4-64. Review Problems

Transcription

1 4-64 Reiew Prblems 4-0 Heat is transferred t a pistn-cylinder deice cntaining air. The expansin wrk is t be determined. Assumptins There is n frictin between pistn and cylinder. Air is an ideal gas. Prperties The gas cntant fr air is 0.87 kj/kg.k (Table A-a). Analysis Nting that the gas cnstant represents the bundary wrk fr a unit mass and a unit temperature change, the expansin wrk is simply determined frm W b m TR (0.5 kg)(5 K)(0.87 kj/kg.k) kj Air 0.5 kg T5 C 4- Slar energy is t be stred as sensible heat using phase-change materials, granite rcks, and water. The amunt f heat that can be stred in a 5-m 5000 L space using these materials as the strage medium is t be determined. Assumptins The materials hae cnstant prperties at the specified alues. N allwance is made fr ids, and thus the alues calculated are the upper limits. Analysis The amunt f energy stred in a medium is simply equal t the increase in its internal energy, which, fr incmpressible substances, can be determined frm U mc T T ). ( (a) The latent heat f glaubers salts is gien t be 9 kj/l. Disregarding the sensible heat strage in this case, the amunt f energy stred is becmes U salt mh if (5000 L)(9 kj/l),645,000 kj This alue wuld be een larger if the sensible heat strage due t temperature rise is cnsidered. (b) The density f granite is 700 kg/m (Table A-), and its specific heat is gien t be c. kj/kg. C. Then the amunt f energy that can be stred in the rcks when the temperature rises by 0 C becmes U rck ρvc T (700 kg/m )(5 m )(. kj/kg. C)(0 C) 66,400 kj (c) The density f water is abut 000 kg/m (Table A-), and its specific heat is gien t be c 4.0 kj/kg. C. Then the amunt f energy that can be stred in the water when the temperature rises by 0 C becmes U rck ρvc T (000 kg/m )(5 m )(4.0 kj/kg. C)(0 C) 400,00 kj Discussin Nte that the greatest amunt f heat can be stred in phase-change materials essentially at cnstant temperature. Such materials are nt withut prblems, hweer, and thus they are nt widely used.

2 The ideal gas in a pistn-cylinder deice is cled at cnstant pressure. The gas cnstant and the mlar mass f this gas are t be determined. Assumptins There is n frictin between pistn and cylinder. Prperties The specific heat rati is gien t be.667 Analysis Nting that the gas cnstant represents the bundary wrk fr a unit mass and a unit temperature change, the gas cnstant is simply determined frm Wb R m T 6.6 kj (0.8 kg)(0 The mlar mass f the gas is Ru M R 8.4 kj/kml.k.075 kj/kg.k The specific heats are determined as c c p.075 kj/kg.k C) kg/kml R.075 kj/kg.k. kj/kg. C k.667 c + R. kj/kg.k kj/kg.k 5.86 kj/kg. C Ideal gas 0.8 kg T0 C 4- Fr a 0 C temperature change f air, the final elcity and final eleatin f air are t be determined s that the internal, kinetic and ptential energy changes are equal. Prperties The cnstant-lume specific heat f air at rm temperature is 0.78 kj/kg. C (Table A-). Analysis The internal energy change is determined frm u c T (0.78 kj/kg. C)(0 C) 7.8 kj/kg Equating kinetic and ptential energy changes t internal energy change, the final elcity and eleatin are determined frm AIR 0 C 0 C 0 m/s Vel 0 m z U KE PE kj/kg u ke ( V V ) 7.8 kj/kg ( V 0 m /s ) V 9.8 m/s 000 m /s kj/kg u pe g( z z) 7.8 kj/kg (9.8 m/s )(z 0 m) z 000 m /s 7.9 m

3 A cylinder equipped with an external spring is initially filled with air at a specified state. Heat is transferred t the air, and bth the temperature and pressure rise. The ttal bundary wrk dne by the air, and the amunt f wrk dne against the spring are t be determined, and the prcess is t be shwn n a P- diagram. Assumptins The prcess is quasi-equilibrium. The spring is a linear spring. Analysis (a) The pressure f the gas changes linearly with lume during this prcess, and thus the prcess cure n a P-V diagram will be a straight line. Then the bundary wrk during this prcess is simply the area under the prcess cure, which is a trapezidal. Thus, W b,ut P + P Area ( V V) ( )kPa kj (0.5 0.)m kpa m 50 kj (b) If there were n spring, we wuld hae a cnstant pressure prcess at P 00 kpa. The wrk dne during this prcess is W Thus, b,ut,n spring W ( ) PdV P V V kj kpa m ( 00 kpa)( ) m /kg 60 kj W kj spring b W b,n spring P (kpa) Air 00 kpa 0. m 0.5 V (m )

4 A cylinder equipped with a set f stps fr the pistn is initially filled with saturated liquid-apr mixture f water a specified pressure. Heat is transferred t the water until the lume increases by 0%. The initial and final temperature, the mass f the liquid when the pistn starts ming, and the wrk dne during the prcess are t be determined, and the prcess is t be shwn n a P- diagram. Assumptins The prcess is quasi-equilibrium. Analysis (a) Initially the system is a saturated mixture at 5 kpa pressure, and thus the initial temperature is T T sat@5 kpa The ttal initial lume is V 06.0 C m f f + mg g m Then the ttal and specific lumes at the final state are Thus, V.V m V m P 00 kpa 4.95 m 5 kg m T /kg m 7.6 C (b) When the pistn first starts ming, P 00 kpa and V V 4.7 m. The specific lume at this state is V /kg 4.7 m m /kg m 5 kg which is greater than g m /kg at 00 kpa. Thus n liquid is left in the cylinder when the pistn starts ming. (c) N wrk is dne during prcess - since V V. The pressure remains cnstant during prcess - and the wrk dne during this prcess is W b kj PdV P kpa m P H O 5 kg ( V V ) ( 00 kpa)( ) m 47.6 kj

5 E A spherical balln is initially filled with air at a specified state. The pressure inside is prprtinal t the square f the diameter. Heat is transferred t the air until the lume dubles. The wrk dne is t be determined. Assumptins Air is an ideal gas. The prcess is quasi-equilibrium. Prperties The gas cnstant f air is R Btu/lbm.R (Table A-E). Analysis The dependence f pressure n lume can be expressed as π π π V V V k kd P D P D D AIR 0 lbm 0 psia 800 R r, 6 PV P V k π Als, 587. V V P P and ( ) R 59 R T P P T T P T P V V V V Thus, ( ) Btu )R R)(59 (0 lbm)( Btu/lbm 5 ) ( 5 ) ( T T mr P P k d k Pd W b V V V V V V V π π

6 E EES Prblem 4-6E is recnsidered. Using the integratin feature, the wrk dne is t be determined and cmpared t the 'hand calculated' result. Analysis The prblem is sled using EES, and the slutin is gien belw. N m0"[lbm]" P_0"[psia]" T_800"[R]" V_*V_ R545"[ft-lbf/lbml-R]"/mlarmass(air)"[ft-lbf/lbm-R]" P_*Cnert(psia,lbf/ft^)*V_m*R*T_ V_4*pi*(D_/)^/"[ft^]" CP_/D_^N (D_/D_)^V_/V_ P_C*D_^N"[psia]" P_*Cnert(psia,lbf/ft^)*V_m*R*T_ PC*D^N*Cnert(psia,lbf/ft^)"[ft^]" V4*pi*(D/)^/"[ft^]" W_bundary_EESintegral(P,V,V_,V_)*cnert(ft-lbf,Btu)"[Btu]" W_bundary_HANDpi*C/(*(N+))*(D_^(N+)-D_^(N+))*cnert(ftlbf,Btu)*cnert(ft^,in^)"[Btu]" N W bundary [Btu] W bundary [Btu] N

7 A cylinder is initially filled with saturated R-4a apr at a specified pressure. The refrigerant is heated bth electrically and by heat transfer at cnstant pressure fr 6 min. The electric current is t be determined, and the prcess is t be shwn n a T- diagram. Assumptins The cylinder is statinary and thus the kinetic and ptential energy changes are negligible. The thermal energy stred in the cylinder itself and the wires is negligible. The cmpressin r expansin prcess is quasi-equilibrium. Analysis We take the cntents f the cylinder as the system. This is a clsed system since n mass enters r leaes. The energy balance fr this statinary clsed system can be expressed as 44 in ut Net energy transfer by heat, wrk, and mass in E + W e,in in E W in + W b, ut e,in m( h + (VI t) m( h Change in internal, kinetic, ptential, etc.energies U Esystem 44 (since KE PE 0) h ) h ) since U + W b H during a cnstant pressure quasiequilibrium prcess. The prperties f R-4a are (Tables A- thrugh A-) Substituting, P 40 kpa h hg@40kpa 47.8 kj/kg sat. apr P 40 kpa h 4.5 kj/kg T 70 C 00,000 VA + ( 0 V)( I )( 6 60 s) ( kg)( ) I.8 A T W e R-4a P cnst. 000 VA kj/kg kj/s

8 A cylinder is initially filled with saturated liquid-apr mixture f R-4a at a specified pressure. Heat is transferred t the cylinder until the refrigerant aprizes cmpletely at cnstant pressure. The initial lume, the wrk dne, and the ttal heat transfer are t be determined, and the prcess is t be shwn n a P- diagram. Assumptins The cylinder is statinary and thus the kinetic and ptential energy changes are negligible. The thermal energy stred in the cylinder itself is negligible. The cmpressin r expansin prcess is quasi-equilibrium. Analysis (a) Using prperty data frm R-4a tables (Tables A- thrugh A-), the initial lume f the refrigerant is determined t be P 00 kpa f , x 0.5 u f 8.8, m /kg g u fg 86. kj/kg u u f f + x + x u fg fg ( ) m /kg kj/kg R-4a 00 kpa V m ( 0. kg)( m /kg) m (b) The wrk dne during this cnstant pressure prcess is P 00 kpa g sat. apr u kpa g@00 kpa m /kg 4.48 kj/kg P W b,ut PdV P ( V V ) ( 0. kg)( 00 kpa)( ).97 kj mp( ) kj m /kg kpa m (c) We take the cntents f the cylinder as the system. This is a clsed system since n mass enters r leaes. The energy balance fr this statinary clsed system can be expressed as E Substituting, 44 in ut Net energy transfer by heat, wrk, and mass E in W b,ut in Change in internal, kinetic, ptential, etc.energies U m( u u ) + W Esystem 44 b,ut in (0. kg)( )kj/kg kj

9 A cylinder is initially filled with helium gas at a specified state. Helium is cmpressed plytrpically t a specified temperature and pressure. The heat transfer during the prcess is t be determined. Assumptins Helium is an ideal gas with cnstant specific heats. The cylinder is statinary and thus the kinetic and ptential energy changes are negligible. The thermal energy stred in the cylinder itself is negligible. 4 The cmpressin r expansin prcess is quasi-equilibrium. Prperties The gas cnstant f helium is R.0769 kpa.m /kg.k (Table A-). Als, c.56 kj/kg.k (Table A-). Analysis The mass f helium and the expnent n are determined t be P V m RT P V PV RT RT ( 50 kpa)( 0.5 m ) (.0769 kpa m /kg K)( 9 K) 0. kg T P 4 K 50 kpa V V 0.5 m T P 9 K 400 kpa n n P V PV P V P V Then the bundary wrk fr this plytrpic prcess can be determined frm W b,in n ( T ) P V P V mr T PdV n n (0. kg)(.0769 kj/kg K)(4 9)K n kj 0.64 m n.56 We take the cntents f the cylinder as the system. This is a clsed system since n mass enters r leaes. Taking the directin f heat transfer t be t the cylinder, the energy balance fr this statinary clsed system can be expressed as Ein Eut E 44 system 44 Net energy transfer by heat, wrk, and mass in + W b,in in Change in internal, kinetic, ptential, etc.energies U m( u m( u mc ( T u ) W u ) b,in T ) W Substituting, in (0. kg)(.56 kj/kg K)(4-9)K - (57. kj) -. kj The negatie sign indicates that heat is lst frm the system. b,in He PV n C 4- A cylinder and a rigid tank initially cntain the same amunt f an ideal gas at the same state. The temperature f bth systems is t be raised by the same amunt. The amunt f extra heat that must be transferred t the cylinder is t be determined. Analysis In the absence f any wrk interactins, ther than the bundary wrk, the H and U represent the heat transfer fr ideal gases fr cnstant pressure and cnstant lume prcesses, respectiely. Thus the extra heat that must be supplied t the air maintained at cnstant pressure is H U mc T mc T m( c c ) T mr T in, extra where R R u M p 8.4 kj / kml K 0.6 kj / kg K 5 kg / kml Substituting, in, extra ( kg)(0.6 kj/kg K)(5 K) 59.9 kj p IDEAL GAS V cnst. IDEAL GAS P cnst.

10 The heating f a passie slar huse at night is t be assisted by slar heated water. The length f time that the electric heating system wuld run that night with r withut slar heating are t be determined. Assumptins Water is an incmpressible substance with cnstant specific heats. The energy stred in the glass cntainers themseles is negligible relatie t the energy stred in water. The huse is maintained at C at all times. Prperties The density and specific heat f water at rm temperature are ρ kg/l and c 4.8 kj/kg C (Table A-). Analysis (a) The ttal mass f water is m w ρv ( kg/l)( 50 0 L) 000 kg Taking the cntents f the huse, including the water as ur system, the energy balance relatin can be written as r, E 44 in ut Net energy transfer by heat, wrk, and mass W& Substituting, It gies W E e,in ut Change in internal, kinetic, ptential, etc.energies U ( U ) ( U ) Esystem 44 water water mc( T e, in t ut [ mc( T T )] water + ( U ) T ) air water water 80 C (5 kj/s) t - (50,000 kj/h)(0 h) (000 kg)(4.8 kj/kg C)( - 80) C t 7,70 s 4.77 h C 50,000 kj/h (b) If the huse incrprated n slar heating, the energy balance relatin abe wuld simplify further t W & t 0 e, in ut Substituting, (5 kj/s) t - (50,000 kj/h)(0 h) 0 It gies t, s 9.6 h 4- An electric resistance heater is immersed in water. The time it will take fr the electric heater t raise the water temperature t a specified temperature is t be determined. Assumptins Water is an incmpressible substance with cnstant specific heats. The energy stred in the cntainer itself and the heater is negligible. Heat lss frm the cntainer is negligible. Prperties The density and specific heat f water at rm temperature are ρ kg/l and c 4.8 kj/kg C (Table A-). Analysis Taking the water in the cntainer as the system, the energy balance can be expressed as Substituting, E E 44 in ut Net energy transfer by heat, wrk, and mass W& W e,in e,in Change in internal, kinetic, ptential, etc.energies ( U ) t mc( T Esystem 44 water T ) water (800 J/s) t (40 kg)(480 J/kg C)(80-0) C Sling fr t gies t 557 s 9.9 min.55 h Resistance Heater Water

11 One tn f liquid water at 80 C is brught int a rm. The final equilibrium temperature in the rm is t be determined. Assumptins The rm is well insulated and well sealed. The thermal prperties f water and air are cnstant. Prperties The gas cnstant f air is R 0.87 kpa.m /kg.k (Table A-). The specific heat f water at rm temperature is c 4.8 kj/kg C (Table A-). Analysis The lume and the mass f the air in the rm are V 4x5x6 0 m³ m PV ( 00 kpa)( 0 m ) ( kpa m /kg K)( 95 K) air RT Taking the cntents f the rm, including the water, as ur system, the energy balance can be written as E 44 Change in internal, kinetic, ptential, etc.energies 4.7 kg ( U ) water + ( ) air in Eut Esystem 0 U U Net energy transfer by heat, wrk, and mass 44 r [ ( T T )] + [ mc ( T T )] 0 Substituting, It gies mc water air 4 m 5 m 6 m Water 80 C ( 000 kg)(4.80 kj/kg C)( T f 80) C + (4.7 kg)(0.78 kj/kg C)( T f ) C 0 T f 78.6 C where T f is the final equilibrium temperature in the rm. ROOM C 00 kpa Heat 4-5 A rm is t be heated by tn f ht water cntained in a tank placed in the rm. The minimum initial temperature f the water is t be determined if it t meet the heating requirements f this rm fr a 4-h perid. Assumptins Water is an incmpressible substance with cnstant specific heats. Air is an ideal gas with cnstant specific heats. The energy stred in the cntainer itself is negligible relatie t the energy stred in water. 4 The rm is maintained at 0 C at all times. 5 The ht water is t meet the heating requirements f this rm fr a 4-h perid. Prperties The specific heat f water at rm temperature is c 4.8 kj/kg C (Table A-). Analysis Heat lss frm the rm during a 4-h perid is lss (8000 kj/h)(4 h) 9,000 kj Taking the cntents f the rm, including the water, as ur system, the energy balance can be written as r Substituting, It gies E E 44 in ut Net energy transfer by heat, wrk, and mass - ut [mc(t - T )] water Esystem 44 Change in internal, kinetic, ptential, etc.energies -9,000 kj (000 kg)(4.8 kj/kg C)(0 - T ) T 65.9 C ut U where T is the temperature f the water when it is first brught int the rm. 0 ( U ) + ( U ) water air water 0 C 8000 kj/h

12 A sample f a fd is burned in a bmb calrimeter, and the water temperature rises by. C when equilibrium is established. The energy cntent f the fd is t be determined. Assumptins Water is an incmpressible substance with cnstant specific heats. Air is an ideal gas with cnstant specific heats. The energy stred in the reactin chamber is negligible relatie t the energy stred in water. 4 The energy supplied by the mixer is negligible. Prperties The specific heat f water at rm temperature is c 4.8 kj/kg C (Table A-). The cnstant lume specific heat f air at rm temperature is c 0.78 kj/kg C (Table A-). Analysis The chemical energy released during the cmbustin f the sample is transferred t the water as heat. Therefre, disregarding the change in the sensible energy f the reactin chamber, the energy cntent f the fd is simply the heat transferred t the water. Taking the water as ur system, the energy balance can be written as Ein Eut Esystem in U Net energy transfer by heat, wrk, and mass Change in internal, kinetic, ptential, etc.energies r in ( U ) water [ mc( T T )] water Substituting, in ( kg)(4.8 kj/kg C)(. C) 40. kj fr a -g sample. Then the energy cntent f the fd per unit mass is Water Reactin chamber 40. kj 000 g T. C 0,060 kj/kg g kg T make a rugh estimate f the errr inled in neglecting the thermal energy stred in the reactin chamber, we treat the entire mass within the chamber as air and determine the change in sensible internal energy: ( ) mc ( T T ) Fd [ ] ( 0.0 kg)( 0.78 kj/kg C)(. C) 0. kj U chamber chamber which is less than % f the internal energy change f water. Therefre, it is reasnable t disregard the change in the sensible energy cntent f the reactin chamber in the analysis. 4-7 A man drinks ne liter f cld water at C in an effrt t cl dwn. The drp in the aerage bdy temperature f the persn under the influence f this cld water is t be determined. Assumptins Thermal prperties f the bdy and water are cnstant. The effect f metablic heat generatin and the heat lss frm the bdy during that time perid are negligible. Prperties The density f water is ery nearly kg/l, and the specific heat f water at rm temperature is C 4.8 kj/kg C (Table A-). The aerage specific heat f human bdy is gien t be.6 kj/kg. C. Analysis. The mass f the water is m ρv kg/l L w ( )( ) kg We take the man and the water as ur system, and disregard any heat and mass transfer and chemical reactins. Of curse these assumptins may be acceptable nly fr ery shrt time perids, such as the time it takes t drink the water. Then the energy balance can be written as Ein Eut E 44 system 44 Net energy transfer by heat, wrk, and mass Change in internal, kinetic, ptential, etc.energies 0 U U + U bdy water r [ mc ( T T )] + [ mc ( T T )] 0 bdy water Substituting ( 68 kg)(.6 kj/kg C)( T f 9) C + ( kg)(4.8 kj/kg C)( T f ) C 0 It gies T f 8.4 C Then T C Therefre, the aerage bdy temperature f this persn shuld drp abut half a degree celsius.

13 A 0.-L glass f water at 0 C is t be cled with ice t 5 C. The amunt f ice r cld water that needs t be added t the water is t be determined. Assumptins Thermal prperties f the ice and water are cnstant. Heat transfer t the glass is negligible. There is n stirring by hand r a mechanical deice (it will add energy). Prperties The density f water is kg/l, and the specific heat f water at rm temperature is c 4.8 kj/kg C (Table A-). The specific heat f ice at abut 0 C is c. kj/kg C (Table A-). The melting temperature and the heat f fusin f ice at atm are 0 C and.7 kj/kg,. Analysis (a) The mass f the water is m w ρv ( kg/l)(0. L) 0. kg Ice cubes 0 C We take the ice and the water as ur system, and disregard any heat and mass transfer. This is a reasnable assumptin since the time perid f the prcess is ery shrt. Then the energy balance can be written as E E 44 in ut Net energy transfer by heat, wrk, and mass Change in internal, kinetic, ptential, etc.energies 0 U 0 U ice Esystem 44 + U water if [ mc( 0 C T ) slid + mh + mc( T 0 C) liquid ] ice + [ mc( T T )] water Nting that T, ice 0 C and T 5 C and substituting gies m[0 +.7 kj/kg + (4.8 kj/kg C)(5-0) C] + (0. kg)(4.8 kj/kg C)(5-0) C 0 m kg 6.4 g (b) When T, ice -8 C instead f 0 C, substituting gies m[(. kj/kg C)[0-(-8)] C +.7 kj/kg + (4.8 kj/kg C)(5-0) C] m kg 4.7 g 0 Water 0 C 0. L + (0. kg)(4.8 kj/kg C)(5-0) C 0 Cling with cld water can be handled the same way. All we need t d is replace the terms fr ice by a term fr cld water at 0 C: Substituting, It gies ( U ) cldwater + ( U ) water 0 [ mc( T T )] + [ mc( T T )] 0 cldwater water [m cld water (4.8 kj/kg C)(5-0) C] + (0. kg)(4.8 kj/kg C)(5-0) C 0 m 0.6 kg 600 g Discussin Nte that this is 7 times the amunt f ice needed, and it explains why we use ice instead f water t cl drinks. Als, the temperature f ice des nt seem t make a significant difference.

14 EES Prblem 4-8 is recnsidered. The effect f the initial temperature f the ice n the final mass f ice required as the ice temperature aries frm -0 C t 0 C is t be inestigated. The mass f ice is t be pltted against the initial temperature f ice. Analysis The prblem is sled using EES, and the slutin is gien belw. "Knwns" rh_water "[kg/l]" V 0. "[L]" T ice 0"[C]" T_ 0"[C]" T_ 5"[C]" C_ice. "[kj/kg-c]" C_water 4.8 "[kj/kg-c]" h_if.7 "[kj/kg]" T CldWater 0"[C]" "The mass f the water is:" m_water rh_water*v "[kg]" "The system is the water plus the ice. Assume a shrt time perid and neglect any heat and mass transfer. The energy balance becmes:" E_in - E_ut DELTAE_sys "[kj]" E_in 0 "[kj]" E_ut 0"[kJ]" DELTAE_sys DELTAU_water+DELTAU_ice"[kJ]" DELTAU_water m_water*c_water*(t_ - T_)"[kJ]" DELTAU_ice DELTAU_slid_ice+DELTAU_melted_ice"[kJ]" DELTAU_slid_ice m_ice*c_ice*(0-t ice) + m_ice*h_if"[kj]" DELTAU_melted_icem_ice*C_water*(T_ - 0)"[kJ]" m_ice_gramsm_ice*cnert(kg,g)"[g]" "Cling with Cld Water:" DELTAE_sys DELTAU_water+DELTAU_CldWater"[kJ]" DELTAU_water m_water*c_water*(t CldWater - T_)"[kJ]" DELTAU_CldWater m_cldwater*c_water*(t CldWater - T CldWater)"[kJ]" m_cldwater_gramsm_cldwater*cnert(kg,g)"[g]" T,ice [g] [C] m ice,grams m ice,grams [g] T,ice [C]

15 A - tn (000 kg) f water is t be cled in a tank by puring ice int it. The final equilibrium temperature in the tank is t be determined. Assumptins Thermal prperties f the ice and water are cnstant. Heat transfer t the water tank is negligible. There is n stirring by hand r a mechanical deice (it will add energy). Prperties The specific heat f water at rm temperature is c 4.8 kj/kg C, and the specific heat f ice at abut 0 C is c. kj/kg C (Table A-). The melting temperature and the heat f fusin f ice at atm are 0 C and.7 kj/kg. Analysis We take the ice and the water as ur system, and disregard any heat transfer between the system and the surrundings. Then the energy balance fr this prcess can be written as Ein Eut E Net energy transfer by heat, wrk, and mass 0 U system Change in internal, kinetic, ptential, etc. energies 0 U + U ice water if [ mc( 0 C T ) slid + mh + mc( T 0 C) liquid ] ice + [ mc( T T )] water Substituting, It gies 0 WATER tn ( 80 kg){(. kj / kg C)[0 (-5)] C +.7 kj / kg + (4.8 kj / kg C)( T 0) C} + ( 000 kg)( 4.8 kj / kg C)( T 0) C 0 T.4 C which is the final equilibrium temperature in the tank. ice -5 C 80 kg

16 An insulated cylinder initially cntains a saturated liquid-apr mixture f water at a specified temperature. The entire apr in the cylinder is t be cndensed isthermally by adding ice inside the cylinder. The amunt f ice that needs t be added is t be determined. Assumptins Thermal prperties f the ice are cnstant. The cylinder is well-insulated and thus heat transfer is negligible. There is n stirring by hand r a mechanical deice (it will add energy). Prperties The specific heat f ice at abut 0 C is c. kj/kg C (Table A-). The melting temperature and the heat f fusin f ice at atm are gien t be 0 C and.7 kj/kg. Analysis We take the cntents f the cylinder (ice and saturated water) as ur system, which is a clsed system. Nting that the temperature and thus the pressure remains cnstant during this phase change prcess and thus W b + U H, the energy balance fr this system can be written as E 44 in ut Net energy transfer by heat, wrk, and mass H ice E W + H b,in water Change in internal, kinetic, ptential, etc.energies U 0 Esystem 44 H 0 if [ mc( 0 C T ) slid + mh + mc( T 0 C) liquid ] ice + [ m( h h )] water The prperties f water at 0 C are (Table A-4) Then, h h h f f , f h f + x + x h 50.8, fg fg h C m 0.89 m /kg g h fg kj.kg ( ) kj/kg 50.8 kj/kg V 0.0 m steam 0.79 m /kg kg Nting that T, ice 0 C and T 0 C and substituting gies m ice 0 C WATER 0.0 m /kg 0 C m[0 +.7 kj/kg + (4.8 kj/kg C)(0-0) C] + ( kg)( ) kj/kg 0 m kg 9.4 g ice

17 The cylinder f a steam engine initially cntains saturated apr f water at 00 kpa. The cylinder is cled by puring cld water utside f it, and sme f the steam inside cndenses. If the pistn is stuck at its initial psitin, the frictin frce acting n the pistn and the amunt f heat transfer are t be determined. Assumptins The deice is air-tight s that n air leaks int the cylinder as the pressure drps. Analysis We take the cntents f the cylinder (the saturated liquid-apr mixture) as the system, which is a clsed system. Nting that the lume remains cnstant during this phase change prcess, the energy balance fr this system can be expressed as E E 44 in ut Net energy transfer by heat, wrk, and mass ut Esystem 44 Change in internal, kinetic, ptential, etc.energies U m( u u ) The saturatin prperties f water at 00 kpa and at 0 C are (Tables A-4 and A-5) P 00 kpa T 0 C Then, and u u P u u f P sat@0 C g@00 kpa g@00 kpa f u f m /kg, g kj/kg, u g.694 m /kg kj/kg f m /kg, g.879 m /kg u f 5.7 kj/kg, u fg 90. kj/kg P kpa sat kpa.694 m /kg kj/kg V 0.05 m m kg.694 m /kg + x u fg x f fg kj/kg Cld water Steam 0.05 m 00 kpa The frictin frce that deelps at the pistn-cylinder interface balances the frce acting n the pistn, and is equal t F A( P P ) (0. m 000 N/m )( )kPa kpa The heat transfer is determined frm the energy balance t be ut m( u u) (0.095 kg)( )kJ/kg 66.8 kj 9575 N

18 Water is biled at sea leel ( atm pressure) in a cffee maker, and half f the water eaprates in 5 min. The pwer rating f the electric heating element and the time it takes t heat the cld water t the biling temperature are t be determined. Assumptins The electric pwer cnsumptin by the heater is cnstant. Heat lsses frm the cffee maker are negligible. Prperties The enthalpy f aprizatin f water at the saturatin temperature f 00 C is h fg 56.4 kj/kg (Table A-4). At an aerage temperature f (00+8)/ 59 C, the specific heat f water is c 4.8 kj/kg. C, and the density is abut kg/l (Table A-). Analysis The density f water at rm temperature is ery nearly kg/l, and thus the mass f L water at 8 C is nearly kg. Nting that the enthalpy f aprizatin represents the amunt f energy needed t aprize a liquid at a specified temperature, the amunt f electrical energy needed t aprize 0.5 kg f water in 5 min is W e W& t mh e fg W& e mh t fg (0.5 kg)(56.4 kj/kg) 0.75 kw (5 60 s) Therefre, the electric heater cnsumes (and transfers t water) 0.75 kw f electric pwer. Water 00 C Heater W e Nting that the specific heat f water at the aerage temperature f (8+00)/ 59 C is c 4.8 kj/kg C, the time it takes fr the entire water t be heated frm 8 C t 00 C is determined t be W e W& t mc T e mc T t W& e ( kg)(4.8 kj/kg C)(00 8) C 456 s 7.60 min 0.75 kj/s Discussin We can als sle this prblem using f data (instead f density), and h f data instead f specific heat. At 00 C, we hae f m /kg and h f 49.7 kj/kg. At 8 C, we hae h f kj/kg (Table A-4). The tw results will be practically the same.

19 Tw rigid tanks that cntain water at different states are cnnected by a ale. The ale is pened and the tw tanks cme t the same state at the temperature f the surrundings. The final pressure and the amunt f heat transfer are t be determined. Assumptins The tanks are statinary and thus the kinetic and ptential energy changes are zer. The tank is insulated and thus heat transfer is negligible. There are n wrk interactins. Analysis We take the entire cntents f the tank as the system. This is a clsed system since n mass enters r leaes. Nting that the lume f the system is cnstant and thus there is n bundary wrk, the energy balance fr this statinary clsed system can be expressed as E E 44 in ut Net energy transfer by heat, wrk, and mass ut ut Change in internal, kinetic, ptential, etc.energies U ( U ) [ U [ m Esystem 44, A+ B,ttal u A U + ( U ), A U ( m u, B ) B A ] (sincew KE PE 0) ( m u The prperties f water in each tank are (Tables A-4 thrugh A-6) Tank A: Tank B: P 400 kpa f x 0.80 u u, A, A u f f + x + x u fg fg P 00 kpa, T 50 C u, m m m, A, B t V A m, A V B, B, A Vt m t , f 604., m 0.5 m kg.989 m /kg 0.7 m 0.77 m kg T 5 C f 0.77 m /kg u g u 7.4 kj/kg fg ) B m /kg 0. m kg m /kg + m, B B B ] kj/kg + [ 0.8 ( ) ] ( ) 6. kj/kg /kg kg , f 04.8, /kg 4.40 m /kg u g fg 04. kj/kg H O 400 kpa A m /kg B H O 00 kpa Thus at the final state the system will be a saturated liquid-apr mixture since f < < g. Then the final pressure must be P P 5 C.7 kpa Als, f x u Substituting, u f fg + x u fg ( ) 4.65 kj/kg ut [(0.957)(4.65) (0.540)(6.) (0.470)(7.4)] 70 kj

20 EES Prblem 4-4 is recnsidered. The effect f the enirnment temperature n the final pressure and the heat transfer as the enirnment temperature aries frm 0 C t 50 C is t be inestigated. The final results are t be pltted against the enirnment temperature. Analysis The prblem is sled using EES, and the slutin is gien belw. "Knwns" Vl_A0. [m^] P_A[]400 [kpa] x_a[]0.8 T_B[]50 [C] P_B[]00 [kpa] Vl_B0.5 [m^] T_final5 [C] "T_final T_surrundings. T d the parametric study r t sle the prblem when _ut 0, place this statement in {}." {_ut0 [kj]} "T determine the surrundings temperature that makes _ut 0, reme the {} and resle the prblem." "Slutin" "Cnseratin f Energy fr the cmbined tanks:" E_in-E_utDELTAE E_in0 E_ut_ut DELTAEm_A*(u_A[]-u_A[])+m_B*(u_B[]-u_B[]) m_avl_a/_a[] m_bvl_b/_b[] Fluid$'Steam_IAPWS' u_a[]intenergy(fluid$,pp_a[], xx_a[]) _A[]lume(Fluid$,PP_A[], xx_a[]) T_A[]temperature(Fluid$,PP_A[], xx_a[]) u_b[]intenergy(fluid$,pp_b[],tt_b[]) _B[]lume(Fluid$,PP_B[],TT_B[]) "At the final state the steam has unifrm prperties thrugh ut the entire system." u_b[]u_final u_a[]u_final m_finalm_a+m_b Vl_finalVl_A+Vl_B _finalvl_final/m_final u_finalintenergy(fluid$,tt_final, _final) P_finalpressure(Fluid$,TT_final, _final) P final [kpa] ut [kj] T final [C]

21 J ] [k t u T final [C] 5 P a] [k a l P fin T final [C]

22 A rigid tank filled with air is cnnected t a cylinder with zer clearance. The ale is pened, and air is allwed t flw int the cylinder. The temperature is maintained at 0 C at all times. The amunt f heat transfer with the surrundings is t be determined. Assumptins Air is an ideal gas. The kinetic and ptential energy changes are negligible, ke pe 0. There are n wrk interactins inled ther than the bundary wrk. Prperties The gas cnstant f air is R 0.87 kpa.m /kg.k (Table A-). Analysis We take the entire air in the tank and the cylinder t be the system. This is a clsed system since n mass crsses the bundary f the system. The energy balance fr this clsed system can be expressed as E 44 in ut Net energy transfer by heat, wrk, and mass E in W b,ut in Change in internal, kinetic, ptential, etc.energies U m( u W b,ut Esystem 44 u ) 0 since u u(t) fr ideal gases, and thus u u when T T. The initial lume f air is PV PV P T 400 kpa Air T 0 C V V (0.4 m ) 0.80 T m T P T 00 kpa The pressure at the pistn face always remains cnstant at 00 kpa. Thus the bundary wrk dne during this prcess is W kj ( ) (00 kpa)( )m PdV P V V kpa m b, ut Therefre, the heat transfer is determined frm the energy balance t be W b, ut in 80 kj 80 kj

23 A well-insulated rm is heated by a steam radiatr, and the warm air is distributed by a fan. The aerage temperature in the rm after 0 min is t be determined. Assumptins Air is an ideal gas with cnstant specific heats at rm temperature. The kinetic and ptential energy changes are negligible. The air pressure in the rm remains cnstant and thus the air expands as it is heated, and sme warm air escapes. Prperties The gas cnstant f air is R 0.87 kpa.m /kg.k (Table A-). Als, c p.005 kj/kg.k fr air at rm temperature (Table A-). Analysis We first take the radiatr as the system. This is a clsed system since n mass enters r leaes. The energy balance fr this clsed system can be expressed as E E 44 in ut Net energy transfer by heat, wrk, and mass ut ut Change in internal, kinetic, ptential, etc.energies U m( u m( u Esystem 44 u ) u ) Using data frm the steam tables (Tables A-4 thrugh A-6), sme prperties are determined t be P 00 kpa m /kg T 00 C u kj/kg P 00 kpa f , (sincew KE PE 0) ( ) u 47.40, u 088. kj/kg x u f u f V 0.05 m m 0.09 kg m /kg f fg + x u.694 m /kg fg g fg kj/kg Substituting, ut (0.09 kg)( )kJ/kg.58 kj The lume and the mass f the air in the rm are V m and m PV ( 00 kpa)( 80 m ) ( kpa m /kg K)( 8 K) air RT The amunt f fan wrk dne in 0 min is W W& t (0.0 kj/s)(0 60 s) fan, in fan,in 6kJ 98.5 kg 0 C 4 m 4 m 5 m Steam radiatr We nw take the air in the rm as the system. The energy balance fr this clsed system is expressed as in + W fan,in E in in E W + W ut b,ut fan,in E U system H mc ( T T ) p since the bundary wrk and U cmbine int H fr a cnstant pressure expansin r cmpressin prcess. It can als be expressed as & + W & ) t mcp ( T T ) Substituting, ( in fan,in,ag (.58 kj) + (6 kj) (98.5 kg)(.005 kj/kg C)(T - 0) C which yields T. C Therefre, the air temperature in the rm rises frm 0 C t. C in 0 min.

24 An insulated cylinder is diided int tw parts. One side f the cylinder cntains N gas and the ther side cntains He gas at different states. The final equilibrium temperature in the cylinder when thermal equilibrium is established is t be determined fr the cases f the pistn being fixed and ming freely. Assumptins Bth N and He are ideal gases with cnstant specific heats. The energy stred in the cntainer itself is negligible. The cylinder is well-insulated and thus heat transfer is negligible. Prperties The gas cnstants and the cnstant lume specific heats are R kpa.m /kg.k is c 0.74 kj/kg C fr N, and R.0769 kpa.m /kg.k is c.56 kj/kg C fr He (Tables A- and A-) Analysis The mass f each gas in the cylinder is m m N He P V RT P V RT N He ( 500 kpa)( m ) ( kpa m /kg K)( 5 K) ( 500 kpa)( m ) (.0769 kpa m /kg K)( 98 K) 4.77 kg kg N m 500 kpa 80 C Taking the entire cntents f the cylinder as ur system, the st law relatin can be written as Substituting, It gies E 44 in ut Net energy transfer by heat, wrk, and mass 0 U 0 [ mc E ( U ) + ( U ) ( T N T )] Change in internal, kinetic, ptential, etc.energies N He + [ mc Esystem 44 ( T T )] He ( 4.77 kg)( 0.74 kj/kg C)( T f 80) C + ( kg)(.56 kj/kg C)( T 5) C 0 T f 57. C where T f is the final equilibrium temperature in the cylinder. f He m 500 kpa 5 C The answer wuld be the same if the pistn were nt free t me since it wuld effect nly pressure, and nt the specific heats. Discussin Using the relatin PV NR u T, it can be shwn that the ttal number f mles in the cylinder is kml, and the final pressure is 50.6 kpa.

25 An insulated cylinder is diided int tw parts. One side f the cylinder cntains N gas and the ther side cntains He gas at different states. The final equilibrium temperature in the cylinder when thermal equilibrium is established is t be determined fr the cases f the pistn being fixed and ming freely. Assumptins Bth N and He are ideal gases with cnstant specific heats. The energy stred in the cntainer itself, except the pistn, is negligible. The cylinder is well-insulated and thus heat transfer is negligible. 4 Initially, the pistn is at the aerage temperature f the tw gases. Prperties The gas cnstants and the cnstant lume specific heats are R kpa.m /kg.k is c 0.74 kj/kg C fr N, and R.0769 kpa.m /kg.k is c.56 kj/kg C fr He (Tables A- and A- ). The specific heat f cpper pistn is c 0.86 kj/kg C (Table A-). Analysis The mass f each gas in the cylinder is m m N He P V RT P V RT N He ( 500 kpa)( m ) ( kpa m /kg K)( 5 K) ( 500 kpa)( m ) (.0769 kpa m /kg K)( 5 K) 4.77 kg kg N m 500 kpa 80 C Taking the entire cntents f the cylinder as ur system, the st law relatin can be written as where Substituting, It gies E 44 in ut Net energy transfer by heat, wrk, and mass 0 U 0 [ mc E ( U ) + ( U ) + ( U ) ( T N T )] Change in internal, kinetic, ptential, etc.energies N He + [ mc Esystem 44 ( T T )] Cu He + [ mc( T T, Cu (80 + 5) / 5.5 C T )] ( 4.77 kg)( 0.74 kj/kg C)( Tf 80) C + ( kg)(.56 kj/kg C)( Tf 5) + ( 5.0 kg)( 0.86 kj/kg C)( T 5.5) C 0 T f 56.0 C where T f is the final equilibrium temperature in the cylinder. f Cu C He m 500 kpa 5 C Cpper The answer wuld be the same if the pistn were nt free t me since it wuld effect nly pressure, and nt the specific heats.

26 EES Prblem 4-9 is recnsidered. The effect f the mass f the cpper pistn n the final equilibrium temperature as the mass f pistn aries frm kg t 0 kg is t be inestigated. The final temperature is t be pltted against the mass f pistn. Analysis The prblem is sled using EES, and the slutin is gien belw. "Knwns:" R_u8.4 [kj/kml-k] V_N[] [m^] C_N0.74 [kj/kg-k] "Frm Table A-(a) at 7C" R_N0.968 [kj/kg-k] "Frm Table A-(a)" T_N[]80 [C] P_N[]500 [kpa] V_He[] [m^] C_He.56 [kj/kg-k] "Frm Table A-(a) at 7C" T_He[]5 [C] P_He[]500 [kpa] R_He.0769 [kj/kg-k] "Frm Table A-(a)" m_pist5 [kg] C_Pist0.86 [kj/kg-k] "Use Cp fr Cpper frm Table A-(b) at 7C" "Slutin:" "mass calculatins:" P_N[]*V_N[]m_N*R_N*(T_N[]+7) P_He[]*V_He[]m_He*R_He*(T_He[]+7) "The entire cylinder is cnsidered t be a clsed system, neglecting the pistn." "Cnseratin f Energy fr the clsed system:" "E_in - E_ut DELTAE_negPist, we neglect DELTA KE and DELTA PE fr the cylinder." E_in - E_ut DELTAE_neglPist E_in 0 [kj] E_ut 0 [kj] "At the final equilibrium state, N and He will hae a cmmn temperature." DELTAE_neglPist m_n*c_n*(t neglpist-t_n[])+m_he*c_he*(t neglpist- T_He[]) "The entire cylinder is cnsidered t be a clsed system, including the pistn." "Cnseratin f Energy fr the clsed system:" "E_in - E_ut DELTAE_withPist, we neglect DELTA KE and DELTA PE fr the cylinder." E_in - E_ut DELTAE_withPist "At the final equilibrium state, N and He will hae a cmmn temperature." DELTAE_withPist m_n*c_n*(t withpist-t_n[])+m_he*c_he*(t withpist- T_He[])+m_Pist*C_Pist*(T withpist-t_pist[]) T_Pist[](T_N[]+T_He[])/ "Ttal lume f gases:" V_ttalV_N[]+V_He[] "Final pressure at equilibrium:" "Neglecting effect f pistn, P_ is:" P neglpist*v_ttaln_ttal*r_u*(t neglpist+7) "Including effect f pistn, P_ is:" N_ttalm_N/mlarmass(nitrgen)+m_He/mlarmass(Helium) P withpist*v_ttaln_ttal*r_u*(t withpist+7)

27 4-90 m Pist [kg] T,neglPist [C] T,withPist [C] Withut Pistn T [C] With Pistn Mass f Pistn [kg]

28 An insulated rigid tank initially cntains saturated liquid water and air. An electric resistr placed in the tank is turned n until the tank cntains saturated water apr. The lume f the tank, the final temperature, and the pwer rating f the resistr are t be determined. Assumptins The tank is statinary and thus the kinetic and ptential energy changes are zer. There are n wrk interactins. Prperties The initial prperties f steam are (Table A-4) T 00 C m /kg x 0 u kj/kg Analysis (a) We take the cntents f the tank as the system. This is a clsed system since n mass enters r leaes. Nting that the lume f the system is cnstant and thus there is n bundary wrk, the energy balance fr this statinary clsed system can be expressed as E E 44 in ut Net energy transfer by heat, wrk, and mass W e,in Esystem 44 Change in internal, kinetic, ptential, etc.energies U m( u u ) The initial water lume and the tank lume are V m (.4 kg)( V tank m (since KE PE 0) m /kg) m m (b) Nw, the final state can be fixed by calculating specific lume V m The final state prperties are (c) Substituting, m m /kg.4 kg m /kg T 7. C x 0.5 kj/kg u W e, in (.4 kg)( )kJ/kg 89 kj Finally, the pwer rating f the resistr is W & e,in We,in t 89 kj.576 kw 0 60 s Air Water.4 kg, 00 C W e

29 A pistn-cylinder deice cntains an ideal gas. An external shaft cnnected t the pistn exerts a frce. Fr an isthermal prcess f the ideal gas, the amunt f heat transfer, the final pressure, and the distance that the pistn is displaced are t be determined. Assumptins The kinetic and ptential energy changes are negligible, W ke pe 0. The frictin between the pistn and the cylinder is negligible. Analysis (a) We take the ideal gas in the cylinder t be the system. This is a clsed system since n mass crsses the system bundary. The energy balance fr this statinary clsed system can be expressed as E 44 in ut Net energy transfer by heat, wrk, and mass W E b,in W ut b,in Change in internal, kinetic, ptential, etc. energies U ut Esystem 44 ideal gas mc ( T T ) ideal gas ) 0 Thus, the amunt f heat transfer is equal t the bundary wrk input in ut W b, 0.kJ (sincet T GAS bar 4 C and KE PE 0) (b) The relatin fr the isthermal wrk f an ideal gas may be used t determine the final lume in the cylinder. But we first calculate initial lume Then, V πd π (0. m) L (0. m) m 4 4 V Wb,in PV ln V V 0.kJ (00 kpa)(0.006 m )ln V m m The final pressure can be determined frm ideal gas relatin applied fr an isthermal prcess P V PV (00 kpa)(0.006 m ) P ( m ) P 55.6 kpa (c) The final psitin f the pistn and the distance that the pistn is displaced are πd π (0. m) V L m L L 4 4 L L L m 7.cm 0.85 m

30 A pistn-cylinder deice with a set f stps cntains superheated steam. Heat is lst frm the steam. The pressure and quality (if mixture), the bundary wrk, and the heat transfer until the pistn first hits the stps and the ttal heat transfer are t be determined. Assumptins The kinetic and ptential energy changes are negligible, ke pe 0. The frictin between the pistn and the cylinder is negligible. Analysis (a) We take the steam in the cylinder t be the system. This is a clsed system since n mass crsses the system bundary. The energy balance fr this statinary clsed system can be expressed as E E 44 in ut Net energy transfer by heat, wrk, and mass W b,in ut Change in internal, kinetic, ptential, etc. energies U Esystem 44 (since KE PE 0) Denting when pistn first hits the stps as state () and the final state as (), the energy balance relatins may be written as W W b,in b,in ut,- ut,- m( u m( u - u - u The prperties f steam at arius states are (Tables A-4 thrugh A-6) T sat@.5 MPa T T + T 4.56 C sat ) ) C P.5 MPa m /kg T C u 67. kj/kg P.5 MPa P m /kg x 0 u kj/kg x m /kg P 555 kpa T 00 C u kj/kg Steam 0.5 kg.5 MPa (b) Nting that the pressure is cnstant until the pistn hits the stps during which the bundary wrk is dne, it can be determined frm its definitin as Wb, in mp ( ) (0.5 kg)(500 kpa)( )m (c) Substituting int energy balance relatins, (d) ut,- ut,- 9.9 kj (0.5 kg)( ) kj/kg 65.7 kj 9.9 kj (0.5 kg)( ) kj/kg 94.8 kj 9.9 kj

31 An insulated rigid tank is diided int tw cmpartments, each cmpartment cntaining the same ideal gas at different states. The tw gases are allwed t mix. The simplest expressin fr the mixture temperature in a specified frmat is t be btained. Analysis We take the bth cmpartments tgether as the system. This is a clsed system since n mass enters r leaes. The energy balance fr this statinary clsed system can be expressed as E 44 in ut Net energy transfer by heat, wrk, and mass ( m E + m ) T and, m m + m Change in internal, kinetic, ptential, etc.energies 0 U 0 m c Esystem 44 ( T (since W KE PE 0) m T + m T Sling fr final temperature, we find m m T + T T m m T ) + m c ( T T ) m T m T 4-45 A relatin fr the explsie energy f a fluid is gien. A relatin is t be btained fr the explsie energy f an ideal gas, and the alue fr air at a specified state is t be ealuated. Prperties The specific heat rati fr air at rm temperature is k.4. Analysis The explsie energy per unit lume is gien as e explsin u u Fr an ideal gas, u - u c (T - T ) and thus Substituting, which is the desired result. c p c R RT P c c R c c p c p / c ( T T ) k c P T eexplsin RT / P k T Using the relatin abe, the ttal explsie energy f 0 m³ f air at 5 MPa and 00 C when the surrundings are at 0 C is determined t be E ( 5000 kpa)( 0 m ) 9 K kj 5,69 kj P V T k T.4 7 K kpa m explsin V eexplsin

32 Using the relatin fr explsie energy gien in the preius prblem, the explsie energy f steam and its TNT equialent at a specified state are t be determined. Assumptins Steam cndenses and becmes a liquid at rm temperature after the explsin. Prperties The prperties f steam at the initial and the final states are (Table A-4 thrugh A-6) P 0 MPa 0.08 m /kg T 500 C u kj/kg T 5 C u u 04.8 kj/kg C Cmp. liquid Analysis The mass f the steam is V 0 m m kg 0.08 m /kg Then the ttal explsie energy f the steam is determined frm E explsie which is equialent t 5 C ( u ) ( kg)( ),79,46 kj m u kj/kg,79,46 kj 55.8 kg f TNT 50 kj/kg f TNT STEAM 0 MPa 500 C

33 4-96 Fundamentals f Engineering (FE) Exam Prblems 4-47 A rm is filled with saturated steam at 00 C. Nw a 5-kg bwling ball at 5 C is brught t the rm. Heat is transferred t the ball frm the steam, and the temperature f the ball rises t 00 C while sme steam cndenses n the ball as it lses heat (but it still remains at 00 C). The specific heat f the ball can be taken t be.8 kj/kg. C. The mass f steam that cndensed during this prcess is (a) 80 g (b) 8 g (c) 99 g (d) 5 g (e) 405 g Answer (c) 99 g Slutin Sled by EES Sftware. Slutins can be erified by cpying-and-pasting the fllwing lines n a blank EES screen. (Similar prblems and their slutins can be btained easily by mdifying numerical alues). m_ball5 "kg" T00 "C" T5 "C" T00 "C" Cp.8 "kj/kg.c" m_ball*cp*(t-t) m_steam*h_fg "kj" h_fenthalpy(steam_iapws, x0,tt) h_genthalpy(steam_iapws, x,tt) h_fgh_g-h_f "Sme Wrng Slutins with Cmmn Mistakes:" Wm_steam*h_g "Using h_g" Wm_steam*4.8*(T-T) "Using m*c*deltat fr water" Wm_steam*h_f "Using h_f" 4-48 A frictinless pistn-cylinder deice and a rigid tank cntain kml f an ideal gas at the same temperature, pressure and lume. Nw heat is transferred, and the temperature f bth systems is raised by 0 C. The amunt f extra heat that must be supplied t the gas in the cylinder that is maintained at cnstant pressure is (a) 0 kj (b) 4 kj (c) 8 kj (d) 0 kj (e) 66 kj Answer (e) 66 kj Slutin Sled by EES Sftware. Slutins can be erified by cpying-and-pasting the fllwing lines n a blank EES screen. (Similar prblems and their slutins can be btained easily by mdifying numerical alues). "Nte that Cp-CR, and thus _diffm*r*dtn*ru*dt" N "kml" Ru8.4 "kj/kml.k" T_change0 _diffn*ru*t_change "Sme Wrng Slutins with Cmmn Mistakes:" W_diff0 "Assuming they are the same" W_diffRu*T_change "Nt using mle numbers" W_diffRu*T_change/N "Diiding by N instead f multiplying" W4_diffN*Rair*T_change; Rair0.87 "using Ru instead f R"

34 The specific heat f a material is gien in a strange unit t be C.60 kj/kg. F. The specific heat f this material in the SI units f kj/kg. C is (a).00 kj/kg. C (b).0 kj/kg. C (c).60 kj/kg. C (d) 4.80 kj/kg. C (e) 6.48 kj/kg. C Answer (e) 6.48 kj/kg. C Slutin Sled by EES Sftware. Slutins can be erified by cpying-and-pasting the fllwing lines n a blank EES screen. (Similar prblems and their slutins can be btained easily by mdifying numerical alues). C.60 "kj/kg.f" C_SIC*.8 "kj/kg.c" "Sme Wrng Slutins with Cmmn Mistakes:" W_CC "Assuming they are the same" W_CC/.8 "Diiding by.8 instead f multiplying" 4-50 A -m rigid tank cntains nitrgen gas at 500 kpa and 00 K. Nw heat is transferred t the nitrgen in the tank and the pressure f nitrgen rises t 800 kpa. The wrk dne during this prcess is (a) 500 kj (b) 500 kj (c) 0 kj (d) 900 kj (e) 400 kj Answer (b) 0 kj Slutin Sled by EES Sftware. Slutins can be erified by cpying-and-pasting the fllwing lines n a blank EES screen. (Similar prblems and their slutins can be btained easily by mdifying numerical alues). V "m^" P500 "kpa" T00 "K" P800 "kpa" W0 "since cnstant lume" "Sme Wrng Slutins with Cmmn Mistakes:" R0.97 W_WV*(P-P) "Using WV*DELTAP" W_WV*P W_WV*P W4_WR*T*ln(P/P)

35 A 0.8-m cylinder cntains nitrgen gas at 600 kpa and 00 K. Nw the gas is cmpressed isthermally t a lume f 0. m. The wrk dne n the gas during this cmpressin prcess is (a) 746 kj (b) 0 kj (c) 40 kj (d) 998 kj (e) 890 kj Answer (d) 998 kj Slutin Sled by EES Sftware. Slutins can be erified by cpying-and-pasting the fllwing lines n a blank EES screen. (Similar prblems and their slutins can be btained easily by mdifying numerical alues). R8.4/8 V0.8 "m^" V0. "m^" P600 "kpa" T00 "K" P*Vm*R*T Wm*R*T* ln(v/v) "cnstant temperature" "Sme Wrng Slutins with Cmmn Mistakes:" W_WR*T* ln(v/v) "Frgetting m" W_WP*(V-V) "Using V*DeltaP" P*V/TP*V/T W_W(V-V)*(P+P)/ "Using P_ae*Delta V" W4_WP*V-P*V "Using WPV-PV" 4-5 A well-sealed rm cntains 60 kg f air at 00 kpa and 5 C. Nw slar energy enters the rm at an aerage rate f 0.8 kj/s while a 0-W fan is turned n t circulate the air in the rm. If heat transfer thrugh the walls is negligible, the air temperature in the rm in 0 min will be (a) 5.6 C (b) 49.8 C (c) 5.4 C (d) 5.5 C (e) 6.4 C Answer (e) 6.4 C Slutin Sled by EES Sftware. Slutins can be erified by cpying-and-pasting the fllwing lines n a blank EES screen. (Similar prblems and their slutins can be btained easily by mdifying numerical alues). R0.87 "kj/kg.k" C0.78 "kj/kg.k" m60 "kg" P00 "kpa" T5 "C" sl0.8 "kj/s" time0*60 "s" Wfan0. "kj/s" "Applying energy balance E_in-E_utdE_system gies" time*(wfan+sl)m*c*(t-t) "Sme Wrng Slutins with Cmmn Mistakes:" Cp.005 "kj/kg.k" time*(wfan+sl)m*cp*(w_t-t) "Using Cp instead f C " time*(-wfan+sl)m*c*(w_t-t) "Subtracting Wfan instead f adding" time*slm*c*(w_t-t) "Ignring Wfan" time*(wfan+sl)/60m*c*(w4_t-t) "Using min fr time instead f s"