SOLUTIONS: HOMEWORK #6

AREN 2110FALL 2006HOMEWORK ASSIGNMENTS 6, 7 and 8SOLUTIONSSOLUTIONS: HOMEWORK #6Chapter 5 Problems5-45 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heatneeds to be removed from the water in order to keep its temperature constant is to be determined.Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperaturebefore and after quenching. 3 The changes in kinetic and potential energies are negligible.Properties The density and specific heat of the brass balls are given to be ρ 8522 kg/m3 and Cp 0.385kJ/kg. C.Analysis We take a single ball as the system. The energy balance for this closedsystem can be expressed asE Eout1in4243 Net energy transferby heat, work, and massΔEsystem14243Brass balls, 120 CChange in internal, kinetic,potential, etc. energies Qout ΔU ball m (u2 u1 )Water bath, 5 CQout mC (T1 T2 )The total amount of heat transfer from a ball ism ρV ρQoutπD 3 (8522 kg / m 3 )π (0.05 m) 3 0.558 kg66 mC(T1 T2 ) (0.558 kg)(0.385 kJ / kg. C)(120 74 ) C 9.88 kJ / ballThen the rate of heat transfer from the balls to the water becomesQ& n& Q (100 balls / min) (9.88 kJ / ball) 988 kJ / mintotalballballTherefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperatureconstant at 50 C since energy input must be equal to energy output for a system whose energy levelremains constant. That is, Ein Eout when ΔEsystem 0 .5-58C It is mostly converted to internal energy as shown by a rise in the fluid temperature.5-59C The kinetic energy of a fluid increases at the expense of the internal energy as evidenced by adecrease in the fluid temperature.

AREN 2110FALL 2006HOMEWORK ASSIGNMENTS 6, 7 and 8SOLUTIONS5-61 Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate, the exit temperature, andthe exit area of the nozzle are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas withconstant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heattransfer is negligible. 5 There are no work interactions.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The specific heat of air at theanticipated average temperature of 450 K is Cp 1.02 kJ/kg. C (Table A-2).&1 m&2 m& . Using the ideal gas relation, theAnalysis (a) There is only one inlet and one exit, and thus mspecific volume and the mass flow rate of air are determined to bev1 RT1 (0.287 kPa m 3 /kg K )(473 K ) 0.4525 m 3 /kgP1300 kPam& 11A1 V1 (0.008m 2 )(30m/s) 0.5304 kg/sv10.4525m 3 /kg(b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energybalance for this steady-flow system can be expressed in the rate form asE& E& out 0ΔE& system Ê0 (steady)1in4243144424443Rate of net energy transferby heat, work, and massRate of change in internal, kinetic,potential, etc. energiesE& in E& out& Δpe 0)m& (h1 V12 / 2) m& (h2 V22 /2) (since Q& W0 h2 h1 V22 V12V 2 V12 0 C p , ave (T2 T1 ) 222Substituting,0 (1.02 kJ/kg K )(T2 200 o C) It yields(180 m/s) 2 (30 m/s) 22 1 kJ/kg 1000 m 2 /s 2 T2 184.6 C(c) The specific volume of air at the nozzle exit isv2 m& RT2 (0.287 kPa m 3 /kg K )(184.6 273 K ) 1.313 m 3 /kgP2100 kPa11A2 V2 0.5304kg/s A2 (180m/s )v21.313m 3 /kgP1 300 kPaT1 200 CV1 30 m/sA1 80 cm2A2 0.00387 m2 38.7 cm25-77C Yes. Because energy (in the form of shaft work) is being added to the air.AIRP2 100 kPaV2 180 m/s

AREN 2110FALL 2006HOMEWORK ASSIGNMENTS 6, 7 and 8SOLUTIONS5-79 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet areaare to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible.Properties From the steam tables (Tables A-4 through 6)P1 10MPa v1 0.02975m 3 /kg T1 450 o C h1 3240.9kJ/kgP1 10 MPaT1 450 CV1 80 m/sandP2 10 kPa h2 h f x 2 h fg 191.83 0.92 2392.8 2393.2kJ/kgx 2 0.92 Analysis (a) The change in kinetic energy is determined fromV 2 V12 (50m/s)2 (80m/s) 2Δke 2 22 1kJ/kg 1000m 2 /s 2 Rate of net energy transferby heat, work, and massΔE& system Ê0 (steady)144424443 0Rate of change in internal, kinetic,potential, etc. energiesE& in E& out& Δpe 0)m& (h1 V12 / 2) W& out m& (h2 V22 /2) (since Q V 2 V12W& out m& h2 h1 22 Then the power output of the turbine is determined by substitution to beW& out (12 kg/s)(2393.2 3240.9 1.95)kJ/kg 10.2 MW(c) The inlet area of the turbine is determined from the mass flow rate relation,m& · 1.95kJ/kg &1 m&2 m& . We take the(b) There is only one inlet and one exit, and thus mturbine as the system, which is a control volume since mass crosses theboundary. The energy balance for this steady-flow system can be expressedin the rate form asE& E& out1in4243STEAM·m 12 kg/sm& v(12 kg/s )(0.02975 m 3 /kg )1A1 V1 A1 1 0.00446 m 2v1V180 m/sWP2 10 kPax2 0.92V2 50 m/s

AREN 2110FALL 2006HOMEWORK ASSIGNMENTS 6, 7 and 8SOLUTIONS5-90 Helium is compressed by a compressor. For a mass flow rate of 90 kg/min, the power input required isto be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Helium is an ideal gas with constant specific heats.Properties The constant pressure specific heat of helium is Cp 5.1926 kJ/kg·K (Table A-2a).&1 m&2 m& . We take the compressor as theAnalysis There is only one inlet and one exit, and thus msystem, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form asE& E& out1in4243ΔE& system Ê0 (steady)144424443 Rate of net energy transferby heat, work, and mass 0P2 700 kPaT2 430 KRate of change in internal, kinetic,potential, etc. energiesE& in E& out& 1 Q& out mh& 2 (since Δke Δpe 0)W& in mh&&& p (T2 T1 )Win Qout m& (h2 h1 ) mCHe·m 90kg/mi·WThus,W& in Q& out m& C p (T2 T1 ) (90/60kg/s)(20 kJ/kg) (90/60kg/s)(5.1926kJ/kg K)(430 310)K 965kWP1 120 kPaT1 310 K5-104 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the massflow rate of cold water is to be determined.Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat lossto the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams arenegligible. 4 Fluid properties are constant. 5 There are no work interactions.Properties Noting that T Tsat @ 250 kPa 127.44 C, the water in all three streams exists as a compressedliquid, which can be approximated as a saturated liquid at the given temperature. Thus,h1 hf @ 80 C 334.91 kJ/kgh2 hf @ 20 C 83.96 kJ/kgh3 hf @ 42 C 175.92 kJ/kgAnalysis We take the mixing chamber as the system, which is a controlvolume. The mass and energy balances for this steady-flow system can beexpressed in the rate form asMass balance:Ê0m& in m& out ΔE& system(steady) 0 m& 1 m& 2 m& 3Energy balance:E& E& out1in4243 Rate of net energy transferby heat, work, and mass 0 (steady)ΔE& system1442443 0T1 80 C· 1 0.5 kg/smRate of change in internal, kinetic,potential, etc. energiesH2O(P 250 kPa)T3 42 CE& in E& outm& 1 h1 m& 2 h2 m& 3 h3 (since Q& W& Δke Δpe 0)T2 20 Cm· 2

AREN 2110FALL 2006HOMEWORK ASSIGNMENTS 6, 7 and 8SOLUTIONS5-104 CONTINUED& 2 givesCombining the two relations and solving for mm& 1h1 m& 2 h2 (m& 1 m& 2 )h3h1 h3&1mh3 h2Substituting, the mass flow rate of cold water stream is determined to be&2 mm& 2 (334.91 175.92)kJ/kg (0.5 kg/s ) 0.864(175.92 83.96)kJ/kgkg/s5-106 Feedwater is heated in a chamber by mixing it with superheated steam. If the mixture is saturatedliquid, the ratio of the mass flow rates of the feedwater and the superheated vapor is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heattransfer is negligible.Properties Noting that T Tsat @ 800 kPa 170.43 C, the cold water stream and the mixture exist as acompressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus,h1 hf @ 50 C 209.33 kJ/kgh3 hf @ 800 kPa 721.11 kJ/kgandP2 800kPa h2 2839.3kJ/kgT2 200 o C Analysis We take the mixing chamber as the system, which is a control volume since mass crosses theboundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as& in m& out Δm& system Ê0mMass balance:Energy balance:E& E& out1in4243 Rate of net energy transferby heat, work, and mass(steady) 0ΔE& system Ê0 (steady)144424443 & in m& outm 0&1 m&2 m&3 mT1 50 C·1mRate of change in internal, kinetic,potential, etc. energiesH2O(P 800 kPa)Sat. liquidE& in E& out& 1h1 m& 2 h2 m& 3h3mCombining the two,& 2 yieldsDividing by m(since Q& W& Δke Δpe 0)m& 1h1 m& 2 h2 (m& 1 m& 2 )h3y h1 h2 ( y 1)h3y Solving for y:h3 h2h1 h3&1 / m& 2 is the desired mass flow rate ratio. Substituting,where y my 721.11 2839.3 4.14209.33 721.11T2 200 Cm· 2

AREN 2110FALL 2006HOMEWORK ASSIGNMENTS 6, 7 and 8SOLUTIONSSOLUTIONS: HOMEWORK #7Chapter 6 Problems6-2C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity.6-3C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to aroom.6-10C Heat engines are cyclic devices that receive heat from a source, convert some of it to work, andreject the rest to a sink.6-12C No. Because 100% of the work can be converted to heat.6-15C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics.6-20 The power output and fuel consumption rate of a power plant are given.The overall efficiency is to be determined.Assumptions The plant operates steadily.Properties The heating value of coal is given to be 30,000 kJ/kg.60 t/hFurnacecoalHEAnalysis The rate of energy supply (in chemical form) to this power plant isQ& H m& coal u coal (60,000 kg/h )(30,000 kJ/kg ) 1.8 10 9 kJ/h 500 MWThen the overall efficiency of the plant becomesη overallW& net ,out 150 MW 0.300 30.0%500 MWQ& Hsink6-40C The difference between the two devices is one of purpose. The purpose of a refrigerator is toremove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm medium.6-47C No. The refrigerator captures energy from a cold medium and carries it to a warm medium. It doesnot create it.