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 eliminatorscooling tower fillcooling tower fill

makeup watermakeup watercooling tower basincooling tower basin

cold watercold water

cooling water returncooling water return

blowdownblowdown(losses)(losses)

UNITSUNITS

blowdownblowdown(losses)(losses)

range = T 1 - T 2approach = T 2 - T 3

to cool a designatedquantity of warm water

with a specifiedwarm water temperature

at a specifiedwet bulb temperature

to a designatedcold 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/Constructionücounterflow/crossflowünatural draft/mechanical draftüvolume - recirculation ratio …

ØInspection & Maintenanceüoperational, mechanical, economical

Ø“Chemical Equipment“üwater treatmentüconditioning of cooling water

ØMechanical Equipmentütower fill, pumps, filtersüexchangers, tubing ..


7

Cooling TowerCooling Tower

8


9


ØØ Flow of air byFlow of air byüNatural 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 flowüCounterflow

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.

1818wet bulb temperaturecooling rangeapproach

1414 1515 1616 1717 1818 1919 20201414 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

wet bulb temperature


Mechanical draft cooling tower -influence of design parameters on the plot area

from GEA-comp.

Basis Plot area factorWet bulb temperature 17 °C 1.0Cooling range 12 K 1.0Approach 5 K 1.0

standard water flow = 1,000 m³/hstandard plot area = 100 m²

Example Plot area factorWet bulb temperature 16.5 °C ( instead 17 from basis) 1.03Cooling range 10 K ( instead 12 from basis) 0.9Approach 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 undrelative Luftfeuchte"

valid for985 to 1020 mbar.

airair / water in / water in counterflowcounterflow air / water in air / water in crossflow crossflow


in comparisonin comparison

counterflow crossflow counterflow 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 distributionA 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 areaof cooling tower fill


20


Water-distribution in cooling tower

Important for design and operation of Important for design and operation of cooling towerscooling towers

ØØ In "VDIIn "VDI--KühlturmregelnKühlturmregeln" defined as "" defined as "LuftzahlLuftzahl""

λGLGLGWGW==

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]

LLGG

liquidliquidgasgas==


22


The operating of a tower is then function The 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 function The 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/m³ 2,36 g/m³

00 4,82 g/m³ 4,82 g/m³ 1010 9,35 g/m³ 9,35 g/m³ 2020 17,15 g/m³ 17,15 g/m³ 30 30 30,10 g/m³ 30,10 g/m³ 4040 50,67 g/m³ 50,67 g/m³ 5050 82,26 g/m³ 82,26 g/m³

0010102020303040405050606070708080

--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 on Depending on ratio of massesratio of masses air / waterair / water

Flow of air = m³/h x densityFlow of water = m³/h x density

0°C 1,292920°C 1,204730°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 inapprox.

8 %reduction in

thermal capability

instead of100 % 100 % 100 %


”Hot Weather Example" ”Hot Weather Example"

ØØ Cooling system:Cooling system:water to tower 1,000 m³/ht35/ 25 °C 10 °C

approach 4 °Cconcentration factor 3.0temperature of makeup 15 °C

operating at: 30°C, 45 % rel.humidity

Questions:Questions:Ø wet bulb temperatureØ evaporationØ makeupØ temperature drop by evaporationØ 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 humidity45 % 45 % relrel./30 °C./30 °C

55

1010

1515

2020

2525

3030 4040 5050 6060 8080 9090 100100

wet bulb temperature [°C]air temperature 30°C





7070

”Hot Weather Example“ - results ”Hot Weather Example“ - results

from charts like shown here) orfrom completepsychrometric 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

3MU = 14,700 x = 22,050 kg/h

3 - 1


water

heatrejection

temp.drop


ØØ Temperature drop by evaporationTemperature drop by evaporation


Water entering tower 1,000,000 kg/hEvaporation losses 14,700 kg/hRemaining 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

33,222,000985,300 33.7kJ/kg

Results in a temperature drop of ~ 8.1 K~ 8.1 K

==

ØØ Consequences of changing Consequences of changing waterflowwaterflow over over

towertowerfor 30 °C, 45 % rel humidity:

Evaporaton losses = 14,700 kg/hRejected 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 K1,000 m³/h 985,300 kg/h 8.1 K1,100 m³/h 1,085,300 kg/h 7.3 K1,200 m³/h 1,185,300 kg/h 6,7 K1,300 m³/h 1,285,300 kg/h 6,2 K


ØØ Consequences of makeupConsequences of makeup--quantity on quantity on

water temperaturewater temperature


985.3 x 25 + 22.05 x 15985.3 + 22.05

985.3 x 25 + 44.1 x 15985.3 + 44.1

985.3 x (26 - 25) = MU x (25 - 15)

MU = 98,5 m³/h

= 24.8 °C

= 24.6 °C

designmakeup

doublemakeup

temperaturesafter mixing25 °C cold water

with 15 °C makeup

makeupfor - 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 on quantity 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

designmakeup

doublemakeup

makeupfor - 1°C

11,4 - timesincrease 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 0,00130,00133,9 3,9 -- 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 relativehumidity of 30 humidity of 30 -- 90 %90 %

Air T Range10°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 "VDI According "VDI KühlturmregelnKühlturmregeln""

Gwo =Gw x cw x ( tw1 - tw2)

i2 - i1x2 - 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 towerx2 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

atsame 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 readingsfor 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 watersystem watermicrobio chemicalsupsets (product leakages ..)

Monitoring performance

corrosion ratestest-heat exchangersdeposit-monitoringexchanger performanceinformation systems economy

Cooling System - MonitoringCooling System - Monitoring

Concentration factor versus makup water andblowdown 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/hvolume 2,000 m³evaporation 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/hvolume 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 isnc = 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 each4.0

blowdown = 20 m³/h blowdown = 40 m³/h 20 m³/h13 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 mW =

sec

kg : mass of liquid( consider specific density if flow is given by volume )

9.81 : gravitym : 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 correctedby 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

Estimation of power-demand for pumps

1,600,000 x 9.81 x 553,600 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 baroperation = 300 kW

1,600,000 x 9.81 x 443,600 x 0.8

4.4 baroperation = 240 kW

200,000 x 9.81 x 253,600 x 0.8

boosterpump = 17.5 kW

Example 3Example 3

Estimation of power-demand for pumps

Pressure drop in cooling water linesIncrease of pressure drop caused by incrustations, example DN 100

1,0

0,50,40,30,2

0,1

0,5 1,0 2,0 3,0water velocity [m/sec]

pressure drop [bar]

14 m³/h29 m³/h

43 m³/h

57 m³/h

72 m³/h

86 m³/hper 100 m

clean1.5 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

Example Example

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/secpressure drop 0.022 bar/100 m

By corrosion products & depositsa 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 mdP 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 kW8,760 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/kg1,500,000

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/kg1,200,000

4.19 kJ/kg for a temperature change of 1 °C ....... 17.5 °C

increase : 3.5 °C

Example 4Example 4


61

ECONOMICSECONOMICS

62

ECONOMICSECONOMICS

Any Any 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 Differenceêê increase of Condensate Temperatureincrease of Condensate Temperatureêê higher Condensate Pressurehigher Condensate Pressureêê a Loss of Efficiencya Loss of Efficiencyêê less Production Outputless Production Output

The Economical Impacts of these ChangesThe Economical Impacts of these Changesare often not quantified !are often not quantified !

Power plant

• Influence on Economy and Efficiency

63

ECONOMICSECONOMICS

Fertilizer plantAmmonia 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 consumptionTower 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%

ActualActual36000 m36000 m33/hr/hr7°C7°C20°C20°C17°C17°C1,041,041500 HP1500 HP18°C18°C26°C26°C33°C33°C9°C9°C252e6 kcal/h252e6 kcal/h290.555 kg/hr290.555 kg/hr90,3%90,3%

Actual to DesignActual to Design36000 m36000 m33/hr/hr10°C10°C27°C27°C22°C22°C1,041,041500 HP1500 HP20°C20°C30,7°C30,7°C40,7°C40,7°C8,7°C8,7°C360e6 kcal/h360e6 kcal/h445.738 kg/hr445.738 kg/hr

64

ECONOMICSECONOMICS

Fertilizer plantAmmonia 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 consumptionTower 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%

ActualActual36000 m36000 m33/hr/hr7°C7°C20°C20°C17°C17°C1,041,041500 HP1500 HP18°C18°C26°C26°C33°C33°C9°C9°C252e6 kcal/h252e6 kcal/h290.555 kg/hr290.555 kg/hr90,3%90,3%

Actual to DesignActual to Design36000 m36000 m33/hr/hr10°C10°C27°C27°C22°C22°C1,041,041500 HP1500 HP20°C20°C30,7°C30,7°C40,7°C40,7°C8,7°C8,7°C360e6 kcal/h360e6 kcal/h445.738 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 / designMoney loss:Money loss:-- FuelFuel

-- 250x0,7x0,09x8700 = $137.000/y 250x0,7x0,09x8700 = $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

Engineering