4 cooling system dynamics

Cooling System
Dynamics
Cooling System
Dynamics
Customer Seminar
November 23-25, 2004;
Vienna

For good efficiency the system has to
Øproduce cold water over the cooling tower
Øabsorb waste heat from the process
at good heat transfer conditions
There is one common denominator for all cooling systems :
The duty is to reject waste heat
PrinciplesPrinciples

3
evaporationevaporation (losses)(losses)
(+ spray(+ spray--losses)losses)
drift eliminatorsdrift eliminators
cooling tower fillcooling tower fill
makeup watermakeup water
cooling tower basincooling tower basin
cold watercold water
cooling water returncooling water return
blowdownblowdown
(losses)(losses)
UNITSUNITS
blowdownblowdown
(losses)(losses)

range = T 1 - T 2
approach = T 2 - T 3
to cool a designated
quantity of warm water
with a specified
warm water temperature
at a specified
wet bulb temperature
to a designated
cold water temperature
T 1
T 3
T 2

5
Maximum Heat Transfer
… is a function of:
Ø air temperature
Ø moisture content of air (wet bulb temperature - WBT)
Ø water distribution
Ø air / water contact
WBT represents the lowest temperature to which water
can theoretically be cooled
Practically, the water temperature approaches the WBT,
but cannot be achieved

ØDesign/Construction
ücounterflow/crossflow
ünatural draft/mechanical draft
üvolume - recirculation ratio …
ØInspection & Maintenance
üoperational, mechanical, economical
Ø“Chemical Equipment“
üwater treatment
üconditioning of cooling water
ØMechanical Equipment
ütower fill, pumps, filters
üexchangers, tubing ..

ØØ Flow of air byFlow of air by
üNatural draft
–by difference of density of air
üMechanical draft
–by fans on top of tower or by impellers on
air inlet
ØØ Direction of air flowDirection of air flow
üCounterflow
üCrossflow
Cooling Tower - DesignCooling Tower - Design

11
Tower size is function of:
ØCooling range (hot-cold water temperature)
ØWet bulb temperature (WBT)
ØApproach to WBT (cold-WB temperature)
ØQuantity of water to be cooled
ØAir velocity through the cell
ØTower height

from GEA-comp.
1818
cooling range
approach
1414 1515 1616 1717 1818 1919 2020
1414 1515 1616 171788 99 1111 1212 131366 77 1010
55 6633 44 8877
0.60.6
0.70.7
0.80.8
0.90.9
1.01.0
1.11.1
1.21.2
1.31.3
1.41.4
1.51.5
plot area factor
approach
cooling range

Mechanical draft cooling tower -
influence of design parameters on the plot area
from GEA-comp.
Basis Plot area factor
Wet bulb temperature 17 °C 1.0
Cooling range 12 K 1.0
Approach 5 K 1.0
standard water flow = 1,000 m³/h
standard plot area = 100 m²
Example Plot area factor
Wet bulb temperature 16.5 °C ( instead 17 from basis) 1.03
Cooling range 10 K ( instead 12 from basis) 0.9
Approach 6.5 K ( instead 5 from basis) 0.79
Total plot area factor 1.03 x 0.9 x 0.79 = 0.732
Required plot area 100 m² x 0.732 = 73.2 m²

14
WBT depends on temperature and humidity of air
5
10
15
20
25
30 40 50 60 70 80 90 100
% relative humidity
wet bulb temperature [°C]
air temperature 30°C
from VDI:
"Kühlgrenze und
relative Luftfeuchte"
valid for
985 to 1020 mbar.

airair / water in/ water in counterflowcounterflow air / water inair / water in crossflowcrossflow

in comparisonin comparison
counterflow crossflowcounterflow crossflow
area demand
investment
operating costs
wet air - recirculation
approach
--
--
--
--
--
++
++
++
++
++

17
For maximum air / water contactFor maximum air / water contact
Splash packing
decks of splash fills -
breaks water into small droplets.
2 - 3 m² surface per m³ of fill
Film packing
water adheres to packing surface -
no droplets formed
60 m² (and more) surface per m³ of fill
Cooling tower fill
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact

18
For maximum air / water contactFor maximum air / water contact
Splash packing
decks of splash fills -
breaks water into small droplets.
2 - 3 m² surface per m³ of fill
Film packing
water adheres to packing surface -
no droplets formed
60 m² (and more) surface per m³ of fill
Cooling tower fill

19
Water distribution
A lot of different systems:
orifices, spray bars, spray nozzles ..
Thermal capability of a cooling tower strongly
depends on
Equal distribution of waterEqual distribution of water
- over the total area
of cooling tower fill

20
Water-distribution
in cooling tower

Important for design and operation ofImportant for design and operation of
cooling towerscooling towers
ØØ In "VDIIn "VDI--KühlturmregelnKühlturmregeln" defined as "" defined as "LuftzahlLuftzahl""
λ
GLGL
GWGW==
Ø according to USA - standards defined as
ratioratio
GL flow of airGL flow of air [kg/h][kg/h]
GW flow of waterGW flow of water [kg/h][kg/h]
LL
GG
liquidliquid
gasgas==

22
The operating of a tower is then functionThe operating of a tower is then function
of the Liquid/Air ratio (of the Liquid/Air ratio (L/GL/G))
Normal limits to the two flows are:
Ø L < 15000 kg/h m2
Above, bad dispersion - big droplets
ØG < 9000 kg/m2 h
Above, high power consumption

23
The operating of a tower is then functionThe operating of a tower is then function
of the Liquid/Air ratio (of the Liquid/Air ratio (L/GL/G))
Normal limits to the two flows are:
Ø L < 15000 kg/h m2
Above, bad dispersion - big droplets
ØG < 9000 kg/m2 h
Above, high power consumption

Ambient Air -
Water Content
at 100 %
Humidity -- 10°C10°C 2,36 g/m³2,36 g/m³
00 4,82 g/m³4,82 g/m³
1010 9,35 g/m³9,35 g/m³
2020 17,15 g/m³17,15 g/m³
3030 30,10 g/m³30,10 g/m³
4040 50,67 g/m³50,67 g/m³
5050 82,26 g/m³82,26 g/m³
00
1010
2020
3030
4040
5050
6060
7070
8080
--1010 00 1010 2020 3030 4040 5050
g H2O per m³ of air
temperature [°C]

Density of Humid Air
1,000
1,100
1,200
1,300
0 10 20 30 40 50
air-density [kg/m³]
temperature [°C]
dry air
50 % rel
100 % rel
valid for 1013 mbar
from VDI

Density of Dry Air
temperature [°C]
1,000
1,100
1,200
1,300
0 10 20 30 40 50
air-density [kg/m³]
dry air
50 % rel
100 % rel
valid for 1013 mbar
from VDI

27
Air/water ratio
Density of air versus temperature:Density of air versus temperature: [kg/m³][kg/m³]
Depending onDepending on ratio of massesratio of masses air / waterair / water
Flow of air = m³/h x density
Flow of water = m³/h x density
0°C 1,2929
20°C 1,2047
30°C 1,1679
(40°C 1,1277)
"loss" on air 0°C 30°C is about 10 %"loss" on air 0°C 30°C is about 10 %

Water- distribution to cooling tower cells
waterflow
of design
90 % 100 % 110 %
results in
approx.
8 %
reduction in
thermal capability
instead of
100 % 100 % 100 %

”Hot Weather Example"”Hot Weather Example"
ØØ Cooling system:Cooling system:
water to tower 1,000 m³/h
t35/ 25 °C 10 °C
approach 4 °C
concentration factor 3.0
temperature of makeup 15 °C
operating at: 30°C, 45 % rel.humidity

Questions:Questions:
Ø wet bulb temperature
Ø evaporation
Ø makeup
Ø temperature drop by evaporation
Ø consequence for temperature drop, when
changing the waterflow to the tower to
900 / 1100 / 1200 / 1300 m³/h
Ø consequence of double makeup quantity
-on concentration factor, on cold water temperature
Ø quantity of makeup to lower cold water
temperature by 1°C
”Hot Weather Example"”Hot Weather Example"

ØØ Wet bulb temperature 21 °CWet bulb temperature 21 °C
% relative humidity
45 %45 % relrel./30 °C./30 °C
55
1010
1515
2020
2525
3030 4040 5050 6060 8080 9090 100100
wet bulb temperature [°C]
7070
”Hot Weather Example“ - results”Hot Weather Example“ - results
from charts like
shown here) or
from complete
psychrometric chart

ØØ Evaporation losses & makeupEvaporation losses & makeup
for 30 °C, 45 % rel humidity:
EV = 10 x [(30 - 1.6667) x 0.0013 +
0.1098]
= 1.47 % RR = 14,700 kg/h
3
MU = 14,700 x = 22,050 kg/h
3 - 1

water
heat
rejection
temp.
drop
ØØ Temperature drop by evaporationTemperature drop by evaporation
Water entering tower 1,000,000 kg/h
Evaporation losses 14,700 kg/h
Remaining cold water 985,300 kg/h
Evaporation (kg/h) x Heat of vaporization (kJ/kg)
14,700 x 2,260 = 33,222,000 kJ/h
33,222,000 kJ/h out of remaining water flow of
985,300 kg/h
33,222,000
985,300 33.7kJ/kg
Results in a temperature drop of ~ 8.1 K~ 8.1 K
==

ØØ Consequences of changingConsequences of changing waterflowwaterflow overover
towertower
Evaporaton losses = 14,700 kg/h
Rejected heat of vaporisation = 33,222,000 kJ/kg
flow remaining temperature
over tower cold water drop
900 m³/h 885,300 kg/h 9.0 K
1,000 m³/h 985,300 kg/h 8.1 K
1,100 m³/h 1,085,300 kg/h 7.3 K
1,200 m³/h 1,185,300 kg/h 6,7 K
1,300 m³/h 1,285,300 kg/h 6,2 K

ØØ Consequences of makeupConsequences of makeup--quantity onquantity on
water temperaturewater temperature
985.3 x 25 + 22.05 x 15
985.3 + 22.05
985.3 x 25 + 44.1 x 15
985.3 + 44.1
985.3 x (26 - 25) = MU x (25 - 15)
MU = 98,5 m³/h
= 24.8 °C
= 24.6 °C
design
makeup
double
makeup
temperatures
after mixing
25 °C cold water
with
15 °C makeup
makeup
for - 1°C
To lower the cold water temperature by 1°C, the
make up quantity would have to increase from
22.05 to 98,5 m³/h

ØØ Consequences of makeupConsequences of makeup--quantity onquantity on
concentrationconcentration--factorfactor
22,05
= 3,0
22,05 - 14,7
44,1
= 1,50
44,1 - 14,7
98,5
= 1,175
98,5 - 14,7
design
makeup
double
makeup
makeup
for - 1°C
11,4 - times
increase in
blowdown

Calculation of Evaporation LossesCalculation of Evaporation Losses
Evaporation constant:Evaporation constant:
Range Relative humidity
< 30 % 30 - 90 % > 90 %
% EV = T x [(T - 1.6667) x km + 0.1098)]
∆∆T = range [°C] (T of warm water - T of cold water )
T = ambient air Temperature [°C] ( dry bulb )
km = evaporation constant
> 7,2 °C> 7,2 °C 0,00130,0013 0,00130,0013 0,00130,0013
3,93,9 -- 7,2 °C7,2 °C 0,00290,0029 0,00190,0019 0,00100,0010
< 3,9 °C< 3,9 °C 0,00580,0058 0,00320,0032 0,00100,0010

38
EvaporationEvaporation
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
-5 0 5 10 15 20 25 30
Evaporation losses
[as % of recirculation rate]
Ambient air temperature [°C]
range = 10 °C
range = 5 °C
valid for a relativevalid for a relative
humidity of 30humidity of 30 -- 90 %90 %
Air T Range
10°C 5°C
30°C 1,47 0,82
25°C 1,40 0,77
20°C 1,34 0,72
15 °C 1,27 0,68
10 °C 1,21 0,63
5°C 1,14 0,58
0°C 1,08 0,53
-5°C 1,01 0,49

ØØ According "VDIAccording "VDI KühlturmregelnKühlturmregeln""
Gwo =
Gw x cw x ( tw1 - tw2)
i2 - i1
x2 - x1
- cw x tw2
Gwo evaporated quantity of waterflow [kg/h]
Gw waterflow (over tower) [kg/h]
cw specific heat of water [kcal/kg °C]
tw1 temperature of water entering the cooling tower [°C]
tw2 temperature of chilled water entering cooling tower basin [°C]
i1 enthalpy of humid air with a content of 1 kg of dry air,
entering cooling tower [kcal/kg]
i2 the same, over water distribution deck [kcal/kg]
x1 content of water vapor, based on 1 kg of dry air, [kg/kg, g/kg]
entering cooling tower
x2 the same, over water distribution deck [kg/kg, g/kg]
Calculation of Evaporation LossesCalculation of Evaporation Losses

40
EvaporationEvaporation
• Outlet-Inlet air
Moisture
difference

Drift losses :
with "old type" eliminators < 0.2 % of recirc. rate
with "high efficiency" eliminators < 0.02 % of recirc. rate
Drift eliminators
Example of high efficiency eliminators :
Cooling Tower – Water LossesCooling Tower – Water Losses

Fan cylinders
("old type")
fan cylinders
"Venturi - Typ"
(velocity regain cylinder)
5 - 8 %
higher air flow
at
same energy demand

Factors to be determined:
Ø flow of water
Ø temperature of cooling water return
Ø temperature of cold water discharge
Ø wet bulb temperature
Ø flow of air
Acceptance testing of
a cooling tower Based on: "VDI Kühlturmregeln", DIN 1947
enough readings
for long enough time !!
(for details see local standards !)
Cooling Tower – Performance TestCooling Tower – Performance Test

44
Typical sketch of the measurement’s locations
m/s at
Cooling Tower – Performance TestCooling Tower – Performance Test

45
Cooling Tower - EconomicsCooling Tower - Economics
The tower performance affect directly the
economics of each producing plant
Main units that suffer for insufficient cold
temperature are condensers, compressors
Higher temperature means more fuel to produce
steam, more work to compress, less final product

46
Cooling Tower - NormsCooling Tower - Norms
Water loading = 5000 - 13000 kg/h m2
Air loading = 6500 - 9000 kg/h m2
L/G ratio = 0,75 - 1,5
Approach = 3 - 5 °C
Tower operating = 80% - 120% of the design
Fan pressure drop < 5 cm
Fan blades pitch = ± 3° (Summer +3°, Winter - 3°)
Air velocity = 1,5 - 2,0 m/s (1,2 - 1,8 natural-draft)
Design WBT = avg. June-September (<5% exceeded)
Minimum contact = 4900 - 7300 kg/(hr)(m2 ground area)
Nd (KaV/L) = 0,5 - 2,5
Drift losses
• old type = < 0,2 % circulating rate
• new = < 0,02 % circulating rate
Air/water contact
• splash fill = 2 - 3 m2
surface per m3
fill
• film = >60 m2
surface per m3
fill

Monitoring by lab
makeup water
system water
microbio
chemicals
upsets (product leakages ..)
Monitoring performance
corrosion rates
test-heat exchangers
deposit-monitoring
exchanger performance
information systems
economy
Cooling System - MonitoringCooling System - Monitoring

Concentration factor versus makup water and
blowdown quantity
0
50
100
150
200
250
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
makeup water [m³/h]
concentration factor
evaporation
blow down
example :
recirculation 5,000 m³/h
volume 2,000 m³
evaporation 0.8 %
Cooling System - OperationCooling System - Operation

Concentration factor versus holdingtime index
0
20
40
60
80
100
120
140
160
1 2 3 4 5
HTI [h]
concentration factor
example :
recirculation 5,000 m³/h
volume 2,000 m³
Cooling System - OperationCooling System - Operation

Water treatment chemicals versus constant /
fluctuating concentration factor
What is the consumption of treatment chemical per year,
based on a concentration factor of nc = 3.0 ?
What is the consumption of treatment chemical per year,
if the concentration factor is
nc = 2,0 over 4 month of the year,
nc = 3,0 over 4 month of the year,
nc = 4,0 over 4 month of the year ???
example :
cooling system, recirculation 5,000 m³/h
evaporation 0,8 %
treatment program 50 ppm
Example 1Example 1

nc = 3.0 (all year) nc = 2.0
3.0 4 month each
4.0
blowdown = 20 m³/h blowdown = 40 m³/h
20 m³/h
13 m³/h
average 24.3 m³/h
treatment chemical treatment chemical
8,760 kg/a 10,660 kg/a
example :
cooling system, recirculation 5,000 m³/h
evaporation 0,8 %
treatment program 50 ppm
of 22 %
Water treatment chemicals versus constant /
fluctuating concentration factor
Example 1Example 1

electrical output 4.4 x105 kW
pressure in condenser (Vacuum) 0.1 bar
specific heat consumption 8.38 x103 kJ/kWh
energy costs 1.0 US$/106 kJ
operating hours 8760 h/a
problem: increase of condensate-temperature 3°C,
increase in specific heat consumption 0.67 %
additional fuel costs: 216,000.- US$ per year
by: James L. Willa
Example: thermal power plant
Example 2Example 2

For estimating the power- demand in Watt :
kg x 9.81 x m
W =
sec
kg : mass of liquid
( consider specific density if flow is given by volume )
9.81 : gravity
m : pressure ( expressed as pumping height )
sec : time ( which is allowed for transport of the given mass
pumping costs !
Estimation of power-demand for pumps
result must be corrected
by given pump-efficiency
( if efficiency is not known, assume ~ 80 % )
Example 3Example 3

Example :
Refinery, vacuum distillation,
consuming 1,600 m³/h cooling water,
200 m³/h out of that for overhead condenser.
cooling water-pressure (ex pumps) : 5.5 bar
Change :
- install a 2.5 bar booster pump for overhead condenser
- decrease cw-pressure ex pumps (total system) to 4.4 bar
Example 3Example 3

1,600,000 x 9.81 x 55
3,600 x 0.8
resulting annual saving of :
( 300 kW - 240 kW) - 17,5 kW = 42.5 kW per operation hour
~ 370,000 kW per year
5.5 bar
operation = 300 kW
1,600,000 x 9.81 x 44
3,600 x 0.8
4.4 bar
operation = 240 kW
200,000 x 9.81 x 25
3,600 x 0.8
booster
pump = 17.5 kW
Example 3Example 3

Pressure drop in cooling water lines
Increase of pressure drop caused by incrustations, example DN 100
1,0
0,5
0,4
0,3
0,2
0,1
0,5 1,0 2,0 3,0
water velocity [m/sec]
pressure drop [bar]
14 m³/h
29 m³/h
43 m³/h
57 m³/h
72 m³/h
86 m³/h
per 100 m
clean
1.5 m/sec, 43 m³/h,
0.28 bar pressure drop/100 m
5 mm incrustation:
same flow rate,
0.45 bar pressure drop/100 m
10 mm incrustation:
same flow rate,
0.7 bar pressure
drop/100 m
ExampleExample

Pressure drop - example
A compressor station for natural gas produces 88,000,000 kJ/h
of waste heat.
It is serviced by a 1,500 m³/h cooling system,
Range over tower : 14 °C,
Cooling water main lines : diameter 600 mm, length 1,000 m
At design flow : water velocity ~ 1.5 m/sec
pressure drop 0.022 bar/100 m
By corrosion products & deposits
a layer of 25 mm has built up in the returnwater line.
Example 4Example 4

Questions:
1) if design flow is kept by higher pumping pressure -
(water velocity will be ~ 1.8 m/sec, pressure drop 0.033 bar/100 m)
What will be the additional pumping costs per year ??
(assume: US$ 0,07/kW, efficiency of pumps = 80 %)
2) if higher pressure drop is not compensated, the water flow will
decline to ~ 1,200 m³/h.
What will be the increase in return water temperature ?
Example 4Example 4

1) additional pumping costs
dP of clean tube: 0.022 bar/100 m ...... 0.22 bar/1000 m
dP of "dirty" tube: 0.033 bar/100 m ..... 0.33 bar/1000 m
Difference: 0.11 bar/1000 m
0.11 bar equals 1.1 m in pumping height
1,500,000 kg/h of water x 1.1 m = 1,650,000 kpm
= 4.5 kW
considering 80 % pump-eff. 5.6 kW
8,760 operating hours / year 49,144 kWh/y
price of US$ 0.07 3,440 US$/y
Example 4Example 4

2) increase in return water temperature
Input of waste heat to cooling water : 88,000,000 kJ/h
Using 1,500 m³/h (1,500,000 kg/h) of water :
88,000,000
= 58.7 kJ/kg
1,500,000
4.19 kJ/kg for a temperature change of 1 °C ....... 14 °C
Using 1,200 m³/h (1,200,000 kg/h) of water :
88,000,000
= 73.3 kJ/kg
1,200,000
4.19 kJ/kg for a temperature change of 1 °C ....... 17.5 °C
increase : 3.5 °C
Example 4Example 4

62
ECONOMICSECONOMICS
AnyAny decrease in the heat transportdecrease in the heat transport (heat exchange)(heat exchange)
of a cooling system results in:of a cooling system results in:
êê change of Temperature Differencechange of Temperature Difference
êê increase of Condensate Temperatureincrease of Condensate Temperature
êê higher Condensate Pressurehigher Condensate Pressure
êê a Loss of Efficiencya Loss of Efficiency
êê less Production Outputless Production Output
The Economical Impacts of these ChangesThe Economical Impacts of these Changes
are often not quantified !are often not quantified !
Power plant
• Influence on Economy and Efficiency

63
ECONOMICSECONOMICS
Fertilizer plant
Ammonia plant designAmmonia plant design
-- 100% production: 1520 tons/d ($27/t)100% production: 1520 tons/d ($27/t)
-- Methanol fuel for steam boiler ($ 0,09/Methanol fuel for steam boiler ($ 0,09/NmcNmc))
+ 1°C CWT = 250+ 1°C CWT = 250 NmcNmc/h over consumption/h over consumption
Tower DesignTower Design
-- RR:RR: 36000 m36000 m33/hr/hr
-- Range:Range: 10°C10°C
-- DBT:DBT: 27°C27°C
-- WBT:WBT: 22°C22°C
-- L/G:L/G: 1,041,04
-- Fan HP:Fan HP: 1500 HP1500 HP
-- TTMUPMUP:: 20°C20°C
-- CWT:CWT: 30°C30°C
-- HWT:HWT: 40°C40°C
-- Approach:Approach: 8°C8°C
-- LoadLoad 360E6 kcal/hr360E6 kcal/hr
-- Evaporation:Evaporation: 445.738 kg/hr445.738 kg/hr
-- Capability:Capability: 100%100%
ActualActual
36000 m36000 m33/hr/hr
7°C7°C
20°C20°C
17°C17°C
1,041,04
1500 HP1500 HP
18°C18°C
26°C26°C
33°C33°C
9°C9°C
252e6 kcal/h252e6 kcal/h
290.555 kg/hr290.555 kg/hr
90,3%90,3%
Actual to DesignActual to Design
36000 m36000 m33/hr/hr
10°C10°C
27°C27°C
22°C22°C
1,041,04
1500 HP1500 HP
20°C20°C
30,7°C30,7°C
40,7°C40,7°C
8,7°C8,7°C
445.738 kg/hr445.738 kg/hr

64
ECONOMICSECONOMICS
Fertilizer plant
Ammonia plant designAmmonia plant design
-- 100% production: 1520 tons/d ($27/t)100% production: 1520 tons/d ($27/t)
-- Methanol fuel for steam boiler ($ 0,09/Methanol fuel for steam boiler ($ 0,09/NmcNmc))
+ 1°C CWT = 250+ 1°C CWT = 250 NmcNmc/h over consumption/h over consumption
Tower DesignTower Design
-- RR:RR: 36000 m36000 m33/hr/hr
-- Range:Range: 10°C10°C
-- DBT:DBT: 27°C27°C
-- WBT:WBT: 22°C22°C
-- L/G:L/G: 1,041,04
-- Fan HP:Fan HP: 1500 HP1500 HP
-- TTMUPMUP:: 20°C20°C
-- CWT:CWT: 30°C30°C
-- HWT:HWT: 40°C40°C
-- Approach:Approach: 8°C8°C
-- LoadLoad 360E6 kcal/hr360E6 kcal/hr
-- Evaporation:Evaporation: 445.738 kg/hr445.738 kg/hr
-- Capability:Capability: 100%100%
ActualActual
36000 m36000 m33/hr/hr
7°C7°C
20°C20°C
17°C17°C
1,041,04
1500 HP1500 HP
18°C18°C
26°C26°C
33°C33°C
9°C9°C
290.555 kg/hr290.555 kg/hr
90,3%90,3%
Actual to DesignActual to Design
36000 m36000 m33/hr/hr
10°C10°C
27°C27°C
22°C22°C
1,041,04
1500 HP1500 HP
20°C20°C
30,7°C30,7°C
40,7°C40,7°C
8,7°C8,7°C
445.738 kg/hr445.738 kg/hr
Actual capability = 90,3%Actual capability = 90,3%
CWT is 0,7°C higher / designCWT is 0,7°C higher / design
Money loss:Money loss:
-- FuelFuel
-- 250x0,7x0,09x8700 = $137.000/y250x0,7x0,09x8700 = $137.000/y
-- Production loss as 0,5%Production loss as 0,5%
-- 76x27x365 = $749.000 / y76x27x365 = $749.000 / y

Know your design !!
cooling tower characteristics,
water flow,
wet bulb temperature, approach,
cold water temperature,
return water temperature,
flow rates,
equipment,
pressure, pressure drop,
temperatures,
materials,
losses & makeup
........
.. and compare it with reality !!

Know your process !!
high temperatures,
high heat flux,
low water velocities,
process contaminants (to water),
technical influence of coolingwater,
economical influence of coolingwater,
specific figures,
bottlenecks,
safety considerations,
importance,
........
.. and watch it !!

Know your partner !!
production,
inspection,
maintenance,
purchasing department,
cost controlling,
environmentalist,
project department,
laboratory,
health & safety,
........
with Nalco – the people you trust to deliver results
.. and help your partner to cooperate !!

4 cooling system dynamics

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