1

Run-around Coil Systems:

• Definition of Run-Around Coil Systems

• Design Factors

• Examples

ISAT 413 - Module V: Industrial Systems

Topic 3: Run-around Coil Systems, Regenerative Heat Exchangers, Pinch Technology

2

Run-around coil system of heat recovery

A run-around coil heat recovery system is the name given to a linking of two recuperative heat exchangers by a third fluid which exchanges heat with each fluid in turn as shown diagrammatically below.

3

Run-around coil system A run-around coil would be used in cases where the two fluids which are required to exchange heat are too far apart to use a conventional direct recuperative heat exchanger. It is also desirable to use such an indirect system if there is a risk of cross-contamination between the two primary fluids (e.g. when a particularly corrosive fluid is involved, or when there is a risk of bacterial contamination as in a hospital).

Advantage would be free choosing of working fluid.

Disadvantage would be low effectiveness of the HX.

Typical applications are recovery of energy from the air leaving a room or building to pre-heat the air entering; and recovery of energy from a corrosive gas for water heating.

4

Run-around coil heat recovery between fluids with the same thermal capacity

; and

i.e.

22112211

212121

CSCSSHSH

HCS

HHCCSSHCS

tttttttt

cmcmcm

ttttttcm

Q

cm

Q

cm

Q

HC cmcm

5

2

as expressed becan recovery heat thehence

since and

2 thus,

2 and

2

:identical are exchangersheat two theAssume

21

21

111111

222

2222

111

1111

C

H

CHH

CCC

CHHCSCSHH

CHS

CSSHCH

CHS

CSSHCH

cm

UAttUA

Q

cm

Qtt

ttUAttUAttUAQ

ttt

tt

Q

tt

QUAUA

ttt

tt

Q

tt

QUAUA

HC cmcm Run-around coil heat recovery between fluids with the same thermal capacity

6

Example A run-around coil heat recovery system similar to that on slide 4 is used for a room in which the presence of bacteria rules out any possibility of air re-circulation or a direct recuperative heat exchanger. Air enters the room at 24oC and leaves at 20oC; the average outside air temperature during the annual period of use is 5oC. Assuming that the mass flow rate of air is 2 kg/s, mean specific heat 1.005 kJ/kg-K, that (UA)H = (UA)C = 4 kW/K, and that the specific heat of the secondary fluid is 2.5 kJ/kg-K, calculate:

(i) the required mass flow rate of secondary fluid;

(ii) the temperature of the air leaving the run-around coil;

(iii) the percentage energy saving by using the run-around coil.

7

Example

39.4%100%

524

5512saving %

:bygiven is savingenergy percentage thehence 24 to5 from

heat bemust air herecovery theat he Without t(iii)

512012

04155 and

0415

0124

2

5204

2 (ii)

804052

012

01200512 (i)

21

21

C

C

oo

o

CCC

C

H

CHH

S

Ss

CHS

cm

.cm

CC

C..

.

cm

Qtt

kW.

.cm

UAttUA

Q

s/kg..

.

c

cmm

K/kW..cmcmcm

8

Round-around coil heat recovery between fluids of different thermal capacity

HC cmcm

9

CSH cmcmcm

Round-around coil heat recovery between fluids of different thermal capacity

10

CSH cmcmcm

Round-around coil heat recovery between fluids of different thermal capacity

11

Idealized run-around coil system

HCCHS

CHCHS UAcmUAcmc

UAUAcmcmm

fluidsecondary Required

12

Example 5.7

(use - NTU method to analyze run-around coil heat recovery system)

13

Example 5.7 A corrosive gas at a flow rate of 30 kg/s from a process at 300oC is to be used to heat 20 kg/s of water entering 10oC using a run-around oil as shown on slides 8 & 12. Calculate using the data given:

(i) the mass flow rate of secondary fluid required;

(ii) the effectiveness of the overall heat transfer;

(iii) the exit temperature of the water;

(iv) the temperatures of the secondary fluid.

Data: Mean specific heat of gases, 1.2 kJ/kg-K; mean specific heat of water, 4.2 kJ/kg-K; mean specific heat of secondary fluid, 3.8 kJ/kg-K; (UA) for the gas to secondary fluid heat exchanger, 40 kW/K; (UA) for the secondary fluid to water heat exchanger, 200 kW/K.

14

Example 5.7

s

kg.

...

..

UAcmUAcmc

UAUAcmcmm

HCCHS

CHCHS

0918200213040242083

2004024202130

bygiven is fluidsecondary of rate flow mass required The (i)

55042901

1

1

1 ess,effectiven HX Hence

429084

36 and

;926036

33333 since

100

3

200

1

40

1111 Since (ii)

57109260

57109260

1

1

.e.

e

Re

e

.cm

cmR

..

cm

UANTU

UAUAUA

..

..

RNTU

RNTUmax

min

min

overall

C_SSHoverall

15

Example 5.7

C.t

.

t..

C.t.

t..

ttcm

ttcm

ttcm

ttcm

ttcm

QQ

oC

H

CC

oH

H

HH

CHH

CCC

CHH

HHH

minmaxmin

CH

478103002130

102420550

5140103002130

3002130550

or

bygiven is esseffectiven HX of definition theSince (iii)

11

22

21

21

21

21

C.ttt.

ttUAttUA

C.t.tt

ttUAttUA

oSSCSH

CSCSHH

oSSCSH

CSCSHH

83110200514040

:2station @

311547820030040

:1station @bygiven isfer heat trans theSince (iv)

222

2222

111

1111

16

Temperature changes for Example 5.7

17

Regenerative Heat Exchangers:

• Definition of Regenerative HX

• Design Factors

• Examples

In a regenerative heat exchanger (sometimes called a capacitance heat exchanger) the hot and cold fluids pass alternately across a matrix of material; the matrix is heated up by the hot fluid then cooled down by the cold fluid so that the process is cyclic.

18

Stationary Regenerative Heat Exchanger

In (a) matrix B is hot and heats up the cold fluid while matrix A is heated by the hot fluid; in (b) the cold fluid is now heated by matrix A while the hot fluid re-heats matrix B; the valves are then switched over and the cycle commences again as in (a) .

19

Rotary Regenerator or Thermal Wheel

A matrix of material is mounted on a wheel which is rotated slowly through the hot and cold fluid streams as shown above. It is known as the thermal wheel, and Ljungstrom rotary regenerator after its Danish inventor.

20

Influence of Matrix Rotational Speed on Effectiveness

MM

M

max

minRNTU

RNTU

c

.

min

M

c

NMcMc

cm

cm

cmR

Re

eE

cm

cmEE

and

;1

1 where,

9

11

1

1

931

21

Example 5.8 (A rotary regenerator)A rotary regenerator is used to recover energy from a gas stream leaving a furnace at 300oC at a mass flow rate of 10 kg/s. Heat is transferred to a mass flow rate of air of 10 kg/s entering at 10oC. The wheel has a diameter 1.5 m, giving an approximate face area of 1.6 m2, and a width of 0.22 m; the matrix has a surface area to volume ratio of 3000 m2/m3 and a mass of 150 kg; the rotational speed of the wheel is 10 rev/min. The heat transfer coefficient for both fluid streams is 30 W/m2-K and the mean specific heats at constant pressure for the gas and air are 1.15 kJ/kg-K and 1.005 kJ/kg-K; the specific heat of matrix material is 0.8 kJ/kg-K.

Calculate:

22

Example 5.8 (A rotary regenerator)(i) the effectiveness of the heat exchanger;

K

kW.cm.

..

cm

cmEE

K

kW.cm.

e.

e

Re

eE

..

.

cm

cmR.

.cm

UANTU

m..V

AVA

K.m

W

hh

hhU

AhAhUA

min..

min

M

c

M..

..

RNTU

RNTU

c

max

min

min

CH

HC

CH

0510 Where61709919

116360

9

11

208015060

10 ;6360

87401

1

1

1

874015110

005110 ;5761

1000005110

105615

1056300022061 area,fer Heat trans

153030

3030111bygiven isct coefficienfer heat trans overall The

931931

12605761

12605761

1

1

2

2

23

Example 5.8 (A rotary regenerator)

(ii) the rate of heat recovery and the temperature of the air at exit;

.C

.t

kWt.ttcmQ

kW..Q

.ttcm

QE

oC

CCCmin

CHmin

1890510

179910 air, of ratureexit tempe

179910005110 Also,

1799103000051106170

6170

esseffectivenexchanger heat of definition theFrom

1

121

21

24

Example 5.8 (A rotary regenerator)(iii) the air temperature at exit if the rotational speed of the wheel is increased to 20 rev/min;

C.t

.ttcm

ttcm

ttcm

QE

..

.

cm

cmEE

..cm

cm

K

kW.NMccm

sec

min

min

.rev

min

.revN

oC

CHmin

CCmin

CHmin

..

min

M

c

min

MMM

19310300631010

6310 Therefore

63109839

116360

9

11

983005110

404080150

60

2060

12020Given

1

21

21

21

931931

25

Example 5.8 (A rotary regenerator)

C..t

.ttcm

ttcm

ttcm

QE

..

.

cm

cmEE

..cm

cm

K

kW.NMccm

sec

min

min

.rev

min

.revN

oC

CHmin

CCmin

CHmin

..

min

M

c

min

MMM

817310300565010

5650 Therefore

565099509

116360

9

11

9950005110

101080150

60

560

155Given

1

21

21

21

931931

(iv) the air temperature at exit if the rotational speed of the wheel is reduced to 5 rev/min;

26

Gas-Fired Regenerative Burners

In (a) the hot gases are fed back through the burner and through a matrix to exhaust; while in (b) air is drawn through the matrix and supplied with gas to the burner where combustion takes place; two burners are used in tandem so that continuous combustion can take place.

27

Magnetic Heat Pump research at Oak Ridge National Laboratory

28

Problem 5.9 Double Accumulator Regenerative Heat Exchanger

A double accumulator as shown above is to be installed to recover energy from the air leaving a building. The air leaves the building at 20oC at a rate of 2 kg/s and the mean outside air temperature for the heating season is 5oC. Calculate the rate of the recovery, etc...

29

Pinch Technology Concepts:

• Basic Concepts

• Design Factors

• Examples

For many years the approach to a large network of heat exchangers was either by ‘rule of thumb’ or a systematic mathematical examination of all possible configurations to try to achieve the best layout.

Another approach to network design is given the name Process Integration, or Pinch Technology.

30

Heat Exchanger Temperature Profiles

The design of the heat exchanger is based on the minimum allowable temperature difference between the two streams being 20K. Additional heating and cooling are required to achieve the desire temperatures.

31

Simple Heat Recovery Scheme

kW

C

Q

C

heating,external

40202 110-130

:heating /ExternalAdditional

kW

C

Q

H

cooling,external

30301 30-60

:cooling /ExternalAdditional

32

Temperature-Heat Load representation of Heat Recovery schemeThe two lines representing the streams are positioned so as to show a region of overlap which represents the action of the HX in transferring 140 KW. The minimum temperature difference occurs where the two lines are nearest together - this point is called the Pinch point.

33

Effect of ‘Moving’ the Cold Stream

The effect of increasing the Pinch temperature difference is twofold; the amount of heat exchange between the two fluids is reduced and the external duties are increased. Note that the slope of the cold stream line is determined by the value of CC, which is 1/CC = 0.5 K/kW.

34

Effect of ‘Moving’ the Hot Stream The same argument could be used for positioning the hot stream line. Thus we can say that the lines can be moved horizontally within the limits of temperature and gradient until the nearest points are separated by the minimum allowable temperature difference, that is the Pinch temperature difference.

Also

To achieve the target for the external cooling duty of the hot stream, there must be no external cooling of the hot stream above the pinch.

A similar argument applies to the external heating of the cold stream below the pinch: there must be no external heating of the cold stream below the Pinch.

35

Stream Netwroks:

• Stream Network Concepts

• Design Factors

• Examples

Considering the design of a system of heat recovery between two (or more) hot streams and two (or more) cold streams to illustrate some fine points of Pinch Technology.

36

Example 6.1 (Pinch Technology)

Stream number

Type Thermal capacity rate, C (kW/K)

Initial temp. (oC)

Final temp. (oC)

Rate of enthalpy increase (CT) (kW)

1 Hot 2 200 60 -280

2 Hot 4 170 70 -400

3 Cold 3 40 175 +405

4 Cold 4.5 100 150 +225

- 50

The heat flow capacities and temperatures of four streams are shown in the table below. For the purpose of definition, a hot stream is defined as one which requires cooling to reach its final temperature and a cold stream is one which requires heating to reach its final temperature. The minimum allowable temperature difference between the streams is 20 K.

37

Composite Stream Heat Flow Capacities(Hot Stream Composite)

38

Composite Stream Heat Flow Capacities(Cold Stream Composite)

39

Hot and Cold Composite Curves

40

Combined Hot and Cold Composite CurvesFrom the graph the following information can be derived:

hot stream temperature at the Pinch: 120oC

cold stream temperature at the Pinch = 100oC

target external heating load = ? kW

target external cooling load = ? kW

Cooling load will exceed the heating load by ? kW.

41

200

40

175

7060

170

120

90

150

148140

100

0

50

100

150

200

0 200 400 600 800

Heat Load (kW)

Te

mp

era

ture

(oC

)Calculations of External Cooling and Heating Loads target external

heating load = 90 kW

target external cooling load = 140 kW

Cooling load will exceed the heating load by 50 kW.