6 Cooling Tower

COOLING TOWER A cooli  is a device commonly used to cool condenser water in power and refrigerating plants.  is cooli ng tower  tower  SYSTEMS USING COOLING TOWER

1. Cooling Towers for Refrigeration 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 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  pumped by the condenser should be cooled with the help of o f cooling towers so that the same water may be re-circulated to the condenser. Water-cooled chillers are normally more energy efficient than air-cooled chillers due to heat to heat rejection to tower water at or near  wet-bulb  wet-bulb temperatures. 2. Industrial Cooling Towers Industrial cooling towers can be used to remove heat from various sources such as machinery or heated  process material. The primary use of large, l arge, industrial ind ustrial cooling towers is to remove the heat absorbed in the circulating cooling circulating cooling water systems used in power in power plants, petroleum refineries, refineries, petrochemical plants, natural gas processing plants, food processing plants, semi-conductor plants, and for other industrial facilities such as in condensers of distillation distillation columns, for cooling liquid in crystallization, etc.

Figure 47 Water cooled chiller system utilizing cooling tower.

Figure 48 Thermal power plant utilizing natural draft hyperboloid cooling tower.

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  ), fresh cool air is drawn “draft ” also known as the chimney or stack eff ect  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 d esign depends on specific site conditions Advantages   ̶   ̶

 No energy consumption for fans operation Low costs of maintenance and spare parts

Disadvantages

Higher investment costs  –  Requires large areas for tower installation  – 

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: 

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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

Figure 53 Induced Draft Cooling Tower

Bottle shaped Induced 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 nozzle. 

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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 W2 = humidity ratio of air leaving, kg/kgd.a

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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 o cooled water temperature shall be 5 to 20 F above the ambient WBT.

ACR =

t4 –  tw1

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Actual Cooling Range, CR A =

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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

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

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 3

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1.) In a cooling tower 28.34 m /min of air at 32 C DB and o o 24 C WB enter the tower and leave saturated at 29 C. (a) To what temperature can the air stream cool a spray o of water which enters at 38 C, 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:

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At td1 = 32 C and tw1 = 24 C, 3 h1  = 72.5 kJ/kg v1 = 0.884 m /kg W1 = 0.0156 kg/kgd.a 3

ma = V1  = 28.34 m /min = 32.06 kg/min 3 v1 0.884 m /kg o

Point 2: At td2 = 29 C and Ø1 = 100% RH (saturated air), h2 = 95 kJ/kg W2 = 0.0256 kg/kgd.a o

Point 3: Using Steam Table No.1: h3 = hf  at 38 C = 159.21 kJ/kg, m3 = 34 kg/min (a) Mass balance:

m3 –  m4 = ma(W2 –  W1) 34 –  m4 = (32.06) (0.0256 –  0.0156) m4 = 33.68 kg/min

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) o

Using steam table No. 1, by interpolation: t4 = tsat at hf  t4  = 33.2 C (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 o o 46 C. Atmospheric air at 16 C DB and 55% RH enters 3 o the tower at 2.85 m /sec and leaves at 32 C saturated. Determine (a) the volume flow rate of water receives by cold water basin, and (b) the exit temperature of the water. Solution: o

Point 1: At td1 = 16 C and Ø1 = 55%, 3 h1  = 32 kJ/kg v1 = 0.828 m /kg W1 = 0.0056 kg/kgd.a 3

ma = V1  = 2.85 m /min = 3.442 kg/min 3 v1 0.828 m /kg o

Point 2: At td2 = 32 C and Ø2 = 100% RH (saturated air),

h2 = 111 kJ/kg W2 = 0.0307 kg/kgd.a

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Point 3: Using Steam Table No.1: h3 = hf  at 46 C = 192.62 kJ/kg, v3 = vf  at 46 C = 0.0010103 m /kg 3

m3 = V3 = (50 gal/min x 1 min/60 sec) (3.7854 L/gal x 1 m /1000L) = 3.122 kg/sec 3 v3 (0.0010103 m /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) Specific volume of water leaving, v4 = vf  at hf(108.51 kJ/kg)

t4 = 25.9oC 3 v4 = 0.0010031 m /kg 3

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Volume flow rate of water leaving = m4v4 = (3.036 kg/sec) (0.0010031 m /kg) = 0.003045 m /sec

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3.) Water at 55 C is cooled in a cooling tower which has an efficiency of 65%. The temperature of the o surrounding air is 32 C 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:

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Point 1: At td1 = 32 C and Ø1 = 70% RH, tw1 = 27.4 C 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

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where: h3 = hf  at 55 C = 230.23 kJ/kg o h4 = hf  at 37.1 C = 155.45 kJ/kg (by interpolation) o -3 3 v4 = vf  at 37.1 C = 1.00674 x 10  m /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)

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where c pw = 4.187 kJ/kg C -3

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Capacity of pump = mw v4 = (8.54 kg/sec) (1.00674 x 10  m /kg) = 0.00859 m3/sec or 8.59 L/s Problems: o

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(1) In a cooling tower water enters at 52 C and leaves at 27 C. Air at 29 C and 47% RH also enters the cooling o tower and leaves at 46 C fully saturated with moisture. It is desired to determine (a) the volume and mass 3 of air necessary to cool one kg of water, and (b) the quantity of water that can be cooled with 142 m /min 3 at atmospheric air. Ans:   (a) 0.5742 m , 0.66 kg; (b) 247.3 kg o

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(2) A cooling tower receives 6 kg/s of water of 60 C. Air enters the tower at 32 C DB and 27 C WB o temperatures and leaves at 50 C 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)

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A mechanical-draft cooling tower receives 115 m /sec of atmospheric air at 103 kPa, 32 C DB o temperature, 55%RH and discharges the air saturated at 36 C. If the tower receives 200 kg/sec of water at o o 40 C, what will be the exit temperature of the cooled water? Ans:   31.2 C