L6 Cooling Tower

8/18/2019 L6 Cooling Tower

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EXPERIMENTAL MANUAL

SOLTEQ® EQUIPMENT FOR ENGINEERING EDUCATION

WATER COOLING

TOWER

WATER COOLING

TOWER


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Table of Contents

Page

List of Figures............................................................................................................................... i

1.0 .INTRODUCTION ……………………………………………………………….…………………… 1

2.0. GENERAL DESCRIPTIONS 2.1 Components of the HE152 Basic Cooling Tower Unit …………………..….............. 2

2.2

The Process Involved in the Operation ………………………………………………... 52.3 Overall Dimensions……………………………………………………………………….. 52.4 General Requirements……………………………………………………………………. 5

3.0 SUMMARY OF THEORY 3.1 Basic Principle ………………………………………………………………............... 63.2 Evaporation from a Wet Surface …………………………….................................. 63.3 Cooling Tower Performance………………………………………………………….. 6

3.4

Thermodynamic Property……………………………………………………………… 73.4.1 Dalton’s and Gibbs Laws…………………………………………………….. 73.4.2 Psychometric Chart…………………………………………………………… 8

3.5 Orifice Calibration…………………………………………………………………………. 93.6 Application of Steady Flow Energy Equation………………………………………….. 103.7 Characteristics Column Study………………………………………………………….. 133.8 Useful Information………………………………………………………………………… 16

4 0 EXPERIMENTAL PROCEDURE


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APPENDIX A Experimental Data Sheets

APPENDIX B Sample Resul ts and Calculation

APPENDIX C Components Properties and Diagram

APPENDIX D Process Flow Diagram


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List of Figures

Page

Figure 1 Parts Identification and Equipment Set-up of Bench Top Cooling Tower 4

Figure 2 System A 10

Figure 3 System B 12

Figure 4 Schematic Representations of the Air and Water Streamsentering and leaving a Block of Packing 13

Figure 5 Graphical Representation of Tower Characteristics 15


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SOLTEQ® BENCH TOP COOLING TOWER UNIT (MODEL: HE152)

1.0 INTRODUCTION

The SOLTEQ®

Basic Cooling Tower Unit (Model: HE152) has been designed todemonstrate students the construction, design and operational characteristics of a moderncooling system. The unit resembles a full size forced draught cooling tower and it isactually an "open system" through which two streams of fluid (in this case air and water)pass and in which there is a mass transfer from one stream to the other. The unit is self-contained supplied with a heating load and a circulating pump. Once energy and massbalances are done, students will then be able to determine the effects on the performanceof the cooling tower by the following parameters:

a) Temperature and flow rate of waterb) Relative Humidity and flow rate of airc) Cooling load

Additionally, a Packing Characteristics Column (optional) is available for SOLTEQ®

Basic

Cooling Tower Unit (Model: HE152). This column is designed to facilitate study of waterand air conditions at three additional stations (I, II and III) within the column. This enables

driving force diagrams to be constructed and the determination of the Characteristic

Equation for the Tower.


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2.0 GENERAL DESCRIPTIONS

2.1 Components of the HE152 Basic Cooling Tower Unit

The unit comes complete with the following main components:

i) Load Tank

The load tank is made of stainless steel having a capacity of approximately 9liters. The tank is fitted with two cartridge heaters, 0.5 kW and 1.0 kW each, toprovide a total of 1.5 kW cooling load. A make-up tank is fixed on top of theload tank. A float type valve at the bottom of the make-up tank is to control theamount of water flowing into the load tank. A centrifugal type pump is suppliedfor circulating the water from the load tank through a flowmeter to the top ofthe column, into the basin and back to the load tank. A temperature sensorand temperature controller is fitted to load tank to prevent overheating. A levelswitch is fitted to the load tank so that when a low level condition occurs, the

heater and the pump will be switched off.

ii) Air Distribution Chamber

The stainless steel air distribution chamber comes with a water collectingbasin and a one-side inlet centrifugal fan. The fan has a capacity ofapproximately 251 CFM of air flow. The air flowrate is adjustable by means ofan intake damper.


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iv) Measurements

Temperature sensors are provided to measure the inlet and outlet watertemperatures as well as the make-up tank water temperature. In addition,temperature sensors have been installed to measure the dry bulb and wet bulbtemperatures of inlet and outlet of the air. The followings show the list of codesassigned to each temperature sensors.

T1 Dry Bulb Temperature of the Inlet AirT2 Wet Bulb Temperature of the Inlet AirT3 Dry Bulb Temperature of the Outlet AirT4 Wet Bulb Temperature of the Outlet AirT5 Inlet Water TemperatureT6 Outlet Water TemperatureT7 Make up Tank TemperatureT8 Hot Water Tank Temperature

A differential pressure transmitter is provided for the measurement of pressuredrop across the packed column. On the other hand, the differential pressuretransmitter and the orifice are also used to determine the air flowrate.

A flowmeter is provided for the measurement of water flowrate. The flowmeteris ranged at 0.4 to 4 LPM.


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1

2

6

4

5

3

7

8

9

10


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2.2 The Process Involved in the Operation

i)

Water Circuit

Water temperature in the load tank will be increased before the water ispumped through a control valve and flow meter to the column cap. Beforeentering the column cap, the inlet temperature of the water is measured andthen the water is uniformly distributed over the top packing deck. This createsa large thin film of water, which is exposed to the air stream. The water getscooled down, while passing downward through the packing, due to theevaporation process. The cooled water falls into the basin below the lowestdeck and return to the load tank where it is re-heated before re-circulation. Theoutlet temperature is measured at a point just before the water flows back intothe load tank. Evaporation causes the water level in the load tank to fall. Theamount of water lost by evaporation will be automatically compensated byequal amount from the make-up tank. At steady state, this compensation rateequals the rate of evaporation plus any small airborne droplets discharged with

the air.

ii) Air Circuit

A one-side inlet centrifugal fan draws the air from the atmosphere into thedistribution chamber. The air flow rate is varied by means of an intake damper.The air passes a dry bulb temperature sensor and wet bulb temperaturesensors before it enters the bottom of the packed column. While the air stream

th h th ki it i t t t i d th t


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3.0 SUMMARY OF THEORY

3.1 Basic Princ iple

First consider an air stream passing over the surface of a warm water droplet orfilm. If we assume that the water is hotter than the air, then the water temperaturewill be cooled down by radiation, conduction and convection, and evaporation. Theradiation effect is normally very small and may be neglected. Conduction andconvection depend on the temperature difference, the surface area, air velocity,etc. The effect of evaporation is the most significant where cooling takes place aswater molecules diffuse from the surface into the surrounding air. During theevaporation process, the water molecules are replaced by others in the liquid fromwhich the required energy is taken.

3.2 Evaporation from a Wet Surface

When considering evaporation from a wet surface into the surrounding air, the rate

is determined by the difference between the vapour pressure at the liquid surfaceand the vapour pressure in the surrounding air. The vapour pressure at the liquidsurface is basically the saturation pressure corresponding with the surfacetemperature, whereas the total pressure of the air and its absolute humiditydetermines the vapour pressure in the surrounding air. Such evaporation processin an enclosed space shall continue until the two vapour pressures are equal. Inother words, until the air is saturated and its temperature equals the surface.

H if t t d i i t tl li d th t f ill h


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The effect of these factors will be studied in depth by varying it. In this way,students will gain an overall view of the operation of cooling tower.

3.4 Thermodynamic Property

In order to understand the working principle and performance of a cooling tower, abasic knowledge of thermodynamic is essential to all students. A brief review onsome of the thermodynamic properties is presented below.

At the triple point (i.e. 0.00602 atm and 0.01°C), the specific enthalpy of saturatedwater is assumed to be zero, which is taken as datum. The specific enthalpy ofsaturated water (h f

) at a range of temperatures above the datum condition can beobtained from thermodynamic tables.

Thespecific enthalpy of compressed liquid is given by

( )satf f ppvhh −+= (1)

The correction for pressure is negligible for the operating condition of the coolingtower; therefore we can see that h ≈ hf

at a given temperature.

Specific heat capacity (Cp) is defined as the rate of change of enthalpy withrespect to temperature (often called the specific heat at constant pressure). For thepurpose of experiment using bench top cooling tower, we may use the followingrelationship:


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same temperature.

The Absolute or Specific Humidity is defined as follows:

Air Dryof Mass

Vapour Water of MassHumiditySpecific =ω , (4)

The Relative Humidity is defined as follows:

etemperatur sametheatVapour Water of essureSaturation

Air theinVapour Water of essurePartialHumiditylative

Pr

Pr ,Re

=

φ (5)

The Percentage Saturation is defined as follows:

TempsametheatVapour Water Satof volsameof Mass Air of VolumegivenainVapour Water of Mass

SaturationPercentage

= (6)

At high humidity conditions, it can be shown that there is not muchdifference between the "Relative Humidity" and the "PercentageSaturation" and thus we shall regard as the same.

To measure the moisture content of the atmosphere, this bench top


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3.5 Orifice Calibration

As mentioned above, the psychometric chart can be used to determine the valueof the specific volume. However, the values given in the chart are for 1 kg of dry airat the stated total pressure.

However, for every 1 kg of dry air, there is w kg of water vapour, yielding the totalmass of 1 + w kg.

Therefore, the actual specific volume of the air/vapor mixture is given by:

ϖ +=1

ba

a

vv (7)

The mass flow rate of air and steam mixture through the orifice is given by

av

x

m 01370

.

=

(8)

Where,m = Mass flow rate of air/vapor mixture vax = Orifice differential in mmH

= Actual specific volume and

2

0.


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3.6 Appl ication of Steady Flow Energy Equation

Consider System A for the cooling tower defined as in Figure 2. It can be seen thatfor this system, indicated by the dotted line,

a) Heat transfer at the load tank and possibly a small quantity to surroundingsb) Work transfer at the pumpc) Low humidity air enters at point Ad) High humidity air leaves at point Be) Make-up enters at point E, the same amount as the moisture increase in the

air stream

A

B

E

am

am

E m


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Note:

a) The mass flow rate of dry air ( am ) through a cooling tower is a constant,

whereas the mass flow rate of moist air increases as the result of evaporationprocess.

b) The term E E hm can usually be neglected since its value is relatively small.

Under steady state conditions, by conservation of mass, the mass flow rate of dryair and of water (as liquid or vapour) must be the same at inlet and outlet to anysystem.

Therefore,

( ) ( ) Ba Aa

mm = (13)

and

( ) ( ) Bs E As mmm

=+ or

( ) ( ) As Bs E

mmm −= (14)

The ratio of steam to air (ϖ ) is known for the initial and final state points on thepsychrometric charts. Therefore,

=


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Say, we re-define the cooling tower system to be as in Figure 3 where the processheat and pump work does not cross the boundary of the system. In this case warm

water enters the system at point C and cool water leaves at point D.

Figure 3: System B

Bam

C

wm

A

am

D

E m E


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3.7 Characteristics Column Study

In order to study the packing characteristics, we define a finite element of thetower (dz) as shown in Figure 4, the energy balances of the water and air streamsin the tower are related to the mass transfer by the following equation:

( )hdVaKdTmC WpW ∆= (20)

where

WpC = Specific heat capacity of water

Wm = Mass flow rate of water per unit plan area of packing

T = Water TemperatureK = Mass Transfer Coefficienta = Area of contact between air and water per unit volume of packingV = Volume occupied by packing per unit plan areah∆

= Difference in specific enthalpy between the saturated boundary layer andthe bulk air

WATER

INLET

AIR

OUTLET

T2

H2

mw

t2

h2

ma


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Integrating Equation 21,

∫ −=2

1

T

T aw

p

W hh

dTCm

VKaW

(22)

The numerical solution to the integral expression Equation 22 using Chebyshevnumerical method gives,

∆+∆+∆+∆

−

=−= ∫ 432112 1111

4

2

1

hhhh

TT

hh

dTCm

VKaT

T aw

p

WW

(23)

Where

Wm

VKa

= Tower Characteristic

1h∆ = value of aw hh − at ( )212 1.0 T T T −+

2

h∆ = value of aw hh − at ( )212 40 TTT −+ .

3h∆ = value of aw hh − at ( )211 4.0 T T T −−

4h∆ = value of aw hh − at ( )211 1.0 T T T −−

Thermodynamics state that the heat removed from the water must be equal to theheat absorbed by the surrounding air. Therefore, the following equation is derived:


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Figure 5:Graphical Representation of Tower Characteristics

The following represents a key to Figure 5:

BA = Initial enthalpy driving force

Air Operating Line

Enthalpy

Driving Force

(hw-ha)

hw2 (Hot water Temp)

hw1 (Cold water Temp)

ha2 (Air out)

ha1 (Air in)

Water Operating Line

C

Approach Range

Temperature

Twb (In) T1 Twb (Out) T2

L/G A

D

B

Saturation Curve

Enthalpy


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3.8 Useful Information

1. Orifice Calibration Formula:

Mass flow rate of air and vapor mixture,

( )

bav

xm

ϖ +=

10137.0

The mass flow rate of dry air,

( )ϖ +=

10137.0

ba

av

xm

Where,

x = orifice differential in mmH2

Ba

v

0,

= specific volume of air at the outlet

ϖ = humidity ratio of the mixture

2. Pump Work Input = 80W (0.08kW)

3. Column Inner Dimension = 150 mm x 150 mm x 600 mm


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4.0 EXPERIMENTAL PROCEDURES

4.1 General Operating Procedures

4.1.1 General Start-up Procedures

1. Check to ensure that valves V1 to V6 are closed and valve V7 ispartially opened.

2. Fill the load tank with distilled or deionised water. It is done by firstremoving the make-up tank and then pouring the water through the

opening at the top of the load tank. Replace the make-up tank ontothe load tank and lightly tighten the nuts. Fill the tank with distilled ordeionised water up to the zero mark on the scale.

3. Add distilled/deionised water to the wet bulb sensor reservoir to thefullest.

4. Connect all appropriate tubing to the differential pressure sensor.5. Install the appropriate cooling tower packing for the experiment.

6. Then, set the temperature set point of temperature controller to 50°C.Switch on the 1.0 kW water heater and heat up the water untilapproximately 40°C.

7. Switch on the pump and slowly open the control valve V1 and set thewater flowrate to 2.0 LPM. Obtain a steady operation where the wateris distributed and flowing uniformly through the packing.

8. Fully open the fan damper, and then switch on the fan. Check that thedifferential pressure sensor is giving reading when the valve manifold


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4.1.2 General Shut-Down Procedure

1. Switch off heaters and let the water to circulate through the coolingtower system for 3-5 minutes until the water cooled down.

2. Switch of the fan and fully close the fan damper.3. Switch off the pump and power supply.4. Retain the water in reservoir tank for the following experiment.5. Completely drain off the water from the unit if it is not in used.

4.2 Experiment 1: General Observation of the Forced Draught Cooling Tower

Objective:

To observe the processes within a forced draught cooling tower

1. Perform the general start-up procedures and observe the forced draughtcooling tower proves.

2. As the warm water enters the top of the tower, it is fed into channels fromwhich it flows via water distribution system onto the packing. The channels aredesigned to distribute the water uniformly over the packing with minimumsplashing.

3. The packing surfaces are easily wetted and the water spreads over thesurfaces to expose a large area to the air stream.

4. The cooled water falls from the lowest packing into the basin and then ispumped to the simulated load in the load tank.


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4.3 Experiment 2: End State Properties of Air and Steady Flow Equations

Objective:To determine the “end state” properties of air and water from tables or chartsTo determine Energy and mass balances using the steady flow equation on theselected systems

Procedure:

1. Prepare and start the cooling tower with according to Section 4.1.1.2. Set the system under the following conditions and allow stabilizing for about 15

minutes.Water flow rate : 2.0 LPM Air Flow : MaximumCooling load : 1.0 kWColumn installed : A

3. Fill up the make-up tank with distilled water up to zero mark at the level scale,and then start the stop watch.

4.

Determine the make-up water supply in an interval of 10 minutes.5. In this 10 minutes interval, record a few sets of the measurements (i.e.temperatures (T1–T7), orifice differential pressure (DP1), water flowrate (FT1)and Heater Power (Q1)), then obtain the mean value for calculation andanalysis.

6. Determine the quantity of make up water that has been supplied during thetime interval by noting the height reduction in the make-up tank.

7. The observation may be repeated at different conditions, i.e. at different water


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4.4 Experiment 3: Investigation of the Effect of Cooling Load on Wet Bulb Approach

Objective:

To investigate the effect of cooling load on “Wet Bulb Approach”

Procedure:


minutes.

Water flow rate : 2.0 LPM Air Flow : MaximumCooling load : 0 kWColumn installed : A

3. After the system stabilizes, record a few sets of measurements (i.e. air inletdry bulb and wet bulb temperature (T1 and T2), water outlet temperature (T6),orifice differential pressure (DP1), water flowrate (FT1) and Heater Power

(Q1)), then obtain the mean value for calculation and analysis.4. Without changes in the conditions, increase the cooling load to 0.5 kW. Whenthe system stabilized, record all data.

5. Similarly, repeat the experiment at 1.0kW and 1.5kW.6. Finally, measure the cross sectional area of the column.7. The four tests may be repeated at another constant airflow.8. The observation may also be repeated at different conditions, i.e. at different

water flow rates, or different air flow rates and with different load.


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4.5 Experiment 4: Investigation of the Effect of Air Velocity on Wet bulb Approach and Pressure Drop through the Pack ing

Objective:

To investigate the effect of air velocity on:(a) Wet Bulb Approach(b) The pressure drop through the packing

Procedure:

1. Prepare and start the cooling tower with according to Section 4.1.1.

2. Set the system under the following conditions and allow stabilizing for about 15minutes.Water flow rate : 2.0 LPM Air flow rate : MaximumCooling load : 1.0 kWColumn installed : A

3. After the system stabilizes, record a few sets of measurements (i.e.temperature (T1-T6), orifice differential pressure (DP1), water flowrate (FT1).heater power (Q1) and pressure drop across packing (DP2)), then obtain themean value for calculation and analysis.

4. Repeat the test with 3 different sets of orifice pressure drop values (75%, 50%and 25% of the maximum value) without changing the water flow rate andcooling loads.

5. Finally, measure the cross sectional area of the column.6. The test may be repeated:


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4.6 Experiment 5: Investigation of the Relationship between Cooling Load andCooling Range

Objective:

To investigate the relationship between cooling load and cooling range

Procedure:


minutes:

Water flow rate : 2.0 LPM Air flow rate : MaximumCooling load : 0.0 kWColumn installed : A

3. After the system stabilized, record a few sets of measurements (i.e.temperature (T1-T6), orifice differential pressure (DP1), water flowrate (FT1)and heater power (Q1)), then obtain the mean value for calculation andanalysis

9. Without changes in the conditions, increase the cooling load to 0.5 kW. Whenthe system stabilized, record all data.

4. Similarly, repeat the experiment at 1.0kW and 1.5kW.5. The tests may be repeated:

i.At other water flow ratesii.At other air flow rate


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4.7 Experiment 6: Investigation of the Effect of Packing Density on thePerformance of the Cooling Tower

Objective:

To investigate the effect of packing density on the performance of the coolingtower

Procedure:


minutes:Water flow rate : 2.0 LPMOrifice differential : MaximumCooling load : 1.0kWColumn installed : A

3. After the system stabilizes, record a few sets of measurements (i.e.temperature (T1-T6), orifice differential pressure (DP1), water flowrate (FT1).heater power (Q1) and pressure drop across packing (DP2)), then obtain themean value for calculation and analysis.

4. Without changing condition, change the column packing to column B. Whenstability is achieved, repeat the observation.

5. Repeat step 4 with column B and C.6. The tests may be repeated:

i. At other water flow ratesii. At other air flow rates


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4.8 Experiment 7: Determination of Characteristic Equation of the PackingCharacteristic Column

Objectives:

To determine the Characteristic Equation of the cooling tower using Packing

Characteristic Column

Procedures:

1.

Install the Packing Characteristic Column (Column E) properly.2. Prepare and start the cooling tower with according to Section 4.1.1.3. Set the system under the following conditions and allow stabilizing for about 15

minutes:Water flow rate : 1.5 LPMOrifice differential : MaximumCooling load : 1.0kWColumn installed : E

4. After the system stabilizes, record a few sets of measurements (i.e.temperature (T1-T6 and T8-T15), orifice differential pressure (DP1), waterflowrate (FT1) and heater power (Q1)), then obtain the mean value forcalculation and analysis.

5. Without changing the air flow rate, and cooling load, change the water flowrate to 2.0 LPM. When stability is achieved, repeat the observation.

6. Similarly, repeat the observations at 2.5 LPM.


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5.0 REFERENCES

Perry, R.H., Green, D.W. and Maloney, J.O., “Perry’s Chemical Engineering Handbook”, 6th

Edition, McGraw Hill, 1984.


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Appendix A


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Experiment 2: End State Properties of Air and Steady Flow Equations

Results:

Column Installed : _______

Initial water level : cm

Final water level : cm

Time Interval : minutes

Packing Density m-1

Air Inlet Dry Bulb, T1 ˚C

Air Inlet Wet Bulb, T2 ˚C

Air Outlet Dry Bulb, T3 ˚C

Air Outlet Wet Bulb, T4 ˚C

˚


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Experiment 4: Investigation of the Effect of Ai r Velocity on Wet bulb Approach and Pressure

Drop through the Packing

Results:

Column Installed : _____

Description Unit

Air Flow

100 % 75 % 50 % 25 %

Packing Density m-1






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Experiment 5: Investigation of the Relationship between Cooling Load and Cooling Range

Results:

Column Installed : ____

Description Unit 0.0kW 0.5kW 1.0kW 1.5kW

Packing Density m-1





Water Inlet Temperature, T5 ˚C


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Experiment 6: Investigation of the Effect of Packing Density on the Performance of the

Cooling Tower

Results:

Description UnitColumn

A B C

Packing Density m-1







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Experiment 7: Determination of Characteristic Equation of the Packing Characteristic

Column

Result:

Description UnitWater Flowrate (LPM)

1.5 2.0 2.5

Top




Station III

Air Dry Bulb, T8˚C


Water Temperature, T14 ˚C

Station II

AirDryBulb T10 ˚C


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Experiment 2: End State Properties of Air and Steady Flow Equations

Results:

Column Installed : A .

Initial water level : 0 cm

Final water level : 4.0 cm

Time Interval : 10 minutes

Descript ion Unit Value

Packing Density m-1 110

Air Inlet Dry Bulb, T1 ˚C 31.6

Air Inlet Wet Bulb, T2 ˚C 27.1

Air Outlet Dry Bulb, T3 ˚C 30.2

˚


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Calculation:

By plotting the air inlet dry bulb and air inlet wet bulb (point A) and air outlet dry bulb and airoutlet wet bulb (point B) on the Psychrometric chart, the enthalpy and humidity of air mixture is

obtained and the result is showed as below:

h A = 88.0 kJ/kg

w A = 0.023 kg water/kg dry air

hB = 96.0kJ/kgwB = 0.0268 kg water/kg dry air

vaB = 0.950 m3/(kg dry air)

From the orifice calibration,

( )baBa

wv xm+

=1

0137.0

x = 88Pa x 1mm H2O / 10.13Pa

= 8.69 mm H2O


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1.600

172.0 −= skg

= 2.867 x 10-4 kg.s-1

Specific enthalpy of make-up,

hE = 126.586kJ/kg (at T7)

Applying the Steady Flow Equation to the system indicated by the chain the (System A)

EKHPQ ∆−∆=−

Now, solving the left side,

( )W W P Q 65883 −−=−

= 948W (Work by pump is estimate –65W)

On the right side of the equation,

0≈∆ E K (change of air velocity is negligible)


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Experiment 3: Investigation of the Effect of Cooling Load on Wet Bulb Approach

Results:



Packing Density m-1 - 110 110 110

Air Inlet Dry Bulb, T1 ˚C - 31.5 31.7 31.8

Air Inlet Wet Bulb, T2 ˚C - 27.1 27.2 27.3

Water Outlet Temperature, T6 ˚C - 29 30.6 32.1

Orifice Differential, DP1 Pa - 89 89 89

Water Flow Rate, FT1 LPM - 2.0 2.0 2.0


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Specific volume at outlet = 0.950 m3/kg

( x = 89 Pa x 1mm H2O / 10.13 Pa = 8.786mm H2O )

am = Bv

x0137.0

=950.0

786.80137.0

= 0.0417 kg/s

Cross sectional area of column (A) = 0.15 x 0.15 m2

= 0.0225m2

Air mass flow per unit area = A

ma

=21

0225.0

0417.0 −−mkgs

= 1.853kg s-1m-2

Water flow rate per unit area = A

mW

min1kg


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Results:

Description 1 2 3

Packing Density, m-1 110 110 110

Air Flow per Unit Area, kgs-1m-2 1.853 1.853 1.853

Total Cooling Load, kW 0.536 0.964 1.424

Approach to Wet Bulb, K 1.9 3.6 4.8


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Experiment 4: Investigation of the Effect of Ai r Velocity on Wet bulb Approach and Pressure

Drop through the Packing

Results:


Description Unit Air Flow

100 % 75 % 50 % 25 %

Packing Density m-1 110 110 110 110

Air Inlet Dry Bulb, T1 ˚C 29.5 29.9 30.2 31.0

Air Inlet Wet Bulb, T2 ˚C 26.3 26.5 26.8 27.2

Air Outlet Dry Bulb, T3 ˚C 29.4 29.8 30.4 31.8

Air Outlet Wet Bulb, T4 ˚C 28.4 28.7 29.5 30.7


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Calculation:

By taking the data obtained when 100% of air flow was employed,Inlet wet bulb temperature = 26.3oC

Outlet water temperature = 30.6 oC

“Approach to wet bulb” = 30.6 – 26.3 K

= 4.3K

Specific volume of air at outlet (by plotting Air Outlet Dry bulb and Air Outlet Wet bulb on the

Psychometric Chart) = 0.890m3kg-1

Air mass flow rate = Bv

x0137.0

(x = 89 Pa x 1mm H2O / 10.13 Pa = 8.786 mm H2O)

=890.0

786.80137.0

= 0.04049 kgs-1

Air volume flow rate =Bvm

v = 0.04049 x 0.893 m3s-1

v =0.0362m3s-1


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Results:

1 2 3 4

Nominal Velocity of Air, ms-1 1.7027 1.609 1.2387 0.9064

Approach to Wet Bulb, K 4.3 3.6 4.5 5.5

Pressure, mm H2O 0.3949 0.8885 1.4808 1.7769


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Experiment 5: Investigation of the Relationship between Cooling Load and Cooling Range

Results:



Packing Density m-1 - 110 110 110

Air Inlet Dry Bulb, T1 ˚C - 30.5 30.8 30.9

Air Inlet Wet Bulb, T2 ˚C - 27.2 27.3 27.4

Air Outlet Dry Bulb, T3 ˚C - 28.6 29.3 30.9

Air Outlet Wet Bulb, T4 ˚C - 28.3 28.7 29.9

Water Inlet Temperature, T5 ˚C - 32.9 35.4 39.4


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Experiment 6: Investigation of the Effect of Packing Density on the Performance of the

Cooling Tower

Results:

Description UnitColumn

A B C

Packing Density m-1 110 77 200

Air Inlet Dry Bulb, T1 ˚C 30.1 30.2 31.0

Air Inlet Wet Bulb, T2 ˚C 27.4 27.3 27.5

Air Outlet Dry Bulb, T3 ˚C 29.7 30.1 30.4

Air Outlet Wet Bulb, T4 ˚C 29.4 29.6 29.7

Water Inlet Temperature, T5 ˚C 37.7 38.6 37.4


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Results:

1 2 3

Packing Density, m-1 77 110 200

Approach to Wet Bulb, K 5 4.2 3.8


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Experiment 7: Determination of Characteristic Equation of the Packing Characteristic

Column

Results:

Description UnitWater Flowrate (LPM)

1.5 2.0 2.5

Top

Air Outlet Dry Bulb, T3 ˚C 29.9 30.3 30.3

Air Outlet Wet Bulb, T4 ˚C 29.1 29.8 29.7

Water Inlet Temperature, T5 ˚C 39.3 38.5 36.2

Station III

Air Dry Bulb, T8˚C 31.6 32.1 31.7

Air Inlet Wet Bulb, T9 ˚C 29.6 30.0 29.8

Water Temperature, T14 ˚C 35.5 35.6 34.2

Station II

AirDryBulb T10 ˚C 297 305 325


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Calculation:

By taking the data obtained when 2.50 LPM of water was used,

s LPM mw

60min

L1kg 50.2 ××=

= 0.04167 kg s-1

vaB = 0.8980 m3/(kg dry air)

wb = 0.0266 kg water/kg dry air

( )baBa

wv

xm

+=

10137.0

x = 49 Pa x 1mmH2O / 10.13Pa = 4.84 mmH2O

10314.0 −= kgsma

For Air Operating Line,

T1 = Water Outlet temperature, T6 (31.5°C)

T2 = Water Inlet temperature, T5 (36.2°C)


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( )212 1.0 T T T −+ = 35.73̊C

( )212 40 TTT −+ . = 34.32̊C

( )211 4.0 T T T −− = 33.38̊C ( )211 1.0 T T T −− = 31.97̊C

Therefore, from the graph,

1h∆ = 15.92 kJ/kg

2h∆ = 14.23 kJ/kg

3h∆ = 12.93 kJ/kg

4h∆ = 10.09 kJ/kg

( )

( )12

12

TT

hh

G

L aa

−

−=

Referring to the Driving Force Diagram Plotted

L/G =110.84 93.96

36.2 31.5

−

−


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Appendix CComponents Properties and Diagrams


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Appendix DProcess Flow Diagram


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T7

V2

T8

TC

Temp

Controller

Heater

Pump

Drain

V8

V7

Hot Water

Tank

Make-up

Tank

V1

FT1

Flowmeter

T4

T2T1

T3

Air Blower

Air In

Receiver Tank

Orifice

Air Out

Water

Distributor

Test Section

(Packed Column)

T5

T6

V3

V4V6 V5

Atm

DP 1

High Low

WATER COOLING TOWER (MODEL: HE152)

L6 Cooling Tower

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Transcript of L6 Cooling Tower