## INTRODUCTION so doing, it cools the warm return

INTRODUCTION
A lab cooling tower is a heat removal mechanism used to study the theory of an actual cooling tower operation and show the heat and mass transfer as well as the mass and energy balance in a system. In this experiment, the heat from the process stream is delivered by the water heater which heats up the water (from load tank) entering at top of the tower. The lab cooling tower allows the speed of the fan which absorb air from the atmosphere and distribute into the bottom of tower column, in so doing, it cools the warm return water and the pump used to return the cooled water to the water heater.

A mass and energy balance on the cooling tower will be presented along with the results of experiments in which mass flowrate of the air steam in the tower was calculated. More details of report about the theory behind the operation of a cooling tower and how the laboratory cooling tower is operated will discussed on the remaining part of the report.

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THEORY
Cooling tower operate by extracting waster heat to the atmosphere at the same time cooling a water stream to a low temperature. This occur by water surface being exposed to air flowing up through the cooling tower, by so doing, a small part of water is cooled to evaporate into the air in order to supply significant cooling to the returning water stream. At this time, heat of evaporation is carried into the bulk air by the water vapour. The heat from the water stream therefore increases the temperature and specific humidity of air and air is released to the atmosphere at the top of the tower. The wet and dry bulb temperature of the air entering and leaving the column are recorded. From these temperatures, a psychometric chart is utilised to find the air inlet and outlet streams parameters such as Relative Humidity, Specific Volume, Humidity Ratio, Dew Point Temperature and Enthalpy.

The principle behind the whole operation of the unit of a cooling tower is the First Law of Thermodynamics which is the conservation of energy. In simpler terms, energy entering the system must exit the system; energy can neither be destroyed nor created, it just transform from one form to another (Mark, 2011). In the cooling tower energy enters in the form of hot water. This hot water was cooled from an inlet temperature (ST-2) to an outlet temperature (ST-7). The water is cooled by the upward flowing air stream through forced convection at the inlet temperature which then gets heated and exits at a temperature than its initial.
PROCEDURE
The water heater was switched on and the water in the tank is allowed to reach its set-point temperature.
Once the set-point temperature is met, the level of water in the reservoir was recorded as well as the dimensions of the tank.

The fan and water pump is then switched on.
A stopwatch was set for 15 minutes as soon as the pump and fan are switched on.
After 15 minutes, the experiment stopped.
The following measurements were recorded:
• Set Point Temperature of the water inlet stream (ST1)
• Dry and Wet Bulb Temperature of Air inlet stream (ST4 and ST3)
• Dry and Wet Bulb Temperature of Air outlet stream (ST6 and ST5)
• Temperature of water inlet stream (ST2)
• Temperature of water outlet stream (ST7)
• Water inlet flowrate (SC-1)
• Air flow rate (AVE-1)
• Level of water in the reservoir after cooling
RESULTS
Measurements Values
ST-1 54.5 °C
ST-2 59.3 °C
ST-3 31.1 °C
ST-4 31.6 °C
ST-5 46.5 °C
ST-6 48.3 °C
ST-7 44.2 °C
SC-1 0.79 L/min
Table 1. Results obtain from the Sensors
Calculations
2.4.1
Parameters
Air inlet stream Air outlet stream
Relative Humidity 97 % 90 %
Specific Volume 0.904 m3/kg dry air 1.013 m3/kg dry air
Humidity Ratio 0.288 kg/kg dry air 0.07 kg/kg dry air
Dew Point Temperature 30.96 °C 46.26 °C
Enthalpy 105.1 kJ /kg dry air 229.69 kJ /kg dry air
2.4.2 ?Hair= Hout-Hin =229.69 kJ/kg dry air-105.1 kJ /kg dry air =124.59kJ/kg dry air More heat energy has been transferred to the bulk air through evaporation.

2.4.3 ?w3=0.79Lmin×1min60sec×0.001m31L×983.64kgm3 =1.295×10-2 kg/sEnthalpies of liquid water
H3 = @ 59.2 °C = 246.97 kJ /kg H2O
H4 = @ 44.2 °C = 184.25 kJ /kg H2O
?Hwater= Hin-Hout = 246.97 kJ /kg H2O – 184.25 kJ /kg H2O
= 62.72 kJ /kg H2O
Mass balance of air
?a1 = ?a2 = Ma
Water mass balance
?w3 + ?a1?1 = ?w4 + ?a2?2
?w3 + Ma*(?1- ?2) = ?w4
Energy Balance
?w3h3 + Mah1 = ?w4h4 + Mah2
Ma*(h1 – h2) = (?w3 + Ma*(?1- ?2))*h4 – ?w3h3
Ma*(h1 – h2) – Ma*(?1- ?2)*h4 = ?w3h4 – ?w3h3
Ma*(h1- h2) – (?1- ?2)*h4 = ?w3*(h4 – h3)
?Ma=M3(h4-h3)(h1-h2)-h4(?1-?2) = 1.295×10-2 (184.25-246.97)105.1-229.69-184.25(0.0288-0.07) = 6.94 x 10-3 kg/s
2.4.4.1. ?w4 = ?w3 + Ma*(?1- ?2)
= 1.295×10-2+6.94×10-3(0.0288-0.07) = 0.0127 kg/s
2.4.5. ?w(evaporated)= ?w3-?w4 = 0.01295-0.0127 = 2.5 x 10-4 kg/s
2.4.6. The purpose of the fill in the cooling tower is to provide a place for efficient heat transfer.

DISCUSSION
The parameter of the air inlet and outlet streams were obtain by the use of wet and dry bulb temperatures in the psychometric chart. The enthalpy difference(?H.) of water and air is obtained from their inlet and outlet enthalpies. Water is equal to 62.72 kJ /kg H2O whereas air is 124.59kJ/kg dry air. ?Hair in the air stream appear to be higher than that of the initial enthalpy of air (105.1 kJ /kg dry air). From these results we deduct that large amount of heat has been transferred via convention into the bulk air.

The inlet and outlet masses of water are determined to be 0.01295 kg/s and 0.0127 kg/s respectively. The slight difference in these result indicate that there was a loss of energy in the water stream.

CONCLUSION
The differences in the heat transfer rates of the air and water flows indicate that the various heat losses in the system are vital and should not be ignored.

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