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Chap 15Heat Exchangers

13-54 Glycerin is heated by ethylene glycol in a thin-walled double-pipe parallel-flow heat exchanger. Therate of heat transfer, the outlet temperature of the glycerin, and the mass flow rate of the ethylene glycolare to be determined.

Assumptions1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer tothe cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There isno fouling. 5 Fluid properties are constant. 6 The thermal resistance of the inner tube is negligible since

the tube is thin-walled and highly conductive.Properties The specific heats of glycerin and ethyleneglycol are given to be 2.4 and 2.5 kJ/kg.C,respectively.

Analysis (a) The temperature differences at thetwo ends are

T T T

T T T T T

h in c in

h out c out h out h out

1

2

60 20

15

= =

= =

, ,

, , , ,(

C C = 40 C

C) = 15 C

and

T

T T

T Tlm =

=

= 1 2

1 2

40 15

40 15255

ln( / ) ln( / ). C

Then the rate of heat transfer becomes

kW19.58==== W584,19C)5.25)(mC)(3.2.W/m240( 22lms TUAQ

(b) The outlet temperature of the glycerin is determined from

C47.2=

+=+==C)kJ/kg.kg/s)(2.43.0(

kW584.19C20)]([ glycerin

pinoutinoutp

Cm

QTTTTCmQ

(c) Then the mass flow rate of ethylene glycol becomes

kg/s3.56=

=

=

=

C]60C15)+C)[(47.2kJ/kg.(2.5

kJ/s584.19

)(

)]([

glycolethylene

glycolethylene

outinp

outinp

TTC

Qm

TTCmQ

13-37

Glycerin20C0.3 kg/s

Hot ethylene

60C3 kg/s

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13-55 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer andthe outlet temperature of the air are to be determined.

Assumptions1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer tothe cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There isno fouling. 5 Fluid properties are constant.

Properties The specific heats of air and combustion gases are given to be 1005 and 1100 J/kg.C,respectively.

Analysis The rate of heat transfer is

kW103=

C)95CC)(180kJ/kg.kg/s)(1.11.1(

)]([ gas.

=

= outinp TTCmQ

The mass flow rate of air is

( ..m

PV

RT= =

=(95 kPa)(0.8 m / s)

kPa. m / kg. K) K kg / s

3

30 287 293

0904

Then the outlet temperature of the air becomes

C133=

+=+=

=

C)J/kg.kg/s)(1005904.0(

W10103C20

)(

3

,,

,,

pincoutc

incoutcp

Cm

QTT

TTCmQ

13-38

Air95 kPa20C

0.8 m3/s

Exhaust gases

1.1 kg/s95C

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13-56 Water is heated by hot oil in a 2-shell passes and 12-tube passes heat exchanger. The heat transfersurface area on the tube side is to be determined.


Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ/kg.C, respectively.

Analysis The rate of heat transfer in this heat exchanger is

[ ( )] ( .Q mC T T p out in= = water kg / s)(4.18 kJ / kg. C)(70 C C) = 940.5 kW4 5 20

The outlet temperature of the hot water is determined from

C129C)kJ/kg.kg/s)(2.310(

kW5.940C170)]([ oil =

===

pinoutoutinp

Cm

QTTTTCmQ

The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are

C109=C20C129

C100=C70C170

,,2

,,1

==

==

incouth

outcinh

TTT

TTT

C105)109/100ln(

109100

)/ln( 21

21, =

=

=

TT

TTT CFlm

0.1

2.1170129

7020

27.017020

170129

12

21

11

12

=

=

=

=

=

=

=

F

tt

TTR

tT

ttP

Then the heat transfer surface area on the tube side becomes

2

m15=

=

==

C)105(C)(1.0).kW/m6.0(

kW5.940

2,

,CFlm

sCFlms TUF

Q

ATFUAQ

13-39

Oil170C10 kg/s

Water20C

4.5 kg/s

70C

(12 tube passes)

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13-57 Water is heated by hot oil in a 2-shell passes and 12-tube passes heat exchanger. The heat transfersurface area on the tube side is to be determined.


Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ/kg.C, respectively.


[ ( )] (Q mC T T p out in= = water kg / s)(4.18 kJ / kg. C)(70 C C) = 418 kW2 20

The outlet temperature of the oil is determined from

C8.151C)kJ/kg.kg/s)(2.310(

kW418C170)]([ oil =

===

pinoutoutinp

Cm

QTTTTCmQ


T T T

T T T

h in c out

h out c in

1

2

170 70

1518 20

= =

= =

, ,

, , .

C C = 100 C

C C = 131.8 C

T

T T

T Tlm CF,

ln( / )

.

ln( / . ).=

=

= 1 2

1 2

100 1318

100 1318115 2 C

}

0.1

7.21708.151

7020

12.017020

1708.151

12

21

11

12

=

=

=

=

=

=

=

F

tt

TTR

tT

ttP


2m6.05=

=

==C)2.115(C)(1.0).kW/m6.0(

kW418

2,, CFlmiiCFlmii TFU

QATFAUQ

13-40

Oil

170C10 kg/s

Water20C2 kg/s

70C

(12 tube passes)

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13-58 Ethyl alcohol is heated by water in a 2-shell passes and 8-tube passes heat exchanger. The heattransfer surface area of the heat exchanger is to be determined.


Properties The specific heats of water and ethyl alcohol are given to be 4.19 and 2.67 kJ/kg.C,respectively.


kW252.3=C)25CC)(70kJ/kg.kg/s)(2.671.2()]([ alcoholethyl == inoutp TTCmQ


T T T

T T T

h in c out

h out c in

1

2

95 70

45 25

= =

= =

, ,

, ,

C C = 25 C

C C = 20 C

T

T T

T Tlm CF,

ln( / ) ln( / ).=

=

= 1 2

1 2

25 20

25 2022 4 C

77.0

9.09545

7025

7.09525

9545

12

21

11

12

=

=

=

=

=

=

=

F

tt

TTR

tT

ttP


2m15.4=

=

==C)4.22(C)(0.77).kW/m950.0(

kW3.2522

,

,

CFlmi

iCFlmiiTFU

QATFAUQ

13-41

Water90C

EthylAlcohol

25C2.1 kg/s

70C

(8 tube passes)

45C

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13-59 Water is heated by ethylene glycol in a 2-shell passes and 12-tube passes heat exchanger. The rateof heat transfer and the heat transfer surface area on the tube side are to be determined.


Properties The specific heats of water and ethylene glycol are given to be 4.18 and 2.68 kJ/kg.C,respectively.

Analysis The rate of heat transfer in this heat exchanger is :

kW160.5=C)22CC)(70kJ/kg.kg/s)(4.188.0()]([ water == inoutp TTCmQ


T T T

T T T

h in c out

h out c in

1

2

110 70

60 22

= =

= =

, ,

, ,

C C = 40 C

C C = 38 C

T

T T

T Tlm CF,

ln( / ) ln( / )=

=

= 1 2

1 2

40 38

40 3839 C

94.0

96.011060

7022

57.011022

11060

12

21

11

12

=

=

=

=

=

=

=

F

tt

TTR

tT

ttP


2m15.6=

=

==C)39(C)(0.94).kW/m28.0(

kW5.1602

,

,

CFlmi

iCFlmiiTFU

QATFAUQ

13-42

Ethylene110C

Water22C

0.8 kg/s

70C

(12 tube passes)

60C

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0.25 0.65 1.05 1.45 1.85 2.25

50

10 0

15 0

20 0

25 0

30 0

35 0

40 0

45 0

5

10

15

20

25

30

35

40

45

mw[kg/s]

Q

[kW]

A

m2

heat

area

13-44

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13-61E Steam is condensed by cooling water in a condenser. The rate of heat transfer, the rate ofcondensation of steam, and the mass flow rate of cold water are to be determined.

Assumptions1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer tothe cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There isno fouling. 5 Fluid properties are constant. 6 The thermal resistance of the inner tube is negligible sincethe tube is thin-walled and highly conductive.

Properties We take specific heat of water are given to be1.0 Btu/lbm.F. The heat of condensation of steam at 90Fis 1043 Btu/lbm.

Analysis (a) The log mean temperature difference isdetermined from

T T T

T T T

h in c out

h out c in

1

2

90 73

90 60

= =

= =

, ,

, ,

F F = 17 F

F F = 30 F

T

T T

T Tlm CF,

ln( / ) ln( / ).=

=

= 1 2

1 2

17 30

17 3022 9 F

The heat transfer surface area is

2ft7.392ft)ft)(548/3(5088 === DLnAs

and

Btu/h105.396 6=== F)9.22)(ftF)(392.7.Btu/h.ft600( 22lms TUAQ

(b) The rate of condensation of the steam is

lbm/s1.44=lbm/h5173=

===Btu/lbm1043

Btu/h10396.5)(

6

fgsteamsteamfg

h

QmhmQ

(c) Then the mass flow rate of cold water becomes

lbm/s115lbm/h104.15 5 ==

=

=

=

F]60FF)(73Btu/lbm.(1.0

Btu/h10396.5

)(

)]([

6

watercold

watercold

inoutp

inoutp

TTC

Qm

TTCmQ

13-45

Steam90F20 lbm/s

60F

Water

73F

90F

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13-62"!PROBLEM 13-62E"

"GIVEN"N_pass=8N_tube=50"T_steam=90 [F], parameter to be varied"h_fg_steam=1043 "[Btu/lbm]"

T_w_in=60 "[F]"T_w_out=73 "[F]"C_p_w=1.0 "[Btu/lbm-F]"D=3/4*1/12 "[ft]"L=5 "[ft]"U=600 "[Btu/h-ft^2-F]"

"ANALYSIS""(a)"DELTAT_1=T_steam-T_w_outDELTAT_2=T_steam-T_w_inDELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2)A=N_pass*N_tube*pi*D*L

Q_dot=U*A*DELTAT_lm*Convert(Btu/h, Btu/s)"(b)"Q_dot=m_dot_steam*h_fg_steam"(c)"Q_dot=m_dot_w*C_p_w*(T_w_out-T_w_in)

Tsteam [F] Q [Btu/s] msteam[lbm/s] mw [lbm/s]

80 810.5 0.7771 62.34

82 951.9 0.9127 73.23

84 1091 1.046 83.89

86 1228 1.177 94.42

88 1363 1.307 104.990 1498 1.436 115.2

92 1632 1.565 125.694 1766 1.693 135.8

96 1899 1.821 146.1

98 2032 1.948 156.3

100 2165 2.076 166.5

102 2297 2.203 176.7

104 2430 2.329 186.9

106 2562 2.456 197.1

108 2694 2.583 207.2110 2826 2.709 217.4

112 2958 2.836 227.5

114 3089 2.962 237.6

116 3221 3.088 247.8

118 3353 3.214 257.9

120 3484 3.341 268

13-46

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80 85 90 95 100 105 110 115 120

50 0

1000

1500

2000

2500

3000

3500

0. 5

1

1. 5

2

2. 5

3

3. 5

Tsteam

[F]

Q

[Btu/s]

msteam

[lbm/s]

heat

msteam

80 85 90 95 100 105 110 115 12050

95

14 0

18 5

23 0

27 5

Tsteam[F ]

mw

[lbm/s]

13-47

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13-63 Glycerin is heated by hot water in a 1-shell pass and 13-tube passes heat exchanger. The mass flowrate of glycerin and the overall heat transfer coefficient of the heat exchanger are to be determined.


Properties The specific heats of water and glycerin are given to be 4.18 and 2.48 kJ/kg.C, respectively.


[ ( )] (5Q mC T T p in out= = water kg / s)(4.18 kJ / kg. C)(100 C C) = 940.5 kW55

The mass flow rate of the glycerin is determined from

kg/s9.5=

=

=

=

C]15CC)[(55kJ/kg.(2.48

kJ/s5.940

)(

)]([

glycerin

glycerin

inoutp

inoutp

TTC

Qm

TTCmQ


T T T

T T T

h in c out

h out c in

1

2

100 55

55 15= = = =

, ,

, ,

C C = 45 C

C C = 40 C

T

T T

T Tlm CF,

ln( / ) ln( / ).=

=

= 1 2

1 2

45 40

45 4042 5 C

77.0

89.010055

5515

53.010015

10055

12

21

11

12

=

=

=

=

=

=

=

F

tt

TTR

tT

ttP

The heat transfer surface area is

2m0.94=m)m)(2015.0(10 == DLnAs

Then the overall heat transfer coefficient of the heat exchanger is determined to be

C.kW/m30.6 2 =

=

==C)5.42)(77.0)(m94.0(

kW5.9402

,,

CFlmsCFlms

TFA

QUTFUAQ

13-48

Glycerin15C

100C

Hot Water5 kg/s

55C

55C

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13-64 Isobutane is condensed by cooling air in the condenser of a power plant. The mass flow rate of airand the overall heat transfer coefficient are to be determined.


Properties The heat of vaporization of isobutane at 75C is given to be hfg = 255.7 kJ/kg and specific heatof air is given to be Cp = 1005 J/kg.C.

Analysis First, the rate of heat transfer is determined from

kW39.690)kJ/kg7.255)(kg/s7.2()( isobutane === fghmQ

The mass flow rate of air is determined from

kg/s98.14=

C)21CC)(28kJ/kg.(1.005

kJ/s39.690

)(

)]([

inoutair

airinout

=

=

=

TTC

Qm

TTCmQ

p

p

The temperature differences between the isobutane andthe air at the two ends of the condenser are

C47=C28C75

C54=C21C75

inc,outh,2

outc,inh,1

==

==

TTT

TTT

and

C4.50)47/54ln(

4754

)/ln( 21

21lm =

=

=

TT

TTT

Then the overall heat transfer coefficient is determined from

C.W/m571 2 == =C)4.50)(m(24W390,690 2lm UUTUAQ s

13-49

Isobutane

75C2.7 kg/s

Air21C

Air28C

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13-65 Water is evaporated by hot exhaust gases in an evaporator. The rate of heat transfer, the exittemperature of the exhaust gases, and the rate of evaporation of water are to be determined.


Properties The heat of vaporization of water at 200C is given to be hfg = 1941 kJ/kg and specific heat ofexhaust gases is given to be Cp = 1051 J/kg.C.

Analysis The temperature differences between the waterand the exhaust gases at the two ends of the evaporator are

C)200(

C350=C200C550

outh,inc,outh,2

outc,inh,1

==

==

TTTT

TTT

and

[ ])200/(350ln)200(350

)/ln( outh,

outh,

21

21lm

=

=

T

T

TT

TTT

Then the rate of heat transfer can be expressed as

[ ])200/(350ln)200(350

)m5.0)(C.kW/m780.1(outh,

outh,22lm

==

T

TTUAQ s

(Eq. 1)

The rate of heat transfer can also be expressed as in the following forms

)CC)(550kJ/kg.51kg/s)(1.025.0()]([ outh,gasesexhaustouth,inh, TTTCmQ p ==

(Eq. 2)

)kJ/kg1941()( waterwater mhmQ fg == (Eq. 3)

We have three equations with three unknowns. Using an equation solver such as EES, the unknowns aredetermined to be

Q

T

m

=

=

=

88.85 kW

211.8 C

0.0458 kg/ s

h,out

water

13-50

Water200C

550F

Exhaustgases

Th,out

200C

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13-66 "!PROBLEM 13-66"

"GIVEN""T_exhaust_in=550 [C], parameter to be varied"C_p_exhaust=1.051 "[kJ/kg-C]"m_dot_exhaust=0.25 "[kg/s]"

T_w=200 "[C]"

h_fg_w=1941 "[kJ/kg]"A=0.5 "[m^2]"U=1.780 "[kW/m^2-C]"

"ANALYSIS"DELTAT_1=T_exhaust_in-T_wDELTAT_2=T_exhaust_out-T_wDELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2)Q_dot=U*A*DELTAT_lmQ_dot=m_dot_exhaust*C_p_exhaust*(T_exhaust_in-T_exhaust_out)Q_dot=m_dot_w*h_fg_w

Texhaust,in [C] Q [kW] Texhaust,out [C] mw [kg/s]

300 25.39 203.4 0.01308

320 30.46 204.1 0.0157

340 35.54 204.7 0.01831

360 40.62 205.4 0.02093

380 45.7 206.1 0.02354

400 50.77 206.8 0.02616420 55.85 207.4 0.02877

440 60.93 208.1 0.03139

460 66.01 208.8 0.03401

480 71.08 209.5 0.03662

500 76.16 210.1 0.03924

520 81.24 210.8 0.04185540 86.32 211.5 0.04447

560 91.39 212.2 0.04709580 96.47 212.8 0.0497

600 101.5 213.5 0.05232

13-51

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300 350 400 450 500 550 600

20

30

40

50

60

70

80

90

10 0

11 0

20 2

20 4

20 6

20 8

21 0

21 2

21 4

Texhaust,in

[C ]

Q

[kW]

T

C

heat

temperature

300 350 400 450 500 550 600

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

0.055

Texhaust,in

[C]

mw

[kg/s]

13-52

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13-67 The waste dyeing water is to be used to preheat fresh water. The outlet temperatures of each fluidand the mass flow rate are to be determined.


Properties The specific heats of waste dyeing water and the fresh water are given to be Cp = 4295 J/kg.Cand Cp = 4180 J/kg.C, respectively.

Analysis The temperature differences between the dyeing waterand the fresh water at the two ends of the heat exchanger are

15

75

outh,inc,outh,2

outc,outc,inh,1

==

==

TTTT

TTTT

and

[ ])15/()75(ln)15()75(

)/ln( outh,outc,

outh,outc,

21

21lm

=

=

TT

TT

TT

TTT

Then the rate of heat transfer can be expressed as

[ ])15/()75(ln)15()75(

)m65.1)(C.kW/m625.0(kW35outh,outc,

outh,outc,22

lm

=

=

TT

TT

TUAQ s

(Eq. 1)

The rate of heat transfer can also be expressed as

)CC)(75kJ/kg.(4.295kW35)]([ outh,waterdyeingouth,inh, TmTTCmQ p ==

(Eq. 2)

C)15C)(kJ/kg.(4.18kW35)]([ outc,waterdyeingouth,inh, == TmTTCmQ p

(Eq. 3)

We have three equations with three unknowns. Using an equation solver such as EES, the unknowns are

determined to be

T

T

m

c,out

h,out

=

=

=

41.4 C

49.3 C

0.317 kg/ s

13-53

Freshwater15C

Dyeingwater

75C

Tc,out

Th,out

heat chap13 054

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