SOLUTIONS: HOMEWORK #6

1 AREN 2110 solutions FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 solutions : HOMEWORK #6 Chapter 5 Problems 5-45 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat needs 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 temperature before 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 = 8522 kg/m3 and Cp = kJ/kg. C. Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as EEEQUmuuQmCTTinoutoutballout = = = = Net energy transferby heat, work, and masssystemChange in internal, kinetic, potential, etc.

2 Energies1243412434 ()()2112 Brass balls, 120 CWater bath, 5 C The total amount of heat transfer from a ball is mVDQmCTTout== === = = 3126852200505580 5580 385120 749 88()(..()(.)(.( ). kg / m m)6 kg kg kJ / kg. C)C kJ / ball33 Then the rate of heat transfer from the balls to the water becomes &&((.)QnQtotal== =ballball balls / min) kJ / ball1009 88988 kJ / minTherefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature constant at 50 C since energy input must be equal to energy output for a system whose energy level remains constant. That is, EEEinout== when system 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 a decrease in the fluid temperature.

3 AREN 2110 solutions FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 &5-61 Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate, the exit temperature, and the 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 with constant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The gas constant of air is (Table A-1). The specific heat of air at the anticipated average temperature of 450 K is Cp = kJ/kg. C (Table A-2). Analysis (a) There is only one inlet and one exit, and thus &&mm m12==. Using the ideal gas relation, the specific volume and the mass flow rate of air are determined to be /kgm 300)K 473)(K/kgmkPa (33111= ==PRTv kg/s )m/s30)( ( (b) We take nozzle as the system, which is a control volume since mass crosses the boundary.

4 The energy balance for this steady-flow system can be expressed in the rate form as &&&&&EEEEE inoutinout ==Rate of net energy transfer by heat, work, and masssystem (steady)Rate of change in internal, kinetic, potential, etc. energies12434124443444 00= ()20200)peW (since /2)+()2/(212212,212212222211 VVVVVV + = + = =+TTChhQhmhmavep&&&& Substituting, + =22222/sm 1000kJ/kg 12)m/s 30()m/s 180()C200)(KkJ/kg (0oT It yields T2 = C (c) The specific volume of air at the nozzle exit is AIR P2 = 100 kPa V2 = 180 m/s P1 = 300 kPa T1 = 200 C V1 = 30 m/s A1 = 80 cm2 /kgm 100)K )(K/kgmkPa (33222=+ ==PRTv ()m/s180 =V& A2 = m2 = cm2 5-77C Yes. Because energy (in the form of shaft work) is being added to the air.

5 AREN 2110 solutions FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 5-79 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A-4 through 6) ==hvTPoSTEAM m = 12 kg/s P1 = 10 MPa T1 = 450 C V1 = 80 m/s P2 = 10 kPa x2 = V2 = 50 m/s W and 102222= +=+= ==fgfhxhhxP Analysis (a) The change in kinetic energy is determined from () = = = 22222122/sm1000kJ/kg12m/s)(80m/s502ke (b) There is only one inlet and one exit, and thus &&mm m12&==. We take the turbine as the system, which is a control volume since mass crosses the boundary.

6 The energy balance for this steady-flow system can be expressed in the rate form as &&&&&EEEEE inoutinout ==Rate of net energy transfer by heat, work, and masssystem (steady)Rate of change in internal, kinetic, potential, etc. energies12434124443444 00= + = +=+20)peQ (since /2)+()2/(212212out222out211 VVVVhhmWhmWhm&&&&&& Then the power output of the turbine is determined by substitution to be MW =kJ/kg) )(kg/s 12(outW&(c) The inlet area of the turbine is determined from the mass flow rate relation, 2m =m/s 80)/kgm )(kg/s 12(13111111vmAAvm&& AREN 2110 solutions FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 & 5-90 Helium is compressed by a compressor. For a mass flow rate of 90 kg/min, the power input required is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time.

7 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats. Properties The constant pressure specific heat of helium is Cp = kJ/kg K (Table A-2a). Analysis There is only one inlet and one exit, and thus &&mm m12==. We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as &&&&&EEEEE inoutinout ==Rate of net energy transfer by heat, work, and masssystem (steady)Rate of change in internal, kinetic, potential, etc. energies12434124443444 00=3He m=90kg/mi P2 = 700 kPa T2 = 430 K P1 = 120 kPa T1 = 310 K W &&&&&&&()&()WmhQ mhWQ mhh mCTTpinoutinout (since kepe0)+=+ = = 122121 Thus, ()kW965= = +=310)KK)(430kJ/kg26kg/s)( (90/60+kJ/kg) kg/s)(20(90/6012 TTCmQWpoutin&&& 5-104 A hot water stream is mixed with a cold water stream.

8 For a specified mixture temperature, the mass flow rate of cold water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat loss to the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There are no work interactions. Properties Noting that T < Tsat @ 250 kPa = C, the water in all three streams exists as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus, h1 hf @ 80 C = kJ/kg h2 hf @ 20 C = kJ/kg h3 hf @ 42 C = kJ/kg Analysis We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance: &&&&&&mmEmmminout == += system (steady) 0120 Energy balance: H2O (P = 250 kPa) T3 = 42 C T1 = 80 C m1 = kg/s T2 = 20 C m2 0)peke (since 0332211energies etc.

9 Potential, kinetic, internal,in change of Rate(steady) 0systemmass and work,heat,by nsferenergy tranet of Rate ===+== = WQhmhmhmEEEEE outinoutin&&&&&&&4434421&43421&& AREN 2110 solutions FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 5-104 CONTINUED Combining the two relations and solving for gives &m2 ()3212211hmmhmhm&&&&+=+ &&mhhhhm213321= Substituting, the mass flow rate of cold water stream is determined to be ()()()kg/s =kg/s 5-106 Feedwater is heated in a chamber by mixing it with superheated steam. If the mixture is saturated liquid, 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 potential energy changes are negligible.

10 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties Noting that T < Tsat @ 800 kPa = C, the cold water stream and the mixture exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus, h1 hf @ 50 C = kJ/kg h3 hf @ 800 kPa = kJ/kg and ==hTPo Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance: && &&& &&&mmmm mmmminoutinout == = += system (steady) 01203 Energy balance: H2O (P = 800 kPa) Sat. liquid T1 = 50 C m1T2 = 200 C m2 &&&&&&& &&&EEEEEmh m hmhQ Winoutinout ===+= Rate of net energy transfer by heat, work, and masssystem (steady)Rate of change in internal, kinetic, potential, etc.