WITHDRAWN: Spray Ponds and Cooling Towers

WITHDRAWN: Spray Ponds and Cooling Towers

CHAPTE R 18 Spray Ponds and Cooling Towers 18.1  Spray Pond When the water required for a condenser is inadequate and expensive, then the same water ...

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CHAPTE R 18

Spray Ponds and Cooling Towers 18.1  Spray Pond When the water required for a condenser is inadequate and expensive, then the same water is recooled and used again and again. Spray ponds and cooling towers are commonly used for this purpose. A schematic diagram of a spray pond is shown in Figure 18.1. The pond can be located either on the ground, on a roof, or at certain height above ground level. The water from the condenser is sprayed through the nozzles to expose the maximum surface of water to the air to allow the droplet surface to evaporate and cool the water. Spray ponds are usually surrounded with wooden walls to prevent the loss of water by wind drift. The major disadvantage with this process is that it requires more floor area per kg of water than any other system used for cooling. The efficiency of cooling ponds is about 40–50%.

18.2  Cooling Towers A cooling tower is a means by which the heat from the condenser is dispersed to the atmosphere. Cooling towers are desired when positive control of water temperature is required. Cooling towers are used for big plants as absorption refrigeration systems, steam jet refrigeration systems, and centrifugal units above 100 TR, because at capacity below 100 TR, the adoption of evaporative condensers with reciprocating compressor is found more economical. Proper siting and orientation of a tower are important. The principle of cooling the water in cooling towers is similar to the evaporative condenser. The water goes into the air in the form of vapor, taking its latent heat from the remaining water and causing a reduction in temperature. Theoretically, the water can be cooled to the wet bulb temperature (WBT) of entering air if the air leaving the tower becomes saturated. Generally, the temperature of water coming out from the cooling tower is 3–5°C above the WBT of the entering air. The temperature difference between WBT of incoming air and the temperature of outgoing water is known as cooling tower “approach.” The higher the quantity of water circulated, the lower will be approach of the tower. The quantity of water circulated economically is also limited by the power requirements of the pump. Engineering for Storage of Fruits and Vegetables. http://dx.doi.org/10.1016/B978-0-12-803365-4.00018-X

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304  Chapter 18

Figure 18.1: Spray Pond.

18.3  Classification of Cooling Towers 1. Cooling towers are divided into two main groups: natural draft cooling towers and forced draft cooling towers.

18.3.1  Natural Draft Atmospheric Spray Towers The arrangement of this type of tower is shown in Figure 18.2. This tower should be located in open space or on the roof of the building where the movement of air is best. The capacity of this cooling tower is nearly 30–75 L per square meter of base area, depending upon the air velocity. They work economically for loads under 200 tons capacity.

18.3.2  Natural Draft Deck–Type Tower The construction of this tower is shown in Figure 18.3.

Figure 18.2: Spray-Type Cooling Tower.

Spray Ponds and Cooling Towers  305

Figure 18.3: Deck-Type Cooling Tower.

It is similar to the atmospheric spray tower except that water-distributing troughs are used, which helps to break the water into small droplets. The object of the decks is to provide additional evaporation area. It gives 20–30% greater efficiency than the atmospheric spray tower for the same size and for the same quantity of water flow.

18.3.3  Forced and Induced Draft Cooling Towers The construction of the forced draft tower is shown in Figure 18.4.

Figure 18.4: Forced Draft Cooling Tower.

306  Chapter 18

Figure 18.5: Induced Draft Cooling Tower.

Water from the condenser is sprayed at the top of the tower and air is forced by the blower from the bottom of the tower as shown in Figure 18.4. The air velocity of 120 m/min is recommended with a flow of 100–130 cmm per min per ton of refrigeration capacity. The construction of induced draft towers is shown in Figure 18.5. The difference lies only in the supply of air. The amount of water usually lost with forced or induced draft cooling tower ranges from 1% to 2% by evaporation and 0.5% to 2% by drift losses. The water for compensating these losses is supplied from external sources. The water must be chemically treated to avoid scale formation of the condenser due to an increase in the concentration of salt in the cooling water. The performance of a cooling tower is measured in terms of its efficiency, which is ­defined  as

η=

twi − twa twi − twha

where twi and twa are the inlet and outlet temperature of water and twha is WBT of inlet air. The efficiencies of different types of cooling systems are: Spray ponds Natural draft towers Mechanical draft towers

40–50% 50–70% 60–80%

Spray Ponds and Cooling Towers  307 Advantages of Mechanical Draft Towers The mechanical draft (MD) towers offer the following advantages over natural draft towers: 1. It requires less space (m2 and m3) per kg of water cooled. 2. Quantity of water required in condenser is reduced due to more effective cooling. 3. There is full control on the temperature of outlet water from the tower because the quantity of air supply can be controlled. 4. Smaller equipment can be used to handle airflow under fairly well-controlled operating conditions. Advantages of Induced Draft Towers 1. The main advantage of induced draft (ID) towers is that the coldest water comes in contact with the driest air, and the warmest water comes in contact with the most humid air. 2. The recalculation is seldom a problem with the tower because the outlet fan discharges the heated and humid air directly away from the air intakes below the tower. 3. Sizes up to 20 m diameter can be used. 4. Lower first cost, require less space and capable of cooling through a wide range. 5. The power consumption per kg of water cooled is less compared with a forced draft fan system. Advantages of Forced Draft Towers 1. Forced draft (FD) towers are more efficient than ID draft because some of the air velocity is converted into static pressure in the tower and recovered in the form of useful work. 2. The vibration and noise are minimal because mechanical equipment is set on solid ­foundations. 3. Because it handles dry air, problems of blade erosion are avoided. 4. It is safer because it is located on the ground level. Disadvantages of Forced Draft Fans 1. The possibility of recirculation of hot air and humid exhaust coming out from the top of the tower through the low pressure air-intake region can cut efficiency as much as 20%. 2. During cold weather, ice forms on nearby equipment and buildings or in the fan housing itself, which can cause fan blade breakage. 3. Fan size is limited to 4 m. 4. The power requirement of the FD system is approximately double that of the ID system for the same capacity.

18.4  Performance of Cooling Towers The factors that decide the performance of the cooling tower are defined here. The schematic and the process of cooling of water in the cooling tower is represented on the psychometric chart shown in Figures 18.6(a) and (b), respectively.

308  Chapter 18

Figure 18.6: (a) Schematic of a Cooling Tower. (b) Cooling Tower Efficiency (Psychrometric Chart).

In the psychrometric chart, Point 1 and Point 2 show the condition of air at the inlet and ­outlet, respectively. twi and two are the inlet and outlet temperature of water passing through the cooling tower (twi > two). twb is the WBT of inlet air. 1. Cooling approach: This is defined as the difference between the water inlet temperature and temperature of the entering air (twi – tab), where tab is the WBT of the inlet air. 2. Cooling range: This is defined as the difference between the inlet (hot) water temperature and outlet (cold) water temperature (twi – two). 3. Cooling tower efficiency: This efficiency is defined as the ratio of actual increase in air temperature (Tao – Tai) to the maximum theoretical drop in air temperature (Tai – Tab), which cannot fall below Tab (its WBT).

η=

Tao – Tai Tai – Tab

where Tao = temperature of the outgoing air, °C Tai = temperature of the incoming air, °C Tab = WBT of the inlet air, °C The desired cooling range (Twi – Two) should be as high as possible, because it reduces the amount of water required in the condenser per kg of refrigerant to be condensed. The factor affecting the cooling range for the cooling tower is mainly the ambient condition. Hotter and drier air will absorb more water vapor and cool the water more effectively.

Spray Ponds and Cooling Towers  309 Example 18.1 Determine the volume of air to be handled and quantity of makeup water in kg/min required by a cooling tower that is used to cool 1000 kg of water per minute from temperature of 35– 30°C. Atmospheric conditions are 35°C dry bulb temperature (DBT) and 25°C WBT, and air leaves the tower at 30°C DBT and 90% relative humidity (RH). Heat lost by water = Heat gained by air = mw1 hw1 − mw2 hw2 =



V ( ha2 − ha1 ) Vs1

(18.1)

where V = volume of free air circulated, m3/min mw1 = mass flow of incoming water, kg/min hw1 = enthalpy of incoming water, kJ/kg mw2 = mass flow of outgoing water, kg/min hw2 = enthalpy of outgoing water, kJ/kg vS1 = specific volume of air at inlet condition, m3/kg of dry air ha2 = enthalpy of outgoing air, kJ/kg ha1 = enthalpy of incoming air, kJ/kg We can also write mw2 = mw1 −



V ( w 2 − w1 ) Vs1

(18.2)

where w2 = specific humidity of incoming air, gm/kg dry air w1 = specific humidity of outgoing air, gm/kg dry air Then, substituting Equation 18.2 in Equation 18.1, we get



  V  V mw1 hw1 −  mw1 − w 2 − w1 )  hw2 = ( ( ha2 − ha1 )  Vs1  Vs1  

(18.3)

Simplified:



mw1 ( hw1 − hw2 ) + hw2

V V w 2 − w1 ) = ( ( ha2 − ha1 ) Vs1 Vs1

(18.4)

We can also write hw1 – hw2 = Cpw(Tw1 – Tw2), and by substituting in Equation 18.4:



(

)

(

V V mw1 Cpw Tw1 − Tw2  + hw 2 ( w 2 − w1 ) = ha2 − ha1   V V si

si

)

(18.5)

310  Chapter 18

Figure 18.7: Psychrometric Chart for Example 18.1.

Equation 18.5 is the final mass balance equation of this problem. The left-hand side of the equation represents heat lost by water and the right-hand side represents the heat gain by the air. The psychrometric chart in Fig. 18.7 provides these values: ha1 = 76.4 kJ/kg

ha2 = 95 kJ/kg

w1 = 16 g/kg. Da

w2 = 24.4 g/kg Da.

Vs1 = 0.895 m /kg 3

(i) Heat lost by water: 1000 × 4.2(35 − 30) + 30 × 4.2 × = 21000 + 1.182 V

V (24.4 − 16) 0.895 1000

(ii) Heat gained by air: V (95 − 76.4 ) = 20.78 V 0.895 ∵ 21000 + 1.182 V = 20.78 V V = 1071.6 m3 / min Makeup water to be added: 1071.6 (24.4 − 16) = 10.05 kg/ min 0.895 × 1000

Spray Ponds and Cooling Towers  311 Example 18.2 Determine the amount of air to be handled and quantity of makeup water required by a cooling tower when 900 kg of water per minute is cooled from 35.5°C to 30.5°C. Atmospheric conditions are 35°C DBT and 25.5°C WBT. Assume that the air leaves the tower 90% saturated and at 32.2°C. Assume that there is no entertainment loss. Solution Using same notations and same equation as in the previous example:   V V mw1 hw1 –  mw1 – ( w 2 – w1 )  hw2 = ( ha2 – ha1 ) Vs1 Vs1   Taking the values from the psychrometric chart: ha1 = 96.6 kJ/kg

ha2 = 122.6 kJ/kg

w1 = 16.8 gm/kg

w2 = 28 gm/kg

Vs1 = 0.886 m3/kg V = volume of atmosphere air supplied, m3/min Substituting the values in the preceding equation: V (28 – 16.8)   900 × 4.2 × 35.5 – 900 – 4.2 × 30.5 0.866 1000   =

V (122.6 – 96.6) 0.866

900 × 4.2 × 35.5 – 900 × 4.2 × 30.5 + ×4.2 × 30.5 =



V 26 0.866

V 11.2 × 0.866 1000

V  11.2 × 4.2 × 30.5  26 –   = 900 × 4.2 × 5 0.866  1000 V (26 – 1.43) = 18900 0.866 V = 681.5m3 /min

Makeup water = evaporated water carried with air =

V 681.5 (28 – 16.8) w 2 – w1 ) = × = 8.6 kg/min ( Vs1 0.886 1000

312  Chapter 18

Figure 18.8: Psychrometric Chart for Example 18.3.

Example 18.3 A cooling tower is to be designed to take the heat load of 200 tons refrigerating plant using F−12 as refrigerant. The heat rejection ratio of the system is 1.2. The rise in temperature allowed in the condenser is 5°C. The atmospheric air condition is 35°C DBT and 25°C WBT. The air leaves the tower at 30°C and 90% RH. Ignoring the heat losses in the system and carryover loss through the cooling tower, find the following: (a) Quantity of air required to pass through a cooling tower per minute. (b) Quantity of makeup water. The temperature of water coming out of the tower is 30°C. Solution The conditions of air at the inlet and outlet of the cooling tower are shown on the psychrometric chart Figure 18.8. The heat carried in the condenser is given by Qc = Heat load of evaporator × Heat rejection ratio = 200 × 3.5 × 1.2 = 840 kJ/sec. The heat carried away in cooling tower must be equal to the heat given out in the condenser. The heat transfer between water and air in the cooling tower is given by the following equation: Qc = mw Cpw (T0 − T1 ) =

V ( ha2 – ha1 ) – ( w 2 – w1 ) Cpw T2   Vs1 

where Ti and T0 are temperatures of water entering and leaving the cooling tower. ha1 and ha2 are the enthalpies of the air at the inlet and exit of the tower. w2 and w1 are the absolute humidities of air at the inlet and exit of the tower. V is the volume of air circulated, and Vsi is the specific volume of air at the inlet of the tower. ∴ From psychrometric chart: ha1 = 76.4 kJ/kg, ha2 = 95 kJ/kg. w1 = 16 gm/kg, w2 = 24.4 gm/kg, Vsi = 0.895 m3/kg

Spray Ponds and Cooling Towers  313 V=

840 × 0.895 = 40.97m3 /s = 14749 m3 /h 18.35

Quantity of makeup water or loss due to evaporation = =

V  w 2 – w1    kg/h Vs1  1000 

14749  24.4 – 16  ×  = 125.25 kg/h 0.895  1000 

Quantity of cooling water circulated through the tower = Quantity of cooling water to the condenser = mw T0 Outlet temperature of water from the tower T1 Inlet temperature of water to the condenser ∴ mw × Cpw (T0 − T1 ) = Q mw =

840 = 40 kg/s 5 × 4.2

18.5  Maintenance of Cooling Towers Modern cooling towers have eliminators, advanced filters, and separators, and new water treatment chemicals have increased the efficiency of the tower and reduced its power consumption. Routine cooling tower services require visual inspection, mechanical maintenance, and a physical cleaning program that will maintain the system cleanup year round. The importance of cooling tower maintenance is well illustrated by its documented benefits. Power savings are reflected in both electrical demand and consumption, and equipment life is significantly increased through reduced corrosion of metal parts and erosion of system components, including tubes and seals, by abrasive particles. Chemical requirements are reduced as a clean system responds better to water treatment. Failures such as unexpected shutdown and severe freeze damage can be avoided along with their associated costs. Problems 1. In an FD cooling tower, 1000 kg of condenser water per minute is cooled from 26°C to 12°C; air enters into the tower at 15°C DBT and 55% RH, and leaves the tower at 24°C and in saturated condition. Calculate the quantity of water supplied in m3/min. Assume that the total pressure throughout the tower is 1.013 bar. 2. Condenser cooling water is supplied to the ID cooling tower at 40°C and is cooled to 3°C of approach temperature while falling through the tower. The air entering the tower is 35°C DBT and 28°C WBT. Determine the quantity of air in m3/min handled by the ID fan

314  Chapter 18 if the air leaves the tower at 38°C and 50% condenser requires 5000 L of cooling water per minute. Also find the makeup water required. Hint: Approach temperature = Temperature of water leaving the tower = WBT of entering air.

Further Readings Arora, S.C., Domkundwar, S., 2004. A Course in Refrigeration and Air Conditioning. Dhanpat and Co. Ltd, New Delhi (Chapter 13). Arora, C.P., 2000. Refrigeration and Air Conditioning. Tata McGraw-Hill Publishing Co. Ltd, New Delhi.