A 8-60
8-79 rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and R134a is allowed to enter the tank. The mass of the R-134a that entered the tank and the exergy destroyed during this process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of refrigerant are (Tables A-11 through A-13) v 1 = v g @ 1.2 MPa = 0.01672 m 3 / kg P1 = 1.2 MPa 1.6 MPa R-134a u1 = u g @ 1.2 MPa = 253.81 kJ/kg 30C sat. vapor s =s 1 g @ 1.2 MPa = 0.91303 kJ/kg K
v 2 = v f @ 1.4 MPa = 0.0009166 m 3 / kg T2 = 1.4 MPa u 2 = u f @ 1.4 MPa = 125.94 kJ/kg sat. liquid s =s 2 f @ 1.4 MPa = 0.45315 kJ/kg K
Pi = 1.6 MPa hi = 93.56 kJ/kg Ti = 30C s i = 0.34554 kJ/kg K Analysis We take the tank as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min - m out = msystem mi = m 2 - m1
R-134a 0.1 m3 1.2 MPa Sat. vapor
Q
Energy balance:
Net energy transfer by heat, work, and mass
E in - E out 1 24 4 3
=
Change in internal, kinetic, potential, etc. energies
E system 1 24 4 3
Qin + mi hi = m 2 u 2 - m1u1 (since W ke pe 0) (a) The initial and the final masses in the tank are V V 0.1 m 3 0.1 m 3 m1 = 1 = = 5.983 kg m2 = 2 = = 109.10 kg v 1 0.01672 m 3 /kg v 2 0.0009166 m 3 /kg Then from the mass balance mi = m 2 - m1 = 109.10 - 5.983 = 103.11 kg The heat transfer during this process is determined from the energy balance to be Qin = -mi hi + m 2 u 2 - m1u1
= -(103.11 kg )(93.56 kJ/kg) + (109.10)(125.94 kJ/kg ) - (5.983 kg )(253.81 kJ/kg ) = 2573 kJ (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy
balance on an extended system that includes the tank and its immediate surroundings so that the boundary temperature of the extended system is the surroundings temperature Tsurr at all times. It gives Q S in - S out + S gen = S system in + mi s i + S gen = S tank = (m 2 s 2 - m1 s1 ) tank 1 24 4 3 { 123 4 4 Tb,in
Net entropy transfer by heat and mass Entropy generation Change in entropy
S gen = m 2 s 2 - m1 s1 - mi s i -
Qin T0
Substituting, the exergy destruction is determined to be Q X destroyed = T0 S gen = T0 m 2 s 2 - m1 s1 - mi s i - in T0 = (318 K)[109.10 0.45315 - 5.983 0.91303 - 103.11 0.34554 - (2573 kJ)/(318 K)]
= 80.3 kJ
8-61
8-80 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, and half of mass in liquid form is withdrawn from the tank. The temperature in the tank is maintained constant. The amount of heat transfer, the reversible work, and the exergy destruction during this process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of water are (Tables A-4 through A-6)
v 1 = v f @170o C = 0.001114 m /kg T1 = 170C u1 = u f @170o C = 718.20 kJ/kg sat. liquid s1 = s f @170o C = 2.0417 kJ/kg K
Te = 170C he = h f @170o C = 719.08 kJ/kg sat. liquid s e = s f @170o C = 2.0417 kJ/kg K
3
H2O 0.6 m3 170C T = const.
Q
me
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: Energy balance:
min - m out = msystem m e = m1 - m 2
Net energy transfer by heat, work, and mass
E in - E out 1 24 4 3
=
Change in internal, kinetic, potential, etc. energies
E system 1 24 4 3
Qin = m e he + m 2 u 2 - m1u1 (since W ke pe 0)
The initial and the final masses in the tank are
m1 = m2 = 0.6 m 3 V = = 538.47 kg v 1 0.001114 m 3 /kg 1 1 m1 = (538.47 kg ) = 269.24 kg = m e 2 2
Now we determine the final internal energy and entropy,
v2 =
x2 =
V
m2
=
0.6 m 3 = 0.002229 m 3 /kg 269.24 kg = 0.002229 - 0.001114 = 0.004614 0.24260 - 0.001114
v 2 -v f v fg
T2 = 170C
u 2 = u f + x 2 u fg = 718.20 + (0.004614 )(1857.5) = 726.77 kJ/kg x 2 = 0.004614 s 2 = s f + x 2 s fg = 2.0417 + (0.004614 )(4.6233) = 2.0630 kJ/kg K
The heat transfer during this process is determined by substituting these values into the energy balance equation,
Qin = m e he + m 2 u 2 - m1u1 = (269.24 kg )(719.08 kJ/kg ) + (269.24 kg )(726.77 kJ/kg ) - (538.47 kg )(718.20 kJ/kg ) = 2545 kJ
8-62
(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on an extended system that includes the tank and the region between the tank and the source so that the boundary temperature of the extended system at the location of heat transfer is the source temperature Tsource at all times. It gives
Net entropy transfer by heat and mass
S in - S out 1 24 4 3
+ S gen = S system { 123 4 4
Entropy generation Change in entropy
Qin - m e s e + S gen = S tank = (m 2 s 2 - m1 s1 ) tank Tb,in S gen = m 2 s 2 - m1 s1 + m e s e - Qin Tsource
Substituting, the exergy destruction is determined to be
Q X destroyed = T0 S gen = T0 m 2 s 2 - m1 s1 + m e s e - in Tsource = (298 K)[269.24 2.0630 - 538.47 2.0417 + 269.24 2.0417 - (2545 kJ)/(523 K)] = 141.2 kJ
For processes that involve no actual work, the reversible work output and exergy destruction are identical. Therefore,
X destroyed = W rev,out - Wact,out W rev,out = X destroyed = 141.2 kJ
8-63
8-81E An insulated rigid tank equipped with an electric heater initially contains pressurized air. A valve is opened, and air is allowed to escape at constant temperature until the pressure inside drops to 30 psia. The amount of electrical work done and the exergy destroys are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of air remains constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats. 5 The environment temperature is given to be 70F. Properties The gas constant of air is R = 0.3704 psia.ft3/lbm.R (Table A-1E). The properties of air are (Table A-17E) Te = 600 R he = 143.47 Btu/lbm, T1 = 600 R u1 = 102.34 Btu/lbm
T2 = 600 R u 2 = 102.34 Btu/lbm Analysis We take the tank as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min - m out = msystem m e = m1 - m 2
Energy balance:
Net energy transfer by heat, work, and mass
E in - E out 1 24 4 3
=
Change in internal, kinetic, potential, etc. energies
E system 1 24 4 3
We,in - m e he = m 2 u 2 - m1u1 (since Q ke pe 0)
The initial and the final masses of air in the tank are PV (75 psia)(150 ft 3 ) m1 = 1 = = 50.62 lbm RT1 (0.3704 psia ft 3 /lbm R)(600 R )
P2V (30 psia)(150 ft 3 ) = = 20.25 lbm RT2 (0.3704 psia ft 3 /lbm R)(600 R ) Then from the mass and energy balances, me = m1 - m2 = 50.62 - 20.25 = 30.37 lbm We,in = m e he + m 2 u 2 - m1u1 m2 =
AIR 150 ft3 75 psia 140F
We
= (30.37 lbm)(143.47 Btu/lbm) + (20.25 lbm)(102.34 Btu/lbm) - (50.62 lbm)(102.34 Btu/lbm) = 1249 Btu (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation Sgen is determined from an entropy balance on
the insulated tank. It gives S in - S out + S gen = S system -m e s e + S gen = S tank = (m 2 s 2 - m1 s1 ) tank 1 24 4 3 { 123 4 4
Net entropy transfer by heat and mass Entropy generation Change in entropy
S gen = m 2 s 2 - m1 s1 + m e s e = m 2 s 2 - m1 s1 + (m1 - m 2 ) s e = m 2 ( s 2 - s e ) - m1 ( s1 - s e )
Assuming a constant average pressure of (75 + 30) / 2 = 52.5 psia for the exit stream, the entropy changes are determined to be
T P 30 psia s 2 - s e = c p ln 2 - R ln 2 = -(0.06855 Btu/lbm R) ln = 0.03836 Btu/lbm R Te Pe 52.5 psia T P 75 psia s1 - s e = c p ln 1 - R ln 1 = -(0.06855 Btu/lbm R) ln = -0.02445 Btu/lbm R Te Pe 52.5 psia Substituting, the exergy destruction is determined to be X destroyed = T0 S gen = T0 [m 2 ( s 2 - s e ) - m1 ( s1 - s e )] = (530 R)[(20.25 lbm)(0.03836 Btu/lbm R) - (50.62 lbm)(-0.02445 Btu/lbm R)] = 1068 Btu
0 0
8-64
8-82 A rigid tank initially contains saturated liquid-vapor mixture of refrigerant-134a. A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank at constant pressure until no liquid remains inside. The final mass in the tank and the reversible work associated with this process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process. It can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of R-134a are (Tables A-11 through A-13)
P1 = 800 kPa v f = 0.0008458 m 3 /kg, v g = 0.025621 m 3 /kg u f = 94.79 kJ/kg, u g = 246.79 kJ/kg s f = 0.35404 kJ/kg.K, s g = 0.91835 kJ/kg.K
R-134a 800 kPa P = const.
Source 60C Q
v 2 = v g @ 800 kPa = 0.02562 m 3 / kg P2 = 800 kPa u 2 = u g @ 800 kPa = 246.79 kJ/kg sat. vapor s =s 2 g @ 800 kPa = 0.91835 kJ/kg K
Pe = 800 kPa he = h f @800 kPa = 95.47 kJ/kg sat. liquid s e = s f @800 kPa = 0.35404 kJ/kg.K
me
Analysis (b) We take the tank as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: Energy balance:
min - m out = msystem m e = m1 - m 2
Net energy transfer by heat, work, and mass
E in - E out 1 24 4 3
=
Change in internal, kinetic, potential, etc. energies
E system 1 24 4 3
Qin = m e he + m 2 u 2 - m1u1 (since W ke pe 0)
The initial mass, initial internal energy, and final mass in the tank are
m1 = m f + m g =
Vf vf
+
Vg vg
=
0.1 0.3 m 3 0.0008458 m 3 / kg
+
0.1 0.7 m 3 0.025621 m 3 / kg
= 35.470 + 2.732 = 38.202 kg
U 1 = m1u1 = m f u f + m g u g = 35.470 94.79 + 2.732 246.79 = 4036.4 kJ S1 = m1 s1 = m f s f + m g s g = 35.470 0.35404 + 2.732 0.91835 = 15.067 kJ/K m2 = 0.1 m 3 V = = 3.903 kg v 2 0.02562 m 3 /kg
Then from the mass and energy balances,
me = m1 - m 2 = 38.202 - 3.903 = 34.299 kg Qin = (34.299 kg)(95.47 kJ/kg) + (3.903 kg)(246.79 kJ/kg) - 4036.4 kJ = 201.2 kJ
(b) This process involves no actual work, thus the reversible work and exergy generation are identical since X destroyed = W rev,out - Wact,out W rev,out = X destroyed . The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on an extended system that includes the tank and the region between the tank and the heat source so that the boundary temperature of the extended system at the location of heat transfer is the source temperature Tsource at all times. It gives
8-65
Net entropy transfer by heat and mass
S in - S out 1 24 4 3
+ S gen = S system { 123 4 4
Entropy generation Change in entropy
Qin - m e s e + S gen = S tank = (m 2 s 2 - m1 s1 ) tank Tb,in S gen = m 2 s 2 - m1 s1 + m e s e - Qin Tsource
Substituting,
Q W rev,out = X destroyed = T0 S gen = T0 m 2 s 2 - m1 s1 + m e s e - in Tsource = (298 K)[3.903 0.91835 - 15.067 + 34.299 0.35404 - 201.2 / 333)] = 16.87 kJ
That is, 16.87 kJ of work could have been produced during this process.
8-66
8-83 A cylinder initially contains helium gas at a specified pressure and temperature. A valve is opened, and helium is allowed to escape until its volume decreases by half. The work potential of the helium at the initial state and the exergy destroyed during the process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process by using constant average properties for the helium leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved other than boundary work. 4 The tank is insulated and thus heat transfer is negligible. 5 Helium is an ideal gas with constant specific heats. Properties The gas constant of helium is R = 2.0769 kPa.m3/kg.K = 2.0769 kJ/kg.K. The specific heats of helium are cp = 5.1926 kJ/kg.K and cv = 3.1156 kJ/kg.K (Table A-2). Analysis (a) From the ideal gas relation, the initial and the final masses in the cylinder are determined to be
m1 = P1V (300 kPa)(0.1 m 3 ) = = 0.0493 kg RT1 (2.0769 kPa m 3 /kg K)(293 K )
m e = m 2 = m1 / 2 = 0.0493 / 2 = 0.0247 kg
The work potential of helium at the initial state is simply the initial exergy of helium, and is determined from the closed-system exergy relation,
1 = m1 = m1 [(u1 - u 0 ) - T0 ( s1 - s 0 ) + P0 (v 1 - v 0 )]
where
v1 =
RT1 (2.0769 kPa m 3 /kg K)(293 K ) = = 2.0284 m 3 /kg P1 300 kPa
RT (2.0769 kPa m 3 /kg K)(293 K ) v0 = 0 = = 6.405 m 3 /kg P0 95 kPa
and
s1 - s 0 = c p ln T1 P - R ln 1 T0 P0
HELIUM 300 kPa 0.1 m3 20C
Q
= (5.1926 kJ/kg K) ln = -2.28 kJ/kg K
293 K 300 kPa - (2.0769 kJ/kg K) ln 293 K 100 kPa
Thus,
1 = (0.0493 kg){(3.1156 kJ/kg K)(20 - 20)C - (293 K)(-2.28 kJ/kg K) + (95 kPa)(2.0284 - 6.405)m 3 /kg[kJ/kPa m 3 ]} = 12.44 kJ
(b) We take the cylinder as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: Energy balance:
min - m out = msystem m e = m1 - m 2
Net energy transfer by heat, work, and mass
E in - E out 1 24 4 3
=
Change in internal, kinetic, potential, etc. energies
E system 1 24 4 3
Qin - m e he + W b,in = m 2 u 2 - m1u1
Combining the two relations gives
8-67
Qin = (m1 - m 2 )he + m 2 u 2 - m1u1 - W b,in = (m1 - m 2 )he + m 2 h2 - m1 h1 = (m1 - m 2 + m 2 - m1 )h1 =0
since the boundary work and U combine into H for constant pressure expansion and compression processes. The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation Sgen can be determined from an entropy balance on the cylinder. Noting that the pressure and temperature of helium in the cylinder are maintained constant during this process and heat transfer is zero, it gives
Net entropy transfer by heat and mass
S in - S out 1 24 4 3
+ S gen = S system { 123 4 4
Entropy generation Change in entropy
- m e s e + S gen = S cylinder = (m 2 s 2 - m1 s1 ) cylinder S gen = m 2 s 2 - m1 s1 + m e s e = m 2 s 2 - m1 s1 + (m1 - m 2 ) s e = (m 2 - m1 + m1 - m 2 ) s1 =0
since the initial, final, and the exit states are identical and thus se = s2 = s1. Therefore, this discharge process is reversible, and
X destroyed = T0 Sgen = 0
8-68
8-84 A rigid tank initially contains saturated R-134a vapor at a specified pressure. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The amount of heat transfer with the surroundings and the exergy destruction are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is from the tank (will be verified). 1.4 MPa Properties The properties of refrigerant are (Tables A-11 through A-13) R-134a 60C u1 = u g @ 1 MPa = 250.68 kJ/kg P1 = 1 MPa s1 = s g @ 1 MPa = 0.91558 kJ/kg K sat.vapor v 1 = v g @ 1 MPa = 0.020313 m 3 / kg R-134a
0.2 m Pi = 1.4 MPa hi = 285.47 kJ/kg Q 1 MPa Sat. vapor Ti = 60C s i = 0.93889 kJ/kg K Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min - m out = msystem mi = m 2 - m1
3
Energy balance:
Net energy transfer by heat, work, and mass
E in - E out 1 24 4 3
=
Change in internal, kinetic, potential, etc. energies
E system 1 24 4 3
mi hi - Qout = m 2 u 2 - m1u1 (since W ke pe 0) The initial and the final masses in the tank are 0.2 m 3 V m1 = = = 9.846 kg v 1 0.020313 m 3 / kg m2 = m f + m g =
Vf vf
+
Vg vg
=
0.1 m 3 0.0008934 m 3 / kg
+
0.1 m 3 0.016715 m 3 / kg
= 111.93 + 5.983 = 117.91 kg
U 2 = m 2 u 2 = m f u f + m g u g = 111.93 116.70 + 5.983 253.81 = 14,581 kJ S 2 = m 2 s 2 = m f s f + m g s g = 111.93 0.42441 + 5.983 0.91303 = 52.967 kJ/K
Then from the mass and energy balances, mi = m 2 - m1 = 117.91 - 9.846 = 108.06 kg The heat transfer during this process is determined from the energy balance to be Qout = mi hi - m 2 u 2 + m1u1 = 108.06 285.47 - 14,581 + 9.846 250.68 = 18,737 kJ (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperature of the extended system is the surroundings temperature Tsurr at all times. It gives Q S in - S out + S gen = S system - out + mi s i + S gen = S tank = (m 2 s 2 - m1 s1 ) tank 1 24 4 3 { 123 4 4 Tb,out
Net entropy transfer by heat and mass Entropy generation Change in entropy
S gen = m 2 s 2 - m1 s1 - mi s i +
Qout T0
Substituting, the exergy destruction is determined to be Q X destroyed = T0 S gen = T0 m 2 s 2 - m1 s1 - mi s i + out T0 = (298 K)[52.967 - 9.846 0.91558 - 108.06 0.93889 + 18,737 / 298] = 1599 kJ
8-69
8-85 An insulated cylinder initially contains saturated liquid-vapor mixture of water. The cylinder is connected to a supply line, and the steam is allowed to enter the cylinder until all the liquid is vaporized. The amount of steam that entered the cylinder and the exergy destroyed are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 The expansion process is quasi-equilibrium. 3 Kinetic and potential energies are negligible. 4 The device is insulated and thus heat transfer is negligible. Properties The properties of steam are (Tables A-4 through A-6)
P1 = 200 kPa h1 = h f + x1 h fg = 504.71 + 0.6 2201.6 = 1825.6 kJ/kg x1 = 9 / 15 = 0.6 s1 = s f + x1 s fg = 1.5302 + 0.6 5.5968 = 4.8883 kJ/kg K P2 = 200 kPa h2 = h g @ 200 kPa = 2706.3 kJ/kg sat.vapor s 2 = s g @ 200 kPa = 7.1270 kJ/kg K Pi = 1 MPa hi = 3264.5 kJ/kg Ti = 400C s i = 7.4670 kJ/kg K
Analysis (a) We take the cylinder as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this unsteady-flow system can be expressed as Mass balance: Energy balance:
min - m out = msystem mi = m 2 - m1
Net energy transfer by heat, work, and mass
H2O 200 kPa P = const.
1 MPa 400C
E in - E out 1 24 4 3
=
Change in internal, kinetic, potential, etc. energies
E system 1 24 4 3
mi hi = W b,out + m 2 u 2 - m1u1 (since Q ke pe 0)
Combining the two relations gives 0 = W b,out - (m 2 - m1 )hi + m 2 u 2 - m1u1 or,
0 = -(m 2 - m1 )hi + m 2 h2 - m1 h1
since the boundary work and U combine into H for constant pressure expansion and compression processes. Solving for m2 and substituting,
m2 = hi - h1 (3264.5 - 1825.6)kJ/kg (15 kg) = 38.66 kg m1 = (3264.5 - 2706.3)kJ/kg hi - h2
Thus,
mi = m 2 - m1 = 38.66 - 15 = 23.66 kg
(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation Sgen is determined from an entropy balance on the insulated cylinder,
Net entropy transfer by heat and mass
S - in S out 1 24 4 3
+ S gen = S system { 123 4 4
Entropy generation Change in entropy
mi s i + S gen = S system = m 2 s 2 - m1 s1 S gen = m 2 s 2 - m1 s1 - mi s i
Substituting, the exergy destruction is determined to be
X destroyed = T0 S gen = T0 [m 2 s 2 - m1 s1 - mi s i ] = (298 K)(38.66 7.1270 - 15 4.8883 - 23.66 7.4670) = 7610 kJ
8-70
8-86 Each member of a family of four take a shower every day. The amount of exergy destroyed by this family per year is to be determined. Assumptions 1 Steady operating conditions exist. 2 The kinetic and potential energies are negligible. 3 & Heat losses from the pipes, mixing section are negligible and thus Q 0. 4 Showers operate at maximum flow conditions during the entire shower. 5 Each member of the household takes a shower every day. 6 Water is an incompressible substance with constant properties at room temperature. 7 The efficiency of the electric water heater is 100%. Properties The density and specific heat of water are at room temperature are = 1 kg/L = 1000 kg/3 and c = 4.18 kJ/kg.C (Table A-3). Analysis The mass flow rate of water at the shower head is
& m = V& = (1 kg/L)(10 L/min) = 10 kg/min
The mass balance for the mixing chamber can be expressed in the rate form as
& & & min - m out = msystem
0 (steady)
& & & & & = 0 min = m out m1 + m 2 = m3
where the subscript 1 denotes the cold water stream, 2 the hot water stream, and 3 the mixture. The rate of entropy generation during this process can be determined by applying the rate form of the entropy balance on a system that includes the electric water heater and the mixing chamber (the Telbow). Noting that there is no entropy transfer associated with work transfer (electricity) and there is no heat transfer, the entropy balance for this steady-flow system can be expressed as
Rate of net entropy transfer by heat and mass
& & S in - S out 1 24 4 3
+
Rate of entropy generation
& S gen {
& = S system 0 (steady) 1442443
Rate of change of entropy
& & & & m1 s1 + m 2 s 2 - m3 s 3 + S gen = 0 (since Q = 0 and work is entropy free) & & & & S gen = m3 s 3 - m1 s1 - m 2 s 2 & & & Noting from mass balance that m1 + m2 = m3 and s2 = s1 since hot water enters the system at the same temperature as the cold water, the rate of entropy generation is determined to be T & & & & & & S gen = m3 s 3 - (m1 + m 2 ) s1 = m3 ( s 3 - s1 ) = m3 c p ln 3 T1 = (10 kg/min)(4.18 kJ/kg.K) ln 42 + 273 = 3.746 kJ/min.K 15 + 273
Noting that 4 people take a 6-min shower every day, the amount of entropy generated per year is
& S gen = ( S gen )t ( No. of people)(No. of days)
= (3.746 kJ/min.K)(6 min/person day)(4 persons)(365 days/year) = 32,815 kJ/K (per year)
The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen ,
X destroyed = T0 S gen = (298 K)(32,815 kJ/K ) = 9,779,000 kJ
Discussion The value above represents the exergy destroyed within the water heater and the T-elbow in the absence of any heat losses. It does not include the exergy destroyed as the shower water at 42C is discarded or cooled to the outdoor temperature. Also, an entropy balance on the mixing chamber alone (hot water entering at 55C instead of 15C) will exclude the exergy destroyed within the water heater.
8-71
8-87 Air is compressed in a steady-flow device isentropically. The work done, the exit exergy of compressed air, and the exergy of compressed air after it is cooled to ambient temperature are to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The process is given to be reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply. 3 The environment temperature and pressure are given to be 300 K and 100 kPa. 4 The kinetic and potential energies are negligible. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The constant pressure specific heat and specific heat ratio of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2). Analysis (a) From the constant specific heats ideal gas isentropic relations,
P T2 = T1 2 P 1
( k -1) / k
1000 kPa = (300 K ) 100 kPa
0.4 / 1.4
= 579.2 K
1 MPa s2 = s1
For a steady-flow isentropic compression process, the work input is determined from
wcomp,in = kRT1 (P2 P1 )(k -1) / k - 1 k -1 (1.4)(0.287kJ/kg K )(300K ) (1000/100) 0.4/1.4 - 1 = 1.4 - 1 = 280.5 kJ/kg
{
}
AIR
{
}
100 kPa 300 K
(b) The exergy of air at the compressor exit is simply the flow exergy at the exit state,
2 = h 2 - h 0 - T0 ( s 2 - s 0 )
= c p (T2 - T0 )
0
V2 + 2 2
0
+ gz 2 (since the proccess 0 - 2 is isentropic)
0
= (1.005 kJ/kg.K)(579.2 - 300)K = 280.6 kJ/kg
which is the same as the compressor work input. This is not surprising since the compression process is reversible. (c) The exergy of compressed air at 1 MPa after it is cooled to 300 K is again the flow exergy at that state,
V2 3 = h3 - h0 - T0 ( s 3 - s 0 ) + 3 2 = c p (T3 - T0 ) = -T0 ( s 3 - s 0 )
0 0
+ gz 3
0
- T0 ( s 3 - s 0 ) (since T3 = T0 = 300 K)
where
T s 3 - s 0 = c p ln 3 T0
0
- R ln
P3 P 1000 kPa = - R ln 3 = -(0.287 kJ/kg K)ln = -0.661 kJ/kg.K P0 P0 100 kPa
Substituting,
3 = -(300 K)(-0.661 kJ / kg.K) = 198 kJ / kg
Note that the exergy of compressed air decreases from 280.6 to 198 as it is cooled to ambient temperature.
8-72
8-88 Cold water is heated by hot water in a heat exchanger. The rate of heat transfer and the rate of exergy destruction within the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 The temperature of the environment is 25C. Properties The specific heats of cold and hot water are given to be 4.18 and 4.19 kJ/kg.C, respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as
Rate of net energy transfer by heat, work, and mass
& & E in - E out 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& E system 0 (steady) 1442443 4
=0
Cold water 15C 0.25 kg/s
Hot water 100C 3 kg/s 45C
& & E in = E out & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mc p (T2 - T1 )
Then the rate of heat transfer to the cold water in this heat exchanger becomes
& & Qin = [mc p (Tout - Tin )] cold water = (0.25 kg/s)(4.18 kJ/kg.C)(45C - 15C) = 31.35 kW
Noting that heat gain by the cold water is equal to the heat loss by the hot water, the outlet temperature of the hot water is determined to be & Q & & Q = [mc p (Tin - Tout )] hot water Tout = Tin - & mc p
= 100C - 31.35 kW = 97.5C (3 kg/s)(4.19 kJ/kg.C)
(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:
Rate of net entropy transfer by heat and mass
& & S in - S out 1 24 4 3
+
Rate of entropy generation
& S gen {
& = S system 0 (steady) 1442443
Rate of change of entropy
& & & & & m1 s1 + m3 s 3 - m 2 s 2 - m3 s 4 + S gen = 0 (since Q = 0) & & & & & m cold s1 + m hot s 3 - m cold s 2 - m hot s 4 + S gen = 0 & & & S gen = m cold ( s 2 - s1 ) + m hot ( s 4 - s 3 )
Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be
T T & & & S gen = m cold c p ln 2 + m hot c p ln 4 T1 T3 = (0.25 kg/s)(4.18 kJ/kg.K)ln 45 + 273 97.5 + 273 + (3 kg/s)(4.19 kJ/kg.K)ln = 0.0190 kW/K 15 + 273 100 + 273
The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen ,
X destroyed = T0 S gen = (298 K)(0.019 kW/K ) = 5.66 kW
8-73
8-89 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and the rate of exergy destruction in the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ/kg.C, respectively. The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as
Rate of net energy transfer by heat, work, and mass
& & E in - E out 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& E system 0 (steady) 1442443 4
=0
& & E in = E out & & & mh1 = Qout + mh2 (since ke pe 0) & & Qout = mC p (T1 - T2 )
Air 95 kPa 20C 0.8 m3/s
Then the rate of heat transfer from the exhaust gases becomes
& & Q = [mc p (Tin - Tout )] gas. = (1.1 kg/s)(1.1 kJ/kg.C)(180C - 95C) = 102.85 kW
Exhaust gases 1.1 kg/s, 95C
The mass flow rate of air is
& m= (95 kPa)(0.8 m 3 /s) PV& = = 0.904 kg/s RT (0.287 kPa.m 3 /kg.K) 293 K
Noting that heat loss by exhaust gases is equal to the heat gain by the air, the air exit temperature becomes & Q 102.85 kW & & Q = mC p (Tout - Tin ) air Tout = Tin + = 20C + = 133.2C & mc p (0.904 kg/s)(1.005 kJ/kg.C)
[
]
The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:
Rate of net entropy transfer by heat and mass
& & S in - S out 1 24 4 3
+
Rate of entropy generation
& S gen {
& = S system 0 (steady) 1442443
Rate of change of entropy
& & & & & m1 s1 + m3 s 3 - m 2 s 2 - m3 s 4 + S gen = 0 (since Q = 0) & & & & & m exhaust s1 + m air s 3 - m exhaust s 2 - m air s 4 + S gen = 0 & & & S gen = m exhaust ( s 2 - s1 ) + m air ( s 4 - s 3 )
Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is
T T & & & S gen = m exhaust c p ln 2 + m air c p ln 4 T1 T3 = (1.1 kg/s)(1.1 kJ/kg.K)ln 95 + 273 133.2 + 273 + (0.904 kg/s)(1.005 kJ/kg.K)ln = 0.0453 kW/K 180 + 273 20 + 273
The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen ,
& & X destroyed = T0 S gen = (293 K)(0.0453 kW/K ) = 13.3 kW
8-74
8-90 Water is heated by hot oil in a heat exchanger. The outlet temperature of the oil and the rate of exergy destruction within the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ/kg.C, respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as
Rate of net energy transfer by heat, work, and mass
& & E in - E out 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& E system 0 (steady) 1442443 4
=0
Oil 170C 10 kg/s 70C Water 20C 4.5
& & E in = E out & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mc p (T2 - T1 )
Then the rate of heat transfer to the cold water in this heat exchanger becomes
(12 tube passes)
& & Q = [mc p (Tout - Tin )] water = (4.5 kg/s)(4.18 kJ/kg.C)(70C - 20C) = 940.5 kW
Noting that heat gain by the water is equal to the heat loss by the oil, the outlet temperature of the hot water is determined from & Q 940.5 kW & & Q = [mc p (Tin - Tout )] oil Tout = Tin - = 170C - = 129.1C & mc p (10 kg/s)(2.3 kJ/kg.C) (b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:
Rate of net entropy transfer by heat and mass
& & S in - S out 1 24 4 3
+
Rate of entropy generation
& S gen {
& = S system 0 (steady) 1442443
Rate of change of entropy
& & & & & m1 s1 + m3 s 3 - m 2 s 2 - m3 s 4 + S gen = 0 (since Q = 0) & & & & & m water s1 + m oil s 3 - m water s 2 - m oil s 4 + S gen = 0 & & & S gen = m water ( s 2 - s1 ) + m oil ( s 4 - s 3 )
Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be
T T & & & S gen = m water c p ln 2 + m oil c p ln 4 T1 T3 = (4.5 kg/s)(4.18 kJ/kg.K) ln 70 + 273 129.1 + 273 + (10 kg/s)(2.3 kJ/kg.K) ln = 0.736 kW/K 20 + 273 170 + 273
The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen ,
& & X destroyed = T0 S gen = (298 K)(0.736 kW/K ) = 219 kW
8-75
8-91E Steam is condensed by cooling water in a condenser. The rate of heat transfer and the rate of exergy destruction within the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 The temperature of the environment is 77F. Properties The specific heat of water is 1.0 Btu/lbm.F (Table A-3E). The enthalpy and entropy of vaporization of water at 120F are 1025.2 Btu/lbm and sfg = 1.7686 Btu/lbm.R (Table A-4E). Analysis We take the tube-side of the heat exchanger where cold water is flowing as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as
Rate of net energy transfer by heat, work, and mass
& & E in - E out 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& E system 0 (steady) 1442443 4
=0
Steam 120F 73F
& & E in = E out & & & Qin + mh1 = mh2 (since ke pe 0) & = mc (T - T ) Qin & p 2 1
Then the rate of heat transfer to the cold water in this heat exchanger becomes & & Q = [mc (T - T )]
p out in water
60F Water 120
= (115.3 lbm/s)(1.0 Btu/lbm.F)(73F - 60F) = 1499 Btu/s
Noting that heat gain by the water is equal to the heat loss by the condensing steam, the rate of condensation of the steam in the heat exchanger is determined from & Q 1499 Btu/s & & & Q = (mh fg ) steam = msteam = = = 1.462 lbm/s h fg 1025.2 Btu/lbm
(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:
Rate of net entropy transfer by heat and mass
& & S in - S out 1 24 4 3
+
Rate of entropy generation
& S gen {
& = S system 0 (steady) 1442443
Rate of change of entropy
& & & & & m1 s1 + m3 s 3 - m 2 s 2 - m 4 s 4 + S gen = 0 (since Q = 0) & & & & & m water s1 + msteam s 3 - m water s 2 - msteam s 4 + S gen = 0 & & & S gen = m water ( s 2 - s1 ) + msteam ( s 4 - s 3 )
Noting that water is an incompressible substance and steam changes from saturated vapor to saturated liquid, the rate of entropy generation is determined to be
T T & & & & & S gen = m water c p ln 2 + msteam ( s f - s g ) = m water c p ln 2 - msteam s fg T1 T1 73 + 460 - (1.462 lbm/s)(1.7686 Btu/lbm.R) = 0.2613 Btu/s.R 60 + 460 The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , = (115.3 lbm/s)(1.0 Btu/lbm.R)ln & & X destroyed = T0 S gen = (537 R)(0.2613 Btu/s.R ) = 140.3 Btu/s
8-76
8-92 Steam expands in a turbine, which is not insulated. The reversible power, the exergy destroyed, the second-law efficiency, and the possible increase in the turbine power if the turbine is well insulated are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. Analysis (a) The properties of the steam at the inlet and exit of the turbine are (Tables A-4 through A-6)
P1 = 12 MPa h1 = 3481.7 kJ/kg T1 = 550C s1 = 6.6554 kJ/kg.K P2 = 20 kPa h2 = 2491.1 kJ/kg x 2 = 0.95 s 2 = 7.5535 kJ/kg.K
Steam 12 MPa 550C, 60 m/s
The enthalpy at the dead state is
T0 = 25C h0 = 104.83 kJ/kg x=0
Turbine
Q
The mass flow rate of steam may be determined from an energy balance on the turbine
V2 & m h1 + 1 2 (60 m/s) 2 1 kJ/kg & m 3481.7 kJ/kg + 2 1000 m 2 /s 2
2 & & = m h2 + V 2 + Qout + W a & 2
20 kPa 130 m/s x=0.95
(130 m/s) 2 1 kJ/kg & = m 2491.1 kJ/kg + 2 1000 m 2 /s 2
& + 150 kW + 2500 kW m = 2.693 kg/s
The reversible power may be determined from
V 2 - V 22 & & W rev = m h1 - h2 - T0 ( s1 - s 2 ) + 1 2 (60 m/s) 2 - (130 m/s) 2 1 kJ/kg = (2.693) (3481.7 - 2491.1) - (298)(6.6554 - 7.5535) + 2 1000 m 2 /s 2 = 3371 kW
(b) The exergy destroyed in the turbine is
& & & X dest = W rev - Wa = 3371 - 2500 = 871 kW
(c) The second-law efficiency is & W 2500 kW II = a = = 0.742 & Wrev 3371 kW (d) The energy of the steam at the turbine inlet in the given dead state is
& & Q = m(h1 - h0 ) = (2.693 kg/s)(3481.7 - 104.83)kJ/kg = 9095 kW
The fraction of energy at the turbine inlet that is converted to power is & W 2500 kW = 0.2749 f = a = & Q 9095 kW Assuming that the same fraction of heat loss from the turbine could have been converted to work, the possible increase in the power if the turbine is to be well-insulated becomes
& & Wincrease = fQout = (0.2749)(150 kW) = 41.2 kW
8-77
8-93 Air is compressed in a compressor that is intentionally cooled. The actual and reversible power inputs, the second law efficiency, and the mass flow rate of cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. 3 Air is an ideal gas with constant specific heats. Properties The gas constant of air is R = 0.287 kJ/kg.K and the specific heat of air at room is cp = 1.005 kJ/kg.K. the specific heat of water at room temperature is cw = 4.18 kJ/kg.K (Tables A-2, A-3). Analysis (a) The mass flow rate of air is
P (100 kPa) & (4.5 m 3 /s) = 5.351 kg/s m = V&1 = 1 V&1 = RT1 (0.287 kJ/kg.K)(20 + 273 K)
900 kPa 60C 80 m/s
Compressor
Q
The power input for a reversible-isothermal process is given by
P 900 kPa & & Wrev = mRT1 ln 2 = (5.351 kg/s)(0.287 kJ/kg.K)(20 + 273 K)ln = 988.8 kW P1 100 kPa
Air 100 kPa 20C
Given the isothermal efficiency, the actual power may be determined from & W 988.8 kW & = 1413 kW Wactual = rev = T 0.70 (b) The given isothermal efficiency is actually the second-law efficiency of the compressor
II = T = 0.70
(c) An energy balance on the compressor gives
V 2 - V 22 & & & Qout = m C p (T1 - T2 ) + 1 + Wactual,in 2 0 - (80 m/s) 2 1 kJ/kg = (5.351 kg/s) (1.005 kJ/kg.C)(20 - 60)C + 2 1000 m 2 /s 2 = 1181 kW + 1413 kW
The mass flow rate of the cooling water is
& mw = & Qout 1181 kW = = 28.25 kg/s c w T (4.18 kJ/kg.C)(10C)
8-78
8-94 Water is heated in a chamber by mixing it with saturated steam. The temperature of the steam entering the chamber, the exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Heat loss from the chamber is negligible. Water Analysis (a) The properties of water are Mixing 15C (Tables A-4 through A-6) chamber 4.6 kg/s T1 = 15C h1 = h0 = 62.98 kJ/kg Mixture x1 = 0 s1 = s 0 = 0.22447 kJ/kg.K 45C T3 = 45C h3 = 188.44 kJ/kg Sat. vap. x1 = 0 s 3 = 0.63862 kJ/kg.K 0.23 kg/s An energy balance on the chamber gives
& & & & & m1h1 + m2 h2 = m3h3 = (m1 + m2 )h3 (4.6 kg/s)(62.98 kJ/kg) + (0.23 kg/s)h2 = (4.6 + 0.23 kg/s)(188.44 kJ/kg) h2 = 2697.5 kJ/kg
The remaining properties of the saturated steam are
h2 = 2697.5 kJ/kg T2 = 114.3C x2 = 1 s 2 = 7.1907 kJ/kg.K
(b) The specific exergy of each stream is
1 = 0 2 = h2 - h0 - T0 ( s2 - s0 ) = (2697.5 - 62.98)kJ/kg - (15 + 273 K)(7.1907 - 0.22447)kJ/kg.K = 628.28 kJ/kg 3 = h3 - h0 - T0 ( s 3 - s 0 ) = (188.44 - 62.98)kJ/kg - (15 + 273 K)(0.63862 - 0.22447)kJ/kg.K = 6.18 kJ/kg
The exergy destruction is determined from an exergy balance on the chamber to be
& & & & & X dest = m1 1 + m 2 2 - (m1 + m 2 ) 3 = 0 + (0.23 kg/s)(628.28 kJ/kg) - (4.6 + 0.23 kg/s)(6.18 kJ/kg) = 114.7 kW
(c) The second-law efficiency for this mixing process may be determined from
II =
& & (m1 + m2 ) 3 (4.6 + 0.23 kg/s)(6.18 kJ/kg) = = 0.207 & & m1 1 + m2 2 0 + (0.23 kg/s)(628.28 kJ/kg)