& out. R-134a 34 C

Transcription

1 Saturated refrigerant-4a vapor at a saturation temperature of T sat 4 C condenses side a tube. Te rate of eat transfer from te refrigerant for te condensate exit temperatures of 4 C and 0 C are to be determed. Assumptions Steady flow conditions exist. Ketic and potential energy canges are negligible. Tere are no work teractions volved. Properties Te properties of saturated refrigerant-4a at 4 C are f kj/kg, are g kj/kg, and are fg 69.7 kj/kg. Te entalpy of saturated liquid refrigerant at 0 C is f 79. kj/kg, (Table A-). Analysis We take te tube and te refrigerant it as te system. Tis is a control volume sce mass crosses te system boundary durg te process. We note tat tere is only one let and one exit, and tus. Notg tat eat is lost from te system, te energy balance for tis steady-flow system can be expressed te rate form as E E & 44 & Rate of net energy transfer E& m & Q& Rate of cange ternal, ketic, potential, etc.energies E& Q& ( 0 (steady) Esystem & m & ) 0 (sce ke pe 0) R-4a 4 C were at te let state g kj/kg. Ten te rates of eat transfer durg tis condensation process for bot cases become Case : T 4 C: f@4 C kj/kg. Q & (0. kg/m)( ) kj/kg Case : T 0 C: f@0 C 79. kj/kg. Q & (0. kg/m)( ) kj/kg 6.9 kg/m 8.9 kg/m Discussion Note tat te rate of eat removal is greater te second case sce te liquid is subcooled tat case. Q

2 E A wterizg project is to reduce te filtration rate of a ouse from. ACH to. ACH. Te resultg cost savgs are to be determed. Assumptions Te ouse is mataed at 7 F at all times. Te latent eat load durg te eatg season is negligible. Te filtratg air is eated to 7 F before it exfiltrates. 4 Air is an ideal gas wit constant specific eats at room temperature. 5 Te canges ketic and potential energies are negligible. 6 Steady flow conditions exist. Properties Te gas constant of air is psia.ft /lbm R (Table A-E). Te specific eat of air at room temperature is 0.4 Btu/lbm F (Table A-E). Analysis Te density of air at te door conditions is Po ρ o RT Te volume of te ouse is V buildg o.5 psia lbm/ft (0.704 psia.ft /lbm.r)(496.5 R) ( Floor area)(heigt) (000 ft )(9 ft) 7,000 ft We can view filtration as a steady stream of air tat is eated as it flows an imagary duct passg troug te ouse. Te energy balance for tis imagary steady-flow system can be expressed te rate form as E E & 44 & Rate of net energy transfer Q& E& + m & Q& Rate of cange ternal, ketic, potential, etc.energies E& 0 (steady) Esystem & m & (sce ke pe 0) mc & ( T T ) ρv & c ( T T ) p p Cold air 6.5 F Warm air 7 F Warm air 7 F Te reduction te filtration rate is... ACH. Te reduction te sensible filtration eat load correspondg to it is Q& filtration, saved ρ oc p ( ACH saved )( V buildg )( Ti To ) (0.074 lbm/ft )(0.4 Btu/lbm. F)(./)(7,000 ft )(7-6.5) F 8,57 Btu/ term/ sce term 00,000 Btu. Te number of ours durg a six mont period is Notg tat te furnace efficiency is 0.65 and te unit cost of natural gas is $.4/term, te energy and money saved durg te 6-mont period are Energy savgs ( Q & filtration, saved (0.857 term/)(40 /year)/ terms/year $50/year )(No.of Cost savgs (Energy savgs)(unit cost of energy) (4 terms/year)($.4/term) ours per year)/efficiency Terefore, reducg te filtration rate by one-alf will reduce te eatg costs of tis omeowner by $50 per year.

3 Outdoors air at -5 C and 90 kpa enters te buildg at a rate of 5 L/s wile te doors is mataed at 0 C. Te rate of sensible eat loss from te buildg due to filtration is to be determed. Assumptions Te ouse is mataed at 0 C at all times. Te latent eat load is negligible. Te filtratg air is eated to 0 C before it exfiltrates. 4 Air is an ideal gas wit constant specific eats at room temperature. 5 Te canges ketic and potential energies are negligible. 6 Steady flow conditions exist. Properties Te gas constant of air is R 0.87 kpa.m /kg K. Te specific eat of air at room temperature is c p.005 kj/kg C (Table A-). Analysis Te density of air at te door conditions is Po ρ o RT o 90 kpa.7 kg/m (0.87 kpa.m /kg.k)( K) We can view filtration as a steady stream of air tat is eated as it flows an imagary duct passg troug te buildg. Te energy balance for tis imagary steady-flow system can be expressed te rate form as E E & 44 & Rate of net energy transfer Q& E& + m & Q& Rate of cange ternal, ketic, potential, etc.energies E& m & mc & ( T 0 (steady) Esystem & 4444 p (sce ke pe 0) T ) 0 Cold air -5 C 90 kpa 5 L/s Warm air Ten te sensible filtration eat load correspondg to an filtration rate of 5 L/s becomes Q & ρ V & ( T filtration o air p i o c (.7 kg/m )(0.05 m /s)(.005 kj/kg. C)[0 - (-5)] C.09 kw T ) Terefore, sensible eat will be lost at a rate of.09 kj/s due to filtration. 0 C Warm air 0 C

4 Te maximum flow rate of a standard sower ead can be reduced from. to 0.5 L/m by switcg to low-flow sower eads. Te ratio of te ot-to-cold water flow rates and te amount of electricity saved by a family of four per year by replacg te standard sower eads by te low-flow ones are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time at any pot and tus mcv 0 and ECV 0. Te ketic and potential energies are negligible, ke pe 0. Heat losses from te system are negligible and tus &Q 0. 4 Tere are no work teractions volved. 5.Sowers operate at maximum flow conditions durg te entire sower. 6 Eac member of te ouseold takes a 5- m sower every day. 7 Water is an compressible substance wit constant properties. 8 Te efficiency of te electric water eater is 00%. Properties Te density and specific eat of water at room temperature are ρ kg/l and c 4.8 kj/kg. C (Table A-). Analysis (a) We take te mixg camber as te system. Tis is a control volume sce mass crosses te system boundary durg te process. We note tat tere are two lets and one exit. Te mass and energy balances for tis steady-flow system can be expressed te rate form as follows: Mass balance: 0 (steady) 0 Energy balance: system E E & 44 & Rate of net energy transfer E& + + Rate of cange ternal, ketic, potential, etc.energies E& 0 (steady) Esystem & 4444 Combg te mass and energy balances and rearrangg, m & + m & ( + ) ( ) ( ) 0 Cold water (sce Q& 0, W& 0, ke pe 0) Ten te ratio of te mass flow rates of te ot water to cold water becomes c( T c( T T ) T T ) T (b) Te low-flow eads will save water at a rate of V& saved saved [(.-0.5) ρv& saved Ten te energy saved per year becomes Energy saved T T (4 5) C.08 (55 4) C ( kg/l)(0,440 L/year) 0,440 kg/year saved 64 kw Mixture L/m](5 m/person.day)(4 persons)(65 days/yr) 0,440 L/year c T (0,440 kg/year)(4.8 kj/kg. C)(4-5) C,07,000 kj/year (scekw 600 kj) Terefore, switcg to low-flow sower eads will save ab 64 kw of electricity per year. Hot water

5 EES Problem 5-79 is reconsidered. Te effect of te let temperature of cold water on te energy saved by usg te low-flow soweread as te let temperature varies from 0 C to 0 C is to be vestigated. Te electric energy savgs is to be plotted agast te water let temperature. Analysis Te problem is solved usg EES, and te results are tabulated and plotted below. "Knowns:" C_P 4.8 [kj/kg-k ] density[kg/l] {T_ 5 [C]} T_ 55 [C] T_ 4 [C] V_dot_old. [L/m] V_dot_new 0.5 [L/m] m_dot_[kg/s] "We can set m_dot_ wit loss of generality." "Analysis:" "(a) We take te mixg camber as te system. Tis is a control volume sce mass crosses te system boundary durg te process. We note tat tere are two lets and one exit. Te mass and energy balances for tis steady-flow system can be expressed te rate form as follows:" "Mass balance:" m_dot_ - m_dot_ DELTAm_dot_sys DELTAm_dot_sys0 m_dot_ m_dot_ + m_dot_ m_dot_ m_dot_ "Te ratio of te mass flow rates of te ot water to cold water is obtaed by settg m_dot_[kg/s]. Ten m_dot_ represents te ratio of m_dot_/m_dot_" "Energy balance:" E_dot_ - E_dot_ DELTAE_dot_sys DELTAE_dot_sys0 E_dot_ m_dot_*_ + m_dot_*_ E_dot_ m_dot_* C_P*T C_P*T C_P*T_ "(b) Te low-flow eads will save water at a rate of " V_dot_saved (V_dot_old - V_dot_new)"L/m"*(5"m/personday")*(4"persons")*(65"days/year") "[L/year]" m_dot_saveddensity*v_dot_saved "[kg/year]" "Ten te energy saved per year becomes" E_dot_savedm_dot_saved*C_P*(T_ - T_)"kJ/year"*convert(kJ,kW) "[kw/year]" "Terefore, switcg to low-flow sower eads will save ab 64 kw of electricity per year. " "Ratio of ot-to-cold water flow rates:" m_ratio m_dot_/m_dot_

6 5-4 E saved [kw/year] T [C] E saved [kw/year] T [C]

7 A fan is powered by a 0.5 p motor, and delivers air at a rate of 85 m /m. Te igest possible air velocity at te fan exit is to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time at any pot and tus m 0 and E 0. Te let velocity and te cange potential energy are negligible, CV V 0 and pe CV 0. Tere are no eat and work teractions oter tan te electrical power consumed by te fan motor. 4 Te efficiencies of te motor and te fan are 00% sce best possible operation is assumed. 5 Air is an ideal gas wit constant specific eats at room temperature. Properties Te density of air is given to be ρ.8 kg/m. Te constant pressure specific eat of air at room temperature is c p.005 kj/kg. C (Table A-). Analysis We take te fan-motor assembly as te system. Tis is a control volume sce mass crosses te system boundary durg te process. We note tat tere is only one let and one exit, and tus. Te velocity of air leavg te fan will be igest wen all of te entire electrical energy drawn by te motor is converted to ketic energy, and te friction between te air layers is zero. In tis best possible case, no energy will be converted to termal energy, and tus te temperature cange of air will be zero, T T. Ten te energy balance for tis steady-flow system can be expressed te rate form as E E & 44 & Rate of net energy transfer W& e, E& + m & Rate of cange ternal, ketic, potential, etc.energies E& ( 0 (steady) Esystem & V /) (scev 0 0 and pe 0) Notg tat te temperature and tus entalpy remas constant, te relation above simplifies furter to were & & W e, mv / ρv & (.8 kg/m )(85 m /m) 00. kg/m.67 kg/s Solvg for V and substitutg gives V W& e, (0.5 p) W m / s.67 kg/s p W. m/s 0.5 p 85 m /m Discussion In reality, te velocity will be less because of te efficiencies of te motor and te fan.

8 Te average air velocity te circular duct of an air-conditiong system is not to exceed 0 m/s. If te fan converts 70 percent of te electrical energy to ketic energy, te size of te fan motor needed and te diameter of te ma duct are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time at any pot and tus mcv 0 and ECV 0. Te let velocity is negligible, V 0. Tere are no eat and work teractions oter tan te electrical power consumed by te fan motor. 4 Air is an ideal gas wit constant specific eats at room temperature. Properties Te density of air is given to be ρ.0 kg/m. Te constant pressure specific eat of air at room temperature is c p.005 kj/kg. C (Table A-). Analysis We take te fan-motor assembly as te system. Tis is a control volume sce mass crosses te system boundary durg te process. We note tat tere is only one let and one exit, and tus. Te cange te ketic energy of air as it is accelerated from zero to 0 m/s at a rate of 80 m /s is ρv & (.0 kg/m )(80 m /m) 6 kg/m.6 kg/s (0 m/s) 0 kj/kg KE & V V (.6 kg/s) 0.8 kw 000 m / s It is stated tat tis represents 70% of te electrical energy consumed by te motor. Ten te total electrical power consumed by te motor is determed to be & & & KE&. kw Wmotor KE Wmotor 0.57 kw Te diameter of te ma duct is 4 4(80 m / m) m V& V VA V ( πd / 4) D & 0.68 m πv π (0 m/s) 60 s 0 m/s 80 m /m Terefore, te motor sould ave a rated power of at least 0.57 kw, and te diameter of te duct sould be at least 6.8 cm 5-8 An evacuated bottle is surrounded by atmosperic air. A valve is opened, and air is allowed to fill te bottle. Te amount of eat transfer troug te wall of te bottle wen termal and mecanical equilibrium is establised is to be determed. Assumptions Tis is an unsteady process sce te conditions wit te device are cangg durg te process, but it can be analyzed as a uniform-flow process sce te state of fluid at te let remas constant. Air is an ideal gas. Ketic and potential energies are negligible. 4 Tere are no work teractions volved. 5 Te direction of eat transfer is to te air te bottle (will be verified). Analysis We take te bottle as te system. It is a control volume sce mass crosses te boundary. Notg tat te microscopic energies of flowg and nonflowg fluids are represented by entalpy and ternal energy u, respectively, te mass and energy balances for tis uniform-flow system can be expressed as Mass balance: m m m m m (sce m m 0) Energy balance: system i itial E 44 Net energy transfer Q + mii Combg te two balances: Q m u E ( Cange ternal, ketic, potential, etc.energies m u ) Esystem 44 (scew E m( cv T i c p T ) But T i T T 0 and c p - c v R. Substitutg, P0 V Q m( cv cp ) T0 mrt RT RT i E 0 0 P0 0 Terefore, Q P 0 V (Heat is lost from te tank) itial V ke pe 0) V Evacuated P 0 T 0

9 An adiabatic air compressor is powered by a direct-coupled steam turbe, wic is also drivg a generator. Te net power delivered to te generator is to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Ketic and potential energy canges are negligible. Te devices are adiabatic and tus eat transfer is negligible. 4 Air is an ideal gas wit variable specific eats. Properties From te steam tables (Tables A-4 troug 6) and P.5 MPa x T 500 C P 0 kpa kj/kg 4 From te air table (Table A-7), T 95 K T 60 K + x f 4 fg kj/kg kj/kg Analysis Tere is only one let and one exit for eiter device, and tus m &. We take eiter te turbe or te compressor as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for eiter steady-flow system can be expressed te rate form as E & 44 & 0 (steady) E Esystem 0 Rate of net energy transfer E& Rate of cange ternal, ketic, potential, etc.energies E& & 4444 For te turbe and te compressor it becomes ( 0.9)( 9.) 9.5 kj/kg Compressor: W& + m & W& ( comp, air air comp, air ) Air comp Turbe: W& W& ( Substitutg, W& W& Terefore, W & comp, turb, net, steam turb, + steam 4 turb, steam 4) ( 0 kg/s)( ) kj/kg 9 kw ( 5 kg/s)( ) kj/kg,777 kw W& W&, ,448 kw turb, comp, 98 kpa 95 K MPa 60 K.5 MPa 500 C Steam turbe 0 kpa

10 Water is eated from 6 C to 4 C by an electric resistance eater placed te water pipe as it flows troug a soweread steadily at a rate of 0 L/m. Te electric power put to te eater, and te money tat will be saved durg a 0-m sower by stallg a eat excanger wit an effectiveness of 0.50 are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time at any pot wit te system and tus m 0 and E 0,. Water is an compressible substance wit constant specific CV CV eats. Te ketic and potential energy canges are negligible, ke pe 0. 4 Heat losses from te pipe are negligible. Properties Te density and specific eat of water at room temperature are ρ kg/l and c 4.8 kj/kg C (Table A-). Analysis We take te pipe as te system. Tis is a control volume sce mass crosses te system boundary durg te process. We observe tat tere is only one let and one exit and tus m &. Ten te energy balance for tis steady-flow system can be expressed te rate form as E E & 44 & Rate of net energy transfer were W& Substitutg, e, + m & m & W& ρv & W & e, Rate of cange ternal, ketic, potential, etc.energies ( 0 (steady) Esystem & 4444 (sce ke pe 0) ) [ c( T 0 T ) + v( P ( kg/l)( 0 L/m) 0 kg/m E& 0 P ) E& o ( 0/60 kg/s)( 4.8 kj/kg C)( 4 6) C 8.8 kw e, Te energy recovered by te eat excanger is Q& saved εq& max εmc & ( Tmax Tm ) o ( kg/s)( 4.8 kj/kg. C)( 9 6) 0.5 0/ kj/s 8.0kW C ] mc & ( T T ) 6 C WATER Terefore, 8.0 kw less energy is needed tis case, and te required electric power tis case reduces to W &, new W&,old Q& saved kw Te money saved durg a 0-m sower as a result of stallg tis eat excanger is ( 8.0 kw)( 0/60 )( 8.5 cents/kw). cents 4 C

11 EES Problem 5-85 is reconsidered. Te effect of te eat excanger effectiveness on te money saved as te effectiveness ranges from 0 percent to 90 percent is to be vestigated, and te money saved is to be plotted agast te effectiveness. Analysis Te problem is solved usg EES, and te results are tabulated and plotted below. "Knowns:" density [kg/l] V_dot 0 [L/m] C 4.8 [kj/kg-c] T_ 6 [C] T_ 4 [C] T_max 9 [C] T_m T_ epsilon 0.5 "eat excanger effectiveness " EleRate 8.5 [cents/kw] "For entrance, one exit, steady flow m_dot_ m_dot_ m_dot_wat er:" m_dot_water density*v_dot /convert(m,s) "Energy balance for te pipe:" W_dot_ele_+ m_dot_water*_m_dot_water*_ "Neglect ke and pe" "For compressible fluid a constant pressure process, te entalpy is:" _ C*T C*T_ "Te energy recovered by te eat excanger is" Q_dot_savedepsilon*Q_dot_max Q_dot_max m_dot_water*c*(t_max - T_m) "Terefore, 8.0 kw less energy is needed tis case, and te required electric power tis case reduces to" W_dot_ele_new W_dot_ele_ - Q_dot_saved "Te money saved durg a 0-m sower as a result of stallg tis eat excanger is" Costs_saved Q_dot_saved*time*convert(m,)*EleRate time0 [m] Costs saved ε [cents] n ts] e [c d e a v s C osts Heat excanger effectiveness ε

12 [Also solved by EES on enclosed CD] Steam expands a turbe steadily. Te mass flow rate of te steam, te exit velocity, and te power put are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Potential energy canges are negligible. Properties From te steam tables (Tables A-4 troug 6) P 0 MPa v m /kg T 550 C 50.0 kj/kg and P 5 kpa v v f + xv fg x 0.95 f + x fg 7.96 Analysis (a) Te mass flow rate of te steam is V A v m /kg ( 0.95)( ) + ( 0.95)( 45.5) 500. kj/kg ( 60 m/s)( 0.05 m ) 5.4 kg/s 5.89 m 0 kj/kg (b) Tere is only one let and one exit, and tus m. Ten te exit velocity is determed from V A v V v A & (5.4 kg/s)(5.89 m 0.4 m /kg) 06 m/s (c) We take te turbe as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady-flow system can be expressed te rate form as E & 44 & 0 (steady) E Esystem 0 Rate of net energy transfer ( + V W& E& / ) Rate of cange ternal, ketic, potential, etc.energies E& W& Q& & Q& + ( + V V + /) V (sce pe 0) Ten te power put of te turbe is determed by substitutg to be W & ( 5.4 0) kj/s ( 5.4 kg/s) 0,0 kw ( 06 m/s) ( 60 m/s) /kg H O kj/kg 000 m /s

13 EES Problem 5-87 is reconsidered. Te effects of turbe exit area and turbe exit pressure on te exit velocity and power put of te turbe as te exit pressure varies from 0 kpa to 50 kpa (wit te same quality), and te exit area to varies from 000 cm to 000 cm is to be vestigated. Te exit velocity and te power put are to be plotted agast te exit pressure for te exit areas of 000, 000, and 000 cm. Analysis Te problem is solved usg EES, and te results are tabulated and plotted below. Fluid$'Steam_IAPWS' A[]50 [cm^] T[]550 [C] P[]0000 [kpa] Vel[] 60 [m/s] A[]400 [cm^] P[]5 [kpa] q_ 0 [kj/kg] m_dot A[]*Vel[]/v[]*convert(cm^,m^) v[]volume(fluid$, TT[], PP[]) "specific volume of steam at state " Vel[]m_dot*v[]/(A[]*convert(cm^,m^)) v[]volume(fluid$, x0.95, PP[]) "specific volume of steam at state " T[]temperature(Fluid$, PP[], vv[]) "[C]" "not required, but good to know" "[conservation of Energy for steady-flow:" "E_dot - E_dot DeltaE_dot" "For steady-flow, DeltaE_dot 0" DELTAE_dot0 "[kw]" "For te turbe as te control volume, neglectg te PE of eac flow steam:" E_dot_E_dot_ []entalpy(fluid$,tt[], PP[]) E_dot_m_dot*([]+ Vel[]^/*Convert(m^/s^, kj/kg)) []entalpy(fluid$,x0.95, PP[]) E_dot_m_dot*([]+ Vel[]^/*Convert(m^/s^, kj/kg))+ m_dot *q_+ W_dot_ PowerW_dot_ Q_dot_m_dot*q_ Power [kw] P [kpa] Vel [m/s]

14 ] W [k e r P o w A 000 cm^ A 000 cm^ A 000 cm^ P[] [kpa] A 000 cm^ A 000 cm^ A 000 cm^ ] s [m / ] V el[ P[] [kpa]

15 5-5-89E Refrigerant-4a is compressed steadily by a compressor. Te mass flow rate of te refrigerant and te exit temperature are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Ketic and potential energy canges are negligible. Te device is adiabatic and tus eat transfer is negligible. Properties From te refrigerant tables (Tables A-E troug A-E) P 5 psia v.55 ft /lbm T 0 F 07.5 Btu/lbm Analysis (a) Te mass flow rate of refrigerant is V &m & v 0 ft /s.07 lbm/s.55 ft /lbm (b) Tere is only one let and one exit, and tus. We take te compressor as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady-flow system can be expressed te rate form as Substitutg, E & 44 & 0 (steady) E Esystem 0 Rate of net energy transfer E& Rate of cange ternal, ketic, potential, etc.energies E& & 4444 W& & & & + m m (sce Q ke pe 0) W& ( ) Btu/s (45 p) p 7.87 Btu/lbm Ten te exit temperature becomes P 00 psia T 7.87 Btu/lbm (.07 lbm/s)( 07.5) 95.7 F Btu/lbm R-4a

16 Air is preeated by te exaust gases of a gas turbe a regenerator. For a specified eat transfer rate, te exit temperature of air and te mass flow rate of exaust gases are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Ketic and potential energy canges are negligible. Tere are no work teractions. 4 Heat loss from te regenerator to te surroundgs is negligible and tus eat transfer from te ot fluid is equal to te eat transfer to te cold fluid. 5 Exaust gases can be treated as air. 6 Air is an ideal gas wit variable specific eats. Properties Te gas constant of air is 0.87 kpa.m /kg.k (Table A-). Te entalpies of air are (Table A- 7) T 550 K T T 800 K K kj/kg 8.95 kj/kg kj/kg Analysis (a) We take te air side of te eat excanger as te system, wic is a control volume sce mass crosses te boundary. Tere is only one let and one exit, and tus. Te energy balance for tis steady-flow system can be expressed te rate form as Substitutg, E E & 44 & Rate of net energy transfer Q& E& + Q& air Rate of cange ternal, ketic, potential, etc.energies E& air air 0 (steady) Esystem & 4444 ( (sce W& ke pe 0) ) 00 ( 800/60 kg/s)( kj/kg) 0 Exaust Gases kj/s kj/kg Ten from Table A-7 we read T 775. K (b) Treatg te exaust gases as an ideal gas, te mass flow rate of te exaust gases is determed from te steady-flow energy relation applied only to te exaust gases, It yields E& exaust Q& E& Q& + exaust 00 kj/s exaust exaust exaust ( ) 4 4 (sce W& ke pe 0) ( )kJ/kg 4.9 kg/s AIR 4

17 Water is to be eated steadily from 0 C to 55 C by an electrical resistor side an sulated pipe. Te power ratg of te resistance eater and te average velocity of te water are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time at any pot wit te system and tus m 0 and E 0. Water is an compressible substance wit constant specific CV CV eats. Te ketic and potential energy canges are negligible, ke pe 0. 4 Te pipe is sulated and tus te eat losses are negligible. Properties Te density and specific eat of water at room temperature are ρ 000 kg/m and c 4.8 kj/kg C (Table A-). Analysis (a) We take te pipe as te system. Tis is a control volume sce mass crosses te system boundary durg te process. Also, tere is only one let and one exit and tus m &. Te energy balance for tis steady-flow system can be expressed te rate form as E E & 44 & Rate of net energy transfer W& e, E& W& e, Rate of cange ternal, ketic, potential, etc.energies E& + m & m & ( 0 (steady) Esystem & 4444 (sce Q& ) [ c( T Te mass flow rate of water troug te pipe is Terefore, ke pe 0) T ) + v P 0 ρv & (000 kg/m )(0.00 m /m) 0 ] mc & 0 kg/m ( T T ) ( ) o o W& e, mc & T T (0/60 kg/s)(4.8 kj/kg C)(55 0) C 7. kw (b) Te average velocity of water troug te pipe is determed from V & & V V 0.00 m /m A πr π(0.05 m) 5. m/m WATER 0 L/m D 5 cm W e

18 Te feedwater of a steam power plant is preeated usg steam extracted from te turbe. Te ratio of te mass flow rates of te extracted seam te feedwater are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Ketic and potential energy canges are negligible. Tere are no work teractions. 4 Heat loss from te device to te surroundgs is negligible and tus eat transfer from te ot fluid is equal to te eat transfer to te cold fluid. Properties Te entalpies of steam and feedwater at are (Tables A-4 troug A-6) and P MPa T 00 C P MPa sat. liquid 88. kj/kg T P.5 MPa f T 50 C P4.5 MPa 4 T4 T 0 70 C 76.5 kj/kg 79.9 C C 09.4 kj/kg C kj/kg Feedwater STEAM Analysis We take te eat excanger as te system, wic is a control volume. Te mass and energy balances for tis steady-flow system can be expressed te rate form as Mass balance (for eac fluid stream): 0 (steady) system 0 s and Energy balance (for te eat excanger): E E & 44 & Rate of net energy transfer E& Rate of cange ternal, ketic, potential, etc. energies E& 0 (steady) Esystem & Combg te two, ( ) ( ) s fw 4 0 (sce Q& W& ke pe 0) 4 fw 4 Dividg by &m fw ( ) and substitutg, s fw kj/kg 76.5 kj/kg ( 88. ) 0.46

19 A buildg is to be eated by a 0-kW electric resistance eater placed a duct side. Te time it takes to raise te terior temperature from 4 C to 4 C, and te average mass flow rate of air as it passes troug te eater te duct are to be determed. Assumptions Steady operatg conditions exist. Air is an ideal gas wit constant specific eats at room temperature. Ketic and potential energy canges are negligible. 4 Te eatg duct is adiabatic, and tus eat transfer troug it is negligible. 5 No air leaks and of te buildg. Properties Te gas constant of air is 0.87 kpa.m /kg.k (Table A-). Te specific eats of air at room temperature are c p.005 and c v 0.78 kj/kg K (Table A-). Analysis (a) Te total mass of air te buildg is PV m RT ( 95 kpa)( 400 m ) ( 0.87 kpa m /kg K)( 87 K) 46. kg. We first take te entire buildg as our system, wic is a closed system sce no mass leaks or. Te time required to raise te air temperature to 4 C is determed by applyg te energy balance to tis constant volume closed system: t W 44 Net energy transfer e, ( W& + W& Q& ) mc ( T T ) e, Solvg for t gives E + W E fan, fan, mcv,avg t W& + W& e, Q ( T T ) fan, Q Cange ternal, ketic, potential, etc. energies U Esystem 44 v,avg (sce KE PE 0) T T + 5 C V 400 m P 95 kpa 4 C 4 C (46. kg)(0.78 kj/kg C)(4 4) C 46 (0 kj/s) + (0.5 kj/s) (450/60 kj/s) o o T s 450 kj/m W e 50 W (b) We now take te eatg duct as te system, wic is a control volume sce mass crosses te boundary. Tere is only one let and one exit, and tus. Te energy balance for tis adiabatic steady-flow system can be expressed te rate form as Tus, E E & 44 & Rate of net energy transfer W& e, + W& W& W& fan, e, e, + W& + W& c T p E& fan, Rate of cange ternal, ketic, potential, etc. energies E& + m & m & fan, ( 0 (steady) Esystem & 4444 (sce Q& ke pe 0) ) mc & p ( T T ) 0 ( ) kj/s 6.0 kg/s o o (.005 kj/kg C)(5 C)

20 [Also solved by EES on enclosed CD] An sulated cylder equipped wit an external sprg itially contas air. Te tank is connected to a supply le, and air is allowed to enter te cylder until its volume doubles. Te mass of te air tat entered and te fal temperature te cylder are to be determed. Assumptions Tis is an unsteady process sce te conditions wit te device are cangg durg te process, but it can be analyzed as a uniform-flow process sce te state of fluid at te let remas constant. Te expansion process is quasi-equilibrium. Ketic and potential energies are negligible. 4 Te sprg is a lear sprg. 5 Te device is sulated and tus eat transfer is negligible. 6 Air is an ideal gas wit constant specific eats. Properties Te gas constant of air is R 0.87 kj/kg K (Table A-). Te specific eats of air at room temperature are c v 0.78 and c p.005 kj/kg K (Table A-a). Also, u c v T and c p T. Analysis We take te cylder as te system, wic is a control volume sce mass crosses te boundary. Notg tat te microscopic energies of flowg and nonflowg fluids are represented by entalpy and ternal energy u, respectively, te mass and energy balances for tis uniform-flow system can be expressed as Mass balance: m m msystem mi m m Energy balance: E E 44 Net energy transfer m W i i Cange ternal, ketic, potential, etc. energies b, Esystem 44 + m u m u Combg te two relations, ( m m ) i Wb, + mu mu F sprg P 00 kpa T C V 0. m Air (sce Q ke pe 0) or, ( m m ) c pti Wb, + mcv T mcv T Te itial and te fal masses te tank are m m P V RT PV RT ( 00 kpa)( 0. m ) ( 0.87 kpa m /kg K)( 95 K) ( 600 kpa)( 0.4 m ) ( 0.87 kpa m /kg K) T T 0.47 kg 86. Ten from te mass balance becomes mi m m 86. T 047. Te sprg is a lear sprg, and tus te boundary work for tis process can be determed from W b P + P Area ( V V ) ( ) kpa ( ) m 80 kj Substitutg to te energy balance, te fal temperature of air T is determed to be It yields Tus, and T T T 44. K m T 44. (.005)( 95) + ( 0.78)( T ) ( 0.47)( 0.78)(95).40 kg m i m - m kg P i 0.8 MPa T i C

21 R-4a is allowed to leave a piston-cylder device wit a pair of stops. Te work done and te eat transfer are to be determed. Assumptions Tis is an unsteady process sce te conditions wit te device are cangg durg te process, but it can be analyzed as a uniform-flow process sce te state of fluid leavg te device is assumed to be constant. Ketic and potential energies are negligible. Properties Te properties of R-4a at various states are (Tables A- troug A-) P 800 kpa m /kg v u kj/kg T 80 C 6.97 kj/kg P 500 kpa v u T 0 C m /kg 4.40 kj/kg 6.46 kj/kg R-4a kg 800 kpa 80 C Analysis (a) We take te tank as te system, wic is a control volume sce mass crosses te boundary. Notg tat te microscopic energies of flowg and nonflowg fluids are represented by entalpy and ternal energy u, respectively, te mass and energy balances for tis uniform-flow system can be expressed as Mass balance: m m msystem me m m Energy balance: 44 Net energy transfer W E b, E Q m e e Cange ternal, ketic, potential, etc.energies m u Esystem 44 m u (sce ke pe 0) Te volumes at te itial and fal states and te mass tat as left te cylder are V m v ( kg)( m /kg) m m V m v m e m (/ ) m v kg (/)( kg)(0.045 m /kg) 0.04 m Te entalpy of te refrigerant witdrawn from te cylder is assumed to be te average of itial and fal entalpies of te refrigerant te cylder e ( / )( + ) (/ )( ) 90. kj/kg Notg tat te pressure remas constant after te piston starts movg, te boundary work is determed from (b) Substitutg, Wb, P ( V V ) (500 kpa)( ) m.6 kj Q 60.7 kj.6 kj ( kg)(90. kj/kg) ( kg)(4.40 kj/kg) ( kg)(90.84 kj/kg) Q Q

22 Air is allowed to leave a piston-cylder device wit a pair of stops. Heat is lost from te cylder. Te amount of mass tat as escaped and te work done are to be determed. Assumptions Tis is an unsteady process sce te conditions wit te device are cangg durg te process, but it can be analyzed as a uniform-flow process sce te state of fluid leavg te device is assumed to be constant. Ketic and potential energies are negligible. Air is an ideal gas wit constant specific eats at te average temperature. Properties Te properties of air are R 0.87 kpa.m /kg.k (Table A-), c v 0.7 kj/kg.k, c p.00 kj/kg.k at te anticipated average temperature of 450 K (Table A-b). Analysis We take te tank as te system, wic is a control volume sce mass crosses te boundary. Notg tat te microscopic energies of flowg and nonflowg fluids are represented by entalpy and ternal energy u, respectively, te mass and energy balances for tis uniform-flow system can be expressed as Mass balance: m m msystem m m m Energy balance: b, E 44 Net energy transfer W E Q m e e Cange ternal, ketic, potential, etc.energies m u Esystem 44 m u or Wb, Q m C T mcv T mc vt e p e e (sce ke pe 0) Te temperature of te air witdrawn from te cylder is assumed to be te average of itial and fal temperatures of te air te cylder. Tat is, T e / )( T + T ) (/ )(47 + ) ( T Air. kg 700 kpa 00 C Te volumes and te masses at te itial and fal states and te mass tat as escaped from te cylder are given by mrt V P (. kg)(0.87 kpa.m /kg.k)( K) 0.7 m (700 kpa) V 0.80V (0.80)(0.7) 0.86 m m m e m PV RT m (600 kpa)(0.86 m ) (0.87 kpa.m /kg.k) T kg T 89.8 kg T Notg tat te pressure remas constant after te piston starts movg, te boundary work is determed from Substitutg, Wb, P ( V V ) (600 kpa)( ) m 89.8 (0.7 kj/kg.k) T T 7.9 kj kj 40 kj. (.00 kj/kg.k)(/)(47 + T T ) (. kg)(0.7 kj/kg.k)(47 K) Te fal temperature may be obtaed from tis equation by a trial-error approac or usg EES to be T 45.0 K Ten, te amount of mass tat as escaped becomes m e kg 45.0 K Q

23 Te pressures across a pump are measured. Te mecanical efficiency of te pump and te temperature rise of water are to be determed. Assumptions Te flow is steady and compressible. Te pump is driven by an external motor so tat te eat generated by te motor is dissipated to te atmospere. Te elevation difference between te let and let of te pump is negligible, z z. 4 Te let and let diameters are te same and tus te let and exit velocities are equal, V V. Properties We take te density of water to be kg/l 000 kg/m and its specific eat to be 4.8 kj/kg C (Table A-). Analysis (a) Te mass flow rate of water troug te pump is ρv & ( kg/l)(50 L/s) 50 kg/s Te motor draws 5 kw of power and is 90 percent efficient. Tus te mecanical (saft) power it delivers to te pump is W & pump, saft η motorw & electric (0.90)(5 kw).5 kw PUMP Pump let Motor 5 kw To determe te mecanical efficiency of te pump, we need to know te crease te mecanical energy of te fluid as it flows troug te pump, wic is P V P V E& mec,fluid E& mec, E& mec, gz gz ρ ρ Simplifyg it for tis case and substitutg te given values, E & P P (00 00)kPa kj (50 kg/s) 000 kg/m kpa m mec, fluid Ten te mecanical efficiency of te pump becomes η pump ρ E& mec,fluid 0 kw % W&.5 kw pump,saft 0 kw (b) Of te.5-kw mecanical power supplied by te pump, only 0 kw is imparted to te fluid as mecanical energy. Te remag.5 kw is converted to termal energy due to frictional effects, and tis lost mecanical energy manifests itself as a eatg effect te fluid, E & mec, loss W& pump,saft E& mec,fluid kw Te temperature rise of water due to tis mecanical efficiency is determed from te termal energy balance, E& u u ) mc & T mec,loss Solvg for T, E& T mc & mec,loss (.5 kw 0.07 C (50 kg/s)(4.8 kj/kg K) Terefore, te water will experience a temperature rise of 0.07 C, wic is very small, as it flows troug te pump. Discussion In an actual application, te temperature rise of water will probably be less sce part of te eat generated will be transferred to te casg of te pump and from te casg to te surroundg air. If te entire pump motor were submerged water, ten te.5 kw dissipated to te air due to motor efficiency would also be transferred to te surroundg water as eat. Tis would cause te water temperature to rise more.

24 Heat is lost from te steam flowg a nozzle. Te exit velocity and te mass flow rate are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Potential energy cange is negligible. Tere are no work teractions. Analysis (a) We take te steam as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady-flow system can be expressed te rate form as Energy balance: E E & 44 & Rate of net energy transfer E& V + Rate of cange ternal, ketic, potential, etc.energies E& or V ( ) q 0 (steady) Esystem & 4444 V + + Q& 0 Te properties of steam at te let and exit are (Table A-6) Substitutg, P 00 kpa T 50 C P 75 kpa v sat. vap. V ( q 769. kj/kg.7 m /kg 66.4 kj/kg ) (b) Te mass flow rate of te steam is 50 C 00 kpa sce W& pe 0) kj/kg ( )kJ/kg 000 m /s AV (0.00 m )(40.7 m/s) 0.8kg/s v.7 m /kg STEAM q 40.7 m/s 75 kpa Sat. vap.

25 Te turbocarger of an ternal combustion enge consistg of a turbe, a compressor, and an aftercooler is considered. Te temperature of te air at te compressor let and te mimum flow rate of ambient air are to be determed. Assumptions All processes are steady sce tere is no cange wit time. Ketic and potential energy canges are negligible. Air properties are used for exaust gases. 4 Air is an ideal gas wit constant specific eats. 5 Te mecanical efficiency between te turbe and te compressor is 00%. 6 All devices are adiabatic. 7 Te local atmosperic pressure is 00 kpa. Properties Te constant pressure specific eats of exaust gases, warm air, and cold ambient air are taken to be c p.06,.008, and.005 kj/kg K, respectively (Table A-b). Analysis (a) An energy balance on turbe gives W& & Turbe Exaust gases Cold air Air Compressor ( T ) (0.0 kg/s)(.06 kj/kg K)(400 50)K.06 kw T mexc p,ex ex, Tex, Aftercooler Tis is also te power put to te compressor sce te mecanical efficiency between te turbe and te compressor is assumed to be 00%. An energy balance on te compressor gives te air temperature at te compressor let W& C c a p,a ( T T.06 kw (0.08 kg/s)(.008 kj/kg K)( T a, a, ) a, 50)K T a, 08.6 C (b) An energy balance on te aftercooler gives te mass flow rate of cold ambient air c a p,a ( T (0.08 kg/s)(.008 kj/kg C)( ) C a, T a, ) c ca ca p,ca ca ( T ca, (.005 kj/kg C)(40 0) C kg/s Te volume flow rate may be determed if we first calculate specific volume of cold ambient air at te let of aftercooler. Tat is, v RT ( ca V& P 0.87 kj/kg K)(0 + 7 K) 00 kpa m /kg T (0.056 kg/s)( m /kg) m /s 44.9 L/s ca v ca ca, )

26 5-44 Fundamentals of Engeerg (FE) Exam Problems 5-00 Steam is accelerated by a nozzle steadily from a low velocity to a velocity of 0 m/s at a rate of. kg/s. If te temperature and pressure of te steam at te nozzle exit are 400 C and MPa, te exit area of te nozzle is (a) 4.0 cm (b) 8.4 cm (c) 0. cm (d) 5 cm (e).0 cm Answer (e).0 cm Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). Vel_0 "m/s" Vel_0 "m/s" m. "kg/s" T400 "C" P000 "kpa" "Te rate form of energy balance is E_dot_ - E_dot_ DELTAE_dot_cv" vvolume(steam_iapws,tt,pp) m(/v)*a*vel_ "A m^" "Some Wrong Solutions wit Common Mistakes:" R0.465 "kj/kg.k" P*videalR*(T+7) m(/videal)*w_a*vel_ "assumg ideal gas" P*videalR*T m(/videal)*w_a*vel_ "assumg ideal gas and usg C" mw_a*vel_ "not usg specific volume" 5-0 Steam enters a diffuser steadily at 0.5 MPa, 00 C, and m/s at a rate of.5 kg/s. Te let area of te diffuser is (a) 5 cm (b) 50 cm (c) 05 cm (d) 50 cm (e) 90 cm Answer (b) 50 cm Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). Vel_ "m/s" m.5 "kg/s" T00 "C" P500 "kpa" "Te rate form of energy balance is E_dot_ - E_dot_ DELTAE_dot_cv" vvolume(steam_iapws,tt,pp) m(/v)*a*vel_ "A m^" "Some Wrong Solutions wit Common Mistakes:" R0.465 "kj/kg.k"

27 5-45 P*videalR*(T+7) m(/videal)*w_a*vel_ "assumg ideal gas" P*videalR*T m(/videal)*w_a*vel_ "assumg ideal gas and usg C" mw_a*vel_ "not usg specific volume" 5-0 An adiabatic eat excanger is used to eat cold water at 5 C enterg at a rate of 5 kg/s by ot air at 90 C enterg also at rate of 5 kg/s. If te exit temperature of ot air is 0 C, te exit temperature of cold water is (a) 7 C (b) C (c) 5 C (d) 85 C (e) 90 C Answer (b) C Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). C_w4.8 "kj/kg-c" Cp_air.005 "kj/kg-c" Tw5 "C" m_dot_w5 "kg/s" Tair90 "C" Tair0 "C" m_dot_air5 "kg/s" "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" m_dot_air*cp_air*(tair-tair)m_dot_w*c_w*(tw-tw) "Some Wrong Solutions wit Common Mistakes:" (Tair-Tair)(W_Tw-Tw) "Equatg temperature canges of fluids" Cv_air0.78 "kj/kg.k" m_dot_air*cv_air*(tair-tair)m_dot_w*c_w*(w_tw-tw) "Usg Cv for air" W_TwTair "Settg let temperature of ot fluid exit temperature of cold fluid" W4_TwTair "Settg exit temperature of ot fluid exit temperature of cold fluid" 5-0 A eat excanger is used to eat cold water at 5 C enterg at a rate of kg/s by ot air at 00 C enterg at rate of kg/s. Te eat excanger is not sulated, and is loosg eat at a rate of 40 kj/s. If te exit temperature of ot air is 0 C, te exit temperature of cold water is (a) 44 C (b) 49 C (c) 9 C (d) 7 C (e) 95 C Answer (c) 9 C Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). C_w4.8 "kj/kg-c" Cp_air.005 "kj/kg-c" Tw5 "C" m_dot_w "kg/s" Tair00 "C" Tair0 "C"

28 5-46 m_dot_air "kg/s" Q_loss40 "kj/s" "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" m_dot_air*cp_air*(tair-tair)m_dot_w*c_w*(tw-tw)+q_loss "Some Wrong Solutions wit Common Mistakes:" m_dot_air*cp_air*(tair-tair)m_dot_w*c_w*(w_tw-tw) "Not considerg Q_loss" m_dot_air*cp_air*(tair-tair)m_dot_w*c_w*(w_tw-tw)-q_loss "Takg eat loss as eat ga" (Tair-Tair)(W_Tw-Tw) "Equatg temperature canges of fluids" Cv_air0.78 "kj/kg.k" m_dot_air*cv_air*(tair-tair)m_dot_w*c_w*(w4_tw-tw)+q_loss "Usg Cv for air" 5-04 An adiabatic eat excanger is used to eat cold water at 5 C enterg at a rate of 5 kg/s by ot water at 90 C enterg at rate of 4 kg/s. If te exit temperature of ot water is 50 C, te exit temperature of cold water is (a) 4 C (b) 47 C (c) 55 C (d) 78 C (e) 90 C Answer (b) 47 C Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). C_w4.8 "kj/kg-c" Tcold_5 "C" m_dot_cold5 "kg/s" Tot_90 "C" Tot_50 "C" m_dot_ot4 "kg/s" Q_loss0 "kj/s" "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" m_dot_ot*c_w*(tot_-tot_)m_dot_cold*c_w*(tcold_-tcold_)+q_loss "Some Wrong Solutions wit Common Mistakes:" Tot_-Tot_W_Tcold_-Tcold_ "Equatg temperature canges of fluids" W_Tcold_90 "Takg exit temp of cold fluidlet temp of ot fluid" 5-05 In a sower, cold water at 0 C flowg at a rate of 5 kg/m is mixed wit ot water at 60 C flowg at a rate of kg/m. Te exit temperature of te mixture will be (a) 4. C (b) 5.0 C (c) 40.0 C (d) 44. C (e) 55. C Answer (a) 4. C Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). C_w4.8 "kj/kg-c" Tcold_0 "C" m_dot_cold5 "kg/m"

29 5-47 Tot_60 "C" m_dot_ot "kg/m" "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" m_dot_ot*c_w*tot_+m_dot_cold*c_w*tcold_(m_dot_ot+m_dot_cold)*c_w*tmix "Some Wrong Solutions wit Common Mistakes:" W_Tmix(Tcold_+Tot_)/ "Takg te average temperature of let fluids" 5-06 In a eatg system, cold door air at 0 C flowg at a rate of 6 kg/m is mixed adiabatically wit eated air at 70 C flowg at a rate of kg/m. Te exit temperature of te mixture is (a) 0 C (b) 40 C (c) 45 C (d) 55 C (e) 85 C Answer (a) 0 C Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). C_air.005 "kj/kg-c" Tcold_0 "C" m_dot_cold6 "kg/m" Tot_70 "C" m_dot_ot "kg/m" "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" m_dot_ot*c_air*tot_+m_dot_cold*c_air*tcold_(m_dot_ot+m_dot_cold)*c_air*tmix "Some Wrong Solutions wit Common Mistakes:" W_Tmix(Tcold_+Tot_)/ "Takg te average temperature of let fluids" 5-07 Hot combustion gases (assumed to ave te properties of air at room temperature) enter a gas turbe at MPa and 500 K at a rate of 0. kg/s, and exit at 0. MPa and 900 K. If eat is lost from te turbe to te surroundgs at a rate of 5 kj/s, te power put of te gas turbe is (a) 5 kw (b) 0 kw (c) 45 kw (d) 60 kw (e) 75 kw Answer (c) 45 kw Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). Cp_air.005 "kj/kg-c" T500 "K" T900 "K" m_dot0. "kg/s" Q_dot_loss5 "kj/s" "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" W_dot_+Q_dot_lossm_dot*Cp_air*(T-T) "Alternative: Variable specific eats - usg EES data" W_dot_variable+Q_dot_lossm_dot*(ENTHALPY(Air,TT)-ENTHALPY(Air,TT)) "Some Wrong Solutions wit Common Mistakes:" W_Wm_dot*Cp_air*(T-T) "Disregardg eat loss" W_W-Q_dot_lossm_dot*Cp_air*(T-T) "Assumg eat ga stead of loss"

30 Steam expands a turbe from 4 MPa and 500 C to 0.5 MPa and 50 C at a rate of 50 kg/. Heat is lost from te turbe at a rate of 5 kj/s durg te process. Te power put of te turbe is (a) 57 kw (b) 07 kw (c) 8 kw (d) 87 kw (e) 46 kw Answer (a) 57 kw Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). T500 "C" P4000 "kpa" T50 "C" P500 "kpa" m_dot50/600 "kg/s" Q_dot_loss5 "kj/s" ENTHALPY(Steam_IAPWS,TT,PP) ENTHALPY(Steam_IAPWS,TT,PP) "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" W_dot_+Q_dot_lossm_dot*(-) "Some Wrong Solutions wit Common Mistakes:" W_Wm_dot*(-) "Disregardg eat loss" W_W-Q_dot_lossm_dot*(-) "Assumg eat ga stead of loss" uintenergy(steam_iapws,tt,pp) uintenergy(steam_iapws,tt,pp) W_W+Q_dot_lossm_dot*(u-u) "Usg ternal energy stead of entalpy" W4_W-Q_dot_lossm_dot*(u-u) "Usg ternal energy and wrong direction for eat" 5-09 Steam is compressed by an adiabatic compressor from 0. MPa and 50 C to 500 MPa and 50 C at a rate of.0 kg/s. Te power put to te compressor is (a) 44 kw (b) 4 kw (c) 48 kw (d) 77 kw (e) 90 kw Answer (a) 44 kw Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). "Note: Tis compressor violates te nd law. Cangg State to 800 kpa and 50C will correct tis problem (it would give 5 kw)" P00 "kpa" T50 "C" P500 "kpa" T50 "C" m_dot.0 "kg/s" Q_dot_loss0 "kj/s" ENTHALPY(Steam_IAPWS,TT,PP) ENTHALPY(Steam_IAPWS,TT,PP) "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" W_dot_-Q_dot_lossm_dot*(-)

31 5-49 "Some Wrong Solutions wit Common Mistakes:" W_W-Q_dot_loss(-)/m_dot "Dividg by mass flow rate stead of multiplyg" W_W-Q_dot_loss- "Not considerg mass flow rate" uintenergy(steam_iapws,tt,pp) uintenergy(steam_iapws,tt,pp) W_W-Q_dot_lossm_dot*(u-u) "Usg ternal energy stead of entalpy" W4_W-Q_dot_lossu-u "Usg ternal energy and ignorg mass flow rate" 5-0 Refrigerant-4a is compressed by a compressor from te saturated vapor state at 0.4 MPa to. MPa and 70 C at a rate of 0.08 kg/s. Te refrigerant is cooled at a rate of.0 kj/s durg compression. Te power put to te compressor is (a) 5.54 kw (b) 7. kw (c) 6.64 kw (d) 7.74 kw (e) 8. kw Answer (d) 7.74 kw Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). P40 "kpa" x P00 "kpa" T70 "C" m_dot0.08 "kg/s" Q_dot_loss.0 "kj/s" ENTHALPY(R4a,xx,PP) ENTHALPY(R4a,TT,PP) "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" W_dot_-Q_dot_lossm_dot*(-) "Some Wrong Solutions wit Common Mistakes:" W_W+Q_dot_lossm_dot*(-) "Wrong direction for eat transfer" W_W m_dot*(-) "Not considerg eat loss" uintenergy(r4a,xx,pp) uintenergy(r4a,tt,pp) W_W-Q_dot_lossm_dot*(u-u) "Usg ternal energy stead of entalpy" W4_W+Q_dot_lossu-u "Usg ternal energy and wrong direction for eat transfer" 5- Refrigerant-4a expands an adiabatic turbe from. MPa and 00 C to 0.8 MPa and 50 C at a rate of.5 kg/s. Te power put of te turbe is (a) 46. kw (b) 66.4 kw (c) 7.7 kw (d) 89. kw (e).0 kw Answer (a) 46. kw Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). P00 "kpa" T00 "C" P80 "kpa"

32 5-50 T50 "C" m_dot.5 "kg/s" Q_dot_loss0 "kj/s" ENTHALPY(R4a,TT,PP) ENTHALPY(R4a,TT,PP) "Te rate form of energy balance for a steady-flow system is E_dot_ E_dot_" -W_dot_-Q_dot_lossm_dot*(-) "Some Wrong Solutions wit Common Mistakes:" -W_W-Q_dot_loss(-)/m_dot "Dividg by mass flow rate stead of multiplyg" -W_W-Q_dot_loss- "Not considerg mass flow rate" uintenergy(r4a,tt,pp) uintenergy(r4a,tt,pp) -W_W-Q_dot_lossm_dot*(u-u) "Usg ternal energy stead of entalpy" -W4_W-Q_dot_lossu-u "Usg ternal energy and ignorg mass flow rate" 5- Refrigerant-4a at.4 MPa and 90 C is trottled to a pressure of 0.6 MPa. Te temperature of te refrigerant after trottlg is (a) C (b) 56 C (c) 8 C (d) 80 C (e) 90.0 C Answer (d) 80 C Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values). P400 "kpa" T90 "C" P600 "kpa" ENTHALPY(R4a,TT,PP) TTEMPERATURE(R4a,,PP) "Some Wrong Solutions wit Common Mistakes:" W_TT "Assumg te temperature to rema constant" W_TTEMPERATURE(R4a,x0,PP) "Takg te temperature to be te saturation temperature at P" uintenergy(r4a,tt,pp) W_TTEMPERATURE(R4a,uu,PP) "Assumg uconstant" vvolume(r4a,tt,pp) W4_TTEMPERATURE(R4a,vv,PP) "Assumg vconstant" 5- Air at 0 C and 5 atm is trottled by a valve to atm. If te valve is adiabatic and te cange ketic energy is negligible, te exit temperature of air will be (a) 0 C (b) 4 C (c) 7 C (d) 0 C (e) 4 C Answer (d) 0 C Solution Solved by EES Software. Solutions can be verified by copyg-and-pastg te followg les on a blank EES screen. (Similar problems and teir solutions can be obtaed easily by modifyg numerical values).