1.0
SUMMARY
This experiment is held to investigate the effect of air velocity on wet bul b approach and pressure drop through the packing. Theoretically, with lower air velocity, the approach on wet bulb in terms of temperature becomes becomes higher. Different with pressure drop, the lower the air flow, fl ow, value of pressure in unit Pa, lowers as well. Shortly, based in Figure 1, the relationship is explained; nominal velocity in function of wet bulb approach, K and packing pressure drops. The performance of cooling tower is basically affected with wet bulb approach; due to the increment of the air velocity made by the fan or blower used used in the experiment experiment which which cause cause the dry air to hold more more water water vapour rather than than it can can hold at high temperature. Meanwhile for pressure drops, speed of the fan or the blower used is high, there will be more pressure drop occur to reduce the temperature of the hot water. As a conclusion, the nominal velocity of the air is inversely proportional to the wet bulb and the pressure drop. It simply means when the air velocity increasing, the wet bulb approach and the pressure drop decreasing.
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2.0
INTRODUCTION
The laboratory cooling tower is a cooling tower unit from a commercial air conditioning system used to study the principles of cooling tower operation. It is used in conjunction with a residential size water heater to simulate a cooling tower used to provide cool water to an industrial process. In the case of the laboratory unit, the industrial process load is provided by the water heater. The laboratory cooling tower allows for complete control of the speed of the fan used in cooling the warm return water and the pump used to return the cooled water to the water heater. Experiments can be conducted which study how adjustment of one or both parameters affect the amount of heat removed from the water provided to the water heater. Cooling towers are heat transfer devices used to remove process waste heat to the atmosphere. Cooling towers may either use the evaporation of water to remove process heat and cool the working fluid to near the wet-bulb air temperature or rely solely on air to cool the working fluid to near the dry-bulb air temperature. Common applications include cooling the circulating water used in oil refineries, chemical plants, and building cooling. The towers vary in size from small roof top units to very large hyperboloid that can be up to 200 meters tall and 100 meters in diameter, or rectangular structures that can be over 40 meters tall and 80 meters long. Smaller towers are normal factor-built, while larger ones are constructed on site. The most widely used in the process industries for employing water by using re-circulated cooling water systems. In the cooling water systems, the processes involved are rejecting the heat from water by evaporation and remove process waste heat into the environment. The cost for this process is inexpensive and very dependable means of removing low grade heat from your process. Environmental considerations, by minimizing consumption of potable water, minimizing the generation and release of contaminated cooling water, and controlling the release into the environment of chemicals from leaking heat exchanger (HX), form the second major reason. It is used to provide lower than ambient water temperatures and are more cost effective and energy efficient than most other alternatives cooling towers are commonly used in many commercial and industrial processes, according to its classifying use. Cooling towers also can be categorized by its air-to-water flow. Cross flow is one of them. Cross flow is a design in which the air flow is directed perpendicular to the water flow. 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. The air continues through the fill and thus past the water flow into an open plenum area. A distribution or hot water basin consisting of a deep pan with holes or nozzles in the bottom uti lized in a cross-flow tower. Gravity distributes the water through the nozzles uniformly across the fill material. The counter flow is another design for cooling tower. It is completely opposite to the above cross flow design. 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. The air continues through the fill and 2
thus past the water flow into an open plenum area. A distribution or hot water basin consisting of a deep pan with holes or nozzles in the bottom is utilized in a cross-flow tower. Gravity distributes the water through the nozzles uniformly across the fill material. There are several formulas involve in this experiment including
)
1. Air mass flowrate, ṁ (
m = 0.0137
√ ℎ
h: orifice differential in mm H2O v: specific volume of air
3
̇
2. Air volumetric flowrate, ( )
̇ = ̇ )
3. Nominal air velocity v ( v =
̇
A: packing area m2
4. Wet bulb approach Wet bulb approach = Outlet water temperature(T6) – inlet air wet bulb temperature(T2)
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3.0
RESULT & DATA
Table A Description
Unit
Air Flow 100%
75%
50%
25%
Packing density
m-1
110
110
110
110
Air inlet dry bulb, T1
˚C
26.65
26.55
26.8
27.2
Air inlet wet bulb, T2
˚C
30.1
30.35
31.3
31.6
Air outlet wet bulb, T3
˚C
29.8
29.2
29.95
30.9
Air outlet dry bulb, T4
˚C
34.8
33.9
34.85
36.2
Water inlet temperature, T5
˚C
28.15
27.4
28.1
29.4
Water outlet temperature, T6
˚C
28.1
28.15
29.4
30.9
Orifice differential, DP1
Pa
89.2
68.8
43.45
22.3
Water flow rate, FT1
LPM
2.0
2.0
2.0
2.0
Heater power, Q1
Watt
1000
1000
1000
1000
Pressure drop across packing, DP2
Pa
40
32
20
13.4
Table B
Air Flow
100%
75%
50%
25%
Nominal Velocity of Air, m/s
3.46×10-8
3.024×10-8
2.443×10-8
1.757×10-8
Wet Bulb Approach, K
-2
-2.2
-1.9
-0.7
Packing Pressure Drops, mm H2O
4.0
3.2
2.0
1.34
4
e r u s s 50 e r p g 40 n i k ) c 30 a O p 2 d H 20 n m a m ) ( K s 10 ( h p c o r 0 o r d p 0 -10 p a p l u b t e w
y = -9.18x + 49.3
y = 0.42x - 2.75 0.5
1
1.5
2
2.5
3
3.5
4
4.5
Norminal velocity of air (m/s) Wet Bulb Approach, K
pressure
Linear (Wet Bulb Approach, K)
Linear (pressure)
Figure 1: Wet bulb approach and packing pressure drops vs nominal velocity
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ANALYSIS & DISCUSSION
Cooling tower is a heat rejection device that is used to transfer process waste heat to the atmosphere. Cooling tower able to lower the temperature of water more than the devices which commonly use only air to reject heat such as radiator. In other word, cooling tower is other kind of specialized heat exchanger in which air and water are brought into direct contact with each other to reduce the water’s temperature. As this occurs, a small volume of water is evaporated, reducing the temperature of the water circulated throughout the tower. However, not all cooling tower are suitable for all applications. Cooling towers are designed and manufactured in several types with numerous sizes. Basically, there are three types of cooling tower which are natural draft, force draft and the induced draft. However, in this experiment, force draft cooling tower was used as the main equipment. The cooling tower is known as the force draft type because it applied fans to produce draft. The fundamental of the cooling tower is quite simple where air was forced by the fan and flowed from the bottom to the top while the hot water flowed in the opposite direction. The hot water, flow from the top of the cooling tower and make a contact with the cold air forced upward by the fan resulting a cold temperature of water collected in the water basin and flow out at the bottom of the machine. Additionally, there are two types of flow that related to the cooling tower which are the cross flow and the counter flow. The cross-flow condition happens when the water flows vertically through the fill while the air flows horizontally across the falling water resulting the does not have to pass through the distribution system, applying the gravity flow of hot water distributions basins mounted at the top of the unit above the fill. Meanwhile, the counter flow condition happens when the air flows vertically upward counter to the falling water in the fill (Hensley, J. C. et all.,1985). Due to the vertically air flow, open, gravity flow basins which is typical in the cross-flow type cannot be use. Basically, this experiment was conducted to investigate the effect of air velocity on wet bulb approach and pressure drop through the packing using the force draft type cooling tower that apply cross flow method. The effect of air velocity on the parameter is based on four different air flows which is 100%, 75%, 50% and 25%. In this experiment, there were 10 data that were measured in this experiment which are air inlet dry bulb T1, air inlet wet bulb T2, air outlet dry bulb T3, air outlet wet bulb T4, water inlet temperature T5, water outlet temperature T6, orifice differential DP1, water flow rate FT1, heater power Q1 and lastly t he pressure drops across the packing DP2.
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From table A, as observed, it can be clearly seen that the value of T1, T2, T3, T5 & T6 is increasing as the percentage of the air flow is reduced. This is due to the reduced amount of the air flow or the forced draft by the fan which cause the increase in the water temperature. For example, the differences can be seen for the 100% air flow, the wet bulb T1 is 26.65 ̊C while it increased to 27.2 ̊C for the 25% air flow. However, there are some fluctuation pattern on the result for the air outlet wet bulb T4 where for the 100% air flow, the value initially 34.8 ˚C. At the 75% air flow, the value of the wet bulb air decreases a little into 33.9 ˚C and then rise again for the 50% air flow into 34.85 ˚C. Meanwhile, for the packing density, water flow rate and the heater power, there were no changes made throughout the experiment where the value is kept constant until the end of the experiment. For the pressure drop across packing DP2, the value is decreasing where the pressure is 40 pascals for 100% air flow and turn into 13.4 pascal for the 25% air flow. Hence, it can be deduced that the pressure drop across packing is decrease as the air flow percentage decrease. From figure 1 which is a graph that has been plotted to show a relationship between wet bulb approach and packing pressure drop against nominal air velocity. Wet bulb approach seems to decrease linearly from -0.7K to -2K as the nominal air increase whereby the packing pressure drop increase linearly from 1.34mmH2O to 4.0mmH2O towards the nominal air velocity. Based from the theory, the wet bulb approach line should be decreases and cross the intersection line of the packing pressure drop that is in increasing. This crossing point shows the limit of the wet bulb approach and packing pressure drop at the highest nominal air velocity or at the 100% air flow. The cooling tower selection and performances is based on the wet bulb approach. As the nominal air velocity increased, the wet bulb approach will decrease. This is due t o the increment of the air velocity made by the fan or blower used in the experiment which cause the dry air to hold more water vapour rather than it can hold at high temperature (Bedekar, S. V., et all.,1998). Meanwhile, the packing pressure drop increase as the nominal air velocity rise. This can be justified by relating to the speed of the fan whereas the speed of the fan or the blower used is high, there will be more pressure drop occur to reduce the temperature of the hot water. In order to ensure the air flow upward through the column, a higher pressure at the bottom of the column is a necessity (Bernier M. A.,1994). Meanwhile, the water flows downward throughout the packing which against the pressure and the flowing air. This is due to the involvement of the gravity pull that makes water denser than the air.
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While running the experiment, there are might be some error that occurred which affect the result obtained when reading was taken. Firstly, it is i mportant to make sure the temperature is 50 ˚C to make it stable enough to get the expected result. A temperature that lower than 50 ˚C could cause error in the temperature reading throughout the experiment. Besides, it is essential to take a break about 10 minutes interval between each air flow percentage reading as it helps the equipment to stabilized to its initial state. Lastly, the air flow percentage blower must be adjusted perfectly. For 100%, it looks easy because the blower cover just need to fully open but when it comes to 75%, 50% and 25%, it is very hard to measure the value correctly. As a result, the temperature from T1 to T6 might be not accurate as the air flow is not following the percentage it should be.
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5.0
CONCLUSION AND RECOMMENDATIONS
The experiment conducted investigating the effect of the air velocity on the wet bulb approach and the pressure drop through the packing. Based on the result obtained, it can be concluded that the nominal velocity of the air is inversely proportional to the wet bulb and the pressure drop. It simply means when the air velocity increasing, the wet bulb approach and the pressure drop decreasing. Moreover, the realationship between the approach to wet bulb and the pressure drop through packing is inversely proportional to each other. This was due to the evaporation process which was the liquid evaporates into the gas phase of another material that was cause from its vapor pressure in the gas phase is lower than the saturated vapor pressure. There were some recommendations to overcome the problem or errors for this experiment. Firstly, keep the cooling tower insulated so the is lesser heat loss to the surrounding. Then, it is recommended for the cooling tower to have the digitalized scalar to adjust the flow rate of air. Thus, the ideal expected results could be achieved. Next , after the pump was switched on, it needs to stablize in about 15 minutes and need to be control before switching the blower button. Besides, this experiment may having some error because in manually control the air blower and it is difficult to ensure the stability of the air velocity taken from appeared the reading on the screen monitor. The manual set up need to be replace by automatic set up to reduce parallax error that making the result not really accurate.. Last but not least, the experiment should be repeated for at least three times to get desired more precise result.
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REFERENCES
Bernier, M. A. (1994). Cooling tower performance: theory and experiments (No. CONF-9406105--). American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc., Atlanta, GA (United States). Bedekar, S. V., Nithiarasu, P., & Seetharamu, K. N. (1998). Experimental investigation of the performance of a counter-flow, packed-bed mechanical cooling tower. Energy, 23(11), 943947. Hensley, J. C. (Ed.). (1985). Cooling tower fundamentals. Marley Cooling Tower Company. Milosavljevic, N., & Heikkilä, P. (2001). A comprehensive approach to cooling tower design. Applied Thermal Engineering , 21(9), 899-915. SHRYOCK, D. R. B. H. A. (1961). A comprehensive approach to the analysis of cooling tower performance. J. Heat Transfer , 83(3), 339-349. Shah, R. K., & London, A. L. (2014). Laminar flow forced convection in ducts: a source book for compact heat exchanger analytical data. Academic press. Whitaker, S. (1972). Forced convection heat transfer correlations for flow in pipes, past flat plates, single cylinders, single spheres, and for flow in packed beds and tube bundles. AIChE Journal , 18(2), 361-371.
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APPENDIX
Calculation at 100% Air Flow: 1. Calculate approach to wet bulb
= Outlet Water Temperature (T 6) – Inlet Air Wet Bulb Temperature (T1) = (28.1 + 273.15) K – (30.1 + 273.15) K = -2 K 2. Check the value of specific volume of air at outlet in the enthalpy-specific volume chart.
T3 = (29.8 + 273.15) K = 302.95 K T4 = (34.8 + 273.15) K = 307.95 ~From the enthalpy-specific volume chart , the value of specific volume is 0.8714 m3/kg dry air. 3. Calculate the air mass flow rate, (kg/s)
h: orifice differential in mm H2O v: specific volume of air (m3/kg) = 4.368 ×
− kg/s
4. Calculate air volumetric flow rate, (m3/s)
10−6 kg/s x 0.8896 m /kg = 3.806×− m /s 3
= 4.368 ×
3
5. Calculate nominal air velocity, v (m/s)
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A = packing area (m2) Calculate packing area:
A = 110 x [0.15 m x 0.15 m x 0.6 m] A = 1.485 m2 v = 3.46×
− m/s
Calculation at 75% Air Flow: 1. Calculate approach to wet bulb
= Outlet Water Temperature (T 6) – Inlet Air Wet Bulb Temperature (T1) = (28.1 + 273.15) K – (30.5 + 273.15) K = -2.2 K 2. Check the value of specific volume of air at outlet in the enthalpy-specific volume chart.
T3 = (29.2 + 273.15) K = 302.35 K T4 = (33.9 + 273.15) K = 307.05 K ~From the enthalpy-specific volume chart , the value of specific volume is 0.8571 m3/kg dry air. 3. Calculate the air mass flow rate, (kg/s)
h: orifice differential in mm H2O v: specific volume of air (m3/kg) = 3.881 ×
− kg/s
4. Calculate air volumetric flow rate, (m3/s)
10−6 kg/s x 0.8896 m /kg
= 4.368 ×
3
12
− m /s 3
= 3.326×
5. Calculate nominal air velocity, v (m/s)
A = packing area (m2) Calculate packing area:
A = 110 x [0.15 m x 0.15 m x 0.6 m] A = 1.485 m2 v = 3.024×
− m/s
Calculation at 25% Air Flow: 1. Calculate approach to wet bulb
= Outlet Water Temperature (T 6) – Inlet Air Wet Bulb Temperature (T1) = (28.129.4 + 273.15) K – (31.3 + 273.15) K = -1.9 K 2. Check the value of specific volume of air at outlet in the enthalpy-specific volume chart.
T3 = (29.95 + 273.15) K = 303.1 K T4 = (34.85+ 273.15) K = 308 K ~From the enthalpy-specific volume chart , the value of specific volume is 0.8857 m3/kg dry air. 3. Calculate the air mass flow rate, (kg/s)
h: orifice differential in mm H2O v: specific volume of air (m3/kg) = 3.034 ×
− kg/s 13
4. Calculate air volumetric flow rate, (m3/s)
10−6 kg/s x 0.8896 m /kg = 2.687×− m /s 3
= 4.368 ×
3
5. Calculate nominal air velocity, v (m/s)
A = packing area (m2) Calculate packing area:
A = 110 x [0.15 m x 0.15 m x 0.6 m] A = 1.485 m2 v = 2.443×
− m/s
Calculation at 25% Air Flow: 1. Calculate approach to wet bulb
= Outlet Water Temperature (T 6) – Inlet Air Wet Bulb Temperature (T1) = (30.9 + 273.15) K – (31.6 + 273.15) K = -0.7 K 2. Check the value of specific volume of air at outlet in the enthalpy-specific volume chart.
T3 = (30.9 + 273.15) K = 304.05 K T4 = (36.2 + 273.15) K = 309.35 K ~From the enthalpy-specific volume chart , the value of specific volume is 0.8929 m3/kg dry air.
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3. Calculate the air mass flow rate, (kg/s)
h: orifice differential in mm H2O v: specific volume of air (m3/kg) = 2.165 ×
− kg/s
4. Calculate air volumetric flow rate, (m3/s)
10−6 kg/s x 0.8896 m /kg = 1.933 × − m /s 3
= 4.368 ×
3
5. Calculate nominal air velocity, v (m/s)
A = packing area (m2) Calculate packing area:
A = 110 x [0.15 m x 0.15 m x 0.6 m] A = 1.485 m2 v = 1.757×
− m/s
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