5.34 E 110 volt⋅:= i 9.7 amp⋅:= Tσ300 K⋅:=
Wdotmech 1.25−hp⋅:= Wdotelect iE⋅:= Wdotelect 1.067 103
×W=
At steady state: Qdot Wdotelect
+Wdotmech
+
t
Ut
d
d
=0=
Qdot
TσSdotG
+
t
St
d
d
=0=
5.33 For the process of cooling the brine:
CP3.5 kJ
kg K⋅
⋅:= ∆T40−K⋅:= mdot 20 kg
sec
⋅:= ηt0.27:=
T1273.15 25+()K⋅:= T1298.15 K=
T2273.15 15−()K⋅:= T2258.15 K=
Tσ273.15 30+()K⋅:= Tσ303.15 K=
134
∆S
R
T1
T2
T
Cp
R
1
T
⌠
⎮
⎮
⌡
dln
P2
P1
⎛
⎜
⎝
⎞
⎠
−=Eq. (5.14)
5.39 (a) T1500K:= P16bar:= T2371K:= P21.2bar:= Cp
7
2R:=
Tσ300K:= Basis: 1 mol n 1mol:=
5.35 Ω25 ohm⋅:= i 10 amp⋅:= Tσ300 K⋅:=
Wdotelect i2Ω⋅:= Wdotelect 2.5 103
×W=
At steady state: Qdot Wdotelect
+
t
Ut
d
d
=0=Qdot Wdotelect
−:=
5.38 mdot 10 kmol
hr
:= T125 273.15+()K:= P110bar:= P21.2bar:=
(b)
135
⎜
⎜
5.41 P12500kPa:= P2150kPa:= Tσ300K:= mdot 20 mol
sec
:=
∆SR−ln P2
P1
⎛
⎜
⎞
⋅:=
ηactual 0.45=ηactual
W
QH
:=
TC298.15 K=TC25 273.15+()K:=
TH523.15 K=TH250 273.15+()K:=W 0.45kJ:=QH1kJ:=
5.42
∆S 0.023 kJ
mol K⋅
=
Wideal Ws
∆S 4.698 J
K
=∆SnC
pln T2
T1
⎛
⎜
⎝
⎞
⎠
⋅Rln P2
P1
⎛
⎜
⎝
⎞
⎠
⋅−
⎛
⎜
⎝
⎞
⎠
⋅:=
4.698 J
K
1409.3J5163−J3753.8−J
(a)
SG
Wlost
136
TC293.15 K=
(b) η0.6 ηmax
⋅:= QdotH
Wdot
η
:= QdotH2.492 109
×W=
River temperature rise: Vdot 165 m3
s
⋅:= ρ 1gm
cm3
⋅:=
5.43 QH150−kJ⋅:= Q150 kJ⋅:= Q2100 kJ⋅:=
TH550 K⋅:= T1350 K⋅:= T2250 K⋅:= Tσ300 K⋅:=
5.44 Wdot 750−MW⋅:= TH315 273.15+()K⋅:= TC20 273.15+()K⋅:=
TH588.15 K=
137
Calculate ideal work using Eqn. (5.26)
Assume air is an Ideal Gas
Tσ70 459.67+( )rankine:=P 1atm:=
T220 459.67+( )rankine:=T170 459.67+( )rankine:=Vdot 100000 ft3
hr
:=
a)
5.47
SG
6
,3.355,0.575 10 3−
⋅, 0,0.016−105
⋅,
()
⋅
…:=
Calculate the rate of entropy generation using Eqn. (5.23)
∆H6
7R⋅ICPH T1T2
,3.355,0.575 10 3−
⋅, 0,0.016−105
⋅,
()
⋅
…:=
First check the First Law using Eqn. (2.33) neglect changes in kinetic and
potential energy.
P21atm:=P15bar:=
T322−273.15+()K:=T227 273.15+()K⋅:=T120 273.15+()K:=
5.46
138
ndotgas
T1
T2
TCpT()
⌠
⎮
⌡d⋅mdotsteam ∆Hv
⋅+ 0=
First apply an energy balance on the boiler to get the ratio of steam flow
rate to gas flow rate.:
a)
Tsteam 212 459.67+( )rankine:=Tσ70 459.67+( )rankine:=
M29
gm
mol
:=∆Hv970 BTU
lbm
:=CpT( ) 3.83 0.000306 T
rankine
⋅+
⎛
⎜
⎝
⎞
⎠R⋅:=
T2300 459.67+( )rankine:=T12000 459.67+( )rankine:=
5.48
Wideal ndot R ICPH T1T2
,3.355,0.575 10 3−
⋅, 0,0.016−105
⋅,
()
⋅
…
⎡
⎢
⎤
⎥
⋅:=
Calculate ideal work using Eqn. (5.26)
ndot 34.064 mol
s
=ndot P Vdot⋅
RT
1
⋅
:=
Assume air is an Ideal Gas
T28−273.15+()K:=T125 273.15+()K:=Vdot 3000 m3
hr
:=
b)
139
∆Hgas
T2
TCpT()
⌠
⎮
⌡d:=
c)
Use ratio to calculate ideal work of steam per lbmol of gas
Calculate lbs of steam generated per lbmol of gas cooled.
Wideal 205.071−BTU
lb
=Wideal ∆Hsteam Tσ∆Ssteam
⋅−
()
:=
∆Ssteam 1.444−BTU
lb rankine⋅
=∆Ssteam
∆Hv
−
Tsteam
:=∆Hsteam ∆Hv
−:=
b)
∆Ssteam 1.444 BTU
lb rankine⋅
=∆Ssteam
∆Hv
Tsteam
:=
SdotG
ndotgas
mdotsteam
ndotgas
∆Ssteam
⋅∆Sgas
+=
Calculate entropy generation per lbmol of gas:
140
∆Ssteam 6.048 103
×J
kg K⋅
=∆Ssteam
∆Hv
Tsteam
:=
SdotG
ndotgas
mdotsteam
ndotgas
∆Ssteam
⋅∆Sgas
+=
Calculate entropy generation per lbmol of gas:
ndotgas
T1
T2
TCpT()
⌠
⎮
⌡d⋅mdotsteam ∆Hv
⋅+ 0=
First appl
y
an ener
gy
balance on the boiler to
g
et the ratio of steam flow rate t
o
gas flow rate.:
a)
Tsteam 100 273.15+()K:=Tσ25 273.15+()K:=
M29
gm
mol
:=∆Hv2256.9 kJ
kg
:=CpT( ) 3.83 0.000551 T
K
⋅+
⎛
⎜
⎝
⎞
⎠R⋅:=
T2150 273.15+()K:=T11100 273.15+()K:=
5.49
141
∆Sethylene 0.09−kJ
mol K⋅
=
∆Sethylene R ICPS T1T2
,1.424,14.394 10 3−
⋅, 4.392−10 6−
⋅, 0,
()
⋅:=
a)
Tσ25 273.15+()K:=T235 273.15+()K:=T1830 273.15+()K:=
5.50
∆Hgas
T1
T2
TCpT()
⌠
⎮
⌡d:=
c)
Calculate lbs of steam generated per lbmol of gas cooled.
Wideal 453.618−kJ
kg
=Wideal ∆Hsteam Tσ∆Ssteam
⋅−
()
:=
∆Ssteam 6.048−103
×J
kg K⋅
=∆Ssteam
∆Hv
−
Tsteam
:=∆Hsteam ∆Hv
−:=
b)
142
∆Sethylene
QC
−0=Solving for QC gives: QCTσ∆Sethylene
⋅:=
143