FACULTY OF ENGINEERINGDEPARTEMENT: CHEMICAL ENGINEERING

PRACTICAL PROGRAMME

CHEMICAL PLANT 3B

REPORT TITLE: COOLING TOWER MASS AND ENERGY BALANCES

Subject Code: CPL301S

Student Full names: Kapiamba Kashala Fabrice

Student Number: 211170283

Lecturer: Mr Andile Thole

Date performed: 04 September 2014

Submission Date: 18 September 2014

Declaration: I, Kapiamba Kashala Fabrice, certify thatthis report is my ownunaided work, except for the assistance

received from the teaching staff. I undertake not to pass this work onto any other student.

Signature:

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Section Possible Mark

Student’s Mark

Title Page 1Structure 2Executive Summary 6Procedure 15Results 38Discussion and Conclusion

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Total 100

Date:

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Table of ContentsList of Symbols....................................................iExecutive Summary.................................................ii

1. Objectives................................12. Procedure.................................1

2.1. Apparatus and Material...................12.2. Procedure................................2

3. Results:..................................23.1. Recorded results.........................2

3.2. Mass balance.............................43.2.1. Mass of water evaporated...............4

3.2.2. Density of air Calculation.............63.2.3. Orifice calculations...................6

3.3. Energy balance...........................64. Discussion................................7

5. Conclusion................................86. References................................9

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List of SymbolsSymbol Description Unit

A Cross sectional area of chamber m2

Ao Combined orifice area m2

h Enthalpy kJ/kgṁ Mass flow rate m3/hrn moles molp Pressure PaPT Total Pressure PaQ Flow rate m3/hrR Ideal gas constant J.K-1.mol−1

T Temperature KVT Total Volume m3

l Height m

Greek Symbols ρ Density Kg/m3

ɛ Efflux coefficient (0.63)

Subscripts

1 Bottom boundary of Column 2 Top boundary of Columna Airs Property of the superheated vapour

at the dry bulb temperaturew Water o Orifice

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Executive SummaryThe purpose of this experiment was to perform mass balance

calculations to determine the mass of water that had evaporated and

to compare with the measured value and to perform energy balance

around the tower to verify if the heat lost by water was gained by

the air.

The calculated mass of water that had evaporated was 0.36 kg/h while

the measured was equal to 0.47 kg/h, the differences was due number

of assumptions such as neglecting the make-up stream for

experimental purposes and due to the mass of water suspended as

bubble was the containers internal walls not counted in the balance

calculations, just to mention a few. From the ideal gas law and

knowing both the dry bulb and wet bulb temperature, the density of

air was calculated and found to be 1.29 kg/m3 which represent a

negligible 7.4 % relative error as compared to the value found using

advanced density calculator at the same conditions (1.201 kg/h).

The energy lost by water obtained was 1245162.24J/h and the energy

gained by air 1083425.68J/h providing a ratio of 1.149 or 13%

relative error, the remaining energy being lost along the way

through conduction and convection from the container’s wall.

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1. ObjectivesThe aim of the experiment to complete mass and energy balances

around the cooling tower in order to determine the mass of water

lost due to evaporation, thus exposing students with the

functioning of a cooling tower.

2. Procedure

2.1. Apparatus and MaterialA computer controlled bench top cooling tower was used and

temperature and pressure were read from the computer and

recorded.

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2.2. Procedure

The level of water in the tank was ensured to be above the

required level.

The Wet Bulb and Dry Bulb of water levels were checked.

A set-point for the water tank was selected.

The water tank was allowed to reach the set point temperature.

The level in the tank was recorded.

An air flow and water flow rate were chosen and allowed to be

reached.

Once steady state was reached the level change was recorded

after 20 minutes.

3. Results:

3.1. Recorded resultsTable 1. Combined Initial and twenty minutes after steady state readings

Initial reading DegreeCelsius

Readingafter 20minutes

water level 135mm 123 mmST-1 ST-1 34.4 34.4ST-2 Temperature of 34.6 34.8

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Figure 1: Schematic of Bench Top Cooling Tower

water inST-3 WB of air entering

the column14.5 14.4

ST-4 DB of air enteringthe column

20.2 20.1

ST-5 WB of air leavingthe column

24.7 24.9

ST-6 DB of air leavingthe column

28.1 28.3

ST-7 Water outlet 26.9 26.7SPD-1

Differentialpressure

23 l/s

SPD Differentialpressure acrosscolumn

0.00

Table 2. Constants and calculated values

OrificecalculationsΔP0 (Pa) 160 PaPatm (Pa) 101325 PaQ (m3/s) 0,0105 m3/sMass of flow air 0,0123 kg/sDry mass flow air 0,012 kg/sMass of vapour out 0,0002 kg/s

0,83 kg/hFor Energy balance

Referencetemperature

0 ºC

Cp water 4200 J/kgKCp dry air 1006 J/kgK

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Cp watervapour

1840 j/kgK

Latent heatofvapoisation

2502000J/kg

Point 2 (Top of thecolumn)

Point 1 (bottom of thecolumn)

T water in 34,6 ºC T water out 26.9ºCAir out DB 28.1ºC Air in DB 20.2ºCAir out WB 24.7ºC Air in WB 14.5ºCWater mass flow 48 kg/h Specific

humidity0,0115 kgwater/kg air

ma2 43,3 kg/h ma1 43,3 kg/hha2 26156 J/kg ha1 21126 J/kgms2 0,83 kg/h ms1 0,5 kg/hhs2 2549840 J/kg hs2 2502000 J/kgl2 0,6 m l1 0 mmw2 48 kg/h mw1 47,7 kg/hhw2 142800 J/kg hw1 117600 J/kg

Table 3. Information related to cooling tower used.

Orifice AreasOrifice 1 diameter 0.78 cmNumber of orifices (1) 16Orifice 2 diameter 1.18 cmNumber of orifices (2) 3Minus steel pipes blocking largeorifice’sDiameter of pipe 0.61 cmNumber 3Chamber dimensions 164 mm by 80 mm

3.2. Mass balance

3.2.1. Mass of water evaporated.

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General equation: Lin+Mw+Gin=Gout+L+B+W (Massconservation).

Assumptions: for experimental purposes the following termswere ignored:

Mw, L, W.

Knowing the outlet conditions (DB=28.1ºC and WB=24.7ºC), thecomposition can be found on the psychrometric chart:

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⇒Lin+Gin=B+G0⇒48+43.2=B+G0⇒B+G0=91.2

kgh

RH=4gramsH2OkgDryAir =

0.004kgH2O1kgDryAir

Mass fraction of water: 0.004

0.004+1=0.00398

Mass of water evaporated (calculated): (0.00398)*91.2=0.36 kg/h

Mass of water out (ma1) = 48-0.36= 47.64 kg/h

From chamber dimensions and water levels recorded after twenty minutes the volume of lost (evaporated) water is:

Volume= (Height 1-Height 2) ˣ Area

V=(135mm−123mm )×164mm×80mmV=157440mm3

Mass of water evaporated(actual)=

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157440mm3

1dm3 1kg 60min =0.47k

g/h20minutes

1000000mm3

1dm3

1 h

3.2.2. Density of air CalculationFrom the ideal gas law PV= nRT, the density of was derived and foundto approximately equal to:

ρ=1+w

(1

Mmair+w

Mmwater )×8.316×273K+TwB

Patm+ΔP0

ρ=1+0.004

(10.029

+0.0040.018

)×8.316×273K+(24.7+273)

101325+160=1.29kg /m3

Online calculator

Literature value at STP: 1.253 kg/m3 (Perry’s handbook, 2008)

3.2.3. Orifice calculationsOrifice calculations are based on Bernoulli’s principle which statesthat there is a relationship between the pressure of the fluid and

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the velocity of the fluid. When the velocity increases, the pressuredecreases and vice versa.

3.3. Energy balance

By applying the equation Q-P = HOutlet - HInlet and assuming that P= 0 and where Q’s value is limited due to heat transfer between the unitand its surrounding. The equation becomes:

HOutlet = HInlet

The derived energy balance below can be used in order to determine if the energy taken in by the air is equal to that supplied by the water

LHS:(43.3×26156−43.3×21126 )+(0.83×2549840−0.5×2502000 )+9.8(0.6(43.3+0.83 )−0)=1083425.68J /h

RHS:

(48×142800−47.7×117600)+9.8(0.6×48−0)=1245162.24J /h

Ratio:RHSLHS

=1245162.241083425.68

=1.149

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Q=ϵA01

√1−(A0

A )2 √2 (p−p0)

ρ

(ma2ha2−ma1ha1)+(ms2hs2−ms1hs1 )+g(l2 (ma2+ ms2)−l1(ma1+ms1))

¿ (mw2hw2−mw1hw1)+g (l2 mw2−l1 mw1 )

4. DiscussionThe experiment aimed to firstly analyse the mini cooling tower at

hand by performing the mass balance to determine the mass of water

that had evaporated during the time interval (20 minutes) under

study and compare the calculated value with the actual value which

was the difference in volumes, having measured the heights before

and after the experiment, secondly to perform energy balance

calculations to make sure the energy gained by air is equal to the

energy supplied by the water.

After mass balance calculations, the mass of water was found to be

equal to 0.36 kg/h while the actual mass, calculated from the volume

difference was found to be equal to 0.47 kg/h. From the general mass

balance equation, the make-up stream was neglected for experimental

reasons and since there was leakage which is often used as a purge

stream to reduce concentration of chemicals.

Considering the above assumptions and the fact chamber dimensions

could not accurately be measured nor the computer readings verified,

and there was water drops suspended on the inner wall of the

container (not counted on mass balance calculations) a significant

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relative error (compounded errors) of 23.4% occurred in calculating

the amount of water that evaporated within twenty minutes.

Knowing the dry bulb and Wet bulb temperature of air entering the

tower, the specific humidity of air was read from the psychrometric

chart and from the ideal gas law, the density of air was calculated

and found to be equal to 1.29 kg/m3, this representing a negligible

relative error of 7.4%, as compared to the calculated from an online

calculator 1.201 kg/m3.

The energy balance calculations revealed that the energy supplied by

water (1245162.24 J/h) was not totally gained by dry air (1083425.68

J/h), representing a ratio of 1.149, this energy amounting to

161736.56 J/h was lost directly to the surrounding through the wall

of the container by conduction then convection to the atmosphere.

5. Conclusion

The aim of the experiment was to complete mass and energy balances

around the cooling tower in order to determine the mass of water

lost due to evaporation, thus exposing students with the functioning

of a cooling tower.

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Referring to the above, the experiment was carried out successfully.

It was found that the mass of the water that had evaporated

calculated by accounting the volume difference after twenty minutes

and knowing the wet bulb and dry bulb temperature was were slightly

different due to minor errors.

The energy provided by hot water stream was not completely gained by

the air, since some heat (energy) was lost to the surrounding from

the container’s wall.

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6. References

Coulson , J.M and Richardson , F.1999. Chemical Engineering “ vol

1 , 3rd edition: BUTTERWORTH HEINEMANN .

Gerald Bowen Hill, E. J. Pring, Peter David Osborn, William

Stanford. (1990). Cooling towers: principles and practice. Technology &

Engineering: Butterworth-Heinemann.

Perry, R. 1997.Perry’s Chemical Engineers’ Handbook, 6th ed. Mc Graw

Hill.

"Chemical Engineering Practical manual",Dep. of Chemical

engineering.

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