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3/48

C3

TABLE OF CONTENTS

Page

Symbols and units 4

H101C fitted to H101 5

Connected for Co-current flow 5

Schematic Counter current flow. 6

Schematic Co-current flow. 7

Description 8

Shell and Tube Heat Exchanger H101C 8

Installation 9

Useful Data 10

Table 1 - Specific Heat capacity Cp of Water in kJ kg1

10

Table 2 - Density of Water in kg Litre-1

11

Capabilities of the Shell and Tube Heat Exchanger H101C 12

1. Demonstration of indirect heating or cooling by transfer of heat from one

fluid stream to another when separated by a solid wall (fluid to fluid heattransfer). 13

2. To perform an energy balance across a Shell and Tube heat exchanger and

calculate the overall efficiency at different fluid flow rates 15

3. To demonstrate the differences between counter-current flow (flows in

opposing directions) and co-current flows (flows in the same direction) and the

effect on heat transferred, temperature efficiencies and temperature profiles

through a shell and tube heat exchanger 18

4. To determine the overall heat transfer coefficient for a shell and tube heat

exchanger using the logarithmic mean temperature difference to perform the

calculations (for counter-current and co-current flows). 24

5. To investigate the effect of changes in hot fluid and cold fluid flow rate on

the temperature efficiencies and overall heat transfer coefficient. 32

6. To investigate the effect of driving force (difference between hot stream and

cold stream temperature) with counter-current and co-current flow 40

Example Data 47


4/48

C4

Symbols and units

Symbol Units

Vcold Cold stream flow rate gram s-1

Vhot Hot stream flow rate gram s-1

T1 Hot fluid inlet temperature C

T2 Hot fluid outlet temperature C

T3 Cold fluid inlet temperature C

T4 Cold fluid outlet temperature C

thot Decrease in hot fluid temperature K

tCold Increase in cold fluid temperature K

dT hot Decrease in hot fluid temperature K

dT cold Increase in cold fluid temperature K

di Inside diameter of hot tube m

do Outside diameter of hot tube m

dmean Mean diameter m

n Number of hot tubes -

L Effective length of hot tube m

Tmean Mean temperature C

Density of stream fluid kg litre

Cp Specific Heat of stream fluid kJkg-1K-1

Q. e Heat flow rate from hot stream Watts

Q.a Heat flow rate to cold stream Watts

Q.f Heat loss to surroundings Watts

LMTD Logarithmic mean temperature difference K

A Heat transfer surface area m2

U Overall heat transfer coefficient Wm-2K-1

Thermal Thermal efficiency %

hot Temperature efficiency hot stream %

cold Temperature efficiency cold stream %

mean Mean temperature efficiency %

L Hot tube effective length m

dTmax Maximum temperature difference across heat exchanger K

dTmin Minimum temperature difference across heat exchanger K


5/48

C5

H101C fitted to H101

Connected for Co-current flow

T1 T3

T2 T4


6/48

C6

Schematic Counter current flow.


7/48

C7

Schematic Co-current flow.


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C8

Description

Shell and Tube Heat Exchanger H101C

The shell and tube heat exchanger is an efficient design and finds application in food, chemical and

refrigeration plant. This type of heat exchanger consists of a number of tubes in parallel enclosed in

a cylindrical shell. Heat is transferred between one fluid flowing through the tube bundle and theother fluid flowing through the cylindrical shell around the tubes. Baffles are often included inside

the shell to increase the velocity and turbulence of the shell side fluid and thereby increasing theheat transfer.

In addition industrial applications often include end plate baffles so that the tube side fluid makesmore than one pass through the tube bundle. This involves greater tube side pumping losses butresults in an increase in the overall heat transfer coefficient. This can result in a smaller heat

exchanger for the same capacity.

For more detailed examination of multi-pass shell and tube heat exchangers the Hilton Steam toWater Heat Exchanger H930is recommended. This allows one two and four-pass operation to beinvestigates and the increased pumping losses to be demonstrated. Details are available from

P.A.Hilton Ltd., or their local agent on request.

The H101C Shell and tube exchanger is a simple model that demonstrates the basic principles ofheat transfer. The H101C is designed for use with the Heat Exchanger Service Unit H101.

The miniature heat exchanger is mounted on the H101 front panel that incorporates two mountingstuds. These locate in the heat exchanger hanging bracket and two plastic thumbnuts retain the

assembly.

The miniature heat exchanger supplied consists of a clear glass shell with end plates through which

pass a bundle of seven equally spaced stainless steel tubes. O ring seals in the end plates allow thestainless steel tubes to be removed for cleaning if necessary. Coupled to the end plates are end capsthat allow hot water from the heater/circulator to pass through all seven tubes and then re-combine

to return to the heater/circulator in a closed loop. Cold water from the mains supply passes throughthe clear glass outer shell and heat is transferred to this from the hot stream. Two baffles are locatedin the shell to promote turbulence and increase the cold fluid velocity.

In normal operation hot water from the heater/circulator passes into the end cap via a stainless steelbraided hose and self-sealing coupling. Its temperature at entry to the heat exchanger is measured

by a thermocouple sensor T1located on a copper tube at the HOT OUTconnection. It then flowsthrough the seven heat exchanger tubes to the opposite end cap and leaves via the braided hoseconnected to HOT RETURN. Its temperature on exit is measured by a similar thermocoupleT2.

The cold water is fed into and out of the heat exchanger via plastic reinforced hoses with self-sealing couplings.

Thermocouples T3 (COLD OUT)and T4 (COLD RETURN)measure the cold water inlet andexit temperatures.

Hot braided hoses terminate with a socket and Cold hoses with a plug to prevent cross-connectionof the hot and cold streams. The flow direction of the cold stream relative to the hot stream can be

reversed by changing the location of the inlet and exit tubes.


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C9

Installation

Heat Exchanger Installation H101C

It is assumed that the basic INSTALLATION AND COMMISSIONINGprocedures for the Heat

Exchanger Service unit H101 have been completed as detailed in the main manual.

Locate the heat exchanger mounting studs on the H101 front panel and remove the plasticthumbnuts. Fit the heat exchanger hanging bracket onto the mounting studs and refit the thumbnuts

to secure.

Temperature SensorsThe four temperature sensors are permanently fitted to copper tubes inside the H101 panel.

Hot Water CircuitThis is alwaysconnected in the following manner.Connect the HOT RETURN braided hose to the lower plug on the heat exchanger.Connect the HOT OUT braided hose to the upper plug on the heat exchanger.

Cold Water Circuit Counter-current FlowIn counter-current flow, the hot and cold-water streams flow in essentially opposing directionsthrough the heat exchanger.Connect the COLD RETURN reinforced hose to the upper socket on the heat exchanger.Connect the COLD OUT reinforced hose to the lower socket on the heat exchanger.

Cold Water Circuit Co-current Flow

In co-current flow, the hot and cold-water streams flow in essentially the same directions

through the heat exchanger.Connect the COLD RETURN reinforced hose to the lower socket on the heat exchanger.Connect the COLD OUT reinforced hose to the upper socket on the heat exchanger.


10/48

C10

Useful Data

Shell and Tube Exchanger H101C

Tube Material Stainless steel

Tube Outside Diameter 0.00476m

Tube Wall Thickness 0.0006mNumber of tubes in bundle 7

Effective length of tube bundle 0.205mEffective heat transfer area 0.0187m

2

Shell Material Clear Borosilicate (Pyrex type glass)

Shell Inside Diameter 0.075mShell Wall Thickness 0.01mNumber of baffles 2

Cold fluid is deflected by two transverse baffles inside the shell before leaving.The effective heat transfer area of the Shell and Tube heat exchanger H101C is close to that of theConcentric Tube heat exchanger H101A and Plate heat exchanger H101B, for direct comparison of

performance.

Table 1 - Specific Heat capacity Cp of Water in kJ kg1

C 0 1 2 3 4 5 6 7 8 9

04.1274 4.2138 4.2104 4.2074 4.2054 4.2019 4.1996 4.1974 4.1954 4.1936

104.1919 4.1904 4.189 4.1877 4.1866 4.1855 4.1864 4.1837 4.1829 4.1822

204.1816 4.181 4.1805 4.1801 4.1797 4.1793 4.1790 4.1787 4.1785 4.1783

304.1782 4.1781 4.1780 4.1780 4.1779 4.1779 4.1780 4.1780 4.1781 4.1782

404.1783 4.1784 4.1786 4.1788 4.1789 4.1792 4.1794 4.1796 4.1799 4.180

504.1804 4.1807 4.1811 4.1814 4.1817 4.1821 4.1825 4.1829 4.1833 4.1837

604.1841 4.1846 4.1850 4.1855 4.1860 4.1865 4.1871 4.1876 4.1882 4.1887

704.1893 4.1899 4.1905 4.1912 4.1918 4.1925 4.1932 4.1939 4.1964 4.1954

To use the table the vertical columns denote whole degrees and the Horizontal rows denote tens ofdegrees. For example the bold value 4.1792 kJ kg-1 is at 40 + 5 = 45 C.

Alternatively the equation C= 6x10

-9

t

4

1.0x10

-6

t

3

+7.0487x10

-5

t

2

2.4403x10

-3

t +4.2113may be used if the data is to be calculated using a spreadsheet.


11/48

C11

Table 2 - Density of Water in kg Litre-1

C 0 2 4 6 8

00.9998 0.9999 0.9999 0.9999 0.9999

100.9997 0.9995 0.9992 0.9989 0.9986

200.9982 0.9978 0.9973 0.9968 0.9962

300.9957 0.9950 0.9944 0.9937 0.9930

400.9922 0.9914 0.9906 0.9898 0.9889

500.9880 0.9871 0.9862 0.9852 0.9842

600.9832 0.9822 0.9811 0.9800 0.9789

70

0.9778 0.9766 0.9754 0.9742 0.9730

To use the table the vertical columns denote degrees and the Horizontal rows denote tens of degrees.For example the bold value 0.9906 kg is at 40 + 4 = 44 C.

Alternatively the equation = -4.582x10-6

t24.0007x10

-5t + 1.004

may be used if the data is to be calculated using a spreadsheet.


12/48

C12

Capabilities of the Shell and Tube Heat Exchanger H101C

When used with the Heat Exchanger Service Unit H101

1.

To demonstrate indirect heating or cooling by transfer of heat from one fluid stream to another

when separated by a solid wall (fluid to fluid heat transfer).

2.

To perform an energy balance across a shell and tube exchanger and calculate the overall

efficiency at different fluid flow rates

3.

To demonstrate the differences between counter-current flow (flows in opposing directions)

and co-current flows (flows in the same direction) and the effect on heat transferred,temperature efficiencies and temperature profiles through a shell and tube heat exchanger.

4. To determine the overall heat transfer coefficient for a shell and tube heat exchanger using thelogarithmic mean temperature difference to perform the calculations (for counter-current andco-current flows).

5.

To investigate the effect of changes in hot fluid and cold fluid flow rate on the temperature

efficiencies and overall heat transfer coefficient.

6.

To investigate the effect of driving force (difference between hot stream and cold streamtemperature) with counter-current and co-current flow.


13/48

C13

1. Demonstration of indirect heating or cooling by transfer of heat from one fluid

stream to another when separated by a solid wall (fluid to fluid heat transfer).

The following procedure demonstrates heat transfer from one fluid stream to another whenseparated by a solid wall.

NOTE that the observations from experiment No 2 may be used for the calculations in thisprocedure in order to save experimental time.

It is assumed that the basic INSTALLATION AND COMMISSIONING procedures for the HeatExchanger Service Unit H101 have been completed as detailed in the main manual.

Procedure

Install the Shell and Tube Exchanger H101C as detailed in INSTALLATION / Heat ExchangerInstallation H101Con page C9 and connect the cold water circuit to give Counter-Current flow asdetailed in the same section.

Follow the OPERATING PROCEDURE detailed in the main manual in order to establish thefollowing operating conditions.

Turn on the MAIN SWITCH and HEATER SWITCHSet the hot water temperature controller to 60C.Set the cold water flow rate Vcoldto 15g/sec

Set the hot water flow rate V hotto 50g/sec.

The procedure may be repeated with different hot and cold flow rates and different hot water inlet

temperature if required.

Monitor the stream temperatures and the hot and cold flow rates to ensure these too remain close to

the original setting. Then record the following:

T1, T2, T3, T4, Vhotand Vcold

Then adjust the cold-water flow valve so that Vcoldis approximately 35g/sec. Maintain the Hot waterflow rate at approximately 50g/sec (the original setting).

Allow the conditions to stabilise and repeat the above observations.

The procedure may be repeated with different hot and cold flow rates and different hot water inlettemperature if required.

OBSERVATIONSFlow Direction: Counter-Current

Sample

No.

T1 T2 T3 T4 Vhot Vcold

---- C C C C G/sec G/sec

1 58.5 53.9 15.4 26.9 48 16

2

3

4

5


14/48

C14

Calculated Data

Sample

No.t hot t cold

--- K K

1 4.6 11.5

CALCULATIONSFor the example result the calculations are as follows.

Reduction in Hot fluid temperature t hot = T1 - T2= 58.5 - 53.9= 4.6 K

Increase in Cold fluid temperature t cold = T4 T3=26.9 - 15.4= 11.5 K

The test results show the effect upon the temperature differences when the flow rates through asimple heat exchanger are varied.

The results from this experiment may also be used in experiment No 2 To perform an energy

balance across a shell and tube heat exchanger and calculate the overall efficiency at different

fluid flow rates.


15/48

C15

2. To perform an energy balance across a Shell and Tube heat exchanger and calculate

the overall efficiency at different fluid flow rates

The following procedure demonstrates heat transfer from one fluid stream to another when

separated by a solid wall and shows that the heat release rate of the hot stream should equal the heatabsorption rate of the cold stream.


It is assumed that the basic INSTALLATION AND COMMISSIONINGprocedures for the HeatExchanger Service Unit H101 have been completed as detailed in the main manual.

Procedure

Install the Shell and Tube Heat Exchanger H101C as detailed in INSTALLATION / HeatExchanger Installation H101Cand connect the cold water circuit to give Counter-Current flow asdetailed in the same section.







Then adjust the cold-water flow valve so that Vcoldis approximately 35g/sec. Maintain the Hot waterflow rate at approximately 50g/sec (the original setting).

Allow the conditions to stabilise and repeat the above observations.

The procedure may be repeated with different hot and cold flow rates and different hot water inlettemperature if required.


SampleNo.


---- C C C C G/sec G/sec1 58.5 53.9 15.4 26.9 48 16.6

2

3

4

5


16/48

C16

Calculated Data

Sample

No.t hot t cold

Q.e Q

.a

Thermal

--- K K W W %

1 4.6 11.5 913 800 87.6

CALCULATIONSFor the example the calculations are as follows.

It is necessary to correct mass flow rates using the conversion factors in table 1 and 2 on pages

C10/11. The water density (kg litre-1) and specific heat capacity Cp (kJ kg-1K-1) is dependantupon the temperature and the mean fluid temperature

is used to calculate the relevant temperature.

For the Hot stream:Tmean= (58.5+53.9) / 2 = 56.2 C

From table 1 and 2 at Tmean= 56.2 C hot = 0.9852 kg litre-1Cp = 4.182 kJ kg-1k-1

Hence the power emitted from the hot stream Q.

e

For the cold stream:Tmean= (26.9 + 15.4) / 2 = 21.2 C

From table 1 and 2 at Tmean= 21.2 C hot = 0.998kg litre-1

Cp = 4.181 kJ kg-1k-1

The power absorbed by the cold stream Q.a

2

TTT

outletinletmean

+=

( )( )

e hot hot HotQ V Cp T1- T2 Watts

48 0.9852 4.182 58.5 53.9

913 Watts

==

=

( )

( )

cold cold coldQa V Cp T4- T3 Watts

16.6 0.998 4.181 26.9 15.4 1000

800Watts

=

=

=


17/48

C17

The overall thermal efficiency

Hence

Note that as the shell is not insulated in order to allow student examination of the components heatwill be lost or gained depending upon the ambient temperature. In extreme cases this can result inan apparentthermal efficiency greater than 100%.

100(%)eQ

aQ Thermal =

%6.87

100(%)913

800 Thermal

=

=


18/48

C18

3. To demonstrate the differences between counter-current flow (flows in opposing

directions) and co-current flows (flows in the same direction) and the effect on heat

transferred, temperature efficiencies and temperature profiles through a shell and

tube heat exchanger

The following procedure demonstrates the effect of changing the direction of fluid flow on the heattransfer and temperature distribution in a shell and tube heat exchanger.

NOTE that the observations from experiment No 4 may be used for the calculations in this

procedure in order to save experimental time.

It is assumed that the basic INSTALLATION AND COMMISSIONING procedures for the Heat

Exchanger Service Unit H101 have been completed as detailed in the main manual.

Procedure

Install the Shell and tube Heat Exchanger H101C as detailed in INSTALLATION / Heat

Exchanger Installation H101C and connect the cold water circuit to give Counter-Currentflow

as detailed in the same section.


Turn on the MAIN SWITCH and HEATER SWITCH

Set the hot water temperature controller to 60C.Set the cold water flow rate Vcoldto 15g/secSet the hot water flow rate V hotto 35g/sec.

Monitor the stream temperatures and the hot and cold flow rates to ensure these too remain close tothe original setting. Then record the following:


This completes the basic Counter-Current flow experiment observations.

Next connect the cold water circuit to give Co-Current flowas detailed in the INSTALLATION /Heat Exchanger Installation H101C Note that there is no need to disconnect the hot water circuitor to turn off the hot water pump during this operation.

Monitor the stream temperatures and the hot and cold flow rates to ensure these remain close to theoriginal setting. Then record the following:


This completes the basic Co-Current flow experiment observations


SampleNo.



1 58.9 53.1 15.5 26.2 35 16.3

2

3

4

5


19/48

C19

Calculated Data

Sample

No.t hot t cold

Q.e Q

.a

Thermal Cold Hot Mean

--- K K W W % % % %

1 5.8 10.7 837 729 87.1 24.6 13.3 18.9

23

4

5

Flow Direction: Co-Current

SampleNo.



1 59.2 52.8 21.7 14.2 32.5 19.6

23

4

5

Calculated Data

SampleNo.

t hot t coldQ.e Q

.a

Thermal Cold Hot Meanl

--- K K W W % % % %

1 6.4 7.5 857 616 71.8 20.0 17.1 18.52

3

4

5

CALCULATIONSFor the examples the calculations are as follows.

Counter-Current FlowIt is necessary to correct mass flow rates using the conversion factors in table 1 and 2 on page

C10/11. The water density (kg litre-1

) and specific heat capacity Cp (kJ kg

-1

K

-1

) is dependantupon the temperature and the mean fluid temperature Tmean


For the Hot Stream:Tmean= (58.9 + 53.1) / 2 = 56.0 C

From table 1 and 2 at Tmean= 56.0 C hot = 0.985kg litre-1Cp = 4.182kJ kg

-1k

-1

2

TTT

outletinletmean

+=


20/48

C20


e

For the Cold stream:Tmean= (26.2 + 15.5) / 2 = 20.9 C

From table 1 and 2 at Tmean= 20.9 C hot = 0.998 kg litre-1Cp = 4.181 kJ kg

-1k

-1

Hence the power absorbed by the cold stream Q.a

Reduction in Hot fluid temperature t hot = T1 - T2= 58.9 - 53.1

= 5.8 K

Increase in Cold fluid temperature t cold = T4 - T3

= 26.2 - 15.5= 10.7 K

A useful measure of the heat exchanger performance is the temperature efficiency.

( )

( )

hot hot HotQe V Cp T1- T2 Watts

35 0.9852 4.180 58.9 53.1

837 Watts

=

=

=

( )

( )

cold cold ColdQa V Cp T4- T3 Watts

16.3 0.998 4.181 26.2 15.5

729 Watts

=

=

=


21/48

C21

The temperature change in each stream (hot and cold) is compared with the maximum temperature

difference between the two streams. This could only occur in a perfect heat exchanger of infinitesize with no external losses or gains.

The temperature efficiency of the hot stream from the above diagram

The temperature efficiency of the cold stream from the above diagram

The mean temperature efficiency

%3.13

%1005.159.58

1.539.58

%100T3-T1

T2-T1Hot

=

=

=

%6.24

%1005.159.585.152.26

%100T3-T1

T3-T4Cold

=

=

=

%9.18

2

6.243.13

2

ColdHotMean

=

+=

+=


22/48

C22

Co-Current FlowFor the co-current flow system the calculation procedure is similar but the formulae are as follows

The power emitted from the hot stream Q.e


Reduction in Hot fluid temperature t hot = T1 - T2 K

Increase in Cold fluid temperature t cold = T4 T3 K



( )hot hot HotQe V Cp T1- T2 Watts=

( )cold cold ColdQa V Cp T4- T3 Watts=

HotT1-T2

100%T1-T3

=

ColdT4-T3

100%

T1-T3

=


23/48

C23


The tabulated and calculated results show the differences between Counter-Current flow and Co-Current flow and the effect upon temperature efficiency and t for the hot and cold stream.

2

ColdHotMean

+=


24/48

C24

4. To determine the overall heat transfer coefficient for a shell and tube heat exchanger

using the logarithmic mean temperature difference to perform the calculations (for

counter-current and co-current flows).

The following procedure demonstrates the effect of changing the direction of fluid flow on theoverall heat transfer coefficient. The Logarithmic mean temperature difference is used to calculate

the overall heat transfer coefficient.



Procedure

Install the Shell and tube Heat Exchanger H101C as detailed in INSTALLATION / HeatExchanger Installation H101C and connect the cold water circuit to give Counter-Currentflowas detailed in the same section.


Turn on the MAIN SWITCH and HEATER SWITCHSet the hot water temperature controller to 60C.

Set the cold water flow rate Vcoldto 15g/secSet the hot water flow rate V hotto 35g/sec.




Next connect the cold water circuit to give Co-Current flowas detailed in the INSTALLATION /

Heat Exchanger Installation H101C Note that there is no need to disconnect the hot water circuit

or to turn off the hot water pump during this operation.

Monitor the stream temperatures and the hot and cold flow rates to ensure these remain close to the

original setting. Then record the following:


This completes the basic Co-Current flow experiment observations


SampleNo.



1 58.9 53.1 15.5 26.2 35 16.3

2

3

4

5


25/48

C25

Calculated Data

Sample

No.t hot t cold

Q.e Q

.a

Thermal Cold Hot Mean

--- K K W W % % % %

1 5.8 10.7 837 729 87.1 24.6 13.3 18.9

23

4

5

Sample

No.

LMTD U

--- K W m2K-1

1 35.09 1275.5

2

3

4

5


SampleNo.



1 59.2 52.8 21.7 14.2 32.5 19.6

2

3

4

5

Calculated Data

SampleNo.

t hot t coldQ.e Q

.a

Thermal Cold Hot Meanl

--- K K W W % % % %

1 6.4 7.5 857 616 71.8 20.0 17.1 18.5

2

3

4

5

Sample

No.

LMTD U

--- K W m2K

-1

1 37.6 1218.8

2

3

4

5


26/48

C26


Counter-Current FlowIt is necessary to correct mass flow rates using the conversion factors in table 1 and 2 on page

C10/11. The water density (kg litre-1) and specific heat capacity Cp (kJ kg-1 K-1) is dependant

upon the temperature and the mean fluid temperature Tmean


For the Hot Stream:Tmean= (58.9 + 53.1) / 2 = 56.0 C

From table 1 and 2 at Tmean= 56.0 C hot = 0.985kg litre-1Cp = 4.182kJ kg-1k-1


e

For the Cold stream:Tmean= (26.2 + 15.5) / 2 = 20.9 C

From table 1 and 2 at Tmean= 20.9 C hot = 0.998 kg litre-1

Cp = 4.181 kJ kg-1

k-1

Hence the power absorbed by the cold stream Q.a

Reduction in Hot fluid temperature t hot = T1 - T2= 58.9 - 53.1= 5.8 K

Increase in Cold fluid temperature t cold = T4 - T3= 26.2 - 15.5

= 10.7 K


2

TTT

outletinletmean

+=

( )

( )


35 0.9852 4.180 58.9 53.1

837 Watts

=

=

=

( )

( )


16.3 0.998 4.181 26.2 15.5

729 Watts

=

=

=


27/48

C27

The temperature change in each stream (hot and cold) is compared with the maximum temperature

difference between the two streams. This could only occur in a perfect heat exchanger of infinitesize with no external losses or gains.




%3.13

%1005.159.58

1.539.58

%100T3-T1

T2-T1Hot

=

=

=

%6.24

%1005.159.58

5.152.26

%100

T3-T1

T3-T4Cold

=

=

=

%9.18

2

6.243.13

2

ColdHotMean

=

+=

+=


28/48

C28

As the temperature difference between the hot and cold fluids vary along the length of the heat

exchanger it is necessary to derive a suitable mean temperature difference that may be used in heattransfer calculations. These calculations are not only of relevance in experimental procedures butalso more importantly to be used in the design of heat exchangers to perform a particular duty.

The derivation and application of the Logarithmic Mean temperature Difference (LMTD) may be

found in most thermodynamics and heat transfer text books.

The LMTD is defined as

Hence from the above Counter-current diagram of temperature distribution

Note that as the temperature measurement points are not fixed on the heat exchanger the LMTD isnot the same formula for both Counter-current flow and Co-current flow.

Hence for the Counter-current example

=

dTmin

dTmaxln

dTmin-dTmaxLMTD

=

T3)-(T2

T4)-(T1ln

T3)-(T2-T4)-(T1LMTD

K09.35

1396.0

9.4

ln(0.8696)

4.9-

15.5)-(53.126.2)-(58.9ln

15.5)-(53.1-26.2)-(58.9LMTD

=

=

=

=


29/48

C29

In order to calculate the overall heat transfer coefficient the following parameters must be used with

consistent units:-

Where

A Heat transfer area of heat exchanger (m2)

Q.e Heat emitted from hot stream (Watts)

LMTD Logarithmic mean temperature difference (K)

The heat transfer area may be calculated from:-

Wheredo Heat transfer tube outside diameter (m)di Heat transfer tube inside diameter (m)

dm Heat transfer tube mean diameter (m)L Heat transfer tube effective length (m)n Number of heat transfer tubes (Non dimensional)

Hence for the heat exchanger from the USEFUL DATAsection on page C10.

Hence for the test conditions the overall heat transfer coefficient:-

LMTDA

QU

e

=

nLdA

And

2

ddd

m

iom

=

+=

( ) ( )m

2

0.00476 0.00476 - (2 0.0006)d

20.00476 0.00356

2

0.00416 m

A 0.00416 0.205 7

0.0187 m

+ =

+=

=

=

=

e

-2 -1

QU

A LMTD

8370.0187 x 35.09

1275.5 Wm K

=

=

=


30/48

C30




Reduction in Hot fluid temperature t hot = T1 - T2 K






HotT1-T2

100%T1-T3

=

Cold

T4-T3

100%T1-T3=


31/48

C31


The logarithmic mean temperature difference LMTD

The Overall heat transfer coefficient U

The tabulated and calculated results show the differences between Counter-Current flow and Co-

Current flow and the effect upon temperature efficiency, t, and the overall heat transfer coefficient.

(T1-T3) - (T2 - T4)LMTD

(T1-T3)ln

(T2 - T4)

=

LMTDA

QU

e

=

2

ColdHotMean

+=


32/48

C32

5. To investigate the effect of changes in hot fluid and cold fluid flow rate on the

temperature efficiencies and overall heat transfer coefficient.

The following procedure demonstrates the effect of changing the flow rate of both the hot and cold

streams on the overall heat transfer coefficient. The Logarithmic mean temperature difference isused to calculate the overall heat transfer coefficient.


Procedure

Install the Shell and Tube Heat Exchanger H101C as detailed in INSTALLATION / HeatExchanger Installation H101Cand connect the cold water circuit to give Counter-Currentflowas detailed in the same section.


Turn on the MAIN SWITCH and HEATER SWITCHSet the hot water temperature controller to 60C.Set the cold water flow rate Vcoldto 15g/secSet the hot water flow rate V hotto 15g/sec.




The hot and cold flows should then be adjusted according to the following suggested values and theprocedure repeated.

Vhot(g/sec) Vcold (g/sec)15 1515 3015

5030 15

30 3030 5050 15

50 3050 50

If time permits the increments between the hot and cold stream flow rates may be made smaller to10g/second, if required.


Next, connect the cold water circuit to give Co-Current flowas detailed in the INSTALLATION /Heat Exchanger Installation H101CNote that there is no need to disconnect the hot water circuitor to turn off the hot water pump during this operation.

Monitor the stream temperatures and the hot and cold flow rates to ensure these remain close to theoriginal setting. Then record the following:


The hot and cold flows should then be adjusted to repeat the procedure used for Counter-Current

above.


33/48

C33


SampleNo.



1 60.5 48.7 14.7 23.1 16.3 15.82

3

4

5

Calculated Data

SampleNo.

t hot t coldQ.e Q

.a

Cold Hot Mean

--- K K W W % % %

1 11.8 8.4 795 555 18.3 25.8 22.1

23

4

5

SampleNo.

LMTD U NominalHot Flow

--- K W m2K

-1 G/sec

1 35.67 1191.8 16

2

3

4

5


SampleNo.



1 60.8 49.5 22.8 14.4 15 15

2

3

4

5

Calculated Data

Sample

No.t hot t cold

Q.e Q

.a

Cold Hot Meanl

--- K K W W % % %

1 11.3 8.4 769 620 22.1 29.7 25.9

2

3

4

5


34/48

C34

SampleNo.

LMTD U

--- K W m2K-1

1 35.64 1153.2

2

3

45


Counter-Current FlowIt is necessary to correct the mass flow rates using the conversion factors in table 1 and 2 on page

C10/11. The water density (kg litre-1) and specific heat capacity Cp (kJ kg-1 K-1) is dependantupon the temperature and the mean fluid temperature Tmean

is used to calculate the relevant temperature of the Hot stream.

For the Hot stream:

Tmean= (60.5 + 48.7 ) / 2 = 54.6 C

From table 1 and 2 at Tmean= 54.6 C

hot = 0.986 kg litre-1

Cp = 4.182 kJ kg-1

k-1


e

For the Cold stream TmeanTmean= (23.1 + 14.7 ) / 2 = 18.9 C

From table 1 and 2 at Tmean= 18.9 C Cold = 0.998 kg litre-1

CpCold = 4.182 kJ kg-1k-1

Hence the power absorbed by the cold stream Q

.

a

Reduction in Hot fluid temperature t hot = T1 T2

= 60.3 48.7= 11.8 K

Increase in Cold fluid temperature t cold = T4 T3

= 23.1 14.7= 8.4 K

( )( )


16.3 0.986 4.182 60.5 48.7

795 Watts

==

=

2

TTT

outletinletmean

+=

( )

( )


15.8 0.998 4.182 23.1 14.7

555 Watts

=

=

=


35/48

C35


The temperature change in each stream (hot and cold) is compared with the maximum temperaturedifference between the two streams. This could only occur in a perfect heat exchanger of infinite

size with no external losses or gains.




%8.25

%1007.145.60

7.485.60

%100T3-T1

T2-T1Hot

=

=

=

%3.18

%1007.145.60

7.141.23

%100T3-T1

T3-T4Cold

=

=

=

%1.22

2

3.188.252

ColdHotMean

=

+=

+=


36/48

C36







Note that as the temperature measurement points are not fixed on the heat exchanger the LMTD isthe not the same formula for both Counter-current flow and Co-current flow.


=

dTmin

dTmaxln

dTmin-dTmaxLMTD

=

T3)-(T2

T4)-(T1ln

T3)-(T2-T4)-(T1LMTD

K67.35

0953.0

4.3

ln(1.1)

3.4

14.7)-(48.7

23.1)-(60.5ln

14.7)-(48.7-23.1)-(60.5LMTD

=

=

=

=


37/48

C37


consistent units: -

Where





Wheredo Heat transfer tube outside diameter (m)

di Heat transfer tube inside diameter (m)dm Heat transfer tube mean diameter (m)L Heat transfer tube effective length (m)

n Number of heat transfer tubes (Non dimensional)



LMTDA

QU

e

=

nLdAAnd

2

ddd

m

iom

=

+=

e

-2 -1

QU

A LMTD

7950.0187 x 35.67

1191.8 Wm K

=

=

=

( ) ( )m

2

0.00476 0.00476 - (2 0.0006)d 2

0.00476 0.00356

2

0.00416 m

A 0.00416 0.205 7

0.0187 m

+ =

+=

=

=

=


38/48

C38

Co-Current FlowFor the co-current flow system the calculation procedure is similar, the formulae are as follows



Reduction in Hot fluid temperature t hot = T1 T2 K






HotT1-T2

100%T1-T3

=

Cold

T4-T3

100%T1-T3=


39/48

C39




In order to see the effect of the volume flow rate on the overall heat transfer coefficient the datashould be plotted with overall heat transfer coefficient on the Y(vertical) axis and the Hot flow rateon the X (horizontal) axis.

2

ColdHotMean

+=

(T1-T3) - (T2 - T4)LMTD

(T1-T3)ln

(T2 - T4)

=

LMTDA

QU

e

=


40/48

C40

6. To investigate the effect of driving force (difference between hot stream and cold

stream temperature) with counter-current and co-current flow

The following procedure demonstrates the effect of changing the temperature difference (or driving

force) between the hot and cold streams. The effect of this on the overall heat transfer coefficientand temperature efficiencies may be investigated for both counter-current and co-current flows.

I It is assumed that the basic INSTALLATION AND COMMISSIONING procedures for the HeatExchanger Service Unit H101 have been completed as detailed in the main manual.

Procedure

Install the Concentric tube Heat Exchanger H101C as detailed in INSTALLATION / HeatExchanger Installation H101Cand connect the cold water circuit to give Counter-Currentflowas detailed in the same section.


Turn on the MAIN SWITCH and HEATER SWITCHSet the hot water temperature controller to 40C.Set the cold water flow rate Vcoldto 15g/secSet the hot water flow rate V hotto 30g/sec.




Repeat the above procedure with the hot water temperature controller set to 50, 60 and 70 C.

This completes the Counter-current observation procedures

Next, connect the cold water circuit to give Co-Current flowas detailed in the INSTALLATION /Heat Exchanger Installation H101CNote that there is no need to disconnect the hot water circuitor to turn off the hot water pump during this operation.






Repeat the above procedure with the hot water temperature controller set to 50, 60 and 70 C.

This completes the Co-current observation procedures


41/48

C41


SampleNo.



1 38.4 35.4 14.6 18.9 30 152 48.5 44.2 14.5 21.5 30 15

3 59.5 52.7 14.4 24.4 30 15

4 69 60.1 14.3 27.4 30 15

5 73 60.9 14.4 27.7 30 15

Calculated Data

SampleNo.

t hot t coldQ.e Q

.a

Cold Hot Mean

--- K K W W % % %

1 3.0 4.3 402 297 18.1 12.6 15.3

2 4.3 7.0592 492 20.6 12.6 16.6

3 6.8 10.0 953 710 22.2 15.1 18.6

4 8.9 13.1 1178 938 23.9 16.3 20.1

5 12.1 13.3 1662 943 22.7 20.6 21.7

SampleNo.

LMTD U

--- K W m2K

-1

1 20.1 1067

2 28.3 1119

3 36.7 1389

4 43.7 1442

5 45.9 1937

Similar observations may be obtained for the Co-current configuration. The calculation proceduresare shown below.

CALCULATIONS

Counter-Current FlowFor sample No 1 above:It is necessary to correct the mass flow rates using the conversion factors in table 1 and 2 on page

C10/11. The water density (kg litre-1) and specific heat capacity Cp (kJ kg-1K-1) is dependantupon the temperature and the mean fluid temperature Tmean

Is used to calculate the relevant temperature of the Hot stream.

For the Hot stream:Tmean= (38.4 + 35.4 ) / 2 = 36.9 C

From table 1 and 2 at Tmean= 36.9 C

hot = 0.993 kg litre-1

Cp = 4.178 kJ kg-1k-1

Hence the power emitted from the hot stream Q.e

( )

( )


32.3 0.993 4.178 38.4 35.4

402 Watts

=

=

=

2

TTT

outletinletmean

+=


42/48

C42


The temperature change in each stream (hot and cold) is compared with the maximum temperaturedifference between the two streams. This could only occur in a perfect heat exchanger of infinite

size with no external losses or gains.




%6.12

%1006.144.38

4.354.38

%100T3-T1

T2-T1Hot

=

=

=

%06.18

%1006.144.38

6.149.18

%100T3-T1

T3-T4Cold

=

=

=

%3.15

206.186.12

2

ColdHotMean

=

+=

+=


43/48

C43







Note that as the temperature measurement points are not fixed on the heat exchanger, the LMTD isnot the same formula for both Counter-current flow and Co-current flow.


=

dTmin

dTmaxln

dTmin-dTmaxLMTD

=

T3)-(T2

T4)-(T1ln

T3)-(T2-T4)-(T1LMTD

K14.20

0645.0

3.1

ln(0.9375)

1.3-

14.6)-(35.418.9)-(38.4ln

14.6)-(35.4-18.9)-(38.4LMTD

=

=

=

=


44/48

C44


consistent units:-

Where





Where

do Heat transfer tube outside diameter (m)di Heat transfer tube inside diameter (m)dm Heat transfer tube mean diameter (m)L Heat transfer tube effective length (m)

n Number of heat transfer tubes (Non dimensional)



LMTDA

QU

e

=

nLdA

And

2

ddd

m

iom

=

+=

e

-2 -1

QU

A LMTD

4020.0187 x 20.14

1067.4 Wm K

=

=

=

( ) ( )m

2

0.00476 0.00476 - (2 0.0006)d2

0.00476 0.00356

2

0.00416 m

A 0.00416 0.205 7

0.0187 m

+ =

+=

=

=

=


45/48

C45




Reduction in Hot fluid temperature t hot = T1 T2 K





HotT1-T2

100%T1-T3

=

ColdT4-T3

100%T1-T3

=



46/48

C46




In order to visualise the effect of temperature difference on the overall heat transfer coefficient the

calculated data may be plotted against logarithmic mean temperature difference.An example of this is given at the end of this section for the Counter-current flow configuration.

2

ColdHotMean

+=

(T1-T3) - (T2 - T4)LMTD

(T1-T3)ln

(T2 - T4)

=

LMTDA

QU

e

=


47/48

C47

Example Data


48/48

C48

Example Data

shell and tube manual

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