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
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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