Chapter9 A

7/30/2019 Chapter9 A

1/14

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


2/14

(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


3/14

(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


4/14

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):

301


5/14

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


6/14

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+:=

303


7/14

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:=

304


8/14

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)]:

305


9/14

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:

306


10/14

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)

307


11/14

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

308


12/14

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

309


13/14

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


14/14

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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Chapter9 A