4 cooling system dynamics
Post on 08-Apr-2017
58 Views
Preview:
TRANSCRIPT
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
PrinciplesPrinciples
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
PrinciplesPrinciples
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 ..
PrinciplesPrinciples
7
Cooling TowerCooling Tower
8
Cooling TowerCooling Tower
9
Cooling TowerCooling Tower
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
Cooling Tower - DesignCooling Tower - Design
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
Cooling Tower - DesignCooling Tower - Design
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
Cooling Tower - DesignCooling Tower - Design
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
air temperature 25°C
air temperature 20°C
air temperature 15°C
air temperature 10°C
Cooling Tower - DesignCooling Tower - Design
from VDI:"Kühlgrenze undrelative Luftfeuchte"
valid for985 to 1020 mbar.
airair / water in / water in counterflowcounterflow air / water in air / water in crossflow crossflow
Cooling Tower - DesignCooling Tower - Design
in comparisonin comparison
counterflow crossflow counterflow crossflow
area demand
investment
operating costs
wet air - recirculation
approach
--
----
--
--
++++
++++
++
Cooling Tower - DesignCooling Tower - Design
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
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
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
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
20
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
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==
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
22
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
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
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
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]
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
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
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
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
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
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 %
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
Water- distribution to cooling tower cells
waterflow
of design
90 % 100 % 110 %
results inapprox.
8 %reduction in
thermal capability
instead of100 % 100 % 100 %
Cooling Tower – Air/Water ContactCooling Tower – Air/Water Contact
”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
air temperature 25°C
air temperature 20°C
air temperature 15°C
air temperature 10°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
”Hot Weather Example“ - results ”Hot Weather Example“ - results
water
heatrejection
temp.drop
”Hot Weather Example“ - results ”Hot Weather Example“ - results
ØØ Temperature drop by evaporationTemperature drop by evaporation
for 30 °C, 45 % rel humidity:
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
”Hot Weather Example“ - results ”Hot Weather Example“ - results
ØØ Consequences of makeupConsequences of makeup--quantity on quantity on
water temperaturewater temperature
for 30 °C, 45 % rel humidity:
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
”Hot Weather Example“ - results ”Hot Weather Example“ - results
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
”Hot Weather Example“ - results ”Hot Weather Example“ - results
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
Pressure drop - example
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
Pressure drop - example
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
Pressure drop - example
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 !!
top related