heat exchangers

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HEAT EXCHANGERS

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Heat Exchangers

• A heat exchanger is used to exchange heat between two fluids of different temperatures, which are separated by a solid wall.

• Heat exchangers are ubiquitous to energy conversion and utilization. They encompass a wide range of flow configurations.

• Applications in heating and air conditioning, power production, waste heat recovery, chemical processing, food processing, sterilization in bio-processes.

• Heat exchangers are classified according to flow arrangement and type of construction.

Heat Exchangers: Heat Transfer Process

• DIRECT CONTACT TYPE – in this type, two immiscible fluids at different temperatures come in direct contact. i.e. jet condensers, desuperheaters, open feed-water heaters, scrubbers and cooling towers.

• TRANSFER TYPE/RECUPERATORS - in this type, the cold and hot fluids flow simultaneously through the device, and the heat is transferred through the wall separating them. This type is commonly used in most fields of engineering.

• REGENERATORS/ STORAGE TYPE – in this type, hot and cold fluids flow alternately on the same surface. This type is used as preheaters for steam power plants, blast furnaces, and oxygen producers.

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Heat Exchangers: Construction Features• TUBULAR HEAT EXCHANGERS – also known as tube in

tube, concentric tube or double-pipe heat exchangers. • SHELL AND TUBE HEAT EXCHANGERS – also known as

surface condensers and are most commonly used for heating, cooling, condensation, or evaporation applications.

• FINNED TUBE TYPE HEAT EXCHANGERS – when a high operating pressure or an enhanced heat transfer rate is required, extended surfaces are used on one side of the heat exchangers. This type are used for liquid to gas heat exchange. i.e. gas turbines, automobiles, airplanes, heat pumps, refrigeration, electronics, cryogenics, air conditioning, humidification.

• COMPACT HEAT EXCHANGERS – used for high area density heat transfer

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5

Finned Exchangers

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Heat Exchanger Applications

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Concentric Tube Construction

• - :• :

Parallel Flow CounterflowParallel Flow Counterflow


Heat Exchangers: Flow Arrangement

• PARALLEL FLOW - also known as concurrent heat exchangers, hot and cold fluids enter through the same point and leaves at the other end.

• COUNTERCURRENT FLOW – with this type, hot and cold fluids enter at opposite ends of the heat exchanger.

• CROSS FLOW – here, two fluids flow at a right angle to each other. For this type, it can further be classified as unmixed flow or mixed flow.

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9

Parallel Flow

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Counter Current Flow

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Heat Exchanger Analysis• Expression for convection heat transfer for flow of a fluid inside a tube:

)( ,, imompconv TTcmq

• For case 3 involving constant surrounding fluid temperature:

lms TAUq

)/ln( io

iolm TT

TTT

12

Heat Exchanger Analysis

In a two-fluid heat exchanger, consider the hot and cold fluids separately:

)(

)(

,,,

,,,

icoccpcc

ohihhphh

TTcmq

TTcmq

lmTUAq

13

Tlm: 1. Parallel-Flow Heat Exchangers

where


lmTUAq

)/ln( 12

12

TT

TTTlm

ocoh

icih

TTT

TTT

,,2

,,1

T1 T2

14

Tlm: 2. Counter-Flow Heat Exchangers

where

lmTUAq

icoh

ocih

TTT

TTT

,,2

,,1


T1 T2

)/ln( 12

12

TT

TTTlm

15

Overall Heat Transfer Coefficient

• For tubular heat exchangers we must take into account the conduction resistance in the wall and convection resistances of the fluids at the inner and outer tube surfaces.

kL

DDR

AhR

AhUA

iocond

oocond

ii

2

111

)/ln(


where inner tube surface

outer tube surface LDA

LDA

oo

ii

ooii AUAUUA

111

16

Shell-and-Tube Heat Exchangers

Baffles are used to establish a cross-flow and to induce turbulent mixing of the shell-side fluid, both of which enhance convection.

The number of tube and shell passes may be varied

One Shell Pass and One Tube Pass

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One Shell Pass,Two Tube Passes

Two Shell Passes,Four Tube Passes

Shell-and-Tube Heat Exchangers

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Multipass and Cross-Flow Heat Exchangers

To account for complex flow conditions in multipass, shell and tube and cross-flow heat exchangers, the log-mean temperature difference can be modified:

CFlmlm TFT ,

where F = correction factor

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Correction Factor

where t is the tube-side fluid temperature

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Fouling

• Heat exchanger surfaces are subject to fouling by fluid impurities, rust formation, or other reactions between the fluid and the wall material. The subsequent deposition of a film or scale on the surface can greatly increase the resistance to heat transfer between the fluids.

• An additional thermal resistance, can be introduced: The Fouling factor, Rf.

Depends on operating temperature, fluid velocity and length of service of heat exchanger. It is variable during heat exchanger operation.

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Overall Heat Transfer Coefficient

•The overall heat transfer coefficient can be written:

ooo

ofcond

i

if

iiooii AhA

RR

A

R

AhAUAUUA

11111

",

",

oofcondo

i

ifo

ii

o

o

hRRA

A

RA

AhA

U1

1

",

",

Need to determine hi and ho

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Determination of ho

• Approach 1: Using correlations Approach 2: Using chart by Kern

Typical values of baffle cuts 20-25% for liquids and 40-45% for vapor

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Determination of tube side film coefficient, hi

• Approach 1: Using correlations Approach 2: Sieder and Tate

24

Determination of Conduction Resistance

• Recall that

)/ln(2

2

)/ln(

ioo

condo

iocond

DDk

DRA

kL

DDR

• or

)/ln(2 io

w

ocondow DD

k

DRAr

25

Example

In a heat exchanger, hot fluid enters at 60OC and leaves at 48OC, where as the cold fluid enters at 35OC and leaves at 44OC. Calculate the mean temperature difference for

a) parallel flow,

b) counter flow,

c) single pass cross flow (both fluids unmixed)

d) single pass cross flow (hot side fluid mixed, cold side fluid unmixed)

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Thi = Tha = 60OC Tho = Thb = 48OCTci = Tca = 35OC Tco = Tcb = 44OC

)/ln( 12

12

TT

TTTlm

ocoh

icih

TTT

TTT

,,2

,,1

)4/25ln(

425 lmT

253560

44448

2

1

T

T

CTlmΟ5.11

Parallel Flow

27


icoh

ocih

TTT

TTT

,,2

,,1

)/ln( 12

12

TT

TTTlm

164460

133548

2

1

T

T)13/16ln(

1316 lmT

CT Olm 45.14

Countercurrent Flow

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Z = Thi – Tho = 60 – 48 = 1.33 Tcb – Tci 44 – 35

ŋH = Tco – Tci = 44 – 35 = 0.36 Thi – Tci 60 – 35


Cross Flow

Single pass cross flow; both fluids unmixed

From Figure 15.7 (b), F = 0.94

LMTD = (0.94)(14.45OC) = 13.583OC

2929

Z = Thi – Tho = 60 – 48 = 1.33 Tcb – Tci 44 – 35

ŋH = Tco – Tci = 44 – 35 = 0.36 Thi – Tci 60 – 35


Cross Flow

Single pass cross flow; hot fluid mixed, cold fluid unmixed

From Figure 15.7 (a), F = 0.98

LMTD = (0.98)(14.45OC) = 14.16OC

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Example

A 1-2 heat exchanger containing one shell pass and two tube passes heats 2.52 kg/s of water from 21.1 to 54.4OC by using hot water under pressure entering at 115.6 and leaving at 48.9OC. The outside surface area of the tubes in the exchanger is Ao = 9.30 m2.

a)Calculate the mean temperature difference LMTD in the exchanger and the overall heat transfer coefficient Uo.

b)For the same temperature but using a 2-4 exchanger, what would be the LMTD?

Thi = Tha = 115.6OC Tho = Thb = 48.9OCTci = Tca = 21.1OC Tco = Tcb = 54.4OC

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q = mCp,c (Tco –Tci) = 2.52 kg/s (4184 J/kg.K)(54.4 – 21.1OC)q = 351104.54 W

LMTD = (115.6 – 54.4) – (48.9 – 21.1) = 42.3261OC ln (115.6 – 54.4) (48.9 – 21.1)

CFlmlm TFT ,For multipass and crossflow heat exchangers

Example

LMTD for crossflow;LMTD = (Tha – Tcb ) – (Thb – Tca)

ln (Tha – Tcb ) (Thb – Tca)

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a) Single pass, shell fluid mixed, other fluids unmixed, use Figure 15.7 a ( 1 – 2 Heat Exchangers)

Z = Thi – Tho = 115.6 – 48.9 = 2.00 Tcb – Tci 54.4 – 21.1

ŋH = Tco – Tci = 54.4 – 21.1 = 0.352 Thi – Tci 115.6 – 21.1

From Figure 15.7 (a), F = 0.74 LMTD = (0.74)( 42.3261) = 31.3213OC

Uo = q = 351104.54 W = 1205.351 W/m2.K Ao (LMTD) (9.30 m2)(31.3213OC)

Example

3333

b) Single pass, both fluids unmixed, use Figure 15.7 b ( 2 – 4 Heat Exchangers)

Z = Thi – Tho = 115.6 – 48.9 = 2.00 Tcb – Tci 54.4 – 21.1

ŋH = Tco – Tci = 54.4 – 21.1 = 0.352 Thi – Tci 115.6 – 21.1

From Figure 15.7 (b), F = 0.94 LMTD = (0.94)( 42.3261) = 33.60693OC

Uo = q = 351104.54 W = 1123.375 W/m2.K Ao (LMTD) (9.30 m2)(33.60693OC)

Example

3434

Example

A stainless steel tube (k = 45 W/m.K) of inner and outer diameters of 22 mm and 27 mm respectively, is used in a cross flow heat exchanger. The fouling factors for the inner and outer surfaces are estimated to be 0.0004 and 0.0002 (m2.K)/W respectively. Determine the overall heat transfer coefficient based on the outside surface area of the tube.

Water at 75OC & 0.5 m/s

Di = 22 mm Do = 27 mm

Air at 15OC & 20 m/s

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Properties of water at 75OCk = 0.6715 W/m.Kν = 0.39 x 10-6 m2/sPr = 2.38

Nu = hiDi = 0.023 Re0.8 Pr0.4 k = 0.023 (28205.13)0.8 (2.38)0.4 = 118.2

hi = 118.2 (0.6715) = 3608 W/m2.K 0.022

Re = u D = 0.5 m/s (0.022 m) = 28205.13 (turbulent) ν 0.39 x 10-6 m2/s

3636

Properties of air at 15OCk = 0.0255 W/m.Kν = 14.16 x 10-6 m2/sPr = 0.704

Nu = hoDo = [0.04 Re0.5 + 0.06 Re0.67] Pr0.4 (μ/μw)0.25 k = [0.04(38135.59)0.5 + 0.06(38135.59)0.67 ](0.704)0.4(1)0.25

= 139.3021

ho = 139.3021 (0.0255) = 131.5631 W/m2.K 0.027

Re = u D = 20 m/s (0.027 m) = 38135.59 ν 14.16 x 10-6 m2/s

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ooo

ofcond

i

if

iiooii AhA

RR

A

R

AhAUAUUA

11111

",

",

oofcondo

i

ifo

ii

o

o

hRRA

A

RA

Ah

AU

1

1

",

",

38

Ro = 1 = 1 = 0.0896 hoAo (131.5631 W/m2.K)(П)(0.027) (1)

Ri = 1 = 1 = 0.00401 hiAi (3608 W/m2.K)(П)(0.022) (1)

Rfi = Fi = 0.0004 = 5.787 x 10 -3

Ai (П)(0.022) (1)

Rfo = Fo = 0.0002 = 2.358 x 10 -3

Ao (П)(0.027) (1)

Rcond = ln Do/Di = ln (0.027/0.022) = 7.24 x 10 -3 2 П k L 2 П (45) (1)

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ooo

ofcond

i

if

iioo AhA

RR

A

R

AhAU

111"

,"

,

0.0896 10 x 2.358 10 x 7.24 10 x 5.78700401.01 3-3-3-

oo AU

0.1024891

oo AU

.K W/m115.0296 2oU

oo

UA

0.102489

1

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A shell and tube heat exchanger with one shell pass and two tube passes is used to heat water (flowing in the tubes) at a rate of 10 kg/s from 30OC to 45OC with steam condensing over the tubes at 160OC. If the overall heat transfer coefficient (based on the outside area) has a value of 2000 W/m2.K, determine the area required. If 20 tubes of 25 mm OD are used. Determine the length of tube required.

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ExampleA counterflow, concentric tube heat exchanger is used to cool the lubricating oil for a large industrial gas turbine engine. The flow rate of cooling water through the inner tube (Di=25 mm) is 0.2 kg/s, while the flow rate of oil through the outer annulus (Do=45 mm) is 0.1 kg/s. The oil and water enter at temperatures of 100 and 30°C respectively. How long must the tube be made if the outlet temperature of the oil is to be 60°C?

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Example A shell-and-tube heat exchanger must be designed to heat 2.5 kg/s of water from 15 to 85°C. The heating is to be accomplished by passing hot engine oil, which is available at 160°C, through the shell side of the exchanger. The oil is known to provide an average convection coefficient of ho=400 W/m2.K on the outside of the tubes. Ten tubes pass the water through the shell. Each tube is thin walled, of diameter D=25 mm, and makes eight passes through the shell. If the oil leaves the exchanger at 100°C, what is the flow rate? How long must the tubes be to accomplish the desired heating?

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Heat Exchanger Effectiveness

Defined as the ratio of actual rate of heat transfer in a given exchanger to the maximum possible amount of heat transfer if an infinite heat transfer area were available.

)(

)(

,,,

,,,

icoccpcc

ohihhphh

TTcmq

TTcmq

ch

ccpc

hhph

CCthen

Ccm

Ccm

;

,

,

The cold fluid undergoes a greater temperature change, hence CC will be Cmin as minimum heat capacity and if there is an infinite area available then Tco = Thi.

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)(

)(

,,

,,

icihc

ohihh

TTC

TTCe

)min(

)max(

,,

,,

icih

ohih

TTC

TTCe

Actual heat transfer

)min( ,, icih TTeCq

45

)min(

)(

,,

,,

icih

ohih

TTC

TTChe


In the case of a single pass, counter current flow

)min(

)(

,,

,,

icih

icoc

TTC

TTCce

)]/()ln[(

)()()(

,,,,

,,,,,,

ocihicoh

ocihicohicoc

TTTT

TTTTUATTCcq

46

e = 1 – exp – UA 1 – Cmin Cmin Cmax

1 – Cmin exp – UA 1 – Cmin Cmax Cmin Cmax

Where: Number of Transfer Units is NTU = UA Cmin

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e = 1 – exp – UA 1 + Cmin Cmin Cmax

1 + Cmin Cmax

For parallel flow

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Example

Water flowing at a rate of 0.667 kg/s enters a counter current heat exchanger at 308 K and is heated by an oil stream entering at 383 K at a rate of 2.85 kg/s (Cp = 1.89 kJ/kg.K). The overall U = 300 W/m2.K and the area A = 15.0 m2. Calculate the heat transfer rate and the exit water temperature.

Assuming Tco = 370 K; Tf = (308 + 370 K) = 339 K Cp,C = 4.192 kJ/kg.K

ṁ Cp,h = Ch = 2.85 kg/s ( 1.89 x 103 J/kg.K) = 5386.5 W/K

ṁ Cp,c = Cc = 0.667 kg/s ( 4.192 x 103 J/kg.K) = 2796.06 W/K

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For counter current flow exchanger, using Figure 15.9 (a) e = 0.71

q = e Cmin (Thi – Tci) = 0.71(2796.06)(383 – 308) = 148890.4 W

q = 148890.4 W = 2796.06 (Tco – 308) Tco = 361.25 K

Cmin = 2796.06 = 0.5191 Cmax 5386.5

NTU = UA = 300 (15.0) = 1.6094 Cmin 2796.06

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A double pipe, parallel flow heat exchanger uses oil (Cp=1.88 kJ/kgOC) at an initial temperature of 205OC to heat water, flowing at 225 kg/h, from 16OC to 44OC. The oil flow rate is 270 kg/h a)what heat exchanger area is required fro an overall heat transfer coefficient of 340 W/m2.K b)determine the number of transfer units c)calculate the effectiveness of HE

Example

(225 kg/h)(4.18 x 103 kJ/kg.OC)(44 – 16) OC = (270 kg/h)(1.88 x103 kJ/kg.OC) (205 – Tho) OC

Tho = 153.12OCLMTD for crossflow;

LMTD = (Tha – Tcb ) – (Thb – Tca) ln (Tha – Tcb ) (Thb – Tca)

51

LMTD for crossflow;LMTD = (205 – 44 ) – (153.12 – 16) = 148. 7406OC

ln (205 – 44 ) (153.12 – 16)

A = 225 (4.18 x 103) (44 – 16) (1/3600) = 0.1446 m2

340 (148.7406OC)

Cmin = (mCp)H = 270 (1.88 x 103) = 507,600 J/h.OC

Cmax = (mCp)C = 225 (4.18 x 103) = 940,500 J/h.OC

NTU = UA = 340 (0.1446) = 0.3487 Cmin 507,600 (1/3600)

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e = 1 – exp – UA 1 + Cmin Cmin Cmax

1 + Cmin Cmax

e = 1 – exp – 0.3487 1 + 507,600 940,500

1 + 507,600

940,500

e = 0.2698 = 26.98 %

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In an open heart surgery, under hypothermic conditions, the patient’s blood is cooled before the surgery and rewarmed afterward. It is proposed that a concentric tube counterflow heat exchanger of 0.5 m length be used for this purpose with a thin-walled inner tube that has a diameter of 55 mm. If the water at 60OC and 0.10 kg/s is used to heat the blood entering the exchanger at 18OC and 0.05 kg/s, what is the temperature of the blood leaving the exchanger? The overall heat exchanger coefficient is 500 W/m2.K and the specific heat of the blood is 3500 J/kg.K.

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A steam condenser condensing at 70OC has to have a capacity of 100 kW. Water at 20OC is used and the outlet water temperature is limited to 45OC. If the overall heat transfer coefficient has a value of 3100 W/m2.K, determine the area required.

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Heat Transfer Coefficients

The heat transfer coefficient hi for the tube side fluid in a shell and tube exchanger can be calculated using equations in chapter 12. the coefficient for the shell side ho cannot be calculated because the direction of flow is partly parallel to the tubes and partly across them. An approximate but general useful equation for predicting shell side coefficients is the Donohue equation.

hoDo = 0.2 DoGe 0.6 Cpμ 0.33 μ 0.14

k μ k μw

Ge = √ Gb Gc Gb = ṁ / Sb Gc = ṁ / Sc

Mass velocity parallel with tubes

Mass velocity for crossflow

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Sb = ƒb π Ds2 – Nb π Do

2

4 4

ƒb = fraction of cross-sectional area of shell occupied by baffle windowDs = inside daimeter of shellNb = number of tubes in baffle windowDo = outside diameter of tubes

Total area of baffle window

Total area occupied by the tubes

Sc = P Ds ( 1 – Do/ƿ)

P = center to center distance between tubesǷ = baffle pitch

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Heat Transfer Coefficients

ho Cpμ 2/3 μw 0.14 = jH = 0.2 DoGe -0.4 CpGe k μ μ

j - factor form

Crossflow Exchangers

hoDo = 0.287 DoG 0.61 Cpμ 0.33 Fa k μ k

58

Example A tubular exchanger with 35-in (889-mm) ID contains eight hundred and twenty eight ¾ -in (19-mm) OD tubes 12 ft (3.66-mm) long on a 1-in (25-mm) square pitch. Standard 25 percent baffles are spaced 12-in (305 mm) apart. Liquid benzene at an average bulk temperature of 60OF (15.6OC) is bing heated in the shell side of the exchanger at the rate of 100,000 lb/h (45,360 kg/h). If the outside surfaces of the tubes are at 140OF (60OC), estimate the individual heat transfer coefficient of the benzene.

Do = 0.75 / 12 = 0.0625 ft Ds = 35 / 12 = 2.9167 ft

Ƿ = 1/ 12 = 0.0833 ft P = 1 ft

Sc = P Ds ( 1 – Do/ƿ)

Sc = 1 x 2.9167 1 – 0.0625 0.0833

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Nb = 0.1955 x 828 = 161.8 ≈ 161 tubes

Sb = ƒb π Ds2 – Nb π Do

2

4 4

Sb = 0.1955 π (2.9167)2 – 161 π (0.0625)2

4 4 Sb = 0.8123 ft2

Ge = √ Gb Gc Gb = ṁ / Sb Gc = ṁ / Sc

Gc = 100,000 = 137,137 lb/ft2.h 0.7292

Gb = 100,000 = 123,107 lb/ft2.h 0.8123

60

Ge = √ (137,137)(123,107) = 129,933 lb/ft2.h

Properties:μ @ 60OF = 0.70 cP μ @ 140OF =m0.38 cPCp = 0.41 Btu/lb.OF k = 0.092 Btu/h.ft.OF

hoDo = 0.2 DoGe 0.6 Cpμ 0.33 μ 0.14

k μ k μw

ho = 0.2 0.0625(129,933) 0.6 (0.41)(0.70)(2.42) 0.33 0.70 0.14 0.092

0.70 (2.42) 0.092 0.38 0.0625

ho = 101 Btu/h.ft2.OF (573 W/m2.OC)

61

HEAT TRANSFER IN AGITATED VESSELS

Heat transfer surfaces, which may be in the form of heating or cooling jackets or coils of pipe immersed in the liquid, are often used in the agitated vessels.

hjDi = 0.76 Da2ŋρ 2/3 Cpμ 1/3 μ 0.24 k μ k μw

hcDc = 0.17 Da2nρ 0.67 Cpμ 0.37 Da

0.1 Dc 0.5 μ 0.24

k μ k Dt Dt μw

With helical coil and turbine impeller

To or from the jacket of a baffled tank

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hjDt = K Da2ŋρ a Cpμ 1/3 μ 0.18 k μ k μw

Anchored agitators for viscous liquid

K = 1.0 a = ½ for 10 < Re < 300

K = 0.36 a = 2/3 for 300 < Re < 40,000

63

Example

A turbine agitated vessel 3 m in diameter contains 6600 kg of a dilute aqueous solution. the agitator is 0.75 m in diameter and turns at 150 rev/min. the vessel is jacketed with steam condensing at 38 psig; the heat transfer area is 16 m2. the steel walls of the vessel are 10-mm thick. If the heat transfer coefficient of the steam is 12 kW/m2.OC, what is the rate of heat transfer between the steam and liquid and the time to heat the vessel contents from 30OC to 90OC?

P = 52.7 psiaTs = 284.216OF = 140.12OC

Ŋ = 150 rev/min δ = 10 mmA = 16 m2 hc = 12 kW/m2.OCTa = 30OC Tb = 90OC

64

hjDi = 0.76 Da2ŋρ 2/3 Cpμ 1/3 μ 0.24 k μ k μw

Properties at Tf = 30 + 90 = 60OC = 140 OF 2

k = 0.378 Btu/ft.h.OFμ = 0.470 cP = 0.47 x 10-3 kg/m.sρ = 61.38 lb/ft3 μw = 0.196 cP

hj (3) = 0.76 0.7522.5(983.65) 2/3 4184(4.7x10-3) 1/3 0.47 0.24 0.6542 4.7 x 10-3 0.6542 0.196

65

hj = 6061.436 W/m2.OC

Ui = 1 1 + 1 3 + 10/1000 3 6061.436 12000 3.02 45 3.01

Ui = 2131.093 W/m2.OC

tT = 6600 (4184) ln (140.12 – 30)/(140.12 – 90) 2131.093 (16)

tT = 637.487 sec = 10.63 min

q = 2131.093 W/m2.OC (16 m2) ( 140.12 – 60)OC

heat exchangers

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