chapter9 a
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S2 0.21868:= P2 138.83:=
State 3, Wet Vapor at TC: Hliq 15.187:= Hvap 104.471:= P3 26.617:=
State 4, Wet Vapor at TC: Sliq 0.03408:= Svap 0.22418:= P4 26.617:=
(a) The pressures in (psia) appear above.
(b) Steps 3--2 and 1--4 (Fig. 8.2) are isentropic, for which S3=S2 and S1=S4.
Thus by Eq. 6.82):
x3S2 Sliq
Svap Sliq:= x3 0.971= x4
S1 Sliq
Svap Sliq:= x4 0.302=
(c) Heat addition, Step 4--3:
H3 Hliq x3 Hvap Hliq ( )+:= H4 Hliq x4 Hvap Hliq ( )+:=
H3 101.888= H4 42.118=
Q43 H3 H4( ):= Q43 59.77= (Btu/lbm)
Chapter 9 - Section A - Mathcad Solutions
9.2 TH 20 273.15+( )K:= TH 293.15K=
TC 20 273.15+( )K:= TC 253.15K=
QdotC 125000kJ
day:=
CarnotTC
TH TC:= (9.3) 0.6 Carnot:= 3.797=
Wdot
QdotC
:=(9.2)
Wdot 0.381kW=
Cost0.08
kW hrWdot:= Cost 267.183
dollars
yr= Ans.
9.4 Basis: 1 lbm of tetrafluoroethane
The following property values are found from Table 9.1:
State 1, Sat. Liquid at TH: H1 44.943:= S1 0.09142:= P1 138.83:=
State 2, Sat. Vapor at TH: H2 116.166:=
298
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(Refrigerator)
By Eq. (5.8): Carnot 1TC
TH:= Carnot 0.43=
By Eq. (9.3): CarnotT'C
T'H T'C
:= Carnot 10.926=
By definition: Wengine
QH=
Q'C
Wrefrig=
But Wengine Wrefrig= Q'C 35kJ
sec:=
Whence QHQ'C
Carnot Carnot:= QH 7.448
kJ
sec= Ans.
Given that: 0.6 Carnot:= 0.6 Carnot:= 6.556=
QHQ'C
:= QH 20.689
kJ
sec= Ans.
(d) Heat rejection, Step 2--1:
Q21 H1 H2( ):= Q21 71.223= (Btu/lbm)
(e) W21 0:= W43 0:=W32 H2 H3( ):= W32 14.278=
W14 H4 H1( ):= W14 2.825=
(f) Q43
W14 W32+:= 5.219=
Note that the first law is satisfied:
Q Q21 Q43+:= W W32 W14+:= Q W+ 0=
9.7 TC 298.15 K:= TH 523.15 K:= (Engine)
T'C 273.15 K:= T'H 298.15 K:=
299
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(isentropic compression)S'3 S2=
H4 37.978Btu
lbm:=T4 539.67 rankine:=
From Table 9.1
for sat. liquid
S2
0.22244
0.22325
0.22418
0.22525
0.22647
Btu
lbm rankine:=H2
107.320
105.907
104.471
103.015
101.542
Btu
lbm:=
QdotC
600
500
400
300
200
Btu
sec
:=
0.79
0.78
0.77
0.76
0.75
:=T2
489.67
479.67
469.67
459.67
449.67
rankine:=
The following vectors contain data for parts (a) through (e). Subscripts
refer to Fig. 9.1. Values of H2 and S2 for saturated vapor come from
Table 9.1.
9.9
or -45.4 degC
Ans.TC 227.75K=
9.8 (a) QC 4kJ
sec:= W 1.5 kW:=
QC
W:= 2.667= Ans.
(b) QH QC W+:= QH 5.5kJ
sec= Ans.
(c) TC
TH TC= TH 40 273.15+( ) K:= TH 313.15K=
TC TH
1+
:=
300
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Ans.Wdot
94.5
100.5
99.2
90.8
72.4
kW=Wdot mdot H23( )
:=
Ans.QdotH
689.6
595.2
494
386.1
268.6
Btu
sec=QdotH mdot H4 H3( )
:=
Ans.mdot
8.653
7.361
6.016
4.613
3.146
lbm
sec
=mdotQdotC
H2 H1
:=
H3
273.711
276.438
279.336
283.026
286.918
kJ
kg=H23
24.084
30.098
36.337
43.414
50.732
kJ
kg=H1 88.337
kJ
kg=
H1 H4:=
H3 H2 H23+:=H23H'3 H2
:=H'3
115.5
116.0
116.5
117.2
117.9
Btu
lbm:=
The saturation pressure at Point 4 from Table 9.1 is 101.37(psia). For
isentropic compression, from Point 2 to Point 3', we must read values for
the enthalpy at Point 3' from Fig. G.2 at this pressure and at the entropy
values S2. This cannot be done with much accuracy. The most
satisfactory procedure is probably to read an enthalpy at S=0.22 (H=114)and at S=0.24 (H=126) and interpolate linearly for intermediate values of
H. This leads to the following values (rounded to 1 decimal):
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H23 402.368kJ
kg=
H1 H4:=H23H'3 H2
:=H'3 2814.7
kJ
kg:=
The saturation pressure at Point 4 from Table F.1 is 5.318 kPa. We must
find in Table F.2 the enthalpy (Point 3') at this pressure and at the
entropy S2. This requires double interpolation. The pressure lies
between entries for pressures of 1 and 10 kPa, and linear interpolation
with P is unsatisfactory. Steam is here very nearly an ideal gas, for
which the entropy is linear in the logarithm of P, and interpolation must
be in accord with this relation. The enthalpy, on the other hand, changes
very little with P and can be interpolated linearly. Linear interpolation
with temperture is satisfactory in either case.
The result of interpolation is
(isentropic compression)S'2 S2=H4 142.4 kJkg:=
S2 9.0526kJ
kg K:=H2 2508.9
kJ
kg:=QdotC 1200
kJ
sec:=
0.76:=T4 34 273.15+( ) K:=T2 4 273.15+( ) K:=
Subscripts in the following refer to Fig. 9.1. All property values come from
Tables F.1 and F.2.
9.10
Ans.Carnot
9.793
7.995
6.71
5.746
4.996
=CarnotTC
TH TC
:=
TH T4:=TC T2:=
Ans.
6.697
5.25
4.256
3.485
2.914
=QdotC
Wdot
:=
302
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H2 Hvap:=Hvap 100.799Btu
lbm:=Hliq 7.505
Btu
lbm:=
At the conditions of Point 2 [t = -15 degF and
P = 14.667(psia)] for sat. liquid and sat. vapor from Table 9.1:
Parts (a) & (b): subscripts refer to Fig. 9.19.11
Ans.Carnot 9.238=CarnotT2
T4 T2:=
Ans. 5.881=QdotC
Wdot:=
Ans.Wdot 204kW=Wdot mdot H23:=
Ans.QdotH 1404kJ
sec=QdotH mdot H4 H3( ):=
Ans.mdot 0.507 kgsec=
mdotQdot
CH2 H1:=
H3 2.911 103
kJ
kg=H3 H2 H23+:=
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mdot 0.0759lbm
sec= Ans.
(c) The sat. vapor from the evaporator is superheated in the heat
exchanger to 70 degF at a pressure of 14.667(psia). Property values
for this state are read (with considerable uncertainty) from Fig. G.2:
H2A 117.5Btu
lbm:= S2A 0.262
Btu
lbm rankine:=
mdotQdotC
H2A H4:= mdot 0.0629
lbm
sec= Ans.
(d) For isentropic compression of the sat. vapor at Point 2,
S3 Svap:= and from Fig. G.2 at this entropy and P=101.37(psia)
H3 118.3Btu
lbm:= Eq. (9.4) may now beapplied to the two cases:
In the first case H1 has the value of H4:
aH2 H4
H3 H2:= a 3.5896= Ans.
Sliq 0.01733Btu
lbm rankine:= Svap 0.22714
Btu
lbm rankine:=
For sat. liquid at Point 4 (80 degF):
H4 37.978Btu
lbm:= S4 0.07892
Btu
lbm rankine:=
(a) Isenthalpic expansion: H1 H4:=
QdotC 5Btu
sec:= mdot
QdotC
H2 H1:= mdot 0.0796
lbm
sec= Ans.
(b) Isentropic expansion: S1 S4:=
x1S1 Sliq
Svap Sliq:= H1 Hliq x1 Hvap Hliq( )+:= H1 34.892
BTU
lbm=
mdotQdotC
H2 H1:=
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mdot 25.634lbm
sec=mdot
QdotC
H2 H1:=QdotC 2000
Btu
sec:=
H1 27.885BTU
lbm=H1 H4 H2A H2+:=
Energy balance, heat exchanger:
S4 0.07892Btu
lbm R:=H4 37.978
Btu
lbm:=
For sat. liquid at Point 4 (80 degF):
S2A 0.2435Btu
lbm rankine:=H2A 116
Btu
lbm:=
At Point 2A we have a superheated vapor at the same pressure and at70 degF. From Fig. G.2:
S2 0.22325Btu
lbm rankine:=H2 105.907
Btu
lbm:=
At the conditions of Point 2 [sat. vapor, t = 20 degF and P = 33.110(psia)]
from Table 9.1:
Subscripts: see figure of the preceding problem.9.12
Ans.c 3.8791=cQdotC
Wdot:=
Wdot 1.289BTU
sec=
Wdot H3 H2A( ) mdot:=H3 138Btu
lbm:= (Last calculated
value of mdot)
In Part (c), compression is at constant entropy of 0.262 to the
final pressure. Again from Fig. G.2:
Ans.b 3.7659=bH2 H1
H3 H2
:=
In the second case H1 has its last calculated value [Part (b)]:
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H1 H4:=H'3
113.3
116.5
119.3
Btu
lbm:=H4
31.239
37.978
44.943
Btu
lbm:=
H values for sat. liquid at Point 4 come from Table 9.1 and H values
for Point 3` come from Fig. G.2. The vectors following give values for
condensation temperatures of 60, 80, & 100 degF at pressures of
72.087, 101.37, & 138.83(psia) respectively.
S'3 S2:=S2 0.22418Btu
lbm R:=H2 104.471
Btu
lbm:=
Subscripts refer to Fig. 9.1.
At Point 2 [sat. vapor @ 10 degF] from Table 9.1:
9.13
Ans.Wdot 418.032 kW=mdot 29.443lbm
sec=
Hcomp 13.457Btu
lbm=Wdot mdot Hcomp:=
HcompH'3 H2
:=H'3 116
Btu
lbm:=mdot
QdotC
H2 H4:=
If the heat exchanger is omitted, then H1 = H4.
Points 2A & 2 coincide, and compression is at a constant entropy of
0.22325 to P = 101.37(psia).
Ans.Wdot 396.66kW=mdot 25.634lbm
sec=
Hcomp 14.667Btu
lbm=Wdot mdot Hcomp:=
HcompH'3 H2A
:= 0.75:=H'3 127Btu
lbm:=
For compression at constant entropy of 0.2435 to the final pressure of
101.37(psia), by Fig. G.2:
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Minimum t = -4.21 degC
Ans.KTC 268.94=
TC Find TC( ):=
Wdot
0.75 TH TC( )TH TC
TH=
Given
(Guess)TC 250:=
Wdot
QdotH
TH TC
TH=
QdotH 0.75 TH TC( )=
Wdot 1.5:=
TH 293.15:=WINTER9.14
Ans.
6.221
4.146
3.011
=H2 H1
H
:=
Eq. (9.4) now becomes
H H3 H2=SinceHH'3 H2
0.75:=(b)
Ans.
8.294
5.528
4.014
=H2 H1
H'3 H2
:=
By Eq. (9.4):(a)
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H41033.5
785.3
kJ
kg:= H9 284.7 kJ
kg:= H15 1186.7
1056.4
kJ
kg:=
By Eq. (9.8): zH4 H15
H9 H15
:= z0.17
0.351
= Ans.
9.17 Advertized combination unit:
TH 150 459.67+( ) rankine:= TC 30 459.67+( ) rankine:=
TH 609.67rankine= TC 489.67rankine=
QC 50000Btu
hr:= WCarnot QC
TH TC
TC:= WCarnot 12253
Btu
hr=
SUMMERTC 298.15:=
QdotC 0.75 TH TC( ):=
Wdot
QdotC
TH TC
TC=
TH 300:= (Guess)
Given
Wdot
0.75 TH TC
( )
TH TC
TC=
TH Find TH( ):=
TH 322.57= K Ans.
Maximum t = 49.42 degC
Data in the following vectors for Pbs. 9.15 and 9.16 come from
Perry's Handbook, 7th ed.9.15 and 9.16
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TC 210:= T'H 260:= T'C 255:= TH 305:=
By Eq. (9.3):
TC
TH TC:= I 0.65
TC
T'H TC:= II 0.65
T'C
TH T'C:=
WCarnotQC
= WI
QC
I= WII
QC
II=
Define r as the ratio of the
actual work, WI + WII, to the
Carnot work:r
1
I
1
II+
:= r 1.477= Ans.
9.19 This problem is just a reworking of Example 9.3 with different values of x.
It could be useful as a group project.
WI 1.5 WCarnot:= WI 18380Btu
hr=
This is the TOTAL power requirement for the advertized combination unit.
The amount of heat rejected at the higher temperature of150 degF is
QH WI QC+:= QH 68380Btu
hr=
For the conventional water heater, this amount of energy must be supplied
by resistance heating, which requires power in this amount.
For the conventional cooling unit,
TH 120 459.67+( ) rankine:=
WCarnot QCTH TC
TC:= WCarnot 9190
Btu
hr=
Work 1.5 WCarnot:= Work 13785Btu
hr=
The total power required is
WII QH Work+:= WII 82165Btu
hr= NO CONTEST
9.18
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Calculate the high and low operating pressures using the given vapor
pressure equation
Guess: PL 1bar:= PH 2bar:=
Given lnPL
bar
45.3274104.67
T1
K
5.146 lnT1
K
615.0
PL
bar
T1
K
2+=
PL Find PL( ):= PL 6.196bar=
Given lnPH
bar
45.3274104.67
T4
K
5.146 lnT4
K
615.0
PH
bar
T4
K
2+=
PH Find PH( ):= PH 11.703bar=
Calculate the heat load
ndottoluene 50kmol
hr
:= T1 100 273.15+( )K:= T2 20 273.15+( )K:=
Using values from Table C.3
QdotC ndottoluene R ICPH T1 T2, 15.133, 6.79 103, 16.35 10 6, 0,( ):=
QdotC 177.536kW=
9.22 TH 290K:= TC 250K:= Ws 0.40kW:=
CarnotTC
TH TC:= Carnot 6.25= 65%Carnot:= 4.063=
Ans.
QC Ws := QC 1.625 103 kg m2 sec-3= QH Ws QC+:= QH 2.025kW=
9.23 Follow the notation from Fig. 9.1
With air at 20 C and the specification of a minimum approach T = 10 C:T1 10 273.15+( )K:= T4 30 273.15+( )K:= T2 T1:=
310
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Vliq 27.112cm
3
mol=
Estimate Hlv at 10C using Watson correlationTrn
Tn
Tc:= Trn 0.591= Tr1
T1
Tc:= Tr1 0.698=
Hlv Hlvn1 Tr11 Trn
0.38
:= Hlv 20.798kJ
mol=
Hliq41 Vliq PH PL( ) R ICPH T1 T4, 22.626, 100.75 103, 192.71 10 6, 0,( )+:=
Hliq41 1.621kJ
mol= x1
Hliq41
Hlv:= x1 0.078=
For the evaporator
H12 H2 H1=
H1vap H1liq x1 Hlv+( )=
1 x1( ) Hlv=
H12 1 x1( ) Hlv:= H12 19.177kJ
mol=
ndotQdotC
H12:= ndot 9.258
mol
sec= Ans.
Since the throttling process is adiabatic: H4 H1=
But: Hliq4 Hliq1 x1 Hlv1+= so: Hliq4 Hliq1 x1 Hlv=
and: Hliq4 Hliq1 Vliq P4 P1( )T1
T4
TCpliq T( )
d+=
Estimate Vliq using the Rackett Eqn.
0.253:= Tc 405.7K:= Pc 112.80bar:=
Zc 0.242:= Vc 72.5cm
3
mol:= Tn 239.7K:= Hlvn 23.34
kJ
mol:=
Tr 20 273.15+( )KTc
:= Tr 0.723=
Vliq Vc Zc1 Tr( )
2
7
:=
311
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