Chambers 5-41
Mixing and Heat Exchangers 5-72C Yes, if the mixing chamber is losing heat to the surrounding medium. 5-73C Under the conditions of no heat and work interactions between the mixing chamber and the surrounding medium. 5-74C Under the conditions of no heat and work interactions between the heat exchanger and the surrounding medium.
5-75 A hot water stream is mixed with a cold water stream. 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 = 127.41C, 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 @ 80C = 335.02 kJ/kg h2 hf @ 20C = 83.915 kJ/kg h3 hf @ 42C = 175.90 kJ/kg
T2 = 20C m2 T1 = 80C m1 = 0.5 kg/s H2O (P = 250 kPa) T3 = 42C
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: Energy balance:
Rate of net energy transfer by heat, work, and mass
& & & min - mout = msystem
0 (steady)
& & & =0 m1 + m2 = m3
& & Ein - Eout 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& Esystem 0 (steady) 1442443
=0
& & Ein = Eout & & & & & m1h1 + m2 h2 = m3h3 (since Q = W = ke pe 0)
& Combining the two relations and solving for m2 gives & & & & m1h1 + m2h2 = (m1 + m2 )h3
& m2 = h1 - h3 & m1 h3 - h2
Substituting, the mass flow rate of cold water stream is determined to be
& m2 =
(335.02 - 175.90) kJ/kg (0.5 kg/s ) = 0.865 (175.90 - 83.915) kJ/kg
kg/s
5-42
5-76 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass flow rate of the steam 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 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties Noting that T < Tsat @ 300 kPa = 133.52C, 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, from steam tables (Tables A-4 through A-6) h1 hf @ 20C = 83.91 kJ/kg h3 hf @ 60C = 251.18 kJ/kg and
T2 = 300C P2 = 300 kPa h2 = 3069.6 kJ/kg
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: Energy balance:
Rate of net energy transfer by heat, work, and mass
& & & min - mout = msystem & Esystem 0 (steady) 1442443
0 (steady)
& & & & & = 0 min = mout m1 + m2 = m3
& & Ein - Eout 1 24 4 3
=
=0
Rate of change in internal, kinetic, potential, etc. energies
T1 = 20C m1 = 1.8 kg/s H2O (P = 300 kPa) T3 = 60C T2 = 300C m2
& & Ein = Eout & & & & & m1h1 + m2 h2 = m3h3 (since Q = W = ke pe 0)
Combining the two,
& Solving for m2 :
& & & & m1h1 + m2h2 = (m1 + m2 )h3 & m2 = h1 - h3 & m1 h3 - h2
Substituting,
& m2 =
(83.91 - 251.18)kJ/kg (1.8 kg/s) = 0.107 kg/s (251.18 - 3069.6)kJ/kg
5-43
5-77 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. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties Noting that T < Tsat @ 1 MPa = 179.88C, 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, from steam tables (Tables A-4 through A-6) h1 hf @ 50C = 209.34 kJ/kg h3 hf @ 1 MPa = 762.51 kJ/kg and
P2 = 1 MPa h2 = 2828.3 kJ/kg T2 = 200C
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: Energy balance:
Rate of net energy transfer by heat, work, and mass
& & & min - mout = msystem & Esystem 0 (steady) 1442443
0 (steady)
& & & & & = 0 min = mout m1 + m2 = m3
& & Ein - Eout 1 24 4 3
=
=0
Rate of change in internal, kinetic, potential, etc. energies
T1 = 50C m1 H2O (P = 1 MPa) Sat. liquid T2 = 200C m2
& & Ein = Eout & & & & & m1h1 + m2 h2 = m3h3 (since Q = W = ke pe 0)
Combining the two,
& Dividing by m2 yields
& & & & m1h1 + m2h2 = (m1 + m2 )h3 y h1 + h2 = ( y + 1)h3 y= h3 - h2 h1 - h3
Solving for y:
& & where y = m1 / m2 is the desired mass flow rate ratio. Substituting,
y= 762.51 - 2828.3 = 3.73 209.34 - 762.51
5-44
5-78E Liquid water is heated in a chamber by mixing it with saturated water vapor. If both streams enter at the same rate, the temperature and quality (if saturated) of the exit stream 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 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From steam tables (Tables A-5E through A-6E), h1 hf @ 50F = 18.07 Btu/lbm h2 = hg @ 50 psia = 1174.2 Btu/lbm 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: min - mout = msystem
0 (steady)
& & & & & & & & & = 0 min = mout m1 + m2 = m3 = 2m m1 = m2 = m
T1 = 50F H2O (P = 50 psia) T3, x3 Sat. vapor m2 = m1
Energy balance:
Rate of net energy transfer by heat, work, and mass
& & Ein - Eout 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& Esystem 0 (steady) 1442443
=0
& & Ein = Eout & & & & & m1h1 + m2 h2 = m3h3 (since Q = W = ke pe 0)
Combining the two gives Substituting,
& & & mh1 + mh2 = 2mh3 or h3 = (h1 + h2 ) / 2
h3 = (18.07 + 1174.2)/2 = 596.16 Btu/lbm At 50 psia, hf = 250.21 Btu/lbm and hg = 1174.2 Btu/lbm. Thus the exit stream is a saturated mixture since hf < h3 < hg. Therefore, T3 = Tsat @ 50 psia = 280.99F and
x3 = h3 - h f h fg = 596.16 - 250.21 = 0.374 924.03
5-45
5-79 Two streams of refrigerant-134a are mixed in a chamber. If the cold stream enters at twice the rate of the hot stream, the temperature and quality (if saturated) of the exit stream are 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 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From R-134a tables (Tables A-11 through A-13), h1 hf @ 12C = 68.18 kJ/kg h2 = h @ 1 MPa,
60C
= 293.38 kJ/kg
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: min - mout = msystem
0 (steady)
& & & & & & & & = 0 min = mout m1 + m2 = m3 = 3m2 since m1 = 2m2
T1 = 12C m1 = 2m2 R-134a (P = 1 MPa) T3, x3 T2 = 60C
Energy balance:
Rate of net energy transfer by heat, work, and mass
& & Ein - Eout 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& Esystem 0 (steady) 1442443
=0
& & Ein = Eout & & & & & m1h1 + m2 h2 = m3h3 (since Q = W = ke pe 0)
Combining the two gives Substituting,
& & & 2m 2 h1 + m 2 h2 = 3m 2 h3 or h3 = (2h1 + h2 ) / 3
h3 = (268.18 + 293.38)/3 = 143.25 kJ/kg At 1 MPa, hf = 107.32 kJ/kg and hg = 270.99 kJ/kg. Thus the exit stream is a saturated mixture since hf < h3 < hg. Therefore, T3 = Tsat @ 1 MPa = 39.37C and
x3 = h3 - h f h fg = 143.25 - 107.32 = 0.220 163.67
5-46
5-80 EES Problem 5-79 is reconsidered. The effect of the mass flow rate of the cold stream of R-134a on the temperature and the quality of the exit stream as the ratio of the mass flow rate of the cold stream to that of the hot stream varies from 1 to 4 is to be investigated. The mixture temperature and quality are to be plotted against the cold-to-hot mass flow rate ratio. Analysis The problem is solved using EES, and the solution is given below. "Input Data" "m_frac = 2" "m_frac =m_dot_cold/m_dot_hot= m_dot_1/m_dot_2" T[1]=12 [C] P[1]=1000 [kPa] T[2]=60 [C] P[2]=1000 [kPa] m_dot_1=m_frac*m_dot_2 P[3]=1000 [kPa] m_dot_1=1 "Conservation of mass for the R134a: Sum of m_dot_in=m_dot_out" m_dot_1+ m_dot_2 =m_dot_3 "Conservation of Energy for steady-flow: neglect changes in KE and PE" "We assume no heat transfer and no work occur across the control surface." E_dot_in - E_dot_out = DELTAE_dot_cv DELTAE_dot_cv=0 "Steady-flow requirement" E_dot_in=m_dot_1*h[1] + m_dot_2*h[2] 0.5 E_dot_out=m_dot_3*h[3] "Property data are given by:" h[1] =enthalpy(R134a,T=T[1],P=P[1]) h[2] =enthalpy(R134a,T=T[2],P=P[2]) T[3] =temperature(R134a,P=P[3],h=h[3]) x_3=QUALITY(R134a,h=h[3],P=P[3]) mfrac 1 1.333 1.667 2 2.333 2.667 3 3.333 3.667 4 T3 [C] 39.37 39.37 39.37 39.37 39.37 39.37 39.37 39.37 39.37 39.37 x3 0.4491 0.3509 0.2772 0.2199 0.174 0.1365 0.1053 0.07881 0.05613 0.03649
0.4
0.3
x3
0.2
0.1
0 1
1.5
2
2.5
3
3.5
4
m
40
frac
38
T[3] [C]
36
34
32
30 1
1.5
2
2.5
3
3.5
4
m
frac
5-47
5-81 Refrigerant-134a is to be cooled by air in the condenser. For a specified volume flow rate of air, the mass flow rate of the refrigerant 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 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant pressure specific heat of air is cp = 1.005 kJ/kgC (Table A-2). The enthalpies of the R-134a at the inlet and the exit states are (Tables A-11 through A-13)
P3 = 1 MPa h3 = 324.64 kJ/kg T3 = 90C P4 = 1 MPa h4 h f @30oC = 93.58 kJ/kg T4 = 30C
R-134a AIR
1 3 4
Analysis The inlet specific volume and the mass flow rate of air are
v1 =
and
& m=
RT1 0.287 kPa m 3 /kg K (300 K ) = = 0.861 m 3 /kg P1 100 kPa
(
)
2
V&1 600 m 3 /min = = 696.9 kg/min v 1 0.861 m 3 /kg
We take the entire heat exchanger 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 ( for each fluid stream):
& & & min - mout = msystem & & Ein - Eout 1 24 4 3
0 (steady)
& & & & & & & & = 0 min = mout m1 = m2 = ma and m3 = m4 = mR & Esystem 0 (steady) 1442443
Energy balance (for the entire heat exchanger):
= =0
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc. energies
& & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0)
Combining the two,
& Solving for mR :
& & m a (h2 - h1 ) = m R (h3 - h4 )
& mR =
c p (T2 - T1 ) h2 - h1 & & ma ma h3 - h4 h3 - h4
Substituting,
& mR =
(1.005 kJ/kg C)(60 - 27)C (696.9 kg/min ) = 100.0 kg/min (324.64 - 93.58) kJ/kg
5-48
5-82E Refrigerant-134a is vaporized by air in the evaporator of an air-conditioner. For specified flow rates, the exit temperature of the air and the rate of heat transfer from the air are 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 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.3704 psia.ft3/lbm.R (Table A-1E). The constant pressure specific heat of air is cp = 0.240 Btu/lbmF (Table A-2E). The enthalpies of the R-134a at the inlet and the exit states are (Tables A-11E through A-13E)
P3 = 20 psia h3 = h f + x3 h fg = 11.445 + 0.3 91.282 = 38.83 Btu/lbm x3 = 0.3 P4 = 20 psia h4 = h g @ 20 psia = 102.73 Btu/lbm sat.vapor
R-134a AIR
1 3 4
Analysis (a) The inlet specific volume and the mass flow rate of air are
v1 =
and
RT1 0.3704 psia ft 3/lbm R (550 R ) = = 13.86 ft 3/lbm P 14.7 psia 1 V&1 200 ft 3/min & m= = = 14.43 lbm/min v1 13.86 ft 3/lbm
(
)
2
We take the entire heat exchanger 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 (for each fluid stream):
& & & min - mout = msystem & & Ein - Eout 1 24 4 3
0 (steady)
& & & & & & & & = 0 min = mout m1 = m2 = ma and m3 = m4 = mR
Energy balance (for the entire heat exchanger):
=
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc. energies
& Esystem 0 (steady) 1442443
=0
& & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0)
Combining the two, Solving for T2 : Substituting,
& & & m R (h3 - h4 ) = m a (h2 - h1 ) = m a c p (T2 - T1 ) & m R (h3 - h4 ) & ma c p
T2 = T1 +
T2 = 90F +
(4 lbm/min )(38.83 - 102.73)Btu/lbm = 16.2F (14.43 Btu/min )(0.24 Btu/lbmo F)
(b) The rate of heat transfer from the air to the refrigerant is determined from the steady-flow energy balance applied to the air only. It yields
& & & - Qair, out = m a (h2 - h1 ) = m a c p (T2 - T1 ) & Qair, out = -(14.43 lbmg/min)(0.24 Btu/lbm F)(16.2 - 90)F = 255.6 Btu/min
5-49
5-83 Refrigerant-134a is condensed in a water-cooled condenser. The mass flow rate of the cooling water 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 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Properties The enthalpies of R-134a at the inlet and the exit states are (Tables A-11 through A-13)
P3 = 700 kPa h3 = 308.33 kJ/kg T3 = 70C P4 = 700 kPa h4 = h f @ 700 kPa = 88.82 kJ/kg sat. liquid
R-134a
Water
1 3 4
Water exists as compressed liquid at both states, and thus (Table A-4) h1 hf @ 15C = 62.98 kJ/kg h2 hf @ 25C = 104.83 kJ/kg Analysis We take the heat exchanger 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 (for each fluid stream):
& & & min - mout = msystem
0 (steady)
2
& & & & & & & & = 0 min = mout m1 = m2 = mw and m3 = m4 = mR & Esystem 0 (steady) 1442443
Energy balance (for the heat exchanger):
Rate of net energy transfer by heat, work, and mass
& & Ein - Eout 1 24 4 3
=
=0
Rate of change in internal, kinetic, potential, etc. energies
& & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0)
Combining the two,
& Solving for mw :
& & m w (h2 - h1 ) = m R (h3 - h4 ) & mw = h3 - h4 & mR h2 - h1
Substituting,
& mw = (308.33 - 88.82)kJ/kg (8 kg/min) = 42.0 kg/min (104.83 - 62.98)kJ/kg
5-50
5-84E [Also solved by EES on enclosed CD] Air is heated in a steam heating system. For specified flow rates, the volume flow rate of air at the inlet 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 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.3704 psia.ft3/lbm.R (Table A-1E). The constant pressure specific heat of air is Cp = 0.240 Btu/lbmF (Table A-2E). The enthalpies of steam at the inlet and the exit states are (Tables A-4E through A-6E)
P3 = 30 psia h3 = 1237.9 Btu/lbm T3 = 400F P4 = 25 psia h4 h f @ 212o F = 180.21 Btu/lbm T4 = 212F
AIR
1
Steam
3 4
Analysis We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance ( for each fluid stream):
& & & min - mout = msystem & & Ein - Eout 1 24 4 3
0 (steady)
2
& & & & & & & & = 0 min = mout m1 = m2 = ma and m3 = m4 = ms
Energy balance (for the entire heat exchanger):
=
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc. energies
& Esystem 0 (steady) 1442443
=0
& & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0)
Combining the two,
& Solving for ma :
& & m a (h2 - h1 ) = m s (h3 - h4 ) & ma = h3 - h4 h3 - h4 & & ms ms c p (T2 - T1 ) h2 - h1
Substituting,
& ma = (1237.9 - 180.21)Btu/lbm (15 lbm/min) = 1322 lbm/min = 22.04 lbm/s (0.240 Btu/lbm F)(130 - 80)F RT1 (0.3704 psia ft 3 /lbm R )(540 R ) = = 13.61 ft 3 /lbm P1 14.7 psia
Also,
v1 =
Then the volume flow rate of air at the inlet becomes
& V&1 = m av 1 = (22.04 lbm/s)(13.61 ft 3 /lbm) = 300 ft 3 /s
5-51
5-85 Steam is condensed by cooling water in the condenser of a power plant. If the temperature rise of the cooling water is not to exceed 10C, the minimum mass flow rate of the cooling water 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 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Liquid water is an incompressible substance with constant specific heats at room temperature. Properties The cooling water exists as compressed liquid at both states, and its specific heat at room temperature is c = 4.18 kJ/kgC (Table A-3). The enthalpies of the steam at the inlet and the exit states are (Tables A-5 and A-6)
P3 = 20 kPa h3 = h f + x3h fg = 251.42 + 0.95 2357.5 = 2491.1 kJ/kg x3 = 0.95
sat. liquid
P4 = 20 kPa h4 hf @ 20 kPa = 251.42 kJ/kg
Analysis We take the heat exchanger 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 (for each fluid stream):
& & & min - mout = msystem & & Ein - Eout 1 24 4 3
0 (steady)
& & & & & & & & = 0 min = mout m1 = m2 = mw and m3 = m4 = ms
Energy balance (for the heat exchanger):
=
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc. Esystem energies
& 0 (steady) 1442443
=0
Steam 20 kPa
& & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0)
Combining the two,
& Solving for mw :
& & m w (h2 - h1 ) = m s (h3 - h4 ) & mw = h3 - h4 h3 - h 4 & & ms ms h2 - h1 c p (T2 - T1 )
Water
Substituting,
& mw =
(2491.1 - 251.42)kJ/kg (4.18 kJ/kg o C)(10C) (20,000/3600 kg/s) = 297.7 kg/s
5-52
5-86 Steam is condensed by cooling water in the condenser of a power plant. The rate of condensation of steam is 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 heat of vaporization of water at 50C is hfg = 2382.0 kJ/kg and specific heat of cold water is cp = 4.18 kJ/kg.C (Tables A-3 and A-4). Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as
Rate of net energy transfer by heat, work, and mass
Steam 50C
27C
& & Ein - Eout 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& Esystem 0 (steady) 1442443
=0
& & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mc p (T2 - T1 )
18C Water 50C
Then the heat transfer rate to the cooling water in the condenser becomes & & Q = [mc (T - T )]
p out in cooling water
= (101 kg/s)(4.18 kJ/kg.C)(27C - 18C) = 3800 kJ/s
The rate of condensation of steam is determined to be & Q 3800 kJ/s & & Q = (mh fg ) steam msteam = & = = 1.60 kg/s h fg 2382.0 kJ/kg
5-53
5-87 EES Problem 5-86 is reconsidered. The effect of the inlet temperature of cooling water on the rate of condensation of steam as the inlet temperature varies from 10C to 20C at constant exit temperature is to be investigated. The rate of condensation of steam is to be plotted against the inlet temperature of the cooling water. Analysis The problem is solved using EES, and the solution is given below. "Input Data" T_s[1]=50 [C] T_s[2]=50 [C] m_dot_water=101 [kg/s] T_water[1]=18 [C] T_water[2]=27 [C] C_P_water = 4.20 [kJ/kg-C] "Conservation of mass for the steam: m_dot_s_in=m_dot_s_out=m_dot_s" "Conservation of mass for the water: m_dot_water_in=lm_dot_water_out=m_dot_water" "Conservation of Energy for steady-flow: neglect changes in KE and PE" "We assume no heat transfer and no work occur across the control surface." E_dot_in - E_dot_out = DELTAE_dot_cv DELTAE_dot_cv=0 "Steady-flow requirement" E_dot_in=m_dot_s*h_s[1] + m_dot_water*h_water[1] E_dot_out=m_dot_s*h_s[2] + m_dot_water*h_water[2] "Property data are given by:" h_s[1] =enthalpy(steam_iapws,T=T_s[1],x=1) "steam data" h_s[2] =enthalpy(steam_iapws,T=T_s[2],x=0) h_water[1] =C_P_water*T_water[1] "water data" h_water[2] =C_P_water*T_water[2] h_fg_s=h_s[1]-h_s[2] "h_fg is found from the EES functions rather than using h_fg = 2305 kJ/kg" ms [kg/s] 3.028 2.671 2.315 1.959 1.603 1.247 Twater,1 [C] 10 12 14 16 18 20
3.5 3
] s / g k [ m
s
2.5 2 1.5 1 10
12
14
16
18
20
Twater[1] [C]
5-54
5-88 Water is heated in a heat exchanger by geothermal water. The rate of heat transfer to the water and the exit temperature of the geothermal water is 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 geothermal fluid are given to be 4.18 and 4.31 kJ/kg.C, respectively. Analysis We take the cold water tubes as the system, which is a control volume. 60C The energy balance for this steady-flow system can be expressed in the rate form as Brine & & & Ein - Eout = Esystem 0 (steady) =0 1 24 4 3 140C 1442443
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies
& & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mc p (T2 - T1 )
Water 25C
Then the rate of heat transfer to the cold water in the heat exchanger becomes & & Q = [mc p (Tout - Tin )]water = (0.2 kg/s)(4.18 kJ/kg.C)(60C - 25C) = 29.26 kW Noting that heat transfer to the cold water is equal to the heat loss from the geothermal water, the outlet temperature of the geothermal water is determined from & Q 29.26 kW & & Q = [mc p (Tin - Tout )]geot.water Tout = Tin - = 140C - = 117.4C & mc p (0.3 kg/s)(4.31 kJ/kg.C)
5-89 Ethylene glycol is cooled by water in a heat exchanger. The rate of heat transfer in the heat exchanger and the mass flow rate of water 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 ethylene glycol are given to be 4.18 and 2.56 kJ/kg.C, respectively. Analysis (a) We take the ethylene glycol 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 & & & Ein - Eout = Esystem 0 (steady) =0 Cold Water 1 24 4 3 1442443 20C
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies
& & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & = mc (T - T ) Qout & p 1 2
Hot Glycol 80C 2 kg/s 40C
Then the rate of heat transfer becomes & & Q = [mc (T - T )] = (2 kg/s)(2.56 kJ/kg.C)(80C - 40C) = 204.8 kW
p in out glycol
(b) The rate of heat transfer from glycol must be equal to the rate of heat transfer to the water. Then,
5-55
& & Q = [mc p (Tout - Tin )]water mwater = &
& Q c p (Tout - Tin )
=
204.8 kJ/s = 1.4 kg/s (4.18 kJ/kg.C)(55C - 20C)
5-56
5-90 EES Problem 5-89 is reconsidered. The effect of the inlet temperature of cooling water on the mass flow rate of water as the inlet temperature varies from 10C to 40C at constant exit temperature) is to be investigated. The mass flow rate of water is to be plotted against the inlet temperature. Analysis The problem is solved using EES, and the solution is given below. "Input Data" {T_w[1]=20 [C]} T_w[2]=55 [C] "w: water" m_dot_eg=2 [kg/s] "eg: ethylene glycol" T_eg[1]=80 [C] T_eg[2]=40 [C] C_p_w=4.18 [kJ/kg-K] C_p_eg=2.56 [kJ/kg-K] "Conservation of mass for the water: m_dot_w_in=m_dot_w_out=m_dot_w" "Conservation of mass for the ethylene glycol: m_dot_eg_in=m_dot_eg_out=m_dot_eg" "Conservation of Energy for steady-flow: neglect changes in KE and PE in each mass steam" "We assume no heat transfer and no work occur across the control surface." E_dot_in - E_dot_out = DELTAE_dot_cv DELTAE_dot_cv=0 "Steady-flow requirement" E_dot_in=m_dot_w*h_w[1] + m_dot_eg*h_eg[1] E_dot_out=m_dot_w*h_w[2] + m_dot_eg*h_eg[2] Q_exchanged =m_dot_eg*h_eg[1] - m_dot_eg*h_eg[2] "Property data are given by:" h_w[1] =C_p_w*T_w[1] "liquid approximation applied for water and ethylene glycol" h_w[2] =C_p_w*T_w[2] h_eg[1] =C_p_eg*T_eg[1] h_eg[2] =C_p_eg*T_eg[2]
mw [kg/s] 1.089 1.225 1.4 1.633 1.96 2.45 3.266
Tw,1 [C] 10 15 20 25 30 35 40
3.5 3
] s / g k [ m
w
2.5 2 1.5 1 10
15
20
25
30
35
40
Tw[1] [C]
5-57
5-91 Oil is to be cooled by water in a thin-walled heat exchanger. The rate of heat transfer in the heat exchanger and the exit temperature of water is 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 2.20 kJ/kg.C, respectively. Analysis We take the oil tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form Hot oil as 150C 0 (steady) & & & Ein - Eout = Esystem =0 2 kg/s 1 24 4 3 1442443
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies
& & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & & Qout = mc p (T1 - T2 )
Cold water 22C 1.5 kg/s
40C
Then the rate of heat transfer from the oil becomes & & Q = [mc (T - T )] = (2 kg/s)(2.2 kJ/kg.C)(150C - 40C) = 484 kW
p in out oil
Noting that the heat lost by the oil is gained by the water, the outlet temperature of the water is determined from & Q 484 kJ/s & & Q = [mc p (Tout - Tin )]water Tout = Tin + = 22C + = 99.2C & mwater c p (1.5 kg/s)(4.18 kJ/kg.C) 5-92 Cold water is heated by hot water in a heat exchanger. The rate of heat transfer and the exit temperature of hot water 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 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 steadyflow system can be expressed in the rate form as Cold Water & & & Ein - Eout = Esystem 0 (steady) =0 1 24 4 3 15C 1442443
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies
Hot water
& & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mc p (T2 - T1 )
100C 3 kg/s
Then the rate of heat transfer to the cold water in this heat exchanger becomes & & Q = [mc (T - T )] = (0.60 kg/s)(4.18 kJ/kg.C)(45C - 15C) = 75.24 kW
p out in cold water
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 75.24 kW & & Q = [mc p (Tin - Tout )]hot water Tout = Tin - = 100C - = 94.0C & mc p (3 kg/s)(4.19 kJ/kg.C)
5-58
5-93 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and the outlet temperature of the air 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. 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 & & & & & Ein - Eout = Esystem 0 (steady) = 0 Ein = Eout 1 24 4 3 1442443
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies
& & & mh1 = Qout + mh2 (since ke pe 0) & = mc (T - T ) Qout & p 1 2
p in out gas.
Air 95 kPa 20C 0.8 m3/s
Then the rate of heat transfer from the exhaust gases becomes & & Q = [mc (T - T )] = (1.1 kg/s)(1.1 kJ/kg.C)(180C - 95C) = 102.85 kW The mass flow rate of air is (95 kPa)(0.8 m 3 /s) PV& & m= = = 0.904 kg/s RT (0.287 kPa.m 3 /kg.K) 293 K
Exhaust gases 1.1 kg/s, 95C
Noting that heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature of the air becomes & Q 102.85 kW & & = 20C + = 133.2C Q = mc p (Tc,out - Tc,in ) Tc,out = Tc,in + & (0.904 kg/s)(1.005 kJ/kg.C) mc p
5-94 Water is heated by hot oil in a heat exchanger. The rate of heat transfer in the heat exchanger and the outlet temperature of oil 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 Oil a control volume. The energy balance for this steady-flow 170C system can be expressed in the rate form as 10 kg/s & & & & & Ein - Eout = Esystem 0 (steady) = 0 Ein = Eout 1 24 4 3 70C 1442443
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies
& & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mc p (T2 - T1 )
Water 20C 4.5 kg/s
Then the rate of heat transfer to the cold water in this heat exchanger becomes & & 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 - = 129.1 C & mc p (10 kg/s)(2.3 kJ/kg.C)
5-59
5-95E Steam is condensed by cooling water in a condenser. The rate of heat transfer in the heat exchanger and the rate of condensation of steam 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 heat of water is 1.0 Btu/lbm.F (Table A-3E). The enthalpy of vaporization of water at 85F is 1045.2 Btu/lbm (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
Steam 85F 73F
& & Ein - Eout 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& Esystem 0 (steady) 1442443
=0
60F Water 85F
& & Ein = Eout & & & 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
& & Q = [mc p (Tout - Tin )] water = (138 lbm/s)(1.0 Btu/lbm.F)(73F - 60F) = 1794 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 1794 Btu/s & & & = = 1.72 lbm/s Q = (mh fg )steam = msteam = h fg 1045.2 Btu/lbm
5-60
5-96 Two streams of cold and warm air are mixed in a chamber. If the ratio of hot to cold air is 1.6, the mixture temperature and the rate of heat gain of the room are 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 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Cold Properties The gas constant of air is 3 air R = 0.287 kPa.m /kg.K. The enthalpies 5C of air are obtained from air table (Table A-17) as Room 24C h1 = h @278 K = 278.13 kJ/kg Warm h2 = h @ 307 K = 307.23 kJ/kg air 34C hroom = h @ 297 K = 297.18 kJ/kg Analysis (a) 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:
& & & min - mout = msystem
0 (steady)
& & & & & & & & = 0 min = mout m1 + 1.6m1 = m3 = 2.6m1 since m2 = 1.6m1
Energy balance:
Rate of net energy transfer by heat, work, and mass
& & Ein - Eout 1 24 4 3
=
Rate of change in internal, kinetic, potential, etc. energies
& Esystem 0 (steady) 1442443
=0
& & Ein = Eout & & & m1h1 + m2 h2 = m3h3 & & (since Q W ke pe 0)
Combining the two gives Substituting,
& & & m1 h1 + 1.6m1 h2 = 2.6m1 h3 or h3 = (h1 + 1.6h2 ) / 2.6
h3 = (278.13 +1.6 307.23)/2.6 = 296.04 kJ/kg From air table at this enthalpy, the mixture temperature is T3 = T @ h = 296.04 kJ/kg = 295.9 K = 22.9C (b) The mass flow rates are determined as follows
RT1 (0.287 kPa m3/kg K)(5 + 273 K) = = 0.7599 m3 / kg 105 kPa P 1.25 m3/s V& & m1 = 1 = = 1.645 kg/s v1 0.7599 m3/kg & & m3 = 2.6m1 = 2.6(1.645 kg/s) = 4.277 kg/s
v1 =
The rate of heat gain of the room is determined from
& & Qcool = m3 (hroom - h3 ) = (4.277 kg/s)(297.18 - 296.04) kJ/kg = 4.88 kW
5-61
5-97 A heat exchanger that is not insulated is used to produce steam from the heat given up by the exhaust gases of an internal combustion engine. The temperature of exhaust gases at the heat exchanger exit and the rate of heat transfer to the water are 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 There are no work interactions. 4 Exhaust gases are assumed to have air properties with constant specific heats. Properties The constant pressure specific heat of the exhaust gases is taken to be cp = 1.045 kJ/kgC (Table A-2). The inlet and exit enthalpies of water are (Tables A-4 and A-5)
Tw,in = 15C hw,in = 62.98 kJ/kg x = 0 (sat. liq.) Pw,out = 2 MPa hw,out = 2798.3 kJ/kg x = 1 (sat. vap.)
Exh. gas 400C
Q Heat exchanger
Analysis We take the entire heat exchanger 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 (for each fluid stream):
& & & min - mout = msystem & & Ein - Eout 1 24 4 3
0 (steady)
2 MPa sat. vap.
Water 15C
& & = 0 min = mout
Energy balance (for the entire heat exchanger):
=
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc. energies
& Esystem 0 (steady) 1442443
=0
& & Ein = Eout & & & & & & mexh hexh,in + mw hw,in = mexh hexh,out + mw hw,out + Qout (since W = ke pe 0) & & & & & mexh c pTexh,in + mw hw,in = mexh c pTexh,out + mw hw,out + Qout & & 15m w (1.045 kJ/kg.C)(400C) + m w (62.98 kJ/kg) & & & = 15m w (1.045 kJ/kg.C)Texh,out + m w (2798.3 kJ/kg) + Qout
or
Noting that the mass flow rate of exhaust gases is 15 times that of the water, substituting gives (1)
The heat given up by the exhaust gases and heat picked up by the water are
& & & Qexh = m exh c p (Texh,in - Texh,out ) = 15m w (1.045 kJ/kg.C)(400 - Texh,out )C & & & Qw = m w (hw,out - hw,in ) = m w (2798.3 - 62.98)kJ/kg
(2) (3)
The heat loss is
& & & Qout = f heat loss Qexh = 0.1Qexh
(4)
The solution may be obtained by a trial-error approach. Or, solving the above equations simultaneously using EES software, we obtain
& & & Texh,out = 206.1 C, Qw = 97.26 kW, m w = 0.03556 kg/s, mexh = 0.5333 kg/s