COOLING TOWER
COOLING TOWER
AAcoocoolili ng ng towetower r is a device commonly used to cool condenser water in power and refrigerating plants. is a device commonly used to cool condenser water in power and refrigerating plants. SYSTEMS USING COOLING TOWER
SYSTEMS USING COOLING TOWER 1.
1. Cooling Towers for RefrigerationCooling Towers for Refrigeration
An important device used in any refrigeration or air conditioning system is a condenser. A condenser An important device used in any refrigeration or air conditioning system is a condenser. A condenser is used in the high pressure side of a refrigeration or air conditioning system to convert the high-pressure is used in the high pressure side of a refrigeration or air conditioning system to convert the high-pressure vapour refrigerant from the compressor into liquid refrigerant. The medium used in a condenser may be vapour refrigerant from the compressor into liquid refrigerant. The medium used in a condenser may be water or air, depending upon the application. In the case of water cooled condensers, the warm water being water or air, depending upon the application. In the case of water cooled condensers, the warm water being pumped
pumped by the by the condenser condenser should should be be cooled cooled with the with the help ohelp of cooling f cooling towers so towers so that the that the same same water water may bemay be re-circulated to the condenser.
re-circulated to the condenser.
Water-cooled chillers are normally more energy efficient than air-cooled chillers due to
Water-cooled chillers are normally more energy efficient than air-cooled chillers due to heat rejection heat rejection toto tower water at or near
tower water at or near wet-bulb temperatures. wet-bulb temperatures. 2.
2.Industrial Cooling TowersIndustrial Cooling Towers
Industrial cooling towers can be used to remove heat from various sources such as machinery or heated Industrial cooling towers can be used to remove heat from various sources such as machinery or heated process material.
process material. The The primary use primary use of lof large, indarge, industrial cooling ustrial cooling towers is towers is to to remove remove the the heat heat absorbed absorbed in in thethe circulating
circulating cooling water cooling water systems used insystems used in power plants, power plants, petroleum refinerpetroleum refineries, petrochemical ies, petrochemical plants, plants, naturalnatural gas
gas processing processing plants, plants, food food processing processing plants, plants, semi-conductor semi-conductor plants, plants, and and for for other other industrial industrial facilitiesfacilities such as in condensers of dis
such as in condensers of distillation columns, tillation columns, for cooling liquid in for cooling liquid in crystallization, etc.crystallization, etc.
Figure
Figure 47 47 Water Water cooled cooled chiller chiller system system utilizing utilizing Figure 48 Figure 48 Thermal Thermal power power plant plant utilizing utilizing natural natural draftdraft cooling
Types of Cooling Tower
1.) Natural Draft Cooling Towers 2.) Mechanical Draft Cooling Towers
Natural Draft Cooling Towers
The natural draft or hyperbolic cooling tower makes use of the difference in temperature between the ambient air and the hotter air inside the tower. As hot air moves upwards through the tower (because hot air rises or of density difference between hot and cool air causes a natural
“draft ” also known as thechimney or stack eff ect ), fresh cool air is drawn
into the tower through an air inlet at the bottom.
Due to the layout of the tower, no fan is required and there is almost no circulation of hot air that could affect the performance. Concrete is used for the tower shell with a height of up to 200 m (and diameter up to 100m). These cooling towers are mostly only for large heat duties (such as in nuclear power plant) because large concrete structures are expensive.
Figure 49 Air flow direction in a natural draft cooling tower.
Figure 50 Cross flow natural draft cooling tower Figure 51 Counter flow natural draft cooling tower
There are two main types of natural draft towers:
Cross flow tower (Figure 50): air is drawn across the falling water and the fill is located
outside the tower
Counter flow tower (Figure 51): air is drawn up through the falling water and the fill is
therefore located inside the tower, although design depends on specific site conditions
Advantages Disadvantages
̶ No energy consumption for fans operation – Higher investment costs
Mechanical Draft Cooling Towers
As the name indicates, air is circulated inside the tower mechanically instead of natural circulation. Propeller fans or centrifugal fans may be used.
Advantages of mechanical draft cooling towers over natural draft cooling towers:
For the same capacity used, the mechanical draft cooling towers are much smaller than the natural draft
cooling towers. This is because of the increase in cooling capacity due to increase in volume of the air being forced out by fan.
Capacity control is possible in mechanical draft cooling tower. By controlling the speed of the fan, the
volume of air can be controlled, which in turn controls the capacity.
The natural draft cooling towers can be located only in open space. As they do not depend upon the
atmospheric air, the mechanical draft cooling towers shall be located even inside the building.
Disadvantages of using mechanical draft cooling towers:
More power is required to run the system,
Increased running cost due to increase in maintenance of the fans, motors and its associated controls,
According to the location of the fan, they are further classified as: 1. Forced draft cooling towers, and
2. Induced draft cooling towers. Forced Draft Cooling Towers
In this system, fan is located near the bottom and on the side. This fan forces the air from bottom to top. An eliminator is used to prevent loss of water droplets along with the forced air.
Induced Draft Cooling Towers
In this system, a centrally located fan at the top, takes suction from the tower and discharges it to the atmosphere. The only difference between the induced draft cooling tower and forced draft cooling tower is that the fan is located at the top in the induced draft cooling tower.
Figure 52 Forced Draft Cooling Tower
Comparison between Induced Draft Counter Flow and Cross Flow Cooling Towers
Figure 54 Induced draft counter flow and cross flow cooling towers.
Counter flow
In a counter flow design, the air flow is directly opposite to the water flow (see Figure 54 upper diagram). Air flow first enters an open area beneath the fill media, and is then drawn up vertically. The water is sprayed through pressurized nozzles near the top of the tower, and then flows downward through the fill, opposite to the air flow.
Advantages of the counter flow design:
Breakup of water in spray makes heat transfer more efficient.
The coldest water comes in contact with the coolest and most dry air, optimizing the heat transfer and
obtaining the maximum performance Disadvantages of the counter flow design:
Typically higher initial and long-term cost, primarily due to pump requirements.
Difficult to use variable water flow, as spray characteristics may be negatively affected.
Cross flow
Cross flow is a design in which the air flow is directed perpendicular to the water flow (see Figure 54 lower diagram). Air flow enters one or more vertical faces of the cooling tower to meet the fill material. Water flows (perpendicular to the air) through the fill by gravity.
A distribution or hot water basin consisting of a deep pan with holes or nozzles in its bottom is located near the top of a cross flow tower. Gravity distributes the water through the nozzles uniformly across the fill material.
Advantages of the cross flow design:
Gravity water distribution allows smaller pumps and maintenance while in use. Typically lower initial and long-term cost, mostly due to pump requirements.
Disadvantages of the cross flow design:
The air flows horizontally and the water falling downwards meets the air at different
temperatures. Therefore the heat transfer is not always optimized
Low pressure head on the distribution pan may encourage orifice clogging and less water breakup at spray
Cooling Towers Parameters and Performance
Figure 53 Diagrammatic cooling tower
Energy balance:
m3h3 + mah1 = m4h4 + mah2
m3h3 – m4h4 = ma(h2 – h1)
where: m3 = mass of water entering
m4 = mass of cooled water received by cold water basin
h3 = enthalpy of water entering, kJ/kg = hf at t3
h4 = enthalpy of water leaving, kJ/kg = hf at t4
h1 = enthalpy of air entering, kJ/kg d.a
h2 = enthalpy of air leaving, kJ/kg d.a
Mass balance:
m3 + mmoist air1 = m4 + mmoist air2
m3 + ma + mv1 = m4 + ma + mv2
m3 + maW1 = m4 + maW2
m3 – m4 = ma (W2 – W1) = mass of water evaporated (or mass of make-up water)
where: W1 = humidity ratio of air entering, kg/kgd.a
Approach Cooling Range, ACR = (also simply Approach) difference between the temperature of the cooled tower water, t4 and the atmospheric wet bulb temperature, tw1. Approach is the most important
indicator of cooling tower performance. It dictates the theoretical limit to the leaving cold-water temperature. The approach temperatures generally fall between 5 and 20ºF implying that the leaving cooled water temperature shall be 5 to 20oF above the ambient WBT.
ACR = t4 – tw1
Actual Cooling Range, CR A = difference between the raw water temperature at the entrance, t3 and
the cooled water at the basin, t4. It is the actual temperature drop of
water. CR A = t3 – t4
Theoretical Cooling Range, CR T = difference between the raw water temperature at the entrance, t3
and the atmospheric wet bulb temperature, tw1. It is the maximum
temperature drop) CR T = t3 – tw1
Cooling Tower Efficiency = Actual Cooling Range x 100% = CR A x 100%
Theoretical Cooling Range CR T
= t3 – t4 x 100%
t3 – tw1
Sample Problems
1.) In a cooling tower 28.34 m3/min of air at 32oC DB and 24oC WB enter the tower and leave saturated at 29oC. (a) To what temperature can the air stream cool a spray of water which enters at 38oC, with a flow of 34 kg/min of water? (b) Cooling tower efficiency and (c) How many kg per hour of make-up water is needed to compensate for the water that is evaporated?
Solution: Point 1: At td1 = 32oC and tw1 = 24oC, h1 = 72.5 kJ/kg v1 = 0.884 m3/kg W1 = 0.0156 kg/kgd.a ma = V1 = 28.34 m3/min = 32.06 kg/min v1 0.884 m3/kg
Point 2: At td2 = 29oC and Ø1 = 100% RH (saturated air), h2 = 95 kJ/kg W2 = 0.0256 kg/kgd.a
Point 3: Using Steam Table No.1: h3 = hf at 38oC = 159.21 kJ/kg, m3 = 34 kg/min
(a) Mass balance: m3 – m4 = ma(W2 – W1)
34 – m4 = (32.06) (0.0256 – 0.0156)
Energy balance:
m3h3 – m4h4 = ma(h2 – h1)
(34) (159.21) – 33.68h4 = (32.06) (95 – 72.5)
h4 = 139.3 kJ/kg = hf at t4
t4 = tsat at hf (139.3 kJ/kg)
Using steam table No. 1, by interpolation: t4 = tsat at hf t4 = 33.2oC
(b) Cooling Tower Efficiency = t3 – t4 x 100% = 38 – 33.2 x 100% = 34.29%
t3 – tw1 38 – 24
(c) Make-up water = ma (W2 – W1) = (32.060) (0.0256 – 0.0156) = 0.3206 kg/min or 19.24 kg/h
2.) Fifty gallons per minute of water enters a cooling tower at 46oC. Atmospheric air at 16oC DB and 55% RH enters the tower at 2.85 m3/sec and leaves at 32oC saturated. Determine (a) the volume flow rate of water receives by cold water basin, and (b) the exit temperature of the water. Solution: Point 1: At td1 = 16oC and Ø1 = 55%, h1 = 32 kJ/kg v1 = 0.828 m3/kg W1 = 0.0056 kg/kgd.a ma = V1 = 2.85 m3/min = 3.442 kg/min v1 0.828 m3/kg
Point 2: At td2 = 32oC and Ø2 = 100% RH (saturated air), h2 = 111 kJ/kg W2 = 0.0307 kg/kgd.a
Point 3: Using Steam Table No.1: h3 = hf at 46oC = 192.62 kJ/kg, v3 = vf at 46oC = 0.0010103 m3/kg
m3 = V3 = (50 gal/min x 1 min/60 sec) (3.7854 L/gal x 1 m3 /1000L) = 3.122 kg/sec
v3 (0.0010103 m3/kg)
(a) Mass balance: m3 – m4 = ma(W2 – W1)
3.122 – m4 = (3.442) (0.0307 – 0.0056) m4 = 3.036 kg/min Energy balance: m3h3 – m4h4 = ma(h2 – h1) (3.122) (192.62) – 3.036h4 = (3.442) (111 – 32) h4 = 108.51 kJ/kg = hf at t4 t4 = tsat at hf (108.51 kJ/kg)
Using steam table No. 1, by interpolation: t4 = tsat at hf(108.51 kJ/kg) t4 = 25.9oC
Specific volume of water leaving, v4 = vf at hf(108.51 kJ/kg) v4 = 0.0010031 m3/kg
3.) Water at 55oC is cooled in a cooling tower which has an efficiency of 65%. The temperature of the surrounding air is 32oC DB and 70% RH. The heat dissipated from the condenser is 2,300,000 kJ/h. Find the capacity in liters per second of the pump used in the cooling tower.
Solution:
Point 1: At td1 = 32oC and Ø1 = 70% RH, tw1 = 27.4oC
Cooling Tower Efficiency = t3 – t4 x 100% = 55 – t4 x 100%
t3 – tw1 55 – 27.4
t4 = 37.1oC
Heat balance about condenser,
mwh4 = mwh3 + Q where: h3 = hf at 55oC = 230.23 kJ/kg
h4 = hf at 37.1oC = 155.45 kJ/kg (by interpolation)
v4 = vf at 37.1oC = 1.00674 x 10-3 m3/kg
Q = mw (h3 – h4) = 2,300,000 kJ/hr = mw (230.23 – 155.45) kJ/kg or
mw= 30,756.89 kg/hr or 8.54 kg/sec
or Q = mw cpw (t3 – t4) where c pw= 4.187 kJ/kgoC
Capacity of pump = mw v4 = (8.54 kg/sec) (1.00674 x 10-3 m3/kg) = 0.00859 m3/sec or8.59 L/s
Problems:
(1) In a cooling tower water enters at 52oC and leaves at 27oC. Air at 29oC and 47% RH also enters the cooling tower and leaves at 46oC fully saturated with moisture. It is desired to determine (a) the volume and mass of air necessary to cool one kg of water, and (b) the quantity of water that can be cooled with 142 m3/min at atmospheric air. Ans: (a) 0.5742 m3, 0.66 kg; (b) 247.3 kg
(2) A cooling tower receives 6 kg/s of water of 60oC. Air enters the tower at 32oC DB and 27oC WB temperatures and leaves at 50oC and 90 per cent relative humidity. The cooling efficiency is 60.6 per cent. Determine (a) the mass flow rate of air entering, and (b) the quantity of make-up water required. Ans: (a) 3.253 kg/s, (b) 0.1818 kg/s
(3) A mechanical-draft cooling tower receives 115 m3/sec of atmospheric air at 103 kPa, 32oC DB temperature, 55%RH and discharges the air saturated at 36oC. If the tower receives 200 kg/sec of water at 40oC, what will be the exit temperature of the cooled water? Ans: 31.2oC
