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.

2 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. 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()(.)

3 ()(.)(.( ). 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.

4 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. 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.

5 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.

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= ()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 ()

7 M/s180 =V& A2 = m2 = cm2 5-77C Yes. Because energy (in the form of shaft work) is being added to the air. 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&==.

8 We take the turbine 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= + = +=+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.

9 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. 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.

10 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.