arts of the system. (Uiformity of temerature). he system uder mechaical ad thermal equilibrium but ot uder chemical equilibrium is said to be uder meta-stable equilibrium. hermodyamic Proerty Ay observable characteristic of the system uder thermodyamic equilibrium is called a roerty. Examles: Pressure, emerature, olume, Desity. Itesive ad Extesive roerty he roerty which is ot deedig o the mass of the system is called itesive or itrisic roerty. Examles: Pressure, emerature, Desity, Secific volume (m 3 /kg), Secific ethaly (J/kg) he roerty which is deedig o the mass of the system is called extesive or extrisic roerty. Examles: olume (m 3 ), Mass (kg), Ethaly (J) osider a system comosed of some fluid. Measure all the roerties of the system, like, ressure ( ), volume ( ), temerature ( ), desity (ρ ), ethaly (H ), mass (m ), secific volume (v ), secific ethaly (h ), etc. Now, disturb the mass by lacig a artitio ad measure all the roerties of a art (,,, ρ, H, m, v, h ). 3

4 It ca be see that, ; ; ; ρ ρ ; H H ; m m ; v v ; h h System hermodyamic State Partitio. State A state is the coditio of the system at a istat of time as described or measured by its roerties. Or each uique coditio of the system is called state. Examle: 8 bar & 0.5 m 3 hermodyamic Path ad Process he successio of states assed through durig a chage of state is called the Path. Whe the ath is comletely secified, the chage of state is called a rocess. Quasi-static rocess Quasi meas almost. A quasi-static rocess is a successio of equilibrium states ad ifiite slowess is its characteristic feature. Stos Pisto Gas P Fial ositio () Weight Iitial ositio () 4

5 osider a system of gas cotaied i a cylider. he force exerted by the gas is balaced by the weight. he gas is i equilibrium i the iitial state. If the weight is removed, the isto will move u util it hits the stos. he system agai comes to equilibrium state. he itermediate states betwee ad are o-equilibrium states ad caot be rereseted by the thermodyamic coordiates. Locus of No-equilibrium states Now if sigle weight o the isto is made u of very small ieces of weights ad these weights are removed oe by oe very slowly from the to of the isto, at ay istat of the uward travel of the isto, the dearture of the state of the system from equilibrium state will be ifiitesimally small. So, each itermediate state will be a equilibrium state. Such a rocess which is locus of all the equilibrium oits is called Quasi-static rocess. x x Locus of equilibrium states x x x hermodyamic ycle A sigle rocess caot take lace cotiuously without ed. o get cotiuous work, a set of rocesses has to be reeated agai ad agai. he set of rocesses which brigs the system to the origial state is called a cycle. otiuous work trasfer is ossible oly with a cycle. Examles: Otto cycle, Diesel cycle, Dual cycle, etc. Work 5

6 Work is oe form of eergy trasfer. Work is said to be doe by a system if the sole effect o thigs exteral to the system ca be reduced to the raisig of weight. he weight may ot actually be raised, but the et effect exteral to the system would be the raisig of a weight. osider a Battery-motor system drivig a fa as show i fig. he system is doig work uo the surroudigs. Whe the fa is relaced by a ulley ad a weight as show, the weight may be raised. he sole effect o the thigs exteral to the system is the raisig of a weight. Work doe by a system Work doe o the system Uit of work d work (or) Dislacemet work ve - ve Nm (or) J osider a cylider-isto arragemet. Let ad be ressure ad volume of the gas i the cylider at iitial ositio. Let the isto moves to a ew fial ositio secified by ressure ad volume. At ay oit i the travel of the isto, let the ressure be ad the volume. Work doe Force x Distace moved dw F x dl A dl d Whe the isto moves out from ositio to ositio with the volume chagig from to, the amout of work W doe by the system will be W d Work trasfer Area uder the curve o a - diagram Electrical work Whe a curret flows through a resistor, there is work trasfer ito the system. his is because the curret ca drive a motor, the motor ca drive a ulley ad the ulley ca raise a weight. 6

7 E Potetial differece (oltage) I urret Shaft work Work trasfer, W E I Whe the shaft is rotated by a motor, there is work trasfer ito the system. his is because the shaft ca rotate a ulley which ca raise a weight. If is the torque alied to the shaft ad ω is the agular velocity, work trasfer ca be writte as W ω Paddle-wheel work or stirrig work As the weight is lowered, ad the addle wheel turs, there is work trasfer ito the fluid system which gets stirred. Sice the volume of the system remais costat, d work is zero. If m is the mass of the weight lowered through a distace dz ad is the torque trasmitted by the shaft i rotatig through a agle of dθ, the work trasfer is give by, W W ' dz d θ W Weight lowered Flow work he flow is sigificat i oe system. his work reresets the eergy trasferred across the system boudary as result of the eergy imarted to the fluid by a um, blower or comressor to make the fluid to flow across the cotrol volume. Flow work is give by, W Work doe i stretchig a wire Let us cosider a wire of legth L, subjected to a tesio force. he ifiitesimal amout of work is doe o the wire which makes the wire to stretch to a legth L dl. W 7 dl

8 L Negative work he work is doe o the wire. Work doe i chagig the area of a surface film A film o the surface of a liquid has a surface tesio, which is a roerty of the liquid ad the surroudigs. he surface tesio acts to make the surface are of the liquid a miimum. he work doe o a homogeeous liquid film i chagig its surface area by a ifiitesimal amout da is W σ da σ Surface tesio (N/m) Magetizatio of a aramagetic solid he work doe er uit volume o a magetic material through which the magetic ad magetizatio fields are uiform is H Field stregth I omoet of the magetizatio field i the directio of the field Heat W I I H Heat is defied as the form of eergy that is trasferred across the boudary by virtue of a temerature differece. he temerature differece is otetial ad heat trasfer er uit area is flux. Zeroth Law of hermodyamics If the bodies A ad B are i thermal equilibrium with the third body, the these two bodies A ad B will be i thermal equilibrium with each other. his law is the basis for measuremet of temerature. di A B If A Alicatio of Zeroth law & B the A B It is the basis for temerature measuremet. Ideal ad Real gases A ideal gas or erfect gas is a hyothetical gas cosistig of idetical articles of zero volume, with o itermolecular forces. Additioally, the costituet atoms or molecules udergo erfectly elastic collisios with the walls of the cotaier. Real gases do ot exhibit these exact roerties. Gases are most ideal at high temeratures ad low ressure. A ideal gas obeys the erfect gas law. he secific heats are costat. 8

9 Perfect gas law, R u m R Pressure i Pa olume i m 3 Amout of gas i kg-mole m / M M Molecular weight R u Uiversal gas costat 834 J/kg-mole K m Amout of gas i kg R haracteristic gas costat i J/kg K emerature of gas i K I reality there is o ideal or erfect gas. At a very low ressure ad at a very high temerature, real gases like itroge, hydroge, oxyge, helium, etc., behave as erfect gases. hese gases are called Semi erfect or Permaet gases. For real gases secific heats vary areciably with temerature ad little with ressure. Iteral eergy ad ethaly Iteral Eergy Eergy ossessed by the system. Joule s law states that the secific iteral eergy of a gas deeds oly o the temerature of the gas ad is ideedet of both ressure ad volume. u f() Ethaly otal heat cotet of the flowig fluid From defiitio of ethaly, h u v But v R herefore, h u R ad h also fuctio of temerature oly. h f() Secific heats Secific heat at costat volume is defied as the amout of heat required to rise the temerature of kg of fluid through o whe the volume is ket costat. he secific heat caacity at costat volume is defied as v u du We kow that u f(). herefore v d Secific heat at costat ressure is defied as the amout of heat required to rise the temerature of kg of fluid through o whe the ressure is ket costat. he secific heat caacity at costat ressure is defied as We kow that h f(). herefore dh d v h 9

10 First Law of thermodyamics I-orollary: For a cycle: Wheever a system udergoes a cyclic chage, the algebraic sum of heat trasfer is roortioal to the algebraic sum of work trasfer. Net heat trasfer Net work trasfer dq dw For a rocess: dq dw du or Q W ΔU ΔU hage i iteral eergy m v Δ m Mass of the workig fluid v Secific heat at costat volume Δ hage i temerature II-orollary: (Law of oservatio of Eergy) I a isolated system, the eergy of the system remais costat. Q W 0 & Δ U 0 III-orollary: (PMM-I) PMM-I is imossible. (Peretual Motio Machie of First kid roduces work cotiuously without ay iut). PMM-I W Q 0 Alicatio of I-Law to No-Flow or losed system Let, m Mass of the workig fluid Iitial ressure of the workig fluid Fial ressure of the workig fluid Iitial temerature of the workig fluid Fial temerature of the workig fluid U Iitial iteral eergy of the fluid 0

11 U Fial iteral eergy of the fluid W Work trasfer Q Heat trasfer Secific heat at costat ressure v Secific heat at costat volume ostat volume rocess (Isometric rocess) I a costat volume rocess the workig fluid is cotaied i a closed vessel. he boudary of the system is immovable ad hece o work trasfer is ossible through boudary of the system. osider a vessel cotaiig m kg of certai gas. Q J of heat is sulied to the gas ad there will be ressure rise, but the volume remais costat. m kg of gas Q Durig costat volume rocess, d 0 Geeral gas equatio For costat volume rocess, Work trasfer W d 0 ; As er I-law for a rocess, Q W Δ U Q Δ U m v ( ) ostat ressure rocess (Isobaric rocess) Pisto m kg of gas ylider Q

12 osider a cylider-isto arragemet as show i fig. he isto is free to move u ad dow. Let the cylider cotai m kg of certai gas. he heat is added to the gas. Sice the isto is free to move, the ressure remais costat ad there is icrease of volume. I this case the boudary of the system is movable. Durig costat ressure rocess, Geeral gas equatio For costat ressure rocess, Work trasfer W d d ( ) As er I-law for a rocess, Q W Δ U Q ( ) m v ( ) m R ( ) m v ( ) m (R v ) ( ) m ( ) ostat temerature rocess (Isothermal rocess or Hyerbolic rocess) A rocess at a costat temerature is called a isothermal rocess. Whe a workig substace i a cylider behid a isto exads from a high ressure to a low ressure there is a tedecy for the temerature to fall. I a isothermal exasio heat must be added cotiuously i order to kee the temerature at the iitial value. Similarly i a isothermal comressio heat must be removed from workig fluid cotiuously durig the rocess. Durig costat temerature rocess, Geeral gas equatio For costat temerature rocess, W Pisto m kg of gas ylider Q d Work trasfer, W d d l

13 3 l l As er I-law for a rocess, Q W Δ U ) ( l m v l 0 l Durig isothermal rocess, Q W Note: he isothermal ad hyerbolic rocesses are idetical oly i the case of a erfect gas ad ot for a vaour. For examle the isothermal exasio of wet steam is ot hyerbolic. ostat etroy rocess (Isetroic rocess) I a isetroic rocess, the heat trasfer betwee the workig fluid ad surroudigs is zero. Durig isetroic rocess, Q 0 he goverig equatio for isetroic or reversible adiabatic rocess is, γ γ Secific heat ratio / v W Pisto ylider m kg of gas Q 0 Work trasfer γ γ d d d W γ γ γ γ γ γ γ γ γ

14 4 γ γ γ γ W Heat trasfer Q W ΔU 0 herefore we ca write, W - ΔU - m v ( ) Polytroic rocess I olytroic rocess, there is heat trasfer betwee the workig fluid ad surroudigs. Pressure, volume ad temerature are variables durig a rocess. he goverig equatio for isetroic or reversible adiabatic rocess is, Polytroic idex W Pisto ylider m kg of gas Q 0 Work trasfer d d d W mr W ) (

15 5 Heat trasfer Q W Δ U ) ( ( m mr Q v R mr v ) ( ) ( R R mr v v ) ) ( v v v v mr ) ) ( v v mr ) ) ( Dividig both umerator ad deomiator by v γ γ ) ) ( mr Q W Q γ γ ) Free exasio rocess (ostat iteral eergy rocess) osider two vessels A ad B itercoected by a short ie with a valve, ad erfectly thermally isulated. Iitially let the vessel A be filled with a fluid at a certai ressure ad let vessel B be comletely evacuated. Whe the valve is oeed, the fluid i vessel A will exad raidly to fill the vessel B. he ressure fially will be lower tha the iitial ressure i the vessel A. his is kow as Free or Urestricted exasio. he rocess is highly irreversible, sice the fluid is eddyig cotiuously durig the rocess. Perfect isulatio I free exasio rocess, W 0; Q 0;

16 As er I-Law, Q W Δ U Δ U 0 U U Steady Flow Eergy Equatio (SFEE) (I-Law for oe system) osider a system as show i fig. Let m kg/s of workig fluid is eterig ad leavig the system. Let, m Mass flow rate of fluid through the system --- kg/s U Iteral eergy of the fluid at ilet --- W elocity of the fluid at ilet --- m/s Pressure of the fluid at ilet -- Pa olume flow rate of fluid at ilet --- m 3 /s Z Height of the ilet sectio from datum --- m Q Heat trasfer through the system --- W W Work trasfer through the system --- W U,,, ad Z are corresodig values at outlet () Z () Datum Lie I additio to iteral eergy, other forms of eergy associated with the mass eterig ad leavig the system is cosidered here. Kietic eergy of fluid m / Potetial eergy of the fluid m g Z Flow eergy of fluid Betwee ilet ad outlet, the eergy equatio ca be writte as, U m mgz Q U m mgz W mu m v m mgz Q mu m v m mgz W Z 6

17 mh m mgz Q mh m mgz W Where, v Secific volume /m u Secific iteral eergy U/m h Secific ethaly u v Alicatio of I-Law to oe system Water turbie I a water turbie water is sulied from a height. he otetial eergy of water is coverted ito kietic eergy whe it eters ito the turbie ad art of it is coverted ito useful work which is used to geerate electricity. SFEE is give by, mu m v m mgz Q mu m v m mgz W Geerally i water turbie, Heat trasfer (Q) 0; Z 0; v v ; or U U Eergy equatio becomes, m v m mgz m v m W () Z Geerator () Z 0 urbie Note: W is ositive sice work is doe by the system. Steam or Gas turbie I a steam or gas turbie, steam or gas is assed through the turbie ad art of its eergy is coverted ito work i the turbie. he outut of the turbie rus a geerator to roduce electricity. Geerator () Steam or gas i () Exhaust steam or gas 7

18 Geerally i steam or gas turbie, Z Z 0; Q 0 (isulated turbie); Q - ve if ot erfectly isulated. Eergy equatio becomes, mh m Q mh m W Note: W is ositive sice work is doe by the system. Q is egative sice the heat is trasferred from the hot casig to low temerature surroudigs. Water um ak () Geerator Pum () Sum A water um draws water from a lower level ad ums it to higher level. Work is required to ru the um ad this may be sulied from a exteral source such as a electric motor or a diesel egie. Geerally i water um, Heat trasfer (Q) 0; v v ; or U U Eergy equatio becomes, m v m mgz m v m mgz W Note: W is egative sice work is doe o the system. etrifugal air comressor A cetrifugal comressor comresses air ad sulies the same at moderate ressure ad i large quatity. Geerator () omressed air out Geerally i cetrifugal comressor, () Air from atmoshere Z Z 0; Q - ve if ot erfectly isulated. 8

19 Eergy equatio becomes, mh m Q mh m W Note: W is egative sice work is doe o the system. Q also is egative sice the heat is trasferred from the hot casig to low temerature surroudigs. Recirocatig air comressor Air i Reservoir ylider Pisto he recirocatig air comressor draws i air from atmoshere ad suly it at relatively higher ressure ad i small quatity. he velocity of air eterig ad leavig the comressor is geerally very small ad is eglected. Geerally i recirocatig comressor, Z Z 0; ; Q - ve if ot erfectly isulated. Eergy equatio becomes, m Q m W Note: W is egative sice work is doe o the system. Q also is egative sice the heat is trasferred from the hot casig to low temerature surroudigs. Heat Exchagers A heat exchager is a device to trasfer heat from oe fluid to aother fluid through walls. here is ethaly chage. Examles: Boiler, odeser, Evaorator. Geerally i heat exchagers, Z Z 0; ; Q ve i boilers ad evaorators Q - ve i evaorators Fluid i () () Fluid out 9

20 Boilers: A boiler roduces high temerature vaour absorbig heat from the exteral source. he heat is trasferred to liquid. Eergy equatio becomes, mh Q mh odesers: A codeser is a device to codese the vaour by rejectig its heat to the coolig medium. Here the heat is rejected by the system. Eergy equatio becomes, mh Q mh Evaorators: A evaorator is device which roduces low temerature vaour by absorbig heat from relatively hot source. Here the heat is absorbed by the system. Eergy equatio becomes, mh Q mh Steam ozzle Steam ozzles are used i steam egies ad steam ower lats. It coverts the ressure eergy ito kietic eergy. Geerally the ozzles are isulated. I steam ozzles the steam is exaded isetroically. Steam i Steam out Geerally i ozzles, Z Z 0; Q 0; W 0; Eergy equatio becomes, mh m mh m he exit velocity of the steam ca be writte as, ( h h ) hrottlig rocess (ostat ethaly rocess) Whe a fluid flows through a costricted assage, like a artially oeed valve, a orifice, or a orous lug, there is a areciable dro i ressure ad the flow is said to be throttled. he fig shows the rocess of throttlig by a orifice. he ie is erfectly isulated. () () Heat trasfer (Q) 0 Work trasfer (W) 0 Orifice ; Z Z Eergy equatio becomes, h h 0

21 he throttlig rocess is commoly used for the followig rocess: (i) (ii) (iii) For determiig the coditio of steam (dryess fractio) For cotrollig the seed of the turbie Used i refrigeratio lat for reducig the ressure of the refrigerat before etry ito the evaorator Path ad Poit fuctio osider a rocess - as show i fig (a). he rocess - may follow a ath A, B or. he area uder a curve i - diagram reresets the work trasfer ad the area uder a curve i a -s diagram reresets the heat trasfer as show i fig (b). Work or heat deeds o the ath of the system, but ot o the ed states. For this reaso, work or heat is called a ath fuctio. W - or Q - is a iexact or imerfect differetial. dw W W hermodyamic roerties are oit fuctios, sice for a give state, there is a defiite value for each roerty. he chage i a thermodyamic roerty of a system i a chage of state is ideedet of the ath of the system ad deeds oly o the ed states of the system. he differetials of oit fuctios are exact or erfect differetials. d Note: I a cyclic rocess, the chage i ay roerty is zero. d 0, d 0, d 0 (a) (b)

22 Iteral eergy A roerty of the system osider a system which chages its state from state to state by followig the ath A ad returs to its origial state followig (i) the ath B ad (ii) the ath. osider a cycle -A--B-. As er I-law for a cycle, dq dw For a rocess -A-, Q A Q B W A W B Q A W A W B Q B () Q A W A ΔU A Q A W A ΔU A () For a rocess -B-, Q B W B ΔU B W B Q B - ΔU B (3) herefore, from (), () ad (3), we ca write, ΔU A - ΔU B (4) osider a cycle -A---. For a rocess --, Q A Q W A W Q A W A W Q (5) Q W ΔU W Q - ΔU (6) herefore, from (), (5) ad (6), we ca write, ΔU A - ΔU (7) From (4) ad (7), ΔU ΔU B From the above, it ca be cocluded that the iteral eergy deeds o ed states ad ideedet of ath. herefore the iteral eergy is a roerty. Molar volume ( m ) Molar volume of a substace is the ratio of the volume of a samle of that substace to the umber of moles i the samle.

23 he molar volume for a ideal gas at 98.5 K ad bar is m³mol - or l/mol. m / Number of moles () m 3 /kmol olume i m 3 M Molar mass i kg/kmol ρ Desity i kg/m 3 m R R u m R u R haracteristic gas costat i J/kg-K R u Uiversal gas costat i J/kmol-K haracteristic ad Uiversal Gas costat haracteristic gas costat (R) v J/kg-K Uiversal gas costat (R u ) (mole) v(mole) J/kmol-K (mole) Molar secific heat at costat ressure M J/kmol-K v(mole) Molar secific heat at costat volume M v J/kmol-K Molar mass ad Molecular mass (M) Molar mass is the mass of oe mole of the substace ad is defied i kg/kmol. Molecular mass (M) is the mass of oe molecule of the substace ad is defied i amu (automic mass uit). Molar mass ad molecular mass differs oly i uits. amu (83) 0 7 kg Note: It is advisable to use aroriate uits while umerical roblems are beig solved. Pressure () N/m (or) Pa emerature () K olume () m 3 olume flow rate ( ) m 3 /s Mass (m) kg Mass flow rate ( m ) elocity () Agular velocity (ω) Seed (N) Eergy Force (F) Power (P) Secific heat ( or v ) haracteristic Gas costat (R) Uiversal gas costat (R u ) kg/s m/s rad/s rm Nm (or) J N J/s (or) W J/kg-K (or) J/kg o J/kg-K 834 J/kmol-K 3

24 Molar volume or Molecular volume ( m ) m 3 /kmol Molecular mass (M) amu Ethaly (H) J Secific ethaly (h) J/kg Etroy (S) J/K Secific etroy (s) J/kg-K Molar secific heat ( (mole) or v(mole) ) J/kmol-K Gravitatioal acceleratio (g) 9.8 m/s Limitatios of I-law of thermodyamics he I-law states that, i carryig out a rocess, heat & work are mutually covertible, a balace of eergy must hold as iteral eergy, the eergy is either gaied or lost i a rocess, it oly trasforms. But, this law does ot lace ay distictio o the directio of the rocess, uder cosideratio. Accordig to the I-law, it is assumed that ay chage of thermodyamic state ca take lace i either directio. But it has bee foud that this is ot the case articularly i the iter-coversio of heat ad work. he rocesses aturally roceed i certai directios ad ot i the oosite directios, eve though the reversal of the rocesses does ot violate the I-law. Examle: If two metal blocks at temeratures & ( > ) are brought ito cotact with each other, the heat flows from the high temerature block to the low temerature block till the temerature of both the blocks are equal. he heat flows from low temerature block to the high temerature block is imossible, i.e., the origial temeratures & caot be restored. (i) (ii) herefore the I-law is a ecessary but ot sufficiet due to the followig restrictios: No restrictio o the directio of eergy flow It does ot deal with the ortio of heat that may be coverted ito useful work 4

25 PROBLEMS. Followig amout of heat trasfer occurs durig a cycle comrisig of four rocesses. alculate the workdoe of the cycle ad idicate about the tye of work. 0 kj, -0 kj, 6 kj ad 4 kj. Give: ycle with four rocesses Heat trasfer durig rocess Q - 0 kj Heat trasfer durig rocess 3 Q -3-0 kj Heat trasfer durig rocess 3 4 Q kj Heat trasfer durig rocess 4 Q 4-4 kj Required: Workdoe ad tye of work Solutio: First law for cycle Net heat trasfer Net work trasfer dq dw dw 0 ( 0) kj he workdoe is ositive, therefore the work is doe by the system.. he followig data refer to a closed system which udergoes a thermodyamics cycle cosistig of four rocesses. Show that the data is cosistet with the I-law of thermodyamics ad calculate, (a) Net rate of work outut i kw ad (b) hage i iteral eergy. Process Heat rasfer (kj/mi) Work rasfer (kj/mi) Nil ,000 Nil 3 4-4,000 6, , Give: Q - 0, Q kj/mi, Q kj/mi, Q kj/mi W kj/mi, W -3 0, W kj/mi, W kj/mi Required: (a) Net work i kw (b) Δ U Solutio: (a) First law for cycle Net heat trasfer Net work trasfer dq dw dw kj / mi 4000 / kw 5

26 dq kj / mi 4000 / kw dq dw herefore, the data is cosistet with the I-law of thermodyamics. (b) hage i iteral eergy durig rocess U U I-law for a rocess Q W Δ U U U Q - W - 0 (-000) 000 kj/mi U 3 U Q -3 W kj/mi U 4 U 3 Q 3-4 W kj/mi U U 4 Q 4- W (-000) -000 kj/mi 3. alculate the workdoe whe the volume chages from 4 m 3 to 8 m 3 through a o-flow quasi-static rocess i which the ressure is give by, (4 5) bar. Give: A rocess Iitial volume ( ) 4 m 3 Fial volume ( ) 8 m 3 (4 5) bar Required: Workdoe Solutio Work trasfer i o-flow rocess is 5 W d ( 4 5)d x he iteral eergy of a certai substace is give by the equatio u 3.56 v 84, where u is give i kj/kg, is i kpa ad v is i m 3 /kg. A system comosed of 3 kg of this substace exads from a iitial ressure of 500 kpa ad a volume of 0. m 3 to a fial ressure of 00 kpa i a rocess i which ressure ad volume are related by.. (a) If the exasio is quasi-static, fid heat trasfer, workdoe ad chage i iteral eergy for the rocess (b) I aother rocess the same system exads accordig to the same ressure-volume relatioshi as i art (a), ad from the same iitial state to the same fial 6 5 [ 4 ( ) / 5 ( )] x 0 5 [4 x (8 4 )/ 5 x (8 4)] x x 0 5 J As

27 state as i art (a), but the heat trasfer i this case is 30 kj. Fid the work trasfer for this rocess (c) Exlai the differece i work trasfer i art (a) ad art (b). Give: Secific iteral eergy u 3.56 v 84 kj/kg Mass of the workig fluid (m) 3 kg Iitial ressure ( ) 500 kpa Iitial volume ( ) 0. m 3 Fial ressure ( ) 00 kpa Idex of exasio (). Required: (a) Q, W, Δ U (b) W (c) Differece W i (a) ad (b) Solutio: (a) Heat trasfer durig a rocess, Q W Δ U u 3.56 v 84 u 3.56 v 84 u 3.56 v 84 o fid Δu 3.56 ( v v ) ΔU m Δu ad m v ΔU 3.56 ( ) / / m 3 ΔU 3.56 (00 x x 0.) kj --- As Process follows the law.. W ( ) / ( ) (00 x 0 3 x x 0 3 x 0.) (.) 9500 J 9.5 kj --- As Q - W - Δ U kj --- As (b) Heat trasfer 30 kj W - Q - Δ U 30 (- 9.04).04 kj --- As (c) he work i art (b) is ot equal to d. he rocess i art (b) is ot quasi-static. 7

28 5. A fluid is cofied i a cylider by a srig loaded, frictioless isto so that the ressure i the fluid is a liear fuctio of the volume ( a b). he iteral eergy of the fluid is give by the equatio, U , where, U is i kj, is i kpa ad is i m 3. If the fluid chages from a iitial state of 70 kpa, 0.03 m 3 to a fial state of 400 kpa, 0.06 m 3, with o work other tha that doe o the isto, fid the directio ad magitude of the work ad heat trasfer. Give: Pressure a b Iteral eergy (U) kj Iitial ressure ( ) 70 kpa Iitial volume ( ) 0.03 m 3 Fial ressure ( ) 400 kpa Fial volume ( ) 0.06 m 3 Required: W ad Q Solutio: U U Δ U 3.5 ( ) 3.5 x (400 x 0 3 x x 0 3 x 0.03) J a b 70 a 0.03 b --- () a b 400 a 0.06 b --- () From () ad () b, b kpa Substitutig i (), a - 60 kpa Work trasfer is W d ( a b ) d [a b /] a ( ) b ( )/ - 60 x ( ) x ( ) / 8.55 kj 8550 J --- As Work is ositive, i.e., Work is doe by the system. Q - W - Δ U J --- As Heat trasfer is ositive, i.e., Heat is sulied to the system. 8

29 6. A steam turbie oerates uder steady flow coditios. It receives 700 kg/h of steam from the boiler. he steam eters the turbie at ethaly of 800 kj/kg, a velocity of 4000 m/mi ad a elevatio of 4 m. he steam leaves the turbie at ethaly of 000 kj/kg, a velocity of 8000 m/mi ad a elevatio of m. Due to radiatio, heat losses from the turbie to the surroudigs amout to 580 kj/h. alculate the outut of the turbie. Give: Mass flow rate of steam (m) 700 kg/h kg/s Iitial secific ethaly (h ) 800 kj/kg Iitial velocity ( ) 4000 m/mi m/s Elevatio of ilet (Z ) m Fial secific ethaly (h ) 000 kj/kg Fial velocity ( ) 8000 m/mi 33.3 m/s Elevatio of outlet (Z ) 4 m Heat losses from turbie (Q) -580 kj/h kj/s Required: Work outut Solutio: SFEE is give by, m (h h ) m ( )/ m g (Z Z ) Q W W ( ) x 0 3 x ( )/ x 9.8 x ( 4) x W --- As 7. Steam eters a ozzle at a ressure of 7 bar ad 0 o (iitial ethaly 850 kj/kg) ad leaves at a ressure of.5 bar. he iitial velocity of steam at the etrace is 40 m/s ad the exit velocity of steam from ozzle is 700 m/s. he mass flow rate through the ozzle is 400 kg/h. he heat loss from the ozzle is 705 kj/h. Determie the fial ethaly of steam ad ozzle area if the secific volume at outlet is.4 m 3 /kg. Give: Iitial ressure ( ) 7 bar Iitial velocity ( ) 40 m/s Iitial ethaly (h ) 850 kj/kg Fial velocity ( ) 700 m/s Secific volume at outlet (v ).4 m 3 /kg Mass flow rate (m) 400 kg/h kg/s Heat loss from ozzle (Q) -05 kj/h J/s Required: h ad A Solutio: SFEE is give by, m (h h ) m ( ) / m g (Z Z ) Q W Assume Z Z 0 if ot give. For ozzle W 0 m (h h ) m ( ) Q x (850 - h ) x x ( ) /

30 h 64.6 kj/kg ---- As Mass flow rate (m) A / v A / v A x 700 /.4 A m --- As 8. Air flows at the rate of 0.5 kg/s through a air comressor, eterig at 7 m/s, 00 kpa ad 0.95 m 3 /kg ad leavig at 5 m/s, 700 kpa, ad 0.9 m 3 /kg. he iteral eergy of air leavig is 90 kj/kg greater tha that of the air eterig. oolig water i the comressor jackets absorbs heat from the air at the rate of 58 kw. (a) omute the rate of shaft work iut to the air i kw (b) Fid the ratio of the ilet ie diameter to outer ie diameter. Give: Mass flow rate (m) 0.5 kg/s Iitial velocity ( ) 7 m/s Iitial secific volume (v ) 0.95 m 3 /kg Iitial ressure ( ) 00 kpa bar Fial velocity ( ) 5 m/s Fial ressure ( ) 700 kpa 7 bar Fial secific volume (v ) 0.9 m 3 /kg u u 90 kj/kg Heat liberated by the air (Q) 58 kj/s ( - ve) Required: (a) W (b) d / d Solutio: (a) SFEE is give by, m (u u ) m ( v v ) m ( ) / m g (Z Z ) Q W Assume Z Z 0 if ot give. 0.5 x (-90) x x ( x 0.95 x x 0.9 x 0 5 ) 0.5 (7 5 ) / W W W Work iut to the comressor 994 W --- As (b) Mass flow rate is give by, m A / v A / v A / A v / ( v ) 5 x 0.95 /(7 x 0.9) 3.57 A / A d / d d / d As 9. I a steam ower statio, steam flows steadily through a 0. diameter ielie from the boiler to the turbie. At the boiler ed, the steam coditios are foud to be 4 MPa, 400 o, secific ethaly 34.6 kj/kg ad secific volume m 3 /kg. At the turbie ed, the coditios are foud to be 3.5 MPa, 39 o, secific ethaly 30.6 kj/kg ad secific volume m 3 /kg. here is a heat loss of 8.5 kj/kg from the ielie. alculate the steam flow rate. 30

31 Give: Diameter of ielie (d) Pressure of the steam at ilet ( ) Iitial temerature ( ) Iitial secific ethaly (h ) Iitial secific volume (v ) Pressure of the steam at outlet ( ) Fial temerature ( ) Fial secific ethaly (h ) Iitial secific volume (v ) Heat losses from turbie (q) Required: Steam flow rate Solutio: his is the roblem of heat exchager. Heat exchager 0. m 4 MPa 40 bar 400 o 673 K 33.6 kj/kg m 3 /kg 3.5 MPa 35 bar 39 o 563 K 30.6 kj/kg m 3 /kg -8.5 kj/kg urbie ed Boiler urbie Boiler ed SFEE is give by, m (h h ) m ( ) / m g (Z Z ) Q W Assume Z Z 0 if ot give. For heat exchager, W 0 ; Mass flow rate (m) A / v A / v A A v / v / () SFEE for heat exchager becomes, m (h h ) m ( ) / m q 0 (h h ) ( ) / q 0 ( ) x 0 3 ((0.869 ) )/ -8.5 x m/s m A / v (π / 4) d / v π x 0. x 4.9 / (4 x 0.084) kg/s --- As 0. A isto ad cylider machie cotais a fluid system which asses through a comlete cycle of four oeratios. Process Q (kj/mi) W (kj/mi) ΔU (kj/mi) a b

32 b c c d d a Durig cycle the sum of all heat trasfers is -70 kj. he system oerates 00 cycles er mi. omlete the followig table showig the method for each item ad comute the et rate if outut i kw. Give: Net heat trasfer -70 kj No of cycles 00 / mi Required: o comlete the table ad to determie the et work, Solutio: ΔU hage i iteral eergy I-law for a cycle, Net work trasfer Net heat trasfer Net work outut - 70 kj - 70 x o of cycles er secod -70 x 00 / kw --- As I-law for a rocess is Q W ΔU Net Heat trasfer, Net work trasfer, For a cycle, sum of ΔU 0 (ΔU) a-b Q a-b W a-b kj/mi --- As (ΔU) b-c Q b-c W b-c kj/mi --- As W c-d Q c-d (ΔU) c-d kj/mi --- As Q d-a Q a-b Q b-c Q c-d Q d-a kj/mi --- As W d-a W a-b W b-c W c-d W d-a kj/mi --- As (ΔU) a-b (ΔU) b-c (ΔU) c-d (ΔU) d-a (ΔU) d-a 0 (ΔU) d-a 7770 kj/mi ---- As Process Q (kj/mi) W (kj/mi) ΔU (kj/mi) a b

33 b c c d d a kg of air er mi is delivered by a cetrifugal comressor. he ilet ad outlet coditios of air are m/s, bar, v 0.5 m 3 /kg ad 90 m/s, 8 bar, v 0.4 m 3 /kg. he icrease i ethaly of air assig through comressor is 50 kj/kg ad heat loss to the surroudigs is 700 kj/mi. Fid motor ower required to drive the comressor ad ratio of ilet ad outlet ie diameters. Assume that ilet ad discharge lies are at the same level. Give: Mass flow rate of air (m) kg/mi 0. kg/s Icrease i ethaly (h h ) 50 kj/kg Heat loss to the surroudigs (Q) 700 kj/mi J/s (-ve) Z Z 0 Required: Power ad Diameter ratio Solutio: SFEE is give by, m (h h ) m ( ) / m g (Z Z ) Q W 0. x (-50 x 0 3 ) 0. x ( 90 ) / W W W Work i W is Power. P W Power required to drive the comressor 46.3 W --- As Mass flow rate (m) A / v A / v A / A v / (v ) 0.5 x 90 / (0.4 x ) d / d (A / A ) As. At the ilet of a certai ozzle, the ethaly of the fluid assig is 3000 kj/kg ad the velocity is 60 m/s. At the discharge ed, the ethaly is 76 kj/kg. he ozzle is horizotal ad there is egligible heat loss from it. (a) Fid the velocity of fluid at exit (b) If the ilet area is 0. m ad the secific volume at ilet is 0.87 m 3 /kg, fid the mass flow rate (c) If the secific volume at the ozzle exit is m 3 /kg, fid the exit area of the ozzle. Give: Nozzle Iitial ethaly (h ) 3000 kj/kg Iitial velocity ( ) 60 m/s Heat trasfer (Q) 0 Required: (a) (b) m (c) A Solutio: (a) I ozzle flow, W 0; Z Z 0 (if ot give); SFEE is give by, m (h h ) m ( ) / m g (Z Z ) Q W 33

34 For give ozzle, m (h h ) m ( ) / 0 (h h ) ( ) / 0 (h h ) 60 x ( ) x m/s --- As (b) A 0. m ; v 0.87 m 3 /kg Mass flow rate (m) A / v A / v 0. x 60 / kg/s --- As (c) v m 3 /kg ; m A / v 3.08 A x 69.5 / A m --- As kj of heat is sulied to a system at costat volume. he system rejects 90 kj of heat at costat ressure ad 0 kj of work is doe o it. he system is brought to its origial state by adiabatic rocess. Determie the adiabatic work. Determie also the values of iteral eergy at all ed states if iitial value is 00 kj. Give: here are three rocesses. ostat volume rocess HS 85 kj 3 ostat ressure rocess HR 90 kj & W -0 kj 3 Adiabatic rocess Iitial iteral eergy (U ) 00 kj Required: W 3-, U & U 3 Solutio: 3 I-law for a rocess, Q W Δ U ostat volume rocess, W - 0 Q - W - U U 85 0 U 00 U 85 kj --- As 3 ostat ressure rocess Q -3 W -3 U 3 U U 3 85 U 3 5 kj --- As 3 Adiabatic rocess, Q 3-0 Q 3- W 3- U U 3 34

35 0 W W 3-5 kj --- As 4. m 3 of hydroge at a ressure of bar ad 0 o is comressed isetroically to 4 bar. he same gas is exaded to origial volume by costat temerature rocess ad reached iitial ressure ad temerature by costat volume heat rejectio rocess. Determie (a) ressure, volume ad temerature at each ed of oeratio (b) the heat trasferred durig the isothermal rocess (c) the heat rejected durig costat volume rocess ad (d) chage i iteral eergy durig each rocess. Assume R 4.06 kj/kg-k ad 4.5 kj/kg-k. Give: here are three rocesses. Isetroic rocess 3 ostat temerature rocess 3 ostat volume rocess Iitial volume of hydroge ( ) m 3 3 Iitial ressure ( ) bar Iitial temerature ( ) 0 o 93 K Pressure after isetroic comressio ( ) 4 bar olume after isothermal exasio ( 3 ) m 3 Required: (a),, 3 Solutio: 3 (a) For adiabatic rocess, R v v kj/kg-k γ / v 4.5 / ( γ ) / γ 93 4 (.49 ) / K --- As 3 / γ 35

36 / ostat temerature rocess m As x x bar --- As (b) Heat trasferred durig isothermal rocess (Q -3 ) Q -3 l[ 3 / ] 4 x 0 5 x x l [/0.753] J --- As (c) Heat rejected durig costat volume rocess (Q 3- ) Q 3- m v ( 3 ) o fid m m R x 0 5 x m x 406 x 93 m 0.63 kg Q x x (93 44.) J --- As (d) U U -W - ( ) / ( γ) - (4 x 0 5 x x 0 5 x ) / (.49) J --- As U 3 U 0 U U 3 Q J ---- As m 3 of air at 30 o ad bar is comressed olytroically to 0.08 m 3. Fid the fial ressure ad temerature ad workdoe, chage i iteral eergy ad ethaly, whe the idex of comressio has the value of.5. ake for air.005 kj/kg-k ad v 0.78 kj/kg-k. Give: Polytroic rocess Iitial volume ( ) 0.5 m 3 Iitial temerature ( ) 30 o 303 K Fial volume ( ) 0.08 m 3 Idex of comressio ().5 Required:,, W, Δ U, Δ H Solutio: ( ) 36

37 303 (. 5) K --- As o fid m bar --- As W - ( ) / ( ) (5.65 x 0 5 x 0.08 x 0 5 x 0.5) / (.5) J ---- As U U m v ( ) m R R v kj/kg-k x 0 5 m x 87 x 303 m kg Δ U x 0.78 x ( ) 87.6 KJ --- As Δ H H H m ( ) x.005 x ( ) 6.6 kj --- As 6. 3 kg of a ideal gas is exaded from a ressure of 8 bar ad volume of.5 m 3 to a ressure of.6 bar ad volume of 4.5 m 3. he chage i iteral eergy is 450 kj. he secific heat at costat volume for the gas is 0.7 kj/kg-k. Determie (a) Gas costat (b) Idex of olytroic exasio (c) Workdoe durig olytroic exasio ad (e) Iitial ad fial temeratures. Give: Mass of gas (m) 3 kg Iitial ressure ( ) 8 bar Iitial olume ( ).5 m 3 Fial ressure ( ).6 bar Fial volume ( ) 4.5 m 3 hage i iteral eergy (U U ) 450 kj Secific heat at costat volume ( v ) 0.7 kj/kg-k Required: (a) R (b) (c) W (d) & Solutio: (a) m R 8 x 0 5 x.5 3 x R 37

38 R () m R.6 x 0 5 x x R R () From () ad () R ( ) / R --- (3) Also U U kj (-ve o exasio) U U m v ( ) x 0.7 x ( ) Substitutig (3), x 0.7 x / R R J/kg-K ---- As (b) l [ / ] / l [ / ] l [8/.6] / l [4.5/.5] As (c) W ( ) / ( ) (.6 x 4.5 x x.5 x 0 5 ) / (.465) 0358 J ---- As (d) m R 8 x 0 5 x.5 3 x x K ---- As m R.6 x 0 5 x x x 3.5 K ---- As 7. A gas mixture obeyig erfect gas law has molar mass of 6.7 kg/kmol. he gas mixture is comressed to a comressio ratio of accordig to the law.5, from iitial coditios of 0.9 bar ad 333 K. Assume a mea molar secific heat at a costat volume of. kj/kmolk, fid er kg of mass, the workdoe ad heat flow across the cylider walls. For the above gas, determie the value of characteristic gas costat, molar secific heat at costat ressure ad ratio of secific heats. Give: Polytroic rocess Idex of comressio ().5 Molar mass (M) 6.7 omressio ratio ( / ) Iitial ressure ( ) 0.9 bar Iitial temerature ( ) 333 K v(mole). kj/kmolk Mass of gas (m) kg Required: W, Q, R, (mole), / v 38

39 Solutio: Work doe durig olytroic rocess is give by, m R W ( ) o fid R Uiversal gas costat (R u ) 834 J/kg-K haracteristic gas costat (R) R u /M 834/ J/kg-K --- As o fid ( ) / Heat trasfer (Q) Gas costat R 333 (. 5)/. 5 [ ] K x x ( ) W J --- As γ Q x W.4.5 x J --- As R v v v(mole) / M (mole) / M v 00 / J/kg-K J/kg-K (mole) 0.64 x J/kmolK ---- As Secific heat ratio / v 0.64/ As 8. A cetrifugal um delivers 750 kg of water er mi from iitial ressure of 0.8 bar absolute to a fial ressure of.8 bar absolute. he suctio is m below ad the delivery is 5 m above the cetre of um. If the suctio ad delivery ies are of 5 cm ad 0 cm diameters resectively, make calculatio for ower required to ru the um. Give: etrifugal um Flow rocess 39

40 Mass flow rate of water (m) 750 kg/mi 750/60 kg/s Iitial ressure ( ) 0.8 bar (abs) Fial ressure ( ).8 bar (abs) Suctio below the cetre of the um m Delivery above the cetre of the um 5 m Suctio ie diameter (d ) 5 cm 0.5 m Delivery ie diameter (d ) 0 cm 0. m Required: W Solutio: SFEE is give by, * * m h g Z Q mh g Z W * * m u v g Z Q mu v g Z W osider datum from suctio (), Z 0 & Z 5 7 m Geerally for ay liquid, u u, v v v ad Q 0 herefore, SFEE is reduced to, * * m v g Z m v g Z W Water Out () Z Pum () Water i o fid ad 40

41 * A A m v v For water ρ 000 kg/m 3 ad v /ρ /000 m 3 /kg π π A d x π π A d x π / 4 x / 60 / m/s π / 4 x / 60 / m/s herefore, (750 / 60) 0.8 x0 x (/000) 9.8x (750 / 60).8 x0 x (/000) (9.8x 7) W W 833 W --- As 9. I a isetroic flow through ozzle, air flows at the rate of 600 kg/h. At the ilet to the ozzle, ressure is MPa ad temerature is 7 o. he exit ressure is 0.5 MPa. Iitial velocity is 300 m/s. Determie, (i) Exit velocity of air ad (ii) Ilet ad exit area of ozzle. Give: Nozzle Flow rocess Fluid Air Mass flow rate of air (m) 600 kg/h 600/3600 kg/s Iitial ressure ( ) MPa x 0 6 Pa Iitial temerature ( ) K Fial ressure ( ) 0.5 MPa 0.5 x 0 6 Pa Iitial velocity ( ) 300 m/s Required: (i) (ii) A & A Solutio: (i) SFEE is give by, * * m h g Z Q mh g Z W 4

42 ake Z Z he flow is isetroic, Q 0 For ozzle, W 0 herefore, SFEE is reduced to, o fid h h h ( h ) ( ) ( γ ) / γ For air, γ.4, R 87 J/kg-K & 005 J/kg-K herefore, (ii) K (.4 ) / x005( ) m/s ---- As A * m v A v o fi v & v v R x 0 6 x v 87 x 400 v m 3 /kg v R 0.5 x 0 6 x v 87 x m 3 /kg A / 3600 x A x 0-5 m --- As 600 / 3600 A x A x 0-5 m --- As 4