Heat Exchangers - ME14mech14.weebly.com/.../me_ecwhr_18_heat_exchangers.pdf · To analyze heat exchange problem, a set of some assumptions are introduced sothat resulting models are

Prof. P. K. Das Energy Conservation and Waste Heat Recovery

Heat Exchangers

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Classification of Heat Exchanger

Indirect contact type Direct contact type

1. Transfer Process

Storage type

Fluidized bed

Single Phase

Multiphase Immiscible fluid

Gas-Liquid

Liquid-VaporDirect transfer type

2.Number of Fluids

Two-fluid Three-fluidN-fluid

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3. Surface Compactness

Gas-to-Fluid Liquid-to-Fluid and Phase change

Noncompact(𝛽 Τ𝑚 𝑚Compact(𝛽 Τ𝑚 𝑚Compact(𝛽 Τ𝑚 𝑚Noncompact(𝛽 Τ𝑚 𝑚

4. Construction

Extended surface Regenerative Plate type Tubular Double-pipe

Shell-and-tube

Spiral

Pipe coils

PHE

Spiral

Plate coil

Printed circuit

Plate-fin

Tube-fin

Rotatory

Fixed-matrix

Rotating Hoods

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Combined convection and

radiation heat transfer

5.Heat Transfer Mechanism

Single-phase

Convection on

both-sides

Two-phase convection

on both side Single-phase convection on one

side , two-phase convection on

other side

6. Flow Arrangement

Single-phase Multi-phase

Counter Parallel Split- flow Divided-flowCounter flow

Extended Surface

Shell-and-tube

Plate

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Double pipe heat exchanger

https://qph.ec.quoracdn.net/main-qimg-

c9e98e0940812c1a290c7e382e81ad7e


41639803193702862731666502e01718

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Shell and tube heat exchanger

Sigle pass

http://classes.engineering.wustl.edu/mase-thermal-lab/htlab-10.jpg

http://www.southwestthermal.com/images/products_

shelltube_drawing_LG.jpg

Multi pass

https://upload.wikimedia.org/wikipedia/commons/5/

59/Straight-tube_heat_exchanger_2-pass.PNG

Exploded view

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http://fchart.com/ees/heat_transfer_library/compact_hx/typehx

_diagram.gif

Finned tube and plate-fin (cross flow) heat exchanger

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Plate & frame and spiral heat exchanger


7cd9a2c31a4fd878142fbac59f1c7194-c

https://4.imimg.com/data4/XV/LF/MY-

13558103/spiral-heat-exchanger-250x250.jpg

http://www.enggcyclopedia.com/wp-content/uploads/2010/12/spiral-

type-heat-exchager-schematic.jpg

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Plate & frame heat exchanger

https://www.youtube.com/watch?v=Jv5p7o-7Pms

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Methods of analysis of Heat Exchangers

To analyze heat exchange problem, a set of some assumptions are introduced so that resulting models are

simple to analyze. The assumptions are:

Heat exchanger operates under steady state condition.

Heat losses to and from surrounding are assumed to be negligible.

There are no thermal energy sources or sinks in the exchanger walls or fluids, such as electric

heating,chemical reaction, or nuclear processes.

Temperature of each fluid is assumed to be uniform in a cross-section.

Wall thermal resistance is distributed uniformly.

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Basic Definitions

The definitions of mean overall heat transfer coefficient and mean temperature difference are introduced

first by reaarging basic equation= − ) 𝒂 𝑨 = 𝜟 𝑨Above equation presented in integral form, ∆ = 𝑨 𝑨Now the definition of mean temperature difference and mean overall heat transfer coefficient as follows:

∆ = ∆=𝑨 𝑨= ∗ 𝑨 ∗ ∆

Here, is mean overall heat transfer coefficient, and ∆ is true mean temperature difference(MTD),

also referred to as mean temperature driving potential or force for heat transfer.

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Temperature distribution in counter flow exchanger of single phase fluids

(Shah,1981).

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Temperature distribution in parallel heat exchanger (Shah,1981).

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Temperature distribution at inlet and outlets of an unmixed cross flow

heat exchanger (Shah,1981)

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Heat Exchanger Analysis

Let ሶ𝑚 =Mass flow rate, Τ𝑔𝐶 = Specific heat of fluid at constant pressure, Τ 𝑔℃,

T= Temperature of fluid, ℃ , and ∆ =Temperature drop or rise of a fluid across the heat exchanger

ℎ,𝑖 ℎ,,𝑖,

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1. F-LMTD

F-LMTD In the analysis of heat exchangers total heat transfer rate is of primary interest. The log-

mean temperature difference (LMTD or ∆ for counter and parallel flow is defined as

LMTD = 𝜵 = ∆ −𝜵Τ𝜵 𝜵

Total heat transfer between hot and cold fluids in a counter flow arrangement= 𝐀 ∆Where 𝛻 = ℎ,𝑖 − , 𝛻 = ℎ, − ,𝑖 ( Counter Flow)𝛻 = ℎ,𝑖 − ,𝑖 𝛻 = ℎ, − , ( Parallel Flow)

LMTD represent maximum temperature potential for heat transfer that can be obtained in counter flow

exchanger.

Special Case: In case of counterflow with ( ሶ𝑚 ℎ= ( ሶ𝑚 , the quantity ∆ is indeterminate. In this

case, by applying L’ Hospital’s rule = 𝐴 ℎ − with ℎ − = ∆ =∆ .

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1. F-LMTD for Multipass and Crossflow Heat Exchangers

The LMTD shown in previous slide is not applicable for heat transfer analysis of crossflow and multipass

flow heat exchangers.

∆ , = , − , − , − ,[ ൗ, − , , − , ]= , − ,, − , = ∆∆ 𝒂𝒙 ( Heat actually transferred to the heat which would be transferred if the same cold fluid

temperature was raised to the hot-fluid inlet temperature. Hence it is temperature effectiveness of heat

exchanger.)= 𝑪𝑪 = , − ,, − , ( Ratio of ሶ𝑚 of cold fluid to that of hot fluid and is called the heat capacity rate ratio.)

Where ∆ , is the LMTD for counter flow arrangement with the same fluid inlet and outlet temperature.

Hence , in crossflow heat is calculated by introducing a non-dimensional number called F.= A F ∆ ,mech14.weebly.com


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F is nondimensional number ; it depends on the temperature effectiveness , the heat capacity rate

ratio , and the flow arrangement. 𝑭 = 𝝓 , , flow arrangement)

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∈ −𝑁 Method

Applicable when inlet or outlet temperatures of the fluid streams are not known.

Method is based on fact inlet or exit temperature difference are function of 𝑈𝐴𝐶𝑐 and Τ𝐶 𝐶ℎ.

Dimensionless Form

Capacity rate ratio: 𝐶∗=𝐶 𝑖𝐶 𝑎 , where 𝑖 and 𝑎 are the smaller and larger of the two magnitudes of

ℎand , respectively , and ∗ . Exchanger heat transfer effectiveness: It is the ratio of actual heat transfer rate in heat exchanger to the

thermodynamically limited maximum possible heat transfer.

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Actual heat transfer is obtained by using energy given off by hot fluid or energy received by cold fluid,= ሶ ( , − , ) =( ሶ ( , − , )If ℎ > , then ( ℎ,𝑖 − ℎ, ) < ( , − ,𝑖If ℎ < ,then ( ℎ,𝑖 − ℎ, ) > ( , − ,𝑖The fluid that might undergo maximum temperature difference is the fluid with minimum heat capacity

rate 𝑖 . Therefore, the maximum possible heat transfer is expressed as

𝒂𝒙=( ሶ ( , − , if < or 𝒂𝒙=( ሶ ( , − ,

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Heat exchanger effectiveness, 𝜖, is therefor written as𝝐 = , − ,𝑪 , − , = , − ,𝑪 , − ,The above equation is valid for all heat exchanger flow arrangements. The value of 𝜖 ranges between

zero and 1.

If the effectiveness is known, Equation provides expression for determination of

= 𝝐 ∗ ሶ *( , − ,

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Number of transfer unit: It depicts the nondimensional heat transfer size of the heat exchanger.

NTU = 𝑨

= 𝑨For counter flow following expression is obtained :

𝝐 = − 𝐞𝐱𝐩[−𝑵 − Τ𝑪 𝐚𝐱 ]− Τ 𝒂𝒙 𝐞𝐱𝐩[−𝑵 − Τ 𝒂𝒙 ]If < ℎ = 𝑖 , ℎ = 𝑎 , the result will be the same.

In case of parallel flow, a similar analysis may be applied to obtain following expression:

𝝐 = − 𝐞𝐱𝐩[−𝑵 + Τ 𝒂𝒙 ]+ Τ 𝒂𝒙

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SPECIAL CASE:

Τ𝐶 𝑖 𝐶 𝑎 equal to 1

𝝐 = 𝑵𝑵 + ( Counter Flow) , 𝜺 = ( − − 𝑵 ( Parallel Flow).

Τ𝐶 𝑖 𝐶 𝑎 equal to 0 as in boiler and condensers for parallel flow and counter flow ,

equation becomes

𝝐 = − 𝐞𝐱𝐩 −𝑵 𝝐 = 𝝓 𝑵 , ∗, flow arrangement)

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Effectiveness- NTU chart: an example

http://userscontent2.emaze.com/images/3f54344d-c988-440d-b903-

539c0c6c9279/Slide21_Pic2_635933982388373132.png

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Exchanger NTU Effectiveness

Automobile Radiator 0.5 40%

Steam Plant Condenser 1 63%

Regenerator for Industrial

Gas Turbine

10 90%

Regenerator for Stirling

Engine

50 98%

Regenerator for LNG

Plant

200 99%

Approximate Values of NTU & Effectivenes for typical

Heat Exchangers

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In a 1-2 TEMA E shell-and-tube exchanger, water enters the shell at 21 ºC at a

rate of 1.4 kg/s. Engine oil flows through the tubes at a rate of 1.0 kg/s. The inlet

and outlet temperatures of the oil are 150 ºC and 90 ºC, respectively. Determine

the surface area of the exchanger by both the MTD and ε–NTU methods if U =

225 W/m2 . K. The specific heats of water and oil are 4.19 and 1.67 J/g . K

respectively.

Fundamentals of heat exchanger design, by R. K.

Shah & D. P. Sekulic, 2003, John Wiley & Sons

Shell & tube heat exchanger: an example with LMTD and ε–NTU

methods

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,𝑖=21 ℃, ሶ𝑚 = 1.4 kg/s

, =90℃

,𝑖=15 ℃, =1.67 J/g . K

Water

Oil, =4.19 J/g . K

U = 225 W/m2. K

Schematic representation of a 1-2 TEMA E shell-and-tube exchanger

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3

3

1.4 / 4.19 10 / . 5866 /

1.0 / 1.67 10 / . 1670 /

s P

s

t P

t

C mc kg s J kg K W K

C mc kg s J kg K W K

The heat capacity rates for the shell fluid (water) and the tube fluid (oil) are:

3

, , 1670 / 150 90 100.2 10t t i t o

q C T T W K C W

The heat transfer rate from the oil is:

Using the energy balance equation, we could find the water outlet temperature: 3

, ,

100.2 1021 38.1

5866 / Ks o s i

s

q WT T C C

C W

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150 38.1 111.9

90 21 69

I

II

T C C C

T C C C

15 ℃38.1℃

9 ℃21℃

x/L0 1

Oil

Water

I II

oil waterC C

oil waterC C

, ,o

1

, s,

s,o s,i

1

, t,o

150 90

150 21

0.4651

1670 /

5866 /

0.2847

t i t

t i i

t

t i s

T T CP

T T C

T T C W KR

T T C W K

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LMTD correction factor F as a function of P1

and R1

for a 1-2 TEMA E shell-and-

tube exchanger with the shell fluid unmixed

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32

2

100.2 105.133

225 / 0.9776 88.74lm

q WA m

UF T W m K K

The heat transfer area from the rate equation is:

The ε–NTU method:

In this problem, t s

C C

1670 /0.2847

5866 /

t

s

C W KC

C W K

Hence,

Using the definition of the effectiveness for the tube side (Cmin side), we get

,

, ,o

, s,

150 900.4651

150 21

t i t

t i i

T T C

T T C

From the figure, F = 0.9776

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2min

2

1670 /0.6916 5.133

225 /

C W KA NTU m

U W m K

Either from the figure or the formula, NTU = 0.6916

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Heat Exchangers Specially Suited for Energy

Conservation & Waste Heat Recovery

Regenerators

Special Recuperators

Heat Pipes

Run Around Coils

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Fixed Twin Bed Regenerator

Hot Gas

Cold Gas

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Rotary Regenerator

https://upload.wikimedia.org/wikipedia/commons/thu

mb/a/a8/Rotary-heat-exchanger.svg/2000px-Rotary-

heat-exchanger.svg.png

http://us.klingenburg.de/fileadmin/_processed_/csm_Rotor-

Schema_HT-US_c759652afb.jpg

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Rotary Regenerator

https://www.youtube.com/watch?v=roImOiIxrjo

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Regenerator with a Rotary Hood

Fundamentals of heat exchanger design, by R. K.

Shah & D. P. Sekulic, 2003, John Wiley & Sons

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Estimation of heat transfer from Regenerator/Heat Wheel

Based on a simple analysis,

Rate of heat transfer in a cycle

1 1 1 1

6

gh gc

h c

h h c c h c

A t tQ

b

h h k

τ - time, the wheel is in contact with a particular gas stream

- temperature of a particular gas stream, h - hot, c - cold

b - length parameter

k - thermal conductivity

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Special types of Recuperators

http://enconi.com/images/recuprater.jpg

Heat recovery for Canadian food and beverage

industries, Agriculture Canada, 1984

Waste Heat Boiler

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http://www.immak.eu/Admin/photos/markabuyuk/63356211

7303593750.jpg

Run around coil

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Run around coil- schematic representation

Hot gas

Th1Th2

Tc2Tc1

Cold gas

Ts2Ts1

1 high temperature

2 low temperaturepump

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Assume

1 2 1 2 1 2h h c c s s

h C s

Q mC T T mC T T mC T T

h C s

mC mC mC

1 2 1 2 1 2

1 1 2 2 1 1 2 2,

h h c c s s

h s h s s c s c

T T T T T T

T T T T T T T T

Three temperature lines are therefore straight

lines and parallel

lm i eT T T

1 1 1 1

h c

h s s c

Q QU A U A

T T T T

Ts1

Th1

Th2

Ts2

Tc1

Tc2

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The temperature line for the secondary fluid must be midway between the temperature

lines for the hot and cold fluids.

1 11

2 22

2

2

h cs

h cs

T TT

T TT

The total heat recovery is

1 1

2

h c

h

T TQ U A

And since1 2c c

c

QT T

mC

1 2

2

h ch

h

c

U A T TQ

U A

mC

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In a shell-and-tube feed water heater, cold water at ℃ flowing at the rate of 180 Kg/h is

preheated to 90℃ by flue gases from 150℃ flowing at the rate of 900 Kg/h. The water flows

inside the copper tubes( 𝑖=25mm, =32mm) having thermal conductivity 𝑘 =381W / m .

K . The heat transfer coefficients on gas and water sides are 120 and 1200 W/m2K,

respectively. The fouling factor on the water side is 0.002 m2.K/W. Determine the flue gas

outlet temperature, the overall heat transfer coefficient based on the outside tube

diameter, and the true mean temperature difference for heat transfer. Consider specific

heats for flue gases and water as 1.05 and 4.19 J/ g .K respectively, and total tube

outside area as 5m2 . There are no fins inside or outside the tubes, and there is no fouling

on the gas side.

Shell & tube heat exchanger: an example with LMTD

Fundamental of heat exchanger design, by R. K.

Shah & D. P. Seculic, 2003, John Wiley & Sons

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SOLUTION

Problem Data and Schematic: Fluid flow rates, inlet temperature , and cold fluid outlet temperature are

provided for a shell-and-tube exchanger of prescribed tube inner and outer diameter. Also, the thermal

conductivity of tube and thermal resistance on the cold fluid side are given. There are no fins on either

side of tubes.

Determine: Hot fluid temperature ℎ, , overall heat transfer coefficient , and true mean temperature

difference ∆ .

Assumptions: Hot-fluids-side fouling is negligible.

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ℎ,𝑖=150 ℃, ሶ𝑚ℎ=900Kg/h, ,ℎ=1.05 J/g.K

, =90 ℃

,𝑖=15 ℃ሶ𝑚 =180 Kg/h, , =4.19J/g.K

ℎ,

𝑖=25mm

=32mm𝑘 =381W/m . K

ℎ 𝑎 =ℎℎ=120 W/m2Kℎ 𝑎 =ℎ =1200 W/m2K, =0.002 m2.K/W𝐴 =5m2

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Analysis: The required heat transfer may be obtained from overall energy balance for the cold

fluid.

𝑞 = 𝐶 , − ,𝑖=(

/ℎ/ℎ )(4.19 Τ𝑔. (1000 Τ𝑔 𝑔 ℃− ℃ = ,Apply the equation on the hot fluid side to find the outlet temperature for flue gas:

ℎ, = ℎ,𝑖 − 𝑝 ሶ ℎSince ሶ𝑚ℎ= /ℎ𝑆/ℎ=0.25 Τ𝑔

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We get

ℎ, = ℃− , 𝑊. Τ ∗ Τ ℃ = . ℃Since U is based on A = 𝐴ℎ= 𝜋 𝑁 ,R = 𝑈ℎ =ℎℎ + ℎℎ,𝑓 + 0ln Τ0 𝑖 + 0ℎ𝑐,𝑓 𝑖 + 0ℎ𝑐 𝑖

= (0.00833+0.00001+0.00256+0.00107) = 0.01197 m2.K/W

Hence, ℎ=85.4W/m2 . K

∆ = 𝑈ℎ𝐴ℎ = 37.6℃mech14.weebly.com


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Fundamental of heat exchanger design, by R. K.

Shah & D. P. Seculic, 2003, John Wiley & Sons

One important design for a radiator design is to cool the engine at 50 km/h on a 7%

grade road. Your responsibility as a design engineer is to make sure that the coolant

(50% water – 50% glycol) at the radiator inlet (top tank) does not exceed 120 ºC

temperature at 100 kPa gauge radiator cap pressure. Determine the radiator top tank

temperature for the following conditions: engine heat rejection rate q = 35 kW, air

flow rate 0.75 kg/s, air inlet temperature 53 ºC, and water-glycol flow rate 1.4 kg/s.

For this radiator, UA =1180 W/K. The specific heats for the air and the water-glycol

mixture are 1009 3664 J/kg.K respectively. What will be the outlet temperature of the

water-glycol mixture? Consider the radiator with both fluids unmixed.

An example

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Water/glycol mixture

Tci

air

Tco

Thi

Tho

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min

min

0.75 / 1009 / . 756.75 /

1.4 / 3664 / . 5129.6 /

756.75 /0.148

5129.6 /

1180 /1.559

756.75 /

air C P

air

liquid h P

liquid

air

liquid

C C mC kg s J kg K W K C

C C mC kg s J kg K W K

C W KC

C W K

UA W KNTU

C W K

From Table for an unmixed-unmixed crossflow exchanger 0.769

min

35 1000 /53 113.1

0.769 756.75 /

35 1000 /113.1 106.3

5129.6 / K

hi ci

ho hi

h

q kW W kWT T C C

C W kW

q kW W kWT T C C

C W

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Heat Exchanger Network Analysis by Pinch Technique

Reference

The Pinch Design Method For Heat Exchanger Networks by B. Linnhoff and E. Hindmarsh

Chemical Engineering Science Vol.38, No.5, pp.745-763, 1983

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Heat Exchanger Network

Process Plant:

Many streams need heating or cooling.

Utilities like hot air , hot oil, steam for heating.

Utilities like air, water, chilled water, refrigerant for cooling.

Efforts to be made for reducing the use of utilities.

This can be done by “ Heat Integration” . “ Heat Integration” implies heat exchange between the hot and cold streams.

This reduces the use of utilities.

However, this increases the number of heat exchangers.

Heat exchanger synthesis means designing a network so that using minimum number of

heat exchangers, the use of utilities can also be minimized

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Synthesis of Heat Exchanger Network (HEN)

There are different techniques of synthesis of HEN.

The pinch technique is very widely used.

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C1 C2 C3

H1

H2

H3

Schematic representation of Heat Exchanger Network

RULE:

Minimum number of Heat

Exchangers ( 𝑖 = 𝑁-1𝑁 =No. of process streams

( N1 ) + No. of utilities(N2)

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Pinch Technique

Logic Based- based on first and second law of thermodynamics.

There is one hot utility and one cold utility of infinite capacity.

Steady state operation.

There is no restriction for matching the streams.

All heat exchangers are without any phase change and without any heat

generation or reaction.

Heat exchangers are assumed to be indirect contact and counter flow.

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STREAM NUMBER &

TYPE

HEAT CAPACITY FLOW RATE 𝑪kW/℃ TS(℃) TT(℃)

(1) HOT 2 150 60

(2) HOT 8 90 60

(3) COLD 2.5 20 125

(4) COLD 3.0 25 100

Table 1. Stream data for test case

𝜟 = ° 𝑪mech14.weebly.com


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℃ ℃ ℃ ℃ ℃ ℃ ℃Temperature

1

2

3

4

Goals of HEN synthesis

1) Minimum utility.

2) Stream matching.

3) No. of heat exchangers needed.

4) End temp of the heat exchangers.

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Hot UtilityCold Utility

T

Q

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SUBN

ETW

ORK

STREAMS AND TEMPERATURE DEFICIT ACCUMULATED HEAT FLOWS

COLD

STREAM

S

T(℃ HOT

STREAMS

INPUT OUTPUT INPUT OUTPUT

150

SN1

145

-10 0 10 107.5/HU 117.5

SN2

120

+12.5 10 -2.5 117.5 105

SN3

70 90

+105 -2.5 -107.5 105 0

SN4 40 60 -135 -107.5 27.5 0 135

SN5 25 +82.5 27.5 -55 135 52.5

SN6 20 +12.5 -55 -67.5 52.5 40/CU

Table 2. The problem table for Test case

3

4

1 2

125

100

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SN1

SN2

SN3

SN4

SN5

SN6

𝐻 =107.5 kW

.

.𝐶 =40 kW

34

3 4

𝐻 =107.5 kW

𝐶 =40 kWFig. 1. (a) Subnetwork heat flow diagram for TC3. (b)Subnetworks combined into hot and cold region

(a) (b)

1

1 2

Cold end

problem

Hot end

problem

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Pinch point divides the HEN into two independent problems to be

solved separately.

No heat flow is allowed across pinch point.

No cold utility to be used in the hot end.

No hot utility to be used in the cold end.

Violation of (2) – (4) gives double penalty

Design of each end should start at the pinch point.

Rules for PINCH ANALYSIS

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1.For hot end 𝑁𝐻 𝑁𝐶 𝑁𝐻- No of Hot stream

For cold end 𝑁𝐶 𝑁𝐻 𝑁𝐶- No. of cold stream

Note: In the hot end, there is some amount of hot utility & hot stream which can only

exchange heat with cold stream.

2. For individual match

Hot end, 𝐶𝐻 𝐶Cold end,𝐶 𝐶𝐻

3.For hot end

(σ 𝑐 𝐶 -σ 𝐶𝐻) σ . 𝑎 ℎ 𝐶 − 𝐶𝐻)

Rules for Pinch End Matching

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Some Examples

4

3

2

1

3

5

2

4

C Difference in C

Composite “C” difference

8-6=2

Summation of match

“C” Diff=(5-4)+(3-2)=2

Solution is possible.

4

3

2

1

C

5

4

5

2

3

1

Difference in C

Composite “C” difference

9-6=3

Summation of match” Diff=(5-4)+(3-2)=2

Solution is possible.

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Hot End Design

Stream Data at Pinch

𝑁𝐻 𝑁𝐶 NoSplit cold stream

Yes

𝐶𝐻 𝐶𝐶 for each match and

cumulative stream

Stream SplittingNo

Yes

Solution

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Stream Matching for Hot End

1150 ℃

125 ℃ 70 ℃100 ℃ 70 ℃

90 ℃118 ℃ .3

4

1150 ℃

125 ℃ 70 ℃100 ℃ 90

.3

4

3082℃

120H

H

H

𝑁𝐻 𝑁𝐶𝐶𝐻 𝐶𝐶(𝐶𝐻)1= 2

(𝐶𝐶)3= 2.5

(𝐶𝐶)4 = 3

All above conditions are satisfied i.e. number of hot

streams number of cold streams

90 ℃70 ℃

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Stream Matching for Cold End

𝑵𝑪 𝑵𝑯Cumulative , 𝐶𝐶 𝐶𝐻

8 3

2 2.5

8 3

2 2.5

8 3

2 2.5

So, we have to split hot stream

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Splitting of Hot Streams

3

2 2.5

87.5

0.5

3

2 2.5

83

5

3

2

0.5

2.5

2.0

87.5

0.5

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So, the restrictions i.e. 𝐶𝐻 𝐶𝐶 just next to pinch point

℃

℃℃ ℃

℃

℃℃ ℃

..

.C

℃

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℃

℃℃ ℃

℃

℃℃ ℃

.℃

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Documents

Heat Exchangers - ME14mech14.weebly.com/.../me_ecwhr_18_heat_exchangers.pdf · To analyze heat exchange problem, a set of some assumptions are introduced sothat resulting models are