3.0 OBJECTIVES The main objectives for this experiment are: 1. To determine the correlation of water to air mass flow ratio with increasing water flow rate. 2. To determine the cooling load effect. 3. To know the effect of different flow rates on the wet bulb approach. 4. To estimate the evaporation rate of water (water loss) for the tower. 4.0 THEORY A cooling tower is a specialized heat exchanger that has been altered in which air and water are brought into direct interaction for the transfer of heat to take effect. In order to accomplish that, it is spraying a flowing mass of water by the spray-filled tower into a rain-like pattern, through which an upward moving mass flow of cool air is encouraged by the action of a fan. The principle of evaporative or „wet-bulb‟ cooling is used in cooling tower in order to level-headed the water. It has some advantages over a conservative heat-exchanger such as it can attain water temperatures below the temperature of the air used to cool it. Besides that, it is also lesser and cheaper for the identical cooling load. The heat increased by the air must equivalent to the heat lost by the water by equilibrium, by ignoring any insignificant amount of sensible heat exchange that may happen through the walls or exterior of the tower. Within the air stream, the rate of heat gain is identified by the expression G (h2 – h1), where: G = Mass flow of dry air through the tower—lb/min. h1 = Enthalpy (total heat content) of entering air—Btu/Ib of dry air. h2 = Enthalpy of leaving air—Btu/Ib of dry air. the rate of heat loss represented by this change in humidity ratio can be expressed as G (H2 . where: L = Mass flow of water entering the tower—lb/min. the total heat balance between air and water.32) where: H1 = Humidity ratio of entering air—lb vapor/lb dry air. the mass flow of water leaving the tower is less than that entering it.Btu/Ib. is: G dh = L dt + G dH (t2 . because of the evaporation that takes place within the tower and an appropriate heat balance must be take into account for this minor difference. stated as a differential equation. t2 = Cold water temperature leaving the tower—°F. (The enthalpy of water is zero at 32°F) Including this loss of heat through evaporation. the rate of heat loss would appear to be L (t1 – t2). t1= Hot water temperature entering the tower—°F.32) (1) . The notation (t2 . Still. This derives from the fact that a Btu (British thermal unit) is the amount of heat gain or loss necessary to change the temperature of 1 pound of water by 1°F.32) equals to an expression of water enthalpy at the cold water temperature .H1) (t2 . H2 = Humidity ratio of leaving air—lb vapor/lb dry air. Since the rate of evaporation must equal the rate of change in the humidity ratio (absolute humidity) of the air stream.Within the water stream. Nonetheless. which establishes two of the parameters required to size a cooling tower—namely. 81⁄3 = Pounds per gallon of water. .000 Btu/min performs best if supplied with 1. Therefore. and is undisturbed with the real hot and cold water temperatures themselves. The heat load established by the process creates a third parameter . we can determine the water temperature elevation through the process. the mere indication of a heat load is meaningless to the Application Engineer attempting to properly size a cooling tower. With a slight transformation of formula (2). More information of a specific nature is required. (2) where: gpm = Water flow rate through process and over tower— gal/min. For example.The expression “L dt” in equation (1) signifies the heat load imposed on the tower by whatever process it is serving. R = “Range” = Difference between hot and cold water temperatures—°F. gpm and cold water temperature. Optimal operation of a process usually occurs within a relatively narrow band of flow rates and cold water temperatures. the hot water temperature coming to the tower would be 85°F + 15°F = 100°F. Therefore. Note from formula (2) that heat load forms only a required temperature differential in the process water.hot water temperature coming to the tower.000 gpm of water at 85°F. let‟s assume that a process developing a heat load of 125. heat load is usually expressed as: Heat Load = gpm x R x 81⁄3 = Btu/min. because pounds of water per unit time are not straightforwardly measured. but air enthalpy is not something that is routinely measured and recorded at any geographic location. Figure 1 . and its relationship to other parameters is as shown in the Figure 1 diagram.Equation (1) would classify enthalpy to be of prime concern. Wet bulb temperature is the only air parameter needed to properly size a cooling tower. 0 LPM. To measure the differential pressure across the orifice. 2. Check that the differential pressure sensor is giving the reading : a. 6. All appropriate tubing to the differential pressure sensor was connected. .0 kW water heaters is switched on and the water is heated up to approximately 40˚C. The temperature set point of temperature controller was set to 45˚C. The water flow rate was set to 2. 9. Refill the make-up tank as required. open valve V4 and V5.0 APPARATUS Water cooling tower MODEL: HE-152 6. 3. 7.0 PROCEDURE General start-up procedure. The 1. Deionised water is added to the wet bulb sensor reservoir to the fullest. b. 4. close valve V3 and V6. Valve V1 to V6 were ensured to be closed while valve V7 was partially closed. The load tank was filled with deionised water. The unit is now ready to use. A steady operation where the water was distributed and flowing uniformly through the packing was obtained. 1. 10. To measure the differential pressure across the column.5. The appropriate cooling tower was installed for the experiment. The make-up tank was filled with deionised water up to zero mark on the scale. The fan damper was fully opened and the fan was switched on. The pump was switched on and the control valve V1 was slowly opened. 5. 8. open valve V3 and V6. The unit was being let to run for 20 minutes for the float valve to correctly adjust the level in the load tank. close valve V4 and V5. 11. 3.Experiment 1 1. Set the water flow rate to 1. Pump and blower is then been switched on. The heater was switched off to let the water to circulate through cooling tower for 3-5 minutes until the water is cooled down. Experiment 2 1. The water cooling tower was being let to operate for 20 minutes. 5. The water flow rate was set to 1. 4. 5. The reading was taken after steady operation achieved. The heater was switched on and set to 0.0 kW heating load.5 LPM. The blower damper was fully opened. Step 1-5 is being repeated with 1. The pump and power supply was switched off. The blower was switched off and the blower damper was fully closed. 3. The unit was being let to run for 20 minutes. 6.0 kW heating load. 6. 4.5 kW. . 2. The blower damper was semi opened. 3. 2.5 LPM. 7. 2. General shut-down procedure 1. 4.5 kW. The reading was taken when the float valve is correctly adjusted. The heater was switched on and set to 0. 5. Step 1-6 is being repeated with 1. The water from the unit was completely drained off. The water in the reservoir tank was retained.