Then the amount of water that flows through the pipe during a differential time interval dt is (1) 4

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1 5-98 Review Problems 5-45 A tank oen to te atmosere is itially filled wit. e tank discarges to te atmosere troug a long ie connected to a valve. e itial discarge velocity from te tank and te time required to emty te tank are to be determed. Assumtions e flow is comressible. e drag ie is orizontal. e tank is considered to be emty wen te level dros to te center of te valve. Analysis (a) Substitutg te known quantities, te discarge velocity can be exressed as gz gz V 0. gz.5 + fl / D (00 m)/(0.0 m) en te itial discarge velocity becomes V 0. 0.(9.8m/s )( m).54 m/s gz were z is te eigt relative to te center of te orifice at tat time. (b) e flow rate of from te tank can be obtaed by multilyg te discarge velocity by te ie cross-sectional area, πd V & AieV 0. gz 4 en te amount of tat flows troug te ie durg a differential time terval dt is πd dv V& dt 0. gzdt () 4 wic, from conservation of mass, must be equal to te decrease te volume of te tank, πd0 dv Atank ( dz) dz () 4 were dz is te cange te level te tank durg dt. (Note tat dz is a negative quantity sce te ositive direction of z is uwards. erefore, we used dz to get a ositive quantity for te amount of discarged). Settg Eqs. () and () equal to eac oter and rearrangg, π D 0.gzdt π 4 D0 dz dt D D 0 0 z dz 0.gz D 4 D 0.g e last relation can be tegrated easily sce te variables are searated. Lettg t f be te discarge time and tegratg it from t 0 wen z z to t t f wen z 0 (comletely draed tank) gives 0 t f D 0 0 / D0 z D0 dt z dz t - 0 f t D 0.g z z D 0.g D 0. z Simlifyg and substitutg te values given, te drag time is determed to be t f 0 D D z 0.g (0 m) (0. m) m 5,940 s 7. 0.(9.8 m/s ) Discussion e drag time can be sortened considerably by stallg a um te ie. D 0 z dz g z D

2 e rate of accumulation of a ool and te rate of discarge are given. e rate suly of to te ool is to be determed. Assumtions Water is sulied and discarged steadily. e rate of evaoration of is negligible. No is sulied or removed troug oter means. Analysis e conservation of mass rcile alied to te ool requires tat te rate of crease te amount of te ool be equal to te difference between te rate of suly of and te rate of discarge. at is, dm ool dt dmool dv mi me mi me V & ool & & & + & i + V& e dt dt sce te density of is constant and tus te conservation of mass is equivalent to conservation of volume. e rate of discarge of is V & e A e Ve ( πd /4) Ve [ π (0.05 m) /4](5 m/s) m e rate of accumulation of te ool is equal to te crosssection of te ool times te rate at wic te level rises, dv ool dt A V cross-section level /s ( m 4 m)(0.05 m/m) 0.8 m /m m /s Substitutg, te rate at wic is sulied to te ool is determed to be dv ool V & i + V& e m dt erefore, is sulied at a rate of 0.08 m /s.8 L/s. /s 5-47 A fluid is flowg a circular ie. A relation is to be obtaed for te average fluid velocity terms of V(r), R, and r. Analysis Coosg a circular rg of area da πrdr as our differential area, te mass flow rate troug a cross-sectional area can be exressed as Solvg for V avg, V ρv ( r) da ρv ( r) π rdr avg A R 0 R V 0 R ()dr r r R r dr

3 Air is accelerated a nozzle. e density of air at te nozzle exit is to be determed. Assumtions Flow troug te nozzle is steady. Proerties e density of air is given to be 4.8 kg/m at te let. Analysis ere is only one let and one exit, and tus m. en, m & ρ A V ρ A V A ρ A V V ρ 0 m/s 80 m/s (4.8 kg/m ).64 & kg/m Discussion Note tat te density of air decreases considerably desite a decrease te cross-sectional area of te nozzle. AIR 5-49 e air a osital room is to be relaced every 5 mutes. e mimum diameter of te duct is to be determed if te air velocity is not to exceed a certa value. Assumtions e volume occuied by te furniture etc te room is negligible. e comg conditioned air does not mix wit te air te room. Analysis e volume of te room is V (6 m)(5 m)(4 m) 0 m o emty tis air 0 m, te volume flow rate must be V 0 m V & 0. m /s t 5 60 s If te mean velocity is 5 m/s, te diameter of te duct is πd V& AV 4 V D 4V & πv 4(0. m /s) π (5 m/s) 0.84 m Hosital Room m 0 bulbs erefore, te diameter of te duct must be at least 0.84 m to ensure tat te air te room is excanged comletely wit 0 m wile te mean velocity does not exceed 5 m/s. Discussion is roblem sows tat engeerg systems are sized to satisfy certa constrats imosed by certa considerations A long roll of large -Mn manganese steel late is to be quenced an oil bat at a secified rate. e mass flow rate of te late is to be determed. Assumtions e late moves troug te bat steadily. Proerties e density of steel late is given to be ρ 7854 kg/m. Analysis e mass flow rate of te seet metal troug te oil bat is Steel late 0 m/m ρv & ρwtv (7854 kg/m )( m)(0.005 m)(0 m/m) 9 kg/m 6.55 kg/s erefore, steel late can be treated conveniently as a flowg fluid calculations.

4 E A study quantifies te cost and benefits of enancg IAQ by creasg te buildg ventilation. e net monetary benefit of stallg an enanced IAQ system to te emloyer er year is to be determed. Assumtions e analysis te reort is alicable to tis work lace. Analysis e reort states tat enancg IAQ creases te roductivity of a erson by $90 er year, and decreases te cost of te resiratory illnesses by $9 a year wile creasg te annual energy consumtion by $6 and te equiment cost by ab $4 a year. e net monetary benefit of stallg an enanced IAQ system to te emloyer er year is determed by addg te benefits and subtractg te costs to be Net benefit otal benefits total cost (90+9) (6+4) $9/year (er erson) e total benefit is determed by multilyg te benefit er erson by te number of emloyees, otal net benefit No. of emloyees Net benefit er erson 0 $9/year $4,80/year Discussion Note tat te unseen savgs roductivity and reduced illnesses can be very significant wen tey are roerly quantified. 5-5 Air flows troug a non-constant cross-section ie. e let and exit velocities of te air are to be determed. Assumtions is is a steady-flow rocess sce tere is no cange wit time. Potential energy cange is negligible. ere are no work teractions. 4 e device is adiabatic and tus eat transfer is negligible. 5 Air is an ideal gas wit constant secific eats. Proerties e gas constant of air is R 0.87 kpa.m /kg.k. Also, c.005 kj/kg.k for air at room temerature (able A-) Analysis We take te ie as te system, wic is a control volume sce mass crosses te boundary. e mass and energy balances for tis steady-flow system can be exressed te rate form as Mass balance: 0 (steady) system Energy balance: E & 44 & 0 ρ AV ρ A V D 00 kpa 50 C Vel P πd V R 4 Rate of cange ternal, ketic, otential, etc.energies V V + c + + & 4444 V + 0 V + P πd V R 4 Air scew& e + q P P D V DV D 50 kpa 40 C Vel or c q Assumg let diameter to be.8 m and te exit diameter to be.0 m, and substitutg given values to mass and energy balance equations 00 kpa 50 kpa (.8 m) V (.0 m) V () K K V kj/kg V kj/kg (.005 kj/kg.k)( K) + (.005 kj/kg.k)( K) + +. kj/kg () 000 m /s 000 m /s ere are two equations and two unknowns. Solvg equations () and () simultaneously usg an equation solver suc as EES, te velocities are determed to be V 8.6 m/s V 0 m/s

5 Geotermal flows troug a flas camber, a searator, and a turbe a geotermal ower lant. e temerature of te steam after te flasg rocess and te ower ut from te turbe are to be determed for different flas camber exit ressures. Assumtions is is a steady-flow rocess sce tere is no cange wit time. Ketic and otential energy canges are negligible. e devices are sulated so tat tere are no eat losses to te surroundgs. 4 Proerties of steam are used for geotermal. Analysis For all comonents, we take te steam as te system, wic is a control volume sce mass crosses te boundary. e energy balance for tis steady-flow system can be exressed te rate form as Energy balance: E & 44 & 0 (steady) 0 Rate of cange ternal, ketic, otential, etc.energies & 4444 For eac comonent, te energy balance reduces to Flas camber: Searator: m & & + & m mliquidliquid urbe: W& ) ( 4 (a) For a flas camber exit ressure of P MPa e roerties of geotermal are 0 C x f kpa kpa kpa 79.9 C C sat. liq. Flas camber searator liquid P 000 kpa 777. kj/kg x P4 0 kpa 4 f + x4 fg (0.05)( 57.5 kj/kg) 49. kj/kg x e mass flow rate of vaor after te flasg rocess is m & x (0.)(50 kg/s) kg/s en, te ower ut from te turbe becomes W & (5.649 kg/s)( ) 66 kw Reeatg te similar calculations for oter ressures, we obta (b) For P 500 kpa, (c) For P 00 kpa, (d) For P 50 kpa, 5.8 C, & 4 kw W 99.6 C, & kw W 8. C, & 7 kw W steam turbe 4 0 kpa x0.95

6 A tank is eated by electricity. e witdrawn from te tank is mixed wit cold a camber. e mass flow rate of ot witdrawn from te tank and te average temerature of mixed are to be determed. Assumtions e rocess te mixg camber is a steady-flow rocess sce tere is no cange wit time. Ketic and otential energy canges are negligible. ere are no work teractions. 4 e device is adiabatic and tus eat transfer is negligible. Proerties e secific eat and density of are taken to be c 4.8 kj/kg.k, ρ 00 kg/m (able A-). Analysis We take te mixg camber as te system, wic is a control volume sce mass crosses te boundary. e energy balance for tis steady-flow system can be exressed te rate form as Energy balance: E & 44 & + Rate of cange ternal, ketic, otential, etc.energies & C m (sce W& ke e m? 0 C 0.06 kg/s ank 80 C 60 C or c + c ( + c () ot tank,ave cold cold ot cold ) mixture Similarly, an energy balance may be written on te tank as were and [ W& + c )] t m c ( ) e, ot ( cold tank,ave tank tank, tank, C tank, tank, tank, ave mtank ρv (000 kg/m )(0.060 m ) Substitutg to Eq. (), [.6 kj/s + (4.8 kj/kg. C)(0 70) ] ot Substitutg to Eq. (), 60 kg () kg/s Mixg camber C (8 60 s) (60 kg)(4.8 kj/kg. C)(60 80) C ( kg/s)(4.8 kj/kg. C)(70 C) + (0.06 kg/s)(4.8 kj/kg. C)(0 C) mixture ot [( )kg/s](4.8 kj/kg. C) 44.5 C mixture mix?

7 Water is boiled at a secified temerature by ot gases flowg troug a staless steel ie submerged. e rate of evaoration of is to be determed. Assumtions Steady oeratg conditions exist. Heat losses from te er surfaces of te boiler are negligible. Proerties e entaly of vaorization of at 50 C is fg.8 kj/kg (able A-4). Analysis e rate of eat transfer to is given to be 74 kj/s. Notg tat te entaly of vaorization reresents te amount of energy needed to vaorize a unit mass of a liquid at a secified temerature, te rate of evaoration of is determed to be evaoration boilg fg 74 kj/s.8 kj/kg kg/s Hot gases Water 50 C Heater 5-56 Cold enters a steam generator at 0 C, and leaves as saturated vaor at sat 50 C. e fraction of eat used to reeat te liquid from 0 C to saturation temerature of 50 C is to be determed. Assumtions Steady oeratg conditions exist. Heat losses from te steam generator are negligible. e secific eat of is constant at te average temerature. Proerties e eat of vaorization of at 50 C is fg.8 kj/kg (able A-4), and te secific eat of liquid is c 4.8 kj/kg. C (able A-). Analysis e eat of vaorization of reresents te amount of eat needed to vaorize a unit mass of liquid at a secified temerature. Usg te average secific eat, te amount of eat transfer needed to reeat a unit mass of from 0 C to 50 C is and q q c (4.8 kj/kg C)(50 0) C 54.4 kj/kg reeatg total qboilg + qreeatg erefore, te fraction of eat used to reeat te is q Fraction t o reeat q reeatg total kj/kg 0.05 (or 0.5% ) Cold 0 C Steam Water 50 C Heater

8 Cold enters a steam generator at 0 C and is boiled, and leaves as saturated vaor at boiler ressure. e boiler ressure at wic te amount of eat needed to reeat te to saturation temerature tat is equal to te eat of vaorization is to be determed. Assumtions Heat losses from te steam generator are negligible. Proerties e entaly of liquid at 0 C is 8.9 kj/kg. Oter roerties needed to solve tis roblem are te eat of vaorization fg and te entaly of saturated liquid at te secified temeratures, and tey can be obtaed from able A-4. Analysis e eat of vaorization of reresents te amount of eat needed to vaorize a unit mass of liquid at a secified temerature, and reresents te amount of eat needed to reeat a unit mass of from 0 C to te saturation temerature. erefore, ( f sat q reeatg f@0 C q ) 8.9 kj/kg boilg fg sat sat sat e solution of tis roblem requires coosg a boilg temerature, readg f and fg at tat temerature, and substitutg te values to te relation above to see if it is satisfied. By trial and error, (able A-4) At 0 C: At kj/kg f sat f sat sat sat kj/kg e temerature tat satisfies tis condition is determed from te two values above by terolation to be 0.6 C. e saturation ressure corresondg to tis temerature is 9.94 MPa. 8.9 kj/kg Cold 0 C Steam Water Heater 5-58 Saturated steam at atm ressure and tus at a saturation temerature of sat 00 C condenses on a vertical late mataed at 90 C by circulatg coolg troug te oter side. e rate of condensation of steam is to be determed. Assumtions Steady oeratg conditions exist. e steam condenses and te condensate dris off at 00 C. (In reality, te condensate temerature will be between 90 and 00, and te coolg of te condensate a few C sould be considered if better accuracy is desired). Proerties e entaly of vaorization of at atm (0.5 kpa) is fg 56.5 kj/kg (able A-5). Analysis e rate of eat transfer durg tis condensation rocess is given to be 80 kj/s. Notg tat te eat of vaorization of reresents te amount of eat released as a unit mass of vaor at a secified temerature condenses, te rate of condensation of steam is determed from condensation fg 80 kj/s 56.5 kj/kg kg/s Plate 90 C Steam 00 C Condensate

9 Water is boiled at sat 00 C by an electric eater. e rate of evaoration of is to be determed. Steam Assumtions Steady oeratg conditions exist. Heat losses from te er surfaces of te tank are negligible. Proerties e entaly of vaorization of at 00 C is fg 56.4 kj/kg (able A-4). Analysis Notg tat te entaly of vaorization reresents te amount of energy needed to vaorize a unit mass of a liquid at a secified temerature, te rate of evaoration of is determed to be evaoration W& e,boilg fg kj/s 56.4 kj/kg 0.00 kg/s 4.79 kg/ Water 00 C Heater 5-60 wo streams of same ideal gas at different states are mixed a mixg camber. e simlest exression for te mixture temerature a secified format is to be obtaed. Analysis e energy balance for tis steady-flow system can be exressed te rate form as E E & 44 & + c + c and, m & + Rate of cange ternal, ketic, otential, etc.energies c Solvg for fal temerature, we fd + Esystem & 4444 (sce W& 0) 0 m, m, Mixg device m,

10 An ideal gas exands a turbe. e volume flow rate at te let for a ower ut of 00 kw is to be determed. Assumtions is is a steady-flow rocess sce tere is no cange wit time. Ketic and otential energy canges are negligible. e device is adiabatic and tus eat transfer is negligible. Proerties e roerties of te ideal gas are given as R 0.0 kpa.m /kg.k, c. kj/kg C, c v 0.8 kj/kg C. Analysis We take te turbe as te system, wic is a control volume sce mass crosses te boundary. e energy balance for tis steady-flow system can be exressed te rate form as system 0 & 44 & & E E E 4444 Rate of cange ternal, ketic, otential, etc.energies E & m & W& + m & (sce ke e wic can be rearranged to solve for mass flow rate W& W& 00 kw 0.54 kg/s ( ) c (. kj/kg.k)(00-700)k e let secific volume and te volume flow rate are R ( 0. kpa m /kg K)( 00 K) v 0.6 m /kg P 600 kpa us, V& v (0.54 kg/s)(0.6 m /kg) 0. m /s P 600 kpa 00 K Ideal gas 00 kw 700 K 5-6 wo identical buildgs Los Angeles and Denver ave te same filtration rate. e ratio of te eat losses by filtration at te two cities under identical conditions is to be determed. Assumtions Bot buildgs are identical and bot are subjected to te same conditions excet te atmoseric conditions. Air is an ideal gas wit constant secific eats at room temerature. Steady flow conditions exist. Analysis We can view filtration as a steady stream of air tat is eated as it flows an imagary duct assg troug te buildg. e energy balance for tis imagary steady-flow system can be exressed te rate form as E & 44 & + m & Rate of cange ternal, ketic, otential, etc.energies & m & (sce ke e mc & ( ) ρv & c ( ) en te sensible filtration eat loss (eat ga for te filtratg air) can be exressed c ) ρ (ACH)( V ) c ( ) filtration air ( i o o, air buildg i o Los Angeles: 0 kpa Denver: 8 kpa were ACH is te filtration volume rate air canges er our. erefore, te filtration eat loss is roortional to te density of air, and tus te ratio of filtration eat losses at te two cities is simly te densities of door air at tose cities, Infiltration eat loss ratio filtration, Los Angeles filtration, Denver ( P0 / R0 ) ( P / R ) 0 Los Angeles 0 Denver ρo, ρ Po, P air, Los Angeles, air, Denver o Los Angeles 0, Denver 0 kpa 8 kpa. erefore, te filtration eat loss Los Angeles will be % iger tan tat Denver under identical conditions.

11 e ventilatg fan of te batroom of an electrically eated buildg San Francisco runs contuously. e amount and cost of te eat vented er mont wter are to be determed. Assumtions We take te atmoseric ressure to be atm 0. kpa sce San Francisco is at sea level. e buildg is mataed at C at all times. e filtratg air is eated to C before it exfiltrates. 4 Air is an ideal gas wit constant secific eats at room temerature. 5 Steady flow conditions exist. Proerties e gas constant of air is R 0.87 kpa.m /kg K (able A-). e secific eat of air at room temerature is c.005 kj/kg C (able A-). Analysis e density of air at te door conditions of atm and C is Po ρ o R o (0. kpa).0 kg/m (0.87 kpa.m /kg.k)( + 7 K) en te mass flow rate of air vented becomes air ρv & air (.0 kg/m )(0.00 m /s) 0.06 kg/s We can view filtration as a steady stream of air tat is eated as it flows an imagary duct assg troug te ouse. e energy balance for tis imagary steady-flow system can be exressed te rate form as E & 44 & + m & Rate of cange ternal, ketic, otential, etc.energies m & mc & ( & 4444 (sce ke e ) 0 0 L/s. C Notg tat te door air vented at C is relaced by filtratg door air at. C, te sensible filtration eat loss (eat ga for te filtratg air) due to ventg by fans can be exressed loss by fan airc ( doors doors ) (0.06 kg/s)(.005 kj/kg. C)(.) C 0.55 kj/s 0.55 kw en te amount and cost of te eat vented er mont ( mont ) becomes Energy loss loss t (0.55 kw)(70 /mont) 56 kw/mont by fan Money loss (Energy loss)(unit cost of energy) (56 kw/mont)($0.09/kw) $.0/mont Discussion Note tat te energy and money loss associated wit ventilatg fans can be very significant. erefore, ventilatg fans sould be used wit care. C

12 Cilled air is to cool a room by removg te eat generated a large sulated classroom by ligts and students. e required flow rate of air tat needs to be sulied to te room is to be determed. Assumtions e moisture roduced by te bodies leave te room as vaor wit any condensg, and tus te classroom as no latent eat load. Heat ga troug te walls and te roof is negligible. 4 Air is an ideal gas wit constant secific eats at room temerature. 5 Steady oeratg conditions exist. Proerties e secific eat of air at room temerature is.005 kj/kg C (able A-). e average rate of sensible eat generation by a erson is given to be 60 W. Analysis e rate of sensible eat generation by te eole te room and te total rate of sensible ternal eat generation are gen, sensible total, sensible q& gen, sensible gen, sensible (No. of eole) (60 W/erson)(50 ersons) 9000 W + ligtg ,000 W Bot of tese effects can be viewed as eat ga for te cilled air stream, wic can be viewed as a steady stream of cool air tat is eated as it flows an imagary duct assg troug te room. e energy balance for tis imagary steady-flow system can be exressed te rate form as E & 44 & + m & Rate of cange ternal, ketic, otential, etc.energies m & & (sce ke e total, sensible mc & ( ) en te required mass flow rate of cilled air becomes air total, sensible c 5 kj/s.49 kg/s (.005 kj/kg C)(5 5) C Cilled air 5 C Return air 5 C Discussion e latent eat will be removed by te air-conditiong system as te moisture condenses side te coolg coils.

13 Cickens are to be cooled by cilled an immersion ciller. e rate of eat removal from te cicken and te mass flow rate of are to be determed. Assumtions Steady oeratg conditions exist. e termal roerties of cickens and are constant. Proerties e secific eat of cicken are given to be.54 kj/kg. C. e secific eat of is 4.8 kj/kg. C (able A-). Analysis (a) Cickens are droed to te ciller at a rate of 500 er our. erefore, cickens can be considered to flow steadily troug te ciller at a mass flow rate of &m cicken (500 cicken / )(. kg / cicken) 00 kg / kg / s akg te cicken flow stream te ciller as te system, te energy balance for steadily flowg cickens can be exressed te rate form as E & 44 & m & Rate of cange ternal, ketic, otential, etc.energies & m & cicken 0 (sce ke e cicken c ( ) en te rate of eat removal from te cickens as tey are cooled from 5 C to ºC becomes Immersion cillg, 0.5 C Cicken 5 C cicken ( mc & ) cicken (0.056 kg/s)(.54 kj/kg.º C)(5 )º C.0 kw e ciller gas eat from te surroundgs at a rate of 00 kj/ kj/s. en te total rate of eat ga by te is cicken eat ga kw Notg tat te temerature rise of is not to exceed ºC as it flows troug te ciller, te mass flow rate of must be at least ( c ).056 kw (4.8 kj/kg.º C)(º C).56 kg/s If te mass flow rate of is less tan tis value, ten te temerature rise of will ave to be more tan C.

14 Cickens are to be cooled by cilled an immersion ciller. e rate of eat removal from te cicken and te mass flow rate of are to be determed. Assumtions Steady oeratg conditions exist. e termal roerties of cickens and are constant. Heat ga of te ciller is negligible. Proerties e secific eat of cicken are given to be.54 kj/kg. C. e secific eat of is 4.8 kj/kg. C (able A-). Analysis (a) Cickens are droed to te ciller at a rate of 500 er our. erefore, cickens can be considered to flow steadily troug te ciller at a mass flow rate of &m cicken (500 cicken / )(. kg / cicken) 00 kg / kg / s akg te cicken flow stream te ciller as te system, te energy balance for steadily flowg cickens can be exressed te rate form as E & 44 & m & Rate of cange ternal, ketic, otential, etc.energies & m & cicken 0 (sce ke e cicken c ( ) en te rate of eat removal from te cickens as tey are cooled from 5 C to ºC becomes cicken ( mc & ) cicken (0.056 kg/s)(.54 kj/kg.º C)(5 )º C.0 kw Immersion cillg, 0.5 C Cicken 5 C Heat ga of te ciller from te surroundgs is negligible. en te total rate of eat ga by te is & Qcicken.0 kw Notg tat te temerature rise of is not to exceed ºC as it flows troug te ciller, te mass flow rate of must be at least ( c ).0 kw (4.8 kj/kg.º C)(º C).56 kg/s If te mass flow rate of is less tan tis value, ten te temerature rise of will ave to be more tan C.

15 A regenerator is considered to save eat durg te coolg of milk a dairy lant. e amounts of fuel and money suc a generator will save er year are to be determed. Assumtions Steady oeratg conditions exist. e roerties of te milk are constant. Proerties e average density and secific eat of milk can be taken to be ρ milk ρ kg/l and c, milk.79 kj/kg. C (able A-). Analysis e mass flow rate of te milk is milk ρv & milk ( kg/l)( L/s) kg/s 4,00 kg/ akg te asteurizg section as te system, te energy balance for tis steady-flow system can be exressed te rate form as E & 44 & + m & Rate of cange ternal, ketic, otential, etc.energies m & & 4444 milk (sce ke e c ( ) 0 Hot milk 7 C Regenerator 4 C Cold milk erefore, to eat te milk from 4 to 7 C as beg done currently, eat must be transferred to te milk at a rate of current [ mc & ( asturization refrigeration )] milk ( kg/s)(.79 kj/kg.ëc)(7 4) C 09 kj/s e roosed regenerator as an effectiveness of ε 0.8, and tus it will save 8 ercent of tis energy. erefore, & saved εqcurrent (. 0 8)( 09 kj / s) 56 kj / s Notg tat te boiler as an efficiency of η boiler 0.8, te energy savgs above corresond to fuel savgs of (56 kj / s) Fuel Saved &Q saved (term) 0.09term / s η (0.8) (05,500 kj) boiler Notg tat year and unit cost of natural gas is $.0/term, te annual fuel and money savgs will be Fuel Saved (0.09 terms/s)( s) 94,400 terms/yr Money saved (Fuel saved)(unit cost of fuel) (94,400 term/yr)($.0/term) $,06,800/yr

16 5-5-68E A refrigeration system is to cool eggs by cilled air at a rate of 0,000 eggs er our. e rate of eat removal from te eggs, te required volume flow rate of air, and te size of te comressor of te refrigeration system are to be determed. Assumtions Steady oeratg conditions exist. e eggs are at uniform temeratures before and after coolg. e coolg section is well-sulated. 4 e roerties of eggs are constant. 5 e local atmoseric ressure is atm. Proerties e roerties of te eggs are given to ρ 67.4 lbm/ft and c 0.80 Btu/lbm. F. e secific eat of air at room temerature c 0.4 Btu/lbm. F (able A-E). e gas constant of air is R sia.ft /lbm.r (able A-E). Analysis (a) Notg tat eggs are cooled at a rate of 0,000 eggs er our, eggs can be considered to flow steadily troug te coolg section at a mass flow rate of egg (0,000 eggs/)(0.4 lbm/egg) 400 lbm/ lbm/s akg te egg flow stream te cooler as te system, te energy balance for steadily flowg eggs can be exressed te rate form as E & 44 & m & Rate of cange ternal, ketic, otential, etc.energies egg & m & egg (sce ke e c ( ) 0 Air 4 F Egg 0.4 lbm en te rate of eat removal from te eggs as tey are cooled from 90 F to 50ºF at tis rate becomes egg ( mc & ) egg (400 lbm/)(0.80 Btu/lbm. F)(90 50) F 44,800 Btu/ (b) All te eat released by te eggs is absorbed by te refrigerated air sce eat transfer troug e walls of cooler is negligible, and te temerature rise of air is not to exceed 0 F. e mimum mass flow and volume flow rates of air are determed to be ρ V & air ( c ) air 44,800 Btu/ (0.4Btu/lbm. F)(0 F) air air air P R ρ air air 8,667 lbm/ lbm/ft³,500 ft³/ 8,667 lbm/ 4.7 sia lbm/ft (0.704 sia.ft /lbm.r)( )R

17 Doug is made wit refrigerated order to absorb te eat of ydration and tus to control te temerature rise durg kneadg. e temerature to wic te city must be cooled before mixg wit flour is to be determed to avoid temerature rise durg kneadg. Assumtions Steady oeratg conditions exist. e doug is at uniform temeratures before and after coolg. e kneadg section is well-sulated. 4 e roerties of and doug are constant. Proerties e secific eats of te flour and te are given to be.76 and 4.8 kj/kg. C, resectively. e eat of ydration of doug is given to be 5 kj/kg. Analysis It is stated tat kg of flour is mixed wit kg of, and tus kg of doug is obtaed from eac kg of. Also, 5 kj of eat is released for eac kg of doug kneaded, and tus 5 45 kj of eat is released from te doug made usg kg of. akg te coolg section of as te system, wic is a steady-flow control volume, te energy balance for tis steadyflow system can be exressed te rate form as E & 44 & m & Rate of cange ternal, ketic, otential, etc.energies & m & (sce ke e c ( ) 0 Water 5 C 5 kj/kg Doug Flour In order for to absorb all te eat of ydration and end u at a temerature of 5ºC, its temerature before enterg te mixg section must be reduced to Q Q doug mc ( ) Q mc 45 kj 5 C 4. C ( kg)(4.8 kj/kg. C) at is, te must be recooled to 4.ºC before mixg wit te flour order to absorb te entire eat of ydration.

18 Glass bottles are wased ot an uncovered rectangular glass wasg bat. e rates of eat and mass tat need to be sulied to te are to be determed. Assumtions Steady oeratg conditions exist. e entire body is mataed at a uniform temerature of 55 C. Heat losses from te er surfaces of te bat are negligible. 4 Water is an comressible substance wit constant roerties. Proerties e secific eat of at room temerature is c 4.8 kj/kg. C. Also, te secific eat of glass is 0.80 kj/kg. C (able A-). Analysis (a) e mass flow rate of glass bottles troug te bat steady oeration is &mbottle mbottle Bottle flow rate (0.50 kg / bottle)(800 bottles / m) 0 kg / m kg / s akg te bottle flow section as te system, wic is a steadyflow control volume, te energy balance for tis steady-flow system can be exressed te rate form as E & 44 & + m & Rate of cange ternal, ketic, otential, etc.energies m & & 4444 bottle (sce ke e c ( ) 0 en te rate of eat removal by te bottles as tey are eated from 0 to 55 C is & ( kg/s)( 0.8 kj/kg.º C)( 55 0) º C 56,000 W bottle mbottlec e amount of removed by te bottles is, ( Flow rate of bottles)( Water removed er bottle) ( 800 bottles/ m)( 0. g/bottle) 60 g/m.67 0 kg/s Notg tat te removed by te bottles is made u by fres enterg at 5 C, te rate of eat removal by te tat sticks to te bottles is Water bat 55 C removed removedc (.67 0 kg/s)(480 J/kg º C)(55 5)º C 446W erefore, te total amount of eat removed by te wet bottles is total, removed + 56, ,446W glass removed removed Discussion In ractice, te rates of eat and removal will be muc larger sce te eat losses from te tank and te moisture loss from te oen surface are not considered.

19 Glass bottles are wased ot an uncovered rectangular glass wasg bat. e rates of eat and mass tat need to be sulied to te are to be determed. Assumtions Steady oeratg conditions exist. e entire body is mataed at a uniform temerature of 50 C. Heat losses from te er surfaces of te bat are negligible. 4 Water is an comressible substance wit constant roerties. Proerties e secific eat of at room temerature is c 4.8 kj/kg. C. Also, te secific eat of glass is 0.80 kj/kg. C (able A-). Analysis (a) e mass flow rate of glass bottles troug te bat steady oeration is &mbottle mbottle Bottle flow rate (0.50 kg / bottle)(800 bottles / m) 0 kg / m kg / s akg te bottle flow section as te system, wic is a steady-flow control volume, te energy balance for tis steady-flow system can be exressed te rate form as E & 44 & + m & Rate of cange ternal, ketic, otential, etc.energies m & & 4444 bottle (sce ke e c ( ) 0 en te rate of eat removal by te bottles as tey are eated from 0 to 50 C is & ( kg/s)( 0.8 kj/kg.º C)( 50 0) º C 48,000W bottle mbottlec e amount of removed by te bottles is, ( Flow rate of bottles)( Water removed er bottle) ( 800 bottles/ m)( 0. g/bottle) 60 g/m.67 0 kg/s Water bat 50 C Notg tat te removed by te bottles is made u by fres enterg at 5 C, te rate of eat removal by te tat sticks to te bottles is removed removedc (.67 0 kg/s)(480 J/kg ºC)(50 5)ºC erefore, te total amount of eat removed by te wet bottles is total, removed + 48, ,9 W glass removed removed 9 W Discussion In ractice, te rates of eat and removal will be muc larger sce te eat losses from te tank and te moisture loss from te oen surface are not considered.

20 Long alumum wires are extruded at a velocity of 0 m/m, and are exosed to atmoseric air. e rate of eat transfer from te wire is to be determed. Assumtions Steady oeratg conditions exist. e termal roerties of te wire are constant. Proerties e roerties of alumum are given to be ρ 70 kg/m and c kj/kg. C. Analysis e mass flow rate of te extruded wire troug te air is m & ρv& ρ( πr0 ) V (70 kg/m ) π (0.005 m) (0 m/m) 0.9 kg/m akg te volume occuied by te extruded wire as te system, wic is a steady-flow control volume, te energy balance for tis steady-flow system can be exressed te rate form as E & 44 & m & Rate of cange ternal, ketic, otential, etc.energies & 4444 wire + m & wire (sce ke e c ( ) 0 en te rate of eat transfer from te wire to te air becomes & mc 50 C Alumum wire 0 m/m [ ( t) ] (0.9 kg/m)(0.896 kj/kg. C)(50 50) C 5. kj/m kw 5-7 Long coer wires are extruded at a velocity of 0 m/m, and are exosed to atmoseric air. e rate of eat transfer from te wire is to be determed. Assumtions Steady oeratg conditions exist. e termal roerties of te wire are constant. Proerties e roerties of coer are given to be ρ 8950 kg/m and c 0.8 kj/kg. C. Analysis e mass flow rate of te extruded wire troug te air is m & ρv& ρ( πr0 ) V (8950 kg/m ) π (0.005 m) (0 m/m) 0.6 kg/m akg te volume occuied by te extruded wire as te system, wic is a steady-flow control volume, te energy balance for tis steady-flow system can be exressed te rate form as E & 44 & m & Rate of cange ternal, ketic, otential, etc.energies & 4444 wire + m & wire (sce ke e c ( ) 0 en te rate of eat transfer from te wire to te air becomes & mc 50 C Coer wire [ ( t) ] (0.6 kg/m)(0.8 kj/kg. C)(50 50) C 7.7 kj/m. kw 0 m/m

21 Steam at a saturation temerature of sat 40 C condenses on te side of a t orizontal tube. Heat is transferred to te coolg tat enters te tube at 5 C and exits at 5 C. e rate of condensation of steam is to be determed. Assumtions Steady oeratg conditions exist. Water is an comressible substance wit constant roerties at room temerature. e canges ketic and otential energies are negligible. Proerties e roerties of at room temerature are ρ 997 kg/m and c 4.8 kj/kg. C (able A-). e entaly of vaorization of at 40 C is fg kj/kg (able A-4). Analysis e mass flow rate of troug te tube is ρvac (997 kg/m )( m/s)[ π (0.0 m) / 4].409 kg/s akg te volume occuied by te cold te tube as te system, wic is a steady-flow control volume, te energy balance for tis steady-flow system can be exressed te rate form as E 0 & 44 & & 4444 Steam Rate of cange ternal, ketic, 40 C + m & m & otential, etc.energies (sce ke e c ( ) Cold Water 5 C en te rate of eat transfer to te and te rate of condensation become & c ( ) (.409 kg/s)(4.8 kj/kg C)(5 5) C 58.9 kw m cond fg cond fg 58.9 kj/s kj/kg kg/s 5 C 5-75E Saturated steam at a saturation ressure of 0.95 sia and tus at a saturation temerature of sat 00 F (able A-4E) condenses on te er surfaces of 44 orizontal tubes by circulatg coolg arranged a square array. e rate of eat transfer to te coolg and te average velocity of te coolg are to be determed. Assumtions Steady oeratg conditions exist. e tubes are isotermal. Water is an comressible substance wit constant roerties at room temerature. 4 e canges ketic and otential energies are negligible. Proerties e roerties of at room temerature are ρ 6. lbm/ft and c.00 Btu/lbm. F (able A-E). e entaly of vaorization of at a saturation ressure of 0.95 sia is fg 06.7 Btu/lbm (able A-4E). Analysis (a) e rate of eat transfer from te steam to te coolg is equal to te eat of vaorization released as te vaor condenses at te secified temerature, fg (6800 lbm/)(06.7 Btu/lbm) 7,049,560 Btu/ 958 Btu/s Saturated steam 0.95 sia (b) All of tis energy is transferred to te cold. erefore, te mass flow rate of cold must be 958 Btu/s c 44.8 lbm/s c (.00 Btu/lbm. F)(8 F) en te average velocity of te coolg troug te 44 tubes becomes 44.8 lbm/s ρav V ρa ρ( nπd / 4) (6.lbm/ft )[44π (/ ft) / 4] 5.0 ft/s Coolg