Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS

Transcription

1 4- Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS Mog Bondary ork 4-C It represents the bondary work for qasi-eqilibrim processes. 4-C Yes. 4-C The area nder the process cre, and ths the bondary work done, is greater the constant pressre case. 4-4C ka m k(n/m m kn m kj 4-5 A piston-cylder deice contas nitrogen gas at a specified state. The bondary work is to be determed for the polytropic expansion of nitrogen. roperties The gas constant for nitrogen is kj/kg.k (Table A-. Analysis The mass and olme of nitrogen at the itial state are V (0 ka(0.07 m m RT (0.968 kj/kg.k(0 + 7 K V kg mrt ( kg(0.968 ka.m /kg.k( K m 00 ka The polytropic dex is determed from n n n N 0 ka 0 C V V (0 ka(0.07 m (00 ka( m n.49 The bondary work is determed from b V V (00 ka( m (0 ka(0.07 m n.49 n.86 kj

2 4-4-6 A piston-cylder deice with a set of stops contas steam at a specified state. Now, the steam is cooled. The compression work for two cases and the fal temperatre are to be determed. Analysis (a The specific olmes for the itial and fal states are (Table A-6 Ma m /kg T 400 C T Ma 0.75 m /kg 50 C Notg that pressre is constant drg the process, the bondary work is determed from b m( (0. kg(000 ka( m /kg.6 kj (b The olme of the cylder at the fal state is 60% of itial olme. Then, the bondary work becomes b Steam 0. kg Ma 400 C m( 0.60 (0. kg(000 ka( m /kg 6.79 kj The temperatre at the fal state is 0.5 Ma T ( m /kg 5.8 C (Table A A piston-cylder deice contas nitrogen gas at a specified state. The fal temperatre and the bondary work are to be determed for the isentropic expansion of nitrogen. roperties The properties of nitrogen are R kj/kg.k, k.4 (Table A-a Analysis The mass and the fal olme of nitrogen are V (0 ka(0.07 m m kg RT (0.968 kj/kg.k(0 + 7 K N 0 ka 0 C k k.4.4 V V (0 ka(0.07 m (00 ka V V m The fal temperatre and the bondary work are determed as T V mr (00 ka( m 64.6 K ( kg(0.968 ka.m /kg.k b V V k (00 ka( m (0 ka(0.07 m.4.64 kj

3 4-4-8 Satrated water apor a cylder is heated at constant pressre ntil its temperatre rises to a specified ale. The bondary work done drg this process is to be determed. Assmptions The process is qasi-eqilibrim. roperties Notg that the pressre remas constant drg this process, the specific olmes at the itial and the fal states are (Table A-4 throgh A-6 00 ka 00 ka m /kg Sat. apor 00 ka m /kg T 00 C Analysis The bondary work is determed from its defition to be b,ot dv ( V V m( kj (5 kg(00 ka( m /kg ka m 65.9 kj (ka Discssion The positie sign dicates that work is done by the system (work otpt. 00 V 4-9 Refrigerant-4a a cylder is heated at constant pressre ntil its temperatre rises to a specified ale. The bondary work done drg this process is to be determed. Assmptions The process is qasi-eqilibrim. roperties Notg that the pressre remas constant drg this process, the specific olmes at the itial and the fal states are (Table A- throgh A- 900 ka 900 ka m /kg Sat. liqid 900 ka m /kg T 70 C Analysis The bondary work is determed from its defition to be and V 0. m m. kg m /kg b,ot dv ( V V m( (ka 900 kj (. kg(900 ka( m /kg ka m 557 kj Discssion The positie sign dicates that work is done by the system (work otpt.

4 EES roblem 4-9 is reconsidered. The effect of pressre on the work done as the pressre aries from 400 ka to 00 ka is to be estigated. The work done is to be plotted erss the pressre. Analysis The problem is soled sg EES, and the soltion is gien below. "Knowns" Vol_L00 [L] x_0 "satrated liqid state" 900 [ka] T_70 [C] "Soltion" Vol_Vol_L*conert(L,m^ "The work is the bondary work done by the R-4a drg the constant pressre process." _bondary*(vol_-vol_ "The mass is:" Vol_m* olme(r4a,,xx_ Vol_m* olme(r4a,,tt_ "lot formation:" []_ []_ [] [] T[]temperatre(R4a,,xx_ T[]T_ C ] [ T R4a 900 ka [m /kg] [ka] bondary [kj] a] [k R4a [m /kg]

5 4-5 J ] [k d ary n o b ka T[] [C] J ] [k d ary n o b T 00 C [ka] J ] [k d ary n o b T 70 C [ka]

6 4-6 4-E Sperheated water apor a cylder is cooled at constant pressre ntil 70% of it condenses. The bondary work done drg this process is to be determed. Assmptions The process is qasi-eqilibrim. roperties Notg that the pressre remas constant drg this process, the specific olmes at the itial and the fal states are (Table A-4E throgh A-6E 40 psia ft /lbm T 600 F 40 psia f + x fg x ( ft /lbm Analysis The bondary work is determed from its defition to be b,ot dv ( V V m( (psia Bt (6 lbm(40 psia( ft /lbm psia ft 48 Bt Discssion The negatie sign dicates that work is done on the system (work pt Air a cylder is compressed at constant temperatre ntil its pressre rises to a specified ale. The bondary work done drg this process is to be determed. Assmptions The process is qasi-eqilibrim. Air is an ideal gas. roperties The gas constant of air is R 0.87 kj/kg.k (Table A-. Analysis The bondary work is determed from its defition to be b,o t V dv Vln mrt ln V 50 ka (.4 kg(0.87 kj/kg K(85 Kln 600 ka 7 kj Discssion The negatie sign dicates that work is done on the system (work pt. T C V

7 Nitrogen gas a cylder is compressed at constant temperatre ntil its pressre rises to a specified ale. The bondary work done drg this process is to be determed. Assmptions The process is qasi-eqilibrim. Nitrogen is an ideal gas. Analysis The bondary work is determed from its defition to be b,ot V dv Vln V ln V 50 ka kj (50 ka(0. m ln 800 ka ka m 50. kj Discssion The negatie sign dicates that work is done on the system (work pt. T 00 K V 4-4 A gas a cylder is compressed to a specified olme a process drg which the pressre changes learly with olme. The bondary work done drg this process is to be determed by plottg the process on a -V diagram and also by tegration. Assmptions The process is qasi-eqilibrim. Analysis (a The pressre of the gas changes learly with olme, and ths the process cre on a -V diagram will be a straight le. The bondary work drg this process is simply the area nder the process cre, which is a trapezoidal. Ths, av + b ( 00 ka/m (0. m + (600 ka 456 ka av + b ( 00 ka/m (0.4 m + (600 ka 96 ka and b,ot + Area ( V V ( ka (0. 0.4m 8.8 kj kj ka m (b The bondary work can also be determed by tegration to be b,ot V V dv ( + + ( av b dv a b V V ( m ( 00 ka/m 8.8 kj 6 (ka + (600 ka(0. 0.4m Discssion The negatie sign dicates that work is done on the system (work pt. 0. av + b 0.4 GAS av + b V (m

8 E A gas a cylder is heated and is allowed to expand to a specified pressre a process drg which the pressre changes learly with olme. The bondary work done drg this process is to be determed. Assmptions The process is qasi-eqilibrim. Analysis (a The pressre of the gas changes learly with olme, and ths the process cre on a -V diagram will be a straight le. The bondary work drg this process is simply the area nder the process cre, which is a trapezoidal. Ths, At state : av + b 5 psia (5 psia/ft b 0 psia (7 ft + b (psia At state : 5 and, av + b 00 psia (5 psia/ft V + ( 0 psia b,ot V 4 ft + Area 8 Bt 00 (00 + 5psia ( V V (4 7ft 7 Bt psia ft Discssion The positie sign dicates that work is done by the system (work otpt. av + b V (ft 4-6 [Also soled by EES on enclosed CD] A gas a cylder expands polytropically to a specified olme. The bondary work done drg this process is to be determed. Assmptions The process is qasi-eqilibrim. Analysis The bondary work for this polytropic process can be determed directly from and, n. V V b,ot 6.5 kj 0.0 m (50 ka 0. m V V dv n ( ka m..74 ka kj ka m (ka 50 Discssion The positie sign dicates that work is done by the system (work otpt. 0.0 V 0. V (m

9 EES roblem 4-6 is reconsidered. The process described the problem is to be plotted on a -V diagram, and the effect of the polytropic exponent n on the bondary work as the polytropic exponent aries from. to.6 is to be plotted. Analysis The problem is soled sg EES, and the soltion is gien below. Fnction Bondork([],V[],[],V[],n "This fnction retrns the Bondary ork for the polytropic process. This fnction is reqired sce the expression for bondary work depens on whether n or n<>" If n<> then Bondork:([]*V[]-[]*V[]/(-n"Use Eqation - when n" else Bondork: []*V[]*ln(V[]/V[] "Use Eqation -0 when n" endif end "Inpts from the diagram wdow" {n. [] 50 [ka] V[] 0.0 [m^] V[] 0. [m^] Gas$'AIR'} "System: The gas enclosed the piston-cylder deice." "rocess: olytropic expansion or compression, *V^n C" []*V[]^n[]*V[]^n "n." "olytropic exponent" "Inpt Data" _b Bondork([],V[],[],V[],n"[kJ]" "If we modify this problem and specify the mass, then we can calclate the fal temperatre of the flid for compression or expansion" m[] m[] "Conseration of mass for the closed system" "Let's sole the problem for m[] 0.05 kg" m[] 0.05 [kg] "Fd the temperatres from the pressre and specific olme." T[]temperatre(gas$,[],V[]/m[] T[]temperatre(gas$,[],V[]/m[] n b [kj] J ] [k b n

10 Nitrogen gas a cylder is compressed polytropically ntil the temperatre rises to a specified ale. The bondary work done drg this process is to be determed. Assmptions The process is qasi-eqilibrim. Nitrogen is an ideal gas. roperties The gas constant for nitrogen is R kj/kg.k (Table A-a Analysis The bondary work for this polytropic process can be determed from b,ot V V mr( T T dv n n ( kg(0.968 kj/kg K(60 00K kj Discssion The negatie sign dicates that work is done on the system (work pt. V n C V 4-9 [Also soled by EES on enclosed CD] A gas whose eqation of state is ( + 0 / RT expands a cylder isothermally to a specified olme. The nit of the qantity 0 and the bondary work done drg this process are to be determed. Assmptions The process is qasi-eqilibrim. Analysis (a The term 0 / mst hae pressre nits sce it is added to. Ths the qantity 0 mst hae the nit ka m 6 /kmol. (b The bondary work for this process can be determed from and b,ot RT 0 RT 0 V / N ( V / N NRT 0N V V NRT 0N V dv dv NR + T ln 0N V V V V V 4 m (0.5 kmol(8.4 kj/kmol K(00 Kln m 6 kj + (0 ka m /kmol (0.5kmol 864 kj 4 m m ka m Discssion The positie sign dicates that work is done by the system (work otpt. T 00 K 4 V

11 4-4-0 EES roblem 4-9 is reconsidered. Usg the tegration featre, the work done is to be calclated and compared, and the process is to be plotted on a -V diagram. Analysis The problem is soled sg EES, and the soltion is gien below. "Inpt Data" N0.5 [kmol] _bar/n "[m^/kmol]" _bar4/n "[m^/kmol]" T00 [K] R_8.4 [kj/kmol-k] "The qation of state is:" _bar*(+0/_bar^r_*t " is ka" "sg the EES tegral fnction, the bondary work, _bees, is" _b_eesn*tegral(,_bar, _bar, _bar,0.0 "e can show that _bhand tegeral of d_bar is (one shold sole for F(_bar and do the tegral 'by hand' for practice." _b_hand N*(R_*T*ln(_bar/_bar +0*(/_bar-/_bar "To plot s _bar, defe _plot f(_bar_plot, T as" {_bar_plot*(_plot+0/_bar_plot^r_*t} " _plot and _bar_bar_plot jst to generate the parametric table for plottg prposes. To plot s _bar for a new temperatre or _bar_plot range, remoe the '{' and '}' from the aboe eqation, and reset the _bar_plot ales the arametric Table. Then press F or select Sole Table from the Calclate men. Next select New lot dow nder the lot men to plot the new data." plot plot plot [ka] s bar plot T 00 K Area bondary [m^/kmol]

12 4-4- CO gas a cylder is compressed ntil the olme drops to a specified ale. The pressre changes drg the process with olme as av. The bondary work done drg this process is to be determed. Assmptions The process is qasi-eqilibrim. Analysis The bondary work done drg this process is determed from b,ot dv (8 ka m 5. kj 6 a V 0. m dv a V 0. m V kj ka m Discssion The negatie sign dicates that work is done on the system (work pt. 0. av V (m 4-E Hydrogen gas a cylder eqipped with a sprg is heated. The gas expands and compresses the sprg ntil its olme dobles. The fal pressre, the bondary work done by the gas, and the work done agast the sprg are to be determed, and a -V diagram is to be drawn. Assmptions The process is qasi-eqilibrim. Hydrogen is an ideal gas. Analysis (a hen the olme dobles, the sprg force and the fal pressre of H becomes F s kx V 5 ft k (5,000 lbf/ft A ft Fs + A 75,000 lbf 75,000 lbf ft (4.7 psia + ft psia (b The pressre of H changes learly with olme drg this process, and ths the process cre on a -V diagram will be a straight le. Then the bondary work drg this process is simply the area nder the process cre, which is a trapezoid. Ths, b,ot + Area ( V V ( psia (0 5ft Bt psia ft 8.7 Bt 5 0 (c If there were no sprg, we wold hae a constant pressre process at 4.7 psia. The work done drg this process wold be Ths, b,ot,no sprg dv ( V V Bt (4.7 psia(0 5 ft psia ft 40.8 sprg b b,no sprg Bt Discssion The positie sign for bondary work dicates that work is done by the system (work otpt. Bt V (ft

13 4-4- ater a cylder eqipped with a sprg is heated and eaporated. The apor expands ntil it compresses the sprg 0 cm. The fal pressre and temperatre, and the bondary work done are to be determed, and the process is to be shown on a -V diagram. Assmptions The process is qasi-eqilibrim. Analysis (a The fal pressre is determed from Fs + A The specific and total olmes at the three states are V 0. m V 0. m m 50 kg + T 5 C f 50 ka kx (00 kn/m(0. m + ka (50 ka A 0. m kn/m V m (50 kg( m /kg 0.05 m V V + x A p (0. m + (0. m(0. m 0. m 5 C m /kg m /kg 450 ka At 450 ka, f m /kg and g 0.49 m /kg. Notg that f < < g, the fal state is a satrated mixtre and ths the fal temperatre is T T sat@450 ka 47.9 C (b The pressre remas constant drg process - and changes learly (a straight le drg process -. Then the bondary work drg this process is simply the total area nder the process cre, b,ot + Area ( V V + (50 ka( m 44.5 kj ( V V ( ka + (0. 0.m kj ka m Discssion The positie sign dicates that work is done by the system (work otpt.

14 EES roblem 4- is reconsidered. The effect of the sprg constant on the fal pressre the cylder and the bondary work done as the sprg constant aries from 50 kn/m to 500 kn/m is to be estigated. The fal pressre and the bondary work are to be plotted agast the sprg constant. Analysis The problem is soled sg EES, and the soltion is gien below. [][]+(Sprg_const*(V[] - V[] "[] is a lear fnction of V[]" "where Sprg_const k/a^, the actal sprg constant diided by the piston face area sqared" "Inpt Data" []50 [ka] m50 [kg] T[]5 [C] [][] V[]0. [m^] A0.[m^] k00 [kn/m] DELTAx0 [cm] Sprg_constk/A^ "[kn/m^5]" V[]m*spol[] spol[]olme(steam_iapws,[],tt[] V[]m*spol[] V[]V[]+A*DELTAx*conert(cm,m V[]m*spol[] "The temperatre at state is:" T[]temperatre(Steam_iapws,[],spol[] "The temperatre at state is:" T[]temperatre(Steam_iapws,[],spol[] net_other 0 _otnet_other + _b+_b _b[]*(v[]-v[] "_b tegral of []*dv[] for Deltax 0 cm and is gien by:" _b[]*(v[]-v[]+sprg_const/*(v[]-v[]^ k [kn/m] [ka] ot [kj]

15 Steam 0 4 a] [k C.4 C 0 5 C [m /kg] a] [k [ ] k [kn/m] J ] [k t o k [kn/m]

16 Seeral sets of pressre and olme data are taken as a gas expands. The bondary work done drg this process is to be determed sg the experimental data. Assmptions The process is qasi-eqilibrim. Analysis lottg the gien data on a -V diagram on a graph paper and ealatg the area nder the process cre, the work done is determed to be 0.5 kj. 4-6 A piston-cylder deice contas nitrogen gas at a specified state. The bondary work is to be determed for the isothermal expansion of nitrogen. roperties The properties of nitrogen are R kj/kg.k, k.4 (Table A-a. Analysis e first determe itial and fal olmes from ideal gas relation, and fd the bondary work sg the relation for isothermal expansion of an ideal gas V V mrt (0.5 kg(0.968 kj/kg.k(0 + 7 K 0.4 m (0 ka mrt (0.5 kg(0.968 kj/kg.k(0 + 7 K 0.96 m (00 ka b V V ln (0 ka(0.4 m V 0.96 m ln 0.4 m 7.65 kj N 0 ka 0 C 4-7 A piston-cylder deice contas air gas at a specified state. The air ndergoes a cycle with three processes. The bondary work for each process and the net work of the cycle are to be determed. roperties The properties of air are R 0.87 kj/kg.k, k.4 (Table A-a. Analysis For the isothermal expansion process: V V mrt (0.5 kg(0.87 kj/kg.k( K 0.04 m (000 ka mrt (0.5 kg(0.87 kj/kg.k( K m (500 ka V V ln (000 ka(0.04 m V m ln 0.04 m b, For the polytropic compression process: 7.8 kj Air Ma 50 C n n.. V V (500 ka( m (000 ka V V m b, V V n For the constant pressre compression process: b (000 ka( m (500 ka( m., ( V V (000 ka( m The net work for the cycle is the sm of the works for each process net b, kj + b, + b, ( ( kj kj

17 4-7 Closed System Energy Analysis 4-8 A rigid tank is itially filled with sperheated R-4a. Heat is transferred to the tank ntil the pressre side rises to a specified ale. The mass of the refrigerant and the amont of heat transfer are to be determed, and the process is to be shown on a - diagram. Assmptions The tank is stationary and ths the ketic and potential energy changes are zero. There are no work teractions. Analysis (a e take the tank as the system. This is a closed system sce no mass enters or leaes. Notg that the olme of the system is constant and ths there is no bondary work, the energy balance for this stationary closed system can be expressed as E E 44 ot Esystem 44 U m( (sce KE E 0 Usg data from the refrigerant tables (Tables A- throgh A-, the properties of R-4a are determed to be 60 ka f , x 0.4 f.09, 0.48 m /kg g fg 90.7kJ/kg R-4a 60 ka f f + x + x fg fg ( m ( kj/kg /kg 700 ka ( kj/kg (Sperheated apor Then the mass of the refrigerant is determed to be V 0.5 m m m 0.0 kg /kg (b Then the heat transfer to the tank becomes m( (0.0 kg( kj/kg 707 kj

18 E A rigid tank is itially filled with satrated R-4a apor. Heat is transferred from the refrigerant ntil the pressre side drops to a specified ale. The fal temperatre, the mass of the refrigerant that has condensed, and the amont of heat transfer are to be determed. Also, the process is to be shown on a - diagram. Assmptions The tank is stationary and ths the ketic and potential energy changes are zero. There are no work teractions. Analysis (a e take the tank as the system. This is a closed system sce no mass enters or leaes. Notg that the olme of the system is constant and ths there is no bondary work, the energy balance for this stationary closed system can be expressed as E E 44 ot ot ot U m( m( Esystem 44 (sce KE E 0 Usg data from the refrigerant tables (Tables A-E throgh A-E, the properties of R-4a are determed to be 60 psia g sat. apor psia 0.96 ft /lbm Bt/lbm 50 psia f ( f 0.05, 4.8, The fal state is satrated mixtre. Ths, ft /lbm g fg Bt/lbm R-4a 60 psia Sat. apor T T 50 psia 40. F (b The total mass and the amont of refrigerant that has condensed are V 0 ft m 0.96 ft x m f f fg ( x 68. lbm /lbm m ( 0.00(68. lbm lbm Also, + x (75.09 f fg Bt/lbm (c Sbstittg, ot m( (68. lbm( Bt/lbm 469 Bt

19 An slated rigid tank is itially filled with a satrated liqid-apor mixtre of water. An electric heater the tank is trned on, and the entire liqid the tank is aporized. The length of time the heater was kept on is to be determed, and the process is to be shown on a - diagram. Assmptions The tank is stationary and ths the ketic and potential energy changes are zero. The deice is well-slated and ths heat transfer is negligible. The energy stored the resistance wires, and the heat transferred to the tank itself is negligible. Analysis e take the contents of the tank as the system. This is a closed system sce no mass enters or leaes. Notg that the olme of the system is constant and ths there is no bondary work, the energy balance for this stationary closed system can be expressed as E E 44 ot e, U m( VI t m( potential, etc. energies Esystem 44 The properties of water are (Tables A-4 throgh A-6 00ka f x , f 47.40, g fg (sce KE E m /kg 088. kj/kg T e H O V const. f f + x + x fg fg [ 0.5 ( ] ( kj/kg 0.44 m /kg 0.44 m /kg 556. kj/kg g@0.44m /kg sat.apor Sbstittg, 000 VA (0 V(8 A t (5 kg( kJ/kg kj/s t 986 s 5. m

20 EES roblem 4-0 is reconsidered. The effect of the itial mass of water on the length of time reqired to completely aporize the liqid as the itial mass aries from kg to 0 kg is to be estigated. The aporization time is to be plotted agast the itial mass. Analysis The problem is soled sg EES, and the soltion is gien below. ROCEDURE X([]:[],x[] Flid$'Steam_IAS' If [] > V_CRIT(Flid$ then []pressre(flid$,[],x x[] else []pressre(flid$,[],x0 x[]0 EndIf End "Knowns" {m5 [kg]} []00 [ka] y0.75 "moistre" Volts0 [V] I8 [amp] t m [m] "Soltion" m [kg] "Conseration of Energy for the closed tank:" E_dot_-E_dot_otDELTAE_dot E_dot dot_ele "[k]" _dot_elevolts*i*convert(j/s,k "[k]" E_dot_ot0 "[k]" DELTAE_dotm*([]-[]/DELTAt_s "[k]" DELTAt_mDELTAt_s*conert(s,m "[m]" "The qality at state is:" Flid$'Steam_IAS' x[]-y []INTENERGY(Flid$,[], xx[] "[kj/kg]" []olme(flid$,[], xx[] "[m^/kg]" T[]temperatre(Flid$,[], xx[] "[C]" "Check to see if state is on the satrated liqid le or satrated apor le:" Call X([]:[],x[] []INTENERGY(Flid$,[], xx[] "[kj/kg]" []olme(flid$,[], xx[] "[m^/kg]" T[]temperatre(Flid$,[], xx[] "[C]" t m [m] m [kg] T [ C] ka 00 ka Steam [m /kg]

21 4-4- One part of an slated tank contas compressed liqid while the other side is eacated. The partition is then remoed, and water is allowed to expand to the entire tank. The fal temperatre and the olme of the tank are to be determed. Assmptions The tank is stationary and ths the ketic and potential energy changes are zero. The tank is slated and ths heat transfer is negligible. There are no work teractions. Analysis e take the entire contents of the tank as the system. This is a closed system sce no mass enters or leaes. Notg that the olme of the system is constant and ths there is no bondary work, the energy balance for this stationary closed system can be expressed as E E 44 ot potential, etc. energies 0 U m( Esystem 44 The properties of water are (Tables A-4 throgh A ka f T 60 C C m (sce KE E 0 /kg 5.6 kj/kg Eacate artition e now assme the fal state the tank is satrated liqid-apor mixtre and determe qality. This assmption will be erified if we get a qality between 0 and. 0 ka f , ( f 9.79, m /kg g fg 45.4 kj/kg H O Ths, and, x f fg T T 0 ka 45.8 C [ ( ] m /kg + x f fg V m (.5 kg( m /kg 0.97 m

22 4-4- EES roblem 4- is reconsidered. The effect of the itial pressre of water on the fal temperatre the tank as the itial pressre aries from 00 ka to 600 ka is to be estigated. The fal temperatre is to be plotted agast the itial pressre. Analysis The problem is soled sg EES, and the soltion is gien below. "Knowns" m.5 [kg] {[]600 [ka]} T[]60 [C] []0 [ka] "Soltion" Flid$'Steam_IAS' "Conseration of Energy for the closed tank:" E_-E_otDELTAE E_0 E_ot0 DELTAEm*([]-[] []INTENERGY(Flid$,[], TT[] []olme(flid$,[], TT[] T[]temperatre(Flid$,[], [] T_T[] []olme(flid$,[], [] V_totalm*[] Steam [ka] T [C] T [ C] ka 0 ka [m /kg] [C] T [] [ka]

23 4-4-4 A cylder is itially filled with R-4a at a specified state. The refrigerant is cooled at constant pressre. The amont of heat loss is to be determed, and the process is to be shown on a T- diagram. Assmptions The cylder is stationary and ths the ketic and potential energy changes are zero. There are no work teractions oled other than the bondary work. The thermal energy stored the cylder itself is negligible. 4 The compression or expansion process is qasi-eqilibrim. Analysis e take the contents of the cylder as the system. This is a closed system sce no mass enters or leaes. The energy balance for this stationary closed system can be expressed as E Eot E 44 system 44 ot b,ot ot U m( m( h h (sce KE E 0 sce U + b H drg a constant pressre qasieqilibrim process. The properties of R-4a are (Tables A- throgh A- 800 ka h kj/kg T 70 C 800 ka h h C 7.4 kj/kg T 5 C Sbstittg, ot - (5 kg( kj/kg 7 kj T R-4a 800 ka 4-5E A cylder contas water itially at a specified state. The water is heated at constant pressre. The fal temperatre of the water is to be determed, and the process is to be shown on a T- diagram. Assmptions The cylder is stationary and ths the ketic and potential energy changes are zero. The thermal energy stored the cylder itself is negligible. The compression or expansion process is qasieqilibrim. Analysis e take the contents of the cylder as the system. This is a closed system sce no mass enters or leaes. The energy balance for this stationary closed system can be expressed as E Eot E 44 system 44 b,ot U m( m( h h (sce KE E 0 sce U + b H drg a constant pressre qasi-eqilibrim process. The properties of water are (Tables A-6E V ft 4 ft m 0.5 lbm /lbm 0 psia 7.0 Bt/lbm h 4 ft /lbm Sbstittg, 00 Bt (0.5 lbm( h 7.0Bt/lbm Then, h 67.0 Bt/lbm 0 psia T 6.4 F h 67.0 Bt/lbm T H O 0 psia

24 A cylder is itially filled with satrated liqid water at a specified pressre. The water is heated electrically as it is stirred by a paddle-wheel at constant pressre. The oltage of the crrent sorce is to be determed, and the process is to be shown on a - diagram. Assmptions The cylder is stationary and ths the ketic and potential energy changes are zero. The cylder is well-slated and ths heat transfer is negligible. The thermal energy stored the cylder itself is negligible. 4 The compression or expansion process is qasi-eqilibrim. Analysis e take the contents of the cylder as the system. This is a closed system sce no mass enters or leaes. The energy balance for this stationary closed system can be expressed as e, E 44 ot + pw, e, E + ( VI t + b,ot pw, pw, U m( h m( h Esystem 44 (sce KE E 0 h h sce U + b H drg a constant pressre qasi-eqilibrim process. The properties of water are (Tables A-4 throgh A-6 Sbstittg, 75 ka h h sat.liqid f 75 ka h x 0.5 V m m 4.7 kg m /kg ka f VI t 485 kj ka + x h fg kj/kg m /kg ( 0.5. VI t + (400kJ (4.7 kg( kJ/kg 485 kj 000 VA V.9 V (8 A(45 60 s kj/s e 59.6 kj/kg H O const. pw

25 A cylder is itially filled with steam at a specified state. The steam is cooled at constant pressre. The mass of the steam, the fal temperatre, and the amont of heat transfer are to be determed, and the process is to be shown on a T- diagram. Assmptions The cylder is stationary and ths the ketic and potential energy changes are zero. There are no work teractions oled other than the bondary work. The thermal energy stored the cylder itself is negligible. 4 The compression or expansion process is qasi-eqilibrim. Analysis e take the contents of the cylder as the system. This is a closed system sce no mass enters or leaes. The energy balance for this stationary closed system can be expressed as E 44 ot E ot b,ot ot U m( m( h Esystem 44 h (sce KE E 0 sce U + b H drg a constant pressre qasieqilibrim process. The properties of water are (Tables A-4 throgh A-6 Ma m /kg T 450 C h 7. kj/kg V.5 m m kg m /kg (b The fal temperatre is determed from Ma T T sat. apor h h sat@ Ma g@ Ma (c Sbstittg, the energy balance gies 79.9 C 777. kj/kg ot - (7.565 kg( kj/kg 4495 kj T H O Ma 450 C

26 [Also soled by EES on enclosed CD] A cylder eqipped with an external sprg is itially filled with steam at a specified state. Heat is transferred to the steam, and both the temperatre and pressre rise. The fal temperatre, the bondary work done by the steam, and the amont of heat transfer are to be determed, and the process is to be shown on a - diagram. Assmptions The cylder is stationary and ths the ketic and potential energy changes are zero. The thermal energy stored the cylder itself is negligible. The compression or expansion process is qasieqilibrim. 4 The sprg is a lear sprg. Analysis e take the contents of the cylder as the system. This is a closed system sce no mass enters or leaes. Notg that the sprg is not part of the system (it is external, the energy balance for this stationary closed system can be expressed as E 44 ot E b,ot U m( m( Esystem 44 + b,ot (sce KE E 0 The properties of steam are (Tables A-4 throgh A-6 00 ka m /kg T 00 C kj/kg V 0.5 m m kg m /kg V m 0.6 m.966 m kg /kg 500 ka T C.966 m /kg 45. kj/kg H O 00 ka 00 C (b The pressre of the gas changes learly with olme, and ths the process cre on a -V diagram will be a straight le. The bondary work drg this process is simply the area nder the process cre, which is a trapezoidal. Ths, b + Area (c From the energy balance we hae ( ka kj ka m ( V V ( m 5 kj (0.468 kg( kJ/kg + 5 kj 808 kj

27 EES roblem 4-8 is reconsidered. The effect of the itial temperatre of steam on the fal temperatre, the work done, and the total heat transfer as the itial temperatre aries from 50 C to 50 C is to be estigated. The fal reslts are to be plotted agast the itial temperatre. Analysis The problem is soled sg EES, and the soltion is gien below. "The process is gien by:" "[][]+k*x*a/a, and as the sprg moes 'x' amont, the olme changes by V[]-V[]." [][]+(Sprg_const*(V[] - V[] "[] is a lear fnction of V[]" "where Sprg_const k/a, the actal sprg constant diided by the piston face area" "Conseration of mass for the closed system is:" m[]m[] "The conseration of energy for the closed system is" "E_ - E_ot DeltaE, neglect DeltaKE and DeltaE for the system" _ - _ot m[]*([]-[] DELTAUm[]*([]-[] "Inpt Data" []00 [ka] V[]0.5 [m^] "T[]00 [C]" []500 [ka] V[]0.6 [m^] Flid$'Steam_IAS' m[]v[]/spol[] spol[]olme(flid$,tt[], [] []tenergy(flid$, TT[], [] spol[]v[]/m[] ot [kj] "The fal temperatre is:" T[]temperatre(Flid$,[],spol[] T[] [C] []tenergy(flid$, [], TT[] net_other 0 _otnet_other + _b "_b tegral of []*dv[] for 0.5<V[]<0.6 and is gien by:" _b[]*(v[]-v[]+sprg_const/*(v[]-v[]^ [kj] T [C] T [C] ot [kj]

28 Steam 0 4 C [ka] C 0 Area b [m /kg] [kj] T[] [C] T[] [C] T[] [C]

29 A cylder eqipped with a set of stops for the piston to rest on is itially filled with satrated water apor at a specified pressre. Heat is transferred to water ntil the olme dobles. The fal temperatre, the bondary work done by the steam, and the amont of heat transfer are to be determed, and the process is to be shown on a - diagram. Assmptions The cylder is stationary and ths the ketic and potential energy changes are zero. There are no work teractions oled other than the bondary work. The thermal energy stored the cylder itself is negligible. 4 The compression or expansion process is qasi-eqilibrim. Analysis e take the contents of the cylder as the system. This is a closed system sce no mass enters or leaes. The energy balance for this stationary closed system can be expressed as E 44 ot E b,ot U m( m( Esystem 44 + b,ot (sce KE E 0 The properties of steam are (Tables A-4 throgh A-6 50 ka g sat. apor V 0.8 m m. kg m /kg V.6 m.475 m /kg m. kg 00 ka.475 m ka g@50 ka T 66 C m /kg 4.4 kj/kg 56.8 kj/kg 00 ka H O 50 ka Sat. Vapor (b The work done drg process - is zero (sce V const and the work done drg the constant pressre process - is b,ot kj dv ( V V (00 ka(.6 0.8m 40 ka m (c Heat transfer is determed from the energy balance, m( + b, ot (. kg( kj/kg + 40 kj kj kj

30 Two tanks itially separated by a partition conta steam at different states. Now the partition is remoed and they are allowed to mix ntil eqilibrim is established. The temperatre and qality of the steam at the fal state and the amont of heat lost from the tanks are to be determed. Assmptions The tank is stationary and ths the ketic and potential energy changes are zero. There are no work teractions. Analysis (a e take the contents of both tanks as the system. This is a closed system sce no mass enters or leaes. Notg that the olme of the system is constant and ths there is no bondary work, the energy balance for this stationary closed system can be expressed as E E 44 ot system 44 ot U A E + U B [ m ] + [ m( ] (sce KE E 0 ( A B TANK A kg Ma 00 C The properties of steam both tanks at the itial state are (Tables A-4 throgh A-6 T, A, A, B, B 000 ka, 00 C, T, B 50 C f x 0.50 f f + x + x fg fg f A A m 79.7 kj/kg , 6.66, g fg /kg The total olme and total mass of the system are V V A m m A + V B + m m B A, A + m + 5 kg m /kg 97.4 kj/kg + [ 0.50 ( ] ( kj/kg B, B Now, the specific olme at the fal state may be determed V.06 m 0.7 m /kg m 5 kg which fixes the fal state and we can determe other properties (b Sbstittg, or T 00 ka x 0.7 m /kg ot U A + U B m /kg TANK B kg 50 C x0.5 ( kg( m /kg + ( kg( m /kg.06 m 00 ka.5 C f g f x 56.+ f fg [ m ] + [ m( ] ( A ( kj/kg ( kg( kJ/kg + ( kg( kJ/kg 959 kj kj 959 ot B

31 4-4-4 A room is heated by an electrical radiator contag heatg oil. Heat is lost from the room. The time period drg which the heater is on is to be determed. Assmptions Air is an ideal gas sce it is at a high temperatre and low pressre relatie to its critical pot ales of -4 C and.77 Ma. The ketic and potential energy changes are negligible, ke pe 0. Constant specific heats at room temperatre can be sed for air. This assmption reslts negligible error heatg and air-conditiong applications. 4 The local atmospheric pressre is 00 ka. 5 The room is air-tight so that no air leaks and ot drg the process. roperties The gas constant of air is R 0.87 ka.m /kg.k (Table A-. Also, c 0.78 kj/kg.k for air at room temperatre (Table A-. Oil properties are gien to be ρ 950 kg/m and c p. kj/kg. C. Analysis e take the air the room and the oil the radiator to be the system. This is a closed system sce no mass crosses the system bondary. The energy balance for this stationary constantolme closed system can be expressed as E 44 ot ( & E & ot t U The mass of air and oil are m m air oil Sbstittg, V RT oil air ρ V oil [ mc air Esystem 44 + U ( T oil T ] air + [ mc p ( T T ] (00 ka(50 m 6. kg (0.87ka m /kg K(0 + 7 K (950 kg/m (0.00 m 8.50 kg t 08 s 4.0 m oil 0 C (sce KE E 0 Radiator Room ( kj/s t (6. kg(0.78 kj/kg C(0 0 C + (8.50 kg(. kj/kg C(50 0 C Discssion In practice, the pressre the room will rema constant drg this process rather than the olme, and some air will leak ot as the air expands. As a reslt, the air the room will ndergo a constant pressre expansion process. Therefore, it is more proper to be conseratie and to sg H stead of se U heatg and air-conditiong applications.

32 4- Specific Heats, and h of Ideal Gases 4-4C It can be sed for any kd of process of an ideal gas. 4-44C It can be sed for any kd of process of an ideal gas. 4-45C The desired reslt is obtaed by mltiplyg the first relation by the molar mass M, Mc p Mc + MR or p c c + R 4-46C Very close, bt no. Becase the heat transfer drg this process is mc p T, and c p aries with temperatre. 4-47C It can be either. The difference temperatre both the K and C scales is the same. 4-48C The energy reqired is mc p T, which will be the same both cases. This is becase the c p of an ideal gas does not ary with pressre. 4-49C The energy reqired is mc p T, which will be the same both cases. This is becase the c p of an ideal gas does not ary with olme. 4-50C For the constant pressre case. This is becase the heat transfer to an ideal gas is mc p T at constant pressre, mc T at constant olme, and c p is always greater than c. 4-5 The enthalpy change of nitrogen gas drg a heatg process is to be determed sg an empirical specific heat relation, constant specific heat at aerage temperatre, and constant specific heat at room temperatre. Analysis (a Usg the empirical relation for c p (T from Table A-c, c p a + bt + ct + dt where a 8.90, b , c , and d Then, h a( T 8.90( ,544 kj/kmol h h M c ( T dt p T + ( b( T 5 [ a + bt + ct + dt ] T,544 kj/kmol 8.0 kg/kmol + c( T ( T ( kj/kg dt + 4 d( T 4 ( (.87 0 T ( (b Usg the constant c p ale from Table A-b at the aerage temperatre of 800 K, c c. kj/kg K p,ag h c p@800 K p,ag ( T T (. kj/kg K( K kj/kg (c Usg the constant c p ale from Table A-a at room temperatre, c c.09 kj/kg K p,ag h c p@00 K p,ag ( T T (.09 kj/kg K( K 45.6 kj/kg

33 4-4-5E The enthalpy change of oxygen gas drg a heatg process is to be determed sg an empirical specific heat relation, constant specific heat at aerage temperatre, and constant specific heat at room temperatre. Analysis (a Usg the empirical relation for c p (T from Table A-Ec, c p a + bt + ct + dt where a 6.085, b , c , and d Then, h a( T T + b( T 6.085( ( Bt/lbmol h h M c ( T dt p [ a + bt + ct + dt ] + T + c( T ( ( ( Bt/lbmol 70. Bt/lbm.999 lbm/lbmol T 4 dt 4 d( T ( T 9 ( (b Usg the constant c p ale from Table A-Eb at the aerage temperatre of 50 R, c c 0.55 Bt/lbm R h c p,ag ( T T (0.55 Bt/lbm R( R 78.5 p,ag p@50 R (c Usg the constant c p ale from Table A-Ea at room temperatre, c c 0.9 Bt/lbm R p,ag h c p@57 R p,ag ( T 4 Bt/lbm T (0.9 Bt/lbm R( R 5. Bt/lbm 4-5 The ternal energy change of hydrogen gas drg a heatg process is to be determed sg an empirical specific heat relation, constant specific heat at aerage temperatre, and constant specific heat at room temperatre. Analysis (a Usg the empirical relation for c p (T from Table A-c and relatg it to c (T, ( a R + bt + ct c ( T c R dt p + where a 9., b , c , and d Then, M ( a R (9. 8.4( c ( T dt ( T T + ( ,487 kj/kmol [( a R + bt + ct + dt ] b( T + T + ( c( T ( ,487 kj/kmol 694 kj/kg.06 kg/kmol T dt + ( ( d( T 4 T ( (b Usg a constant c p ale from Table A-b at the aerage temperatre of 500 K, c c 0.89 kj/kg K,ag K,ag ( T T (0.89 kj/kg K(800 00K 6 kj/kg (c Usg a constant c p ale from Table A-a at room temperatre, c c 0.8 kj/kg K,ag K,ag ( T T (0.8 kj/kg K(800 00K 60 kj/kg

34 4-4 Closed System Energy Analysis: Ideal Gases 4-54C No, it isn't. This is becase the first law relation - U redces to 0 this case sce the system is adiabatic ( 0 and U 0 for the isothermal processes of ideal gases. Therefore, this adiabatic system cannot receie any net work at constant temperatre. 4-55E The air a rigid tank is heated ntil its pressre dobles. The olme of the tank and the amont of heat transfer are to be determed. Assmptions Air is an ideal gas sce it is at a high temperatre and low pressre relatie to its critical pot ales of -4 C and.77 Ma. The ketic and potential energy changes are negligible, pe ke 0. Constant specific heats at room temperatre can be sed for air. This assmption reslts negligible error heatg and air-conditiong applications. roperties The gas constant of air is R psia.ft /lbm.r (Table A-E. Analysis (a The olme of the tank can be determed from the ideal gas relation, mrt V (0 lbm(0.704 psia ft /lbm R(540 R psia (b e take the air the tank as or system. The energy balance for this stationary closed system can be expressed as E E 44 ot U The fal temperatre of air is Esystem 44 m( mc ( T T V V T T (540 R 080 R T T The ternal energies are (Table A-7E Sbstittg, 9.04 Bt / 540 R 86.9 Bt / 080 R (0 lbm( bt/lbm 898 Bt ft Air 0 lbm 50 psia 80 F Alternatie soltions The specific heat of air at the aerage temperatre of T ag ( / 80 R 50 F is, from Table A-Eb, c,ag 0.75 Bt/lbm.R. Sbstittg, (0 lbm( 0.75 Bt/lbm.R( R 890 Bt Discssion Both approaches reslted almost the same soltion this case.

35 The hydrogen gas a rigid tank is cooled ntil its temperatre drops to 00 K. The fal pressre the tank and the amont of heat transfer are to be determed. Assmptions Hydrogen is an ideal gas sce it is at a high temperatre and low pressre relatie to its critical pot ales of -40 C and.0 Ma. The tank is stationary, and ths the ketic and potential energy changes are negligible, ke pe 0. roperties The gas constant of hydrogen is R 4.4 ka.m /kg.k (Table A-. The constant olme specific heat of hydrogen at the aerage temperatre of 450 K is, c,ag 0.77 kj/kg.k (Table A-. Analysis (a The fal pressre of hydrogen can be determed from the ideal gas relation, V V T 50 K (50 ka 59. ka T T T 550 K (b e take the hydrogen the tank as the system. This is a closed system sce no mass enters or leaes. The energy balance for this stationary closed system can be expressed as where E E 44 ot ot ot U Esystem 44 U m( mc ( T T V (50 ka(.0 m m 0.07 kg RT (4.4 ka m /kg K(550 K Sbstittg to the energy balance, ot (0.07 kg(0.77 kj/kg K(550-50K 686. kj H 50 ka 550 K

36 A resistance heater is to raise the air temperatre the room from 7 to C with 5 m. The reqired power ratg of the resistance heater is to be determed. Assmptions Air is an ideal gas sce it is at a high temperatre and low pressre relatie to its critical pot ales of -4 C and.77 Ma. The ketic and potential energy changes are negligible, ke pe 0. Constant specific heats at room temperatre can be sed for air. This assmption reslts negligible error heatg and air-conditiong applications. 4 Heat losses from the room are negligible. 5 The room is air-tight so that no air leaks and ot drg the process. roperties The gas constant of air is R 0.87 ka.m /kg.k (Table A-. Also, c 0.78 kj/kg.k for air at room temperatre (Table A-. Analysis e take the air the room to be the system. This is a closed system sce no mass crosses the system bondary. The energy balance for this stationary constant-olme closed system can be expressed as or, E E 44 ot The mass of air is & e, Esystem 44 U mc,ag ( T T (sce KE E 0 ( T e, t mc,ag T V m V (00 ka(0 m m 49. kg RT (0.87 ka m /kg K(80 K Sbstittg, the power ratg of the heater becomes & e, (49. kg(0.78 kj/kg C( 7 C.9 k 5 60 s o o e m 7 C Discssion In practice, the pressre the room will rema constant drg this process rather than the olme, and some air will leak ot as the air expands. As a reslt, the air the room will ndergo a constant pressre expansion process. Therefore, it is more proper to be conseratie and to se H stead of sg U heatg and air-conditiong applications. AIR

37 A room is heated by a radiator, and the warm air is distribted by a fan. Heat is lost from the room. The time it takes for the air temperatre to rise to 0 C is to be determed. Assmptions Air is an ideal gas sce it is at a high temperatre and low pressre relatie to its critical pot ales of -4 C and.77 Ma. The ketic and potential energy changes are negligible, ke pe 0. Constant specific heats at room temperatre can be sed for air. This assmption reslts negligible error heatg and air-conditiong applications. 4 The local atmospheric pressre is 00 ka. 5 The room is air-tight so that no air leaks and ot drg the process. roperties The gas constant of air is R 0.87 ka.m /kg.k (Table A-. Also, c 0.78 kj/kg.k for air at room temperatre (Table A-. Analysis e take the air the room to be the system. This is a closed system sce no mass crosses the system bondary. The energy balance for this stationary constant-olme closed system can be expressed as or, 44 ot E + E fan, The mass of air is ot Esystem 44 U mc,ag ( & + & fan, & ot t mc,ag( T T V m ( T T (sce KE E 0 V (00 ka(40 m m 7.4 kg RT (0.87 ka m /kg K(8 K Usg the c ale at room temperatre, It yields Steam pw ROOM o o [( 0,000 5,000 /600 kj/s + 0. kj/s] t (7.4 kg(0.78 kj/kg C(0 0 C t 8 s 4m 5m 7m 5,000 kj/h 0,000 kj/h Discssion In practice, the pressre the room will rema constant drg this process rather than the olme, and some air will leak ot as the air expands. As a reslt, the air the room will ndergo a constant pressre expansion process. Therefore, it is more proper to be conseratie and to sg H stead of se U heatg and air-conditiong applications.

38 A stdent lig a room trns her 50- fan on the morng. The temperatre the room when she comes back 0 h later is to be determed. Assmptions Air is an ideal gas sce it is at a high temperatre and low pressre relatie to its critical pot ales of -4 C and.77 Ma. The ketic and potential energy changes are negligible, ke pe 0. Constant specific heats at room temperatre can be sed for air. This assmption reslts negligible error heatg and air-conditiong applications. 4 All the doors and wdows are tightly closed, and heat transfer throgh the walls and the wdows is disregarded. roperties The gas constant of air is R 0.87 ka.m /kg.k (Table A-. Also, c 0.78 kj/kg.k for air at room temperatre (Table A-. Analysis e take the room as the system. This is a closed system sce the doors and the wdows are said to be tightly closed, and ths no mass crosses the system bondary drg the process. The energy balance for this system can be expressed as E E 44 ot The mass of air is e, e, U V m Esystem 44 m( mc ( T T V (00 ka(44 m m 74. kg RT (0.87 ka m /kg K(88 K The electrical work done by the fan is & t (0.5 kj / s(0 600 s 5400 kj e e Sbstittg and sg the c ale at room temperatre, 5400 kj (74. kg(0.78 kj/kg C(T - 5 C T 58. C ROOM Fan 4m 6m 6m Discssion Note that a fan actally cases the ternal temperatre of a confed space to rise. In fact, a 00- fan spplies a room with as mch energy as a 00- resistance heater.

39 E A paddle wheel an oxygen tank is rotated ntil the pressre side rises to 0 psia while some heat is lost to the srrondgs. The paddle wheel work done is to be determed. Assmptions Oxygen is an ideal gas sce it is at a high temperatre and low pressre relatie to its critical pot ales of -8 F and 76 psia. The ketic and potential energy changes are negligible, ke pe 0. The energy stored the paddle wheel is negligible. 4 This is a rigid tank and ths its olme remas constant. roperties The gas constant and molar mass of oxygen are R 0.5 psia.ft /lbm.r and M lbm/lbmol (Table A-E. The specific heat of oxygen at the aerage temperatre of T ag (75+540/ 68 R is c,ag 0.60 Bt/lbm.R (Table A-E. Analysis e take the oxygen the tank as or system. This is a closed system sce no mass enters or leaes. The energy balance for this system can be expressed as E E 44 ot pw, ot pw, U ot ot Esystem 44 + m( + mc ( T T The fal temperatre and the mass of oxygen are V V T T T Sbstittg, V (4.7 psia(0 ft m 0.8 lbm RT (0.5 psia ft /lbmol R(540 R T 0 psia (540 R 75 R 4.7 psia pw, (0 Bt + (0.8 lbm(0.60 Bt/lbm.R( R 45. Bt O 4.7 psia 80 F 0 Bt 4-6 One part of an slated rigid tank contas an ideal gas while the other side is eacated. The fal temperatre and pressre the tank are to be determed when the partition is remoed. Assmptions The ketic and potential energy changes are negligible, ke pe 0. The tank is slated and ths heat transfer is negligible. Analysis e take the entire tank as the system. This is a closed system sce no mass crosses the bondaries of the system. The energy balance for this system can be expressed as Therefore, E Eot E U m( system potential, etc. energies 0 T T 50 C Sce (T for an ideal gas. Then, V V V (800 ka 400 ka T T V IDEAL GAS 800 ka 50 C Eacated

40 A cylder eqipped with a set of stops for the piston to rest on is itially filled with helim gas at a specified state. The amont of heat that mst be transferred to raise the piston is to be determed. Assmptions Helim is an ideal gas with constant specific heats. The ketic and potential energy changes are negligible, ke pe 0. There are no work teractions oled. 4 The thermal energy stored the cylder itself is negligible. roperties The specific heat of helim at room temperatre is c.56 kj/kg.k (Table A-. Analysis e take the helim gas the cylder as the system. This is a closed system sce no mass crosses the bondary of the system. The energy balance for this constant olme closed system can be expressed as E Eot E 44 system 44 U m( m( mc ( T T The fal temperatre of helim can be determed from the ideal gas relation to be V V 500 ka T T (98 K 490 K T T 00 ka Sbstittg to the energy balance relation gies (0.5 kg(.56 kj/kg K(490-98K 857 kj 500 ka He 00 ka 5 C 4-6 An slated cylder is itially filled with air at a specified state. A paddle-wheel the cylder stirs the air at constant pressre. The fal temperatre of air is to be determed. Assmptions Air is an ideal gas with ariable specific heats. The cylder is stationary and ths the ketic and potential energy changes are zero. There are no work teractions oled other than the bondary work. 4 The cylder is well-slated and ths heat transfer is negligible. 5 The thermal energy stored the cylder itself and the paddle-wheel is negligible. 6 The compression or expansion process is qasi-eqilibrim. roperties The gas constant of air is R 0.87 ka.m /kg.k (Table A-. Also, c p.005 kj/kg.k for air at room temperatre (Table A-. The enthalpy of air at the itial temperatre is h K 98.8 kj/kg (Table A-7 Analysis e take the air the cylder as the system. This is a closed system sce no mass enters or leaes. The energy balance for this stationary closed system can be expressed as E Eot E 44 system 44 pw, b,ot U pw, m( h h sce U + b H drg a constant pressre qasi-eqilibrim process. The mass of air is V (400 ka(0. m m kg RT (0.87 ka m /kg K(98 K pw AIR const. Sbstittg to the energy balance, 5 kj (0.468 kg(h kj/kg h 0. kj/kg From Table A-7, T 9.9 K Alternatie soltion Usg specific heats at room temperatre, c p.005 kj/kg. C, the fal temperatre is determed to be m h h mcp( T T 5 kj (0.468 kg(.005 kj/kg. C(T - 5 C which gies pw, ( T 56.9 C