CHE504 Lab 2 Cooling Tower PDF

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Lab 2 – Cooling TowerEH2204E: Group 2AHMAD AIMAN ASYRAAF BIN AHMAD SEZALI (2019268738)MOHAMAD SYAFIQ HAFIZUDDIN BIN ABDULLAH (2019230338)NURUL MIZA NADIA BINTI MOHD BAHAROM (2019229824)NUR AINA FASIHAH BINTI KHAIRUL AMIR (2019294926)ABSTRACTAs we know, the experiment was conducted to evaluate and st...

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Lab 2 – Cooling Tower EH2204E: Group 2 AHMAD AIMAN ASYRAAF BIN AHMAD SEZALI (2019268738) MOHAMAD SYAFIQ HAFIZUDDIN BIN ABDULLAH (2019230338) NURUL MIZA NADIA BINTI MOHD BAHAROM (2019229824) NUR AINA FASIHAH BINTI KHAIRUL AMIR (2019294926) ABSTRACT

process heat and cool the working fluid to near the wet-

As we know, the experiment was conducted to evaluate

bulb air temperature or, in the case of closed circuit dry

and study the performance of cooling tower effected by

cooling towers, rely solely on air to cool the working

following parameters such as cooling load and air velocity.

fluid to near the dry-bulb air temperature. In this

Cooling tower were primarily used for heating, ventilation,

experiment, we use the SOLTEQ® Basic Cooling Tower

air conditioning (HVAC) and industrial purposes. Cooling

Unit (Model: HE152) which has been designed to

towers provided a cost-effective and energy efficient

demonstrate students the construction, design and

operation of systems in need of cooling. There were two

operational characteristics of a modern cooling system.

experiments been conducted with each experiment had 3

The unit resembles a full size forced draught cooling

runs respectively. Experiment 1 had different heater power

tower and it is actually an "open system" through which

for each run while Experiment 2 had different blower

two streams of fluid (in this case air and water) pass and

opening for each run. As the result, temperature (T1-T6),

in which there is a mass transfer from one stream to the

orifices differential pressure (DP1), water flowrate (FT1)

other. The unit is self-contained supplied with a heating

and heater power (Q1) were obtained from Experiment 1

load and a circulating pump.

while temperature (T1-T6), orifices differential pressure (DP1), pressure drop across packing (DP2) were obtained OBJECTIVES

from Experiment 2. Then, we were assigned to calculate

wet bulb approach and total cooling load in Experiment 1 The objective for Experiment 1 is to study the effect of while nominal velocity of air and approach to wet bulb in different heater power towards the cooling range. The Experiment 2. In Experiment 1, the wet bulb approach in objective for Experiment 2 is to study the effect of blower the 3 runs were -2.20K, -3.00K and -2.00K respectively opening towards wet bulb approach and pressure drop while the total cooling loads in the 3 runs were 1.5kW, through packing. 2.0kW and 2.5kW respectively. In Experiment 2, the nominal velocity of air in the 3 runs were 1.4622 (fully opened), 1.3786 (partially opened), and 0 (fully closed) respectively while the approach to wet bulb in the 3 runs were -2 K, 7.6 K and 10.8 K respectively. INTRODUCTION

THEORY A cooling tower is a specialized heat exchanger in which water and air are brought into direct contact in order to lower the temperature of the water. The process is happen when a small volume of water is evaporated, so it reducing the temperature of water that circulated through the tower. Cooling towers can either use water evaporation

A cooling tower is a heat rejection device that rejects to extract heat and cool the working fluid to near wet-bulb waste heat to the atmosphere through the cooling of a air temperature, or they can use air to cool the working water stream to a lower temperature. Cooling towers fluid to near dry-bulb air temperature in closed circuit dry may either use the evaporation of water to remove cooling towers. In this heat exchanger, the terms cross flow

and counter flow are used to explain how air flowing as the surface. The wet surface will reach the equilibrium through a cooling tower interacts with the process water temperature if unsaturated air is constantly circulated. So being cooled, as well as the variations between them. Air that, the heat is transferred mainly in the load tank where flows horizontally through the direction of falling water in the water is heated to the feed temperature and from this a cross flow tower, while air travels in the opposite principle, small amount of heat will lost to the surrounding. direction of falling water in a counter flow tower.

From the experiment, pumps work on the water and

According to the First Law of Thermodynamics, heat transferring the energy and mass loss as dry air enters and is a source of energy, and thermodynamic processes are humid air exits. The energy balance of the system can be thus subject to the principle of energy conservation. From defined as; this statement, we can say that heat energy cannot be created nor destroyed but it can be transferred from one Q + P = Hexit – Hentry

location to another location and transformed into and out

of other sources of energy. For this experiment, cooling Where ; tower is operated according to this law. When energy Q – the rate of heating added to the system enters the system, it must leave before it can diffuse P – rate work done by the pump on the water through the system. From this experiment, the hot water Hexit – rate of enthalpy loss in the exiting vapor which is entering the cooling water in form of energy was Hentry – rate of enthalpy gain due entering air and entering cooled down from temperature T1 to temperature T2. The water from make up tank hot water was cooled by convection, which involved blowing ambient air over the hot water at T1 and exiting

The wet bulb temperature is the temperature of

the cooling tower at some temperature T2. Enthalpy, adiabatic saturation where the temperature is indicated by a which is the sum of the internal energy (U) and the moistened thermometer bulb that exposed to the air flow. It product of pressure and volume (PV) is the primary is possible to measure it with a thermometer and a wet component in energy balance. The equation of energy muslin bulb. The rate of evaporation from the wet bandage on the bulb as well as the temperature difference between

balance is: H = U + PV

the wet and dry bulbs, are also affected by air humidity.

Which;

The evaporation is reduced when the air contains a lot of

H is enthalpy

water vapor. The temperature of the wet bulb is always

U is internal energy

between that of the dry bulb and that of the dew point.

P is pressure

There is a complex equilibrium for the wet bulb between

V is the volume

heat obtained because it is cooler than the ambient air and

In the cooling tower, heat is transferred mainly in the heat lost due to evaporation. The wet bulb temperature is load tank, where water is heated to feed temperature. The the maximum temperature that can be reached by difference between the vapor pressure at the liquid surface, evaporative cooling, assuming good air flow and constant the saturation pressure that corresponds to the surface ambient air temperature. The condition of the humid air is temperature, and the vapor pressure in the ambient air determined by combining the dry bulb and wet bulb determines the rate of evaporation from a wet surface into temperatures in a psychrometric diagram or Mollier map. the surrounding air. It is also influenced by the air's overall In the Psychrometric Chart, lines of constant wet bulb pressure and absolute humidity. Evaporation will continue temperatures extend diagonally from upper left to lower in an enclosed space until the two vapor pressures are right. equal and the air is saturated and at the same temperature

PROCEDURE

until the water cooled down.

Before Experiment:

2. Switch off the fan and fully close the fan damper.

General Start-up Procedure

3. Switch off the pump and power supply.

1. Check to ensure that valves V1 to V6 are closed and

4. Retain the water in reservoir tank for the following

valve V7 is partially opened. 2. Fill the load tank with distilled or deionised water. It is done by first removing the make-up tank and then

experiment. 5. Completely drain off the water from the unit if it is not used.

pouring the water through the opening at the top of the load bank. Replace the make-up tank onto the

Experiment 1 Procedure

load tank and lightly tighten the nuts. Fill the tank with distilled or deionised water up to the zero mark on the scale. 3. Add distilled/deionised water to the wet bulb sensor reservoir to the fullest. 4. Connect all appropriate tubing to the differential pressure sensor. 5. Install the appropriate cooling tower packing for the experiment. 6. Then, set the temperature set point of temperature

1) Prepare and perform the start-up procedure as shown above. 2) Set the system under the following conditions and allow stabilizing for about 15 minutes: Water flow rate : 1 LPM Air flow rate

: Maximum

Cooling load

: 0 kW

Column installed

: A (cross sectional area is

0.0225 m2)

controller to 50°C. Switch on the 1.0 kW water

3) After the system stabilizes, record a few sets of

heater and heat up the water until approximately

measurements (i.e. temperature (T1-T6), orifices

40°C.

differential pressure (DP1), water flowrate (FT1) and

7. Switch on the pump and slowly open the control valve V1 and set the water flowrate to 2.0 LPM.

heater power (Q1)), then obtain the mean value for calculation and analysis.

Obtain a steady operation where the water is

4) Without changes in the conditions, increase the

distributed and flowing uniformly through the

cooling load to 0.5kW. When the system stabilized,

packing. 8. Fully open the fan damper, and then

record all the data.

switch on the fan. Check that the differential pressure

5) Similarly, repeat the experiment at 1.0kW and 1.5kW.

sensor is giving reading when the valve manifold is switched to measure the orifice differential pressure. 8. Let the unit run for about 20 minutes, for the float valve to correctly adjust the level in the load tank. Refill the makeup tank as required. 9. Now, the unit is ready for use.

Experiment 2 Procedure 1) Prepare and perform the start-up procedure as shown above. 2) Set the system under the following conditions and allow stabilizing for about 15 minutes:

General Shut-down Procedure 1. Switch off heaters and let the water to circulate through the cooling tower system for 3-5 minutes

Water flow rate : 1 LPM Air flow rate

: Maximum (fully opened)

Cooling load

: 1 kW

Column installed is 0.0225

: A (cross sectional area

m2)

3) After the system stabilizes, record a few sets of measurements (i.e. temperature (T1-T6), orifices

Experiment 2: Manipulated variable

: Blower opening

Constant variable

: Water flow rate at 1 LPM and heater power of 1 kW

differential pressure (DP1), pressure drop across packing (DP2)), then obtain the mean value for Table 2.2 Differences of Blower Opening

calculation and analysis. 4) Similarly, repeat the experiment for different blower opening (i.e. partially opened and fully closed).

RESULTS

Blower

Fully

Partially

Fully

opening

opened

opened

closed

T1 (°C)

28.5

29.5

32.3

T2 (°C)

28.9

29.9

30.4

T3 (°C)

28.5

29.1

43.2

T4 (°C)

29.9

31.2

39.9

T5 (°C)

35.7

36.5

48.5

T6 (°C)

26.9

37.5

41.2

Differential

68

60

0

57

54

0

Experiment 1: Manipulated variable

:

Heater power

Constant variable

:

Water flowrate at 2 LPM

Blower’s Opening

:

Fully opened

Table 2.1 Differences of Heater Power Heater power (kW)

0.5

1.0

1.5

T1 (°C)

27.7

28.0

28.5

T2 (°C)

27.4

28.9

28.9

pressure orifice (Pa) Differential

T3 (°C)

25.2

26.6

28.5

T4 (°C)

26.5

27.7

29.9

T5 (°C)

29.1

32.1

35.7

pressure Column (Pa)

CALCULATION T6 (°C)

25.2

25.9

26.9 Experiment 1

Differential pressure

67

70

68

Orifice (Pa) Differential pressure Column (Pa)

Water flow remains constant = 2 LPM The variable is the heater power = 0.5 kW, 1.0 kW and 1.5

56

59

57

Kw The pump transfers approximately 100W to the water, and this should be added to the load imposed in the load tank. Total cooling load = Applied load + Pump input = 1.0 +

0.1 kW = 1.1 kW Approach to Wet Bulb = T6 – T2

ΔTcr= T5 – T6 = 305.25–299.05 K = 6.2 K

Cooling Range = T5 – T6 ΔTwb= T6– T2 = 299.05– 302.05 K At power = 0.5 kW = -3.00 K Total cooling load = Applied load + Pump input = 1.0 + 0.5 kW = 1.5 kW From data obtained when 100% flow air employed, T5 = 29.10 +273.15 K = 302.25 K T6 = 25.20 + 273.15 = 298.35 K T2 = 27.40 +273.15 K = 300.55 K ΔTcr= T5– T6 302.25– 298.35

At power = 1.5 kW Total cooling load = Applied load + Pump input = 1.0 + 1.5 kW = 2.5 kW T5 = 35.70 +273.15 K = 308.85 K T6 = 26.90 + 273.15 = 300.05 K T2 = 28.90 +273.15 K = 302.05 K

= 3.90 K ΔTcr= T5 – T6 ΔTwb =T6– T2 = 298.35 – 300.55

308.85 -300.05 K = 8.8 K

= -2.20 K ΔTwb =T6– T2 = 300.05 – 302.05 At power = 1.0 kW Total cooling load = Applied load + Pump input = 1.0 + 1.0kW = 2.0 kW T5 = 32.1 +273.15 K = 305.25 K T6 = 25.90 + 273.15 = 299.05 K T2 = 28.90 +273.15 K = 302.05 K

= -2.00 K

Outlet water temperature, T6 = 37.5°C Experiment 2

37.50 + 273.15

Fully opened

= 310.65 K

From data obtained when 100% flow air employed,

Approach to wet bulb

= 310.65 – 303.05 K

Inlet wet bulb temperature, T2 = 28.9°C

= 7.6 K

28.9 + 273. 15 = 302.05 K Outlet water temperature, T6 = 26.9 °C 26.90 + 273.15

Specific volume of air at outlet (by plotting air outlet dry bulb and air outlet wet bulb on psychometric chart) = 0.864 m3/kg

= 300.05 Approach to wet bulb

= 300.05 – 302.05 K = -2 K

Specific volume of air at outlet (by plotting air outlet dry bulb and air outlet wet bulb on psychometric chart) = 0.856

m3/kg

Air mass flow rate = 0.0137 �/�� x= 60 Pa x 1mm H2O / 10.13 Pa = 5.9230 mm H2O 0.0137 5.9230/0.864 ma= 0.0359kg/s Air volume rate = ṁvB

Air mass flow rate = 0.0137 �/��

vB=0.0359 x 0.864

x= 68 Pa x 1mm H2O / 10.13 Pa = 6.7127 mm H2O

= 0.0310 m3s-1

0.0137 6.7127/0.856

Cross sectional area of empty tower, A = 0.0225 m2

ma= 0.0384 kg/s

Air velocity,

Air volume rate = ṁvB

�� �

0.0310

= 0.0225

= 1.3786 ms-1

vB= 0.0384 x 0.856 = 0.0329 m3s-1 Cross sectional area of empty tower, A = 0.0225 m2 Air velocity,

�� �

Fully closed From data obtained when 100% flow air employed,

0.0329

= 0.0225

Inlet wet bulb temperature, T2 = 30.4°C

= 1.4622 ms-1

30.40 + 273.15 = 303.55 K Outlet water temperature, T6 = 41.2°C

Partially opened From data obtained when 100% flow air employed,

41.20 + 273.15

Inlet wet bulb temperature, T2 = 29.9°C

= 314.35 K

29.90 + 273.15 = 303.05 K

Approach to wet bulb

= 314.35 – 303.55 = 10.8 K

Specific volume of air at outlet (by plotting air outlet dry bulb and air outlet wet bulb on psychometric chart) = 0.895 m3/kg

DISCUSSION SOLTEQ Water Cooling Tower (Model HE152 Unit) was being used in this experiment. As we know, a cooling tower is a specialized heat exchanger in which

Air mass flow rate = 0.0137 �/��

water and air are brought into direct contact in order to

x= 0 Pa x 1mm H2O / 10.13 Pa = 0 mm H2O

lower the temperature of the water. The process is happen

0.0137 0/0.856

when a small volume of water is evaporated, so it reducing the temperature of water that circulated through the tower.

= 0kg/s

When warm water and colder air come together, the latent

Air volume rate = ṁvB

heat of vaporization is released and causing the water to cool. In this experiment, two experiments were done as

vB=0 x 0.856

cooling tower unit was being used. Experiment 1 was

= 0 m3s-1

carried out to determine the effect of different heater power

Cross sectional area of empty tower, A = 0.0225 Air velocity,

�� �

m2

at power 0.5 kW, 1.0 kW and 1.5kW towards the cooling range of the cooling tower. In Experiment 2, the

0

= 0.0225

experiment was conducted to investigate the impact of a

= 0 ms-1

cooling tower with varying blower opening at fully opened,

Table 2.3: Relationship of Nominal Velocity Air and Approach to Wet Bulb and Pressure Drop.

partially opened and fully closed. A load tank with a total of 2.5 kW electric heater, an air distribution chamber, a make-up tank, and a test column

Fully opened

Partially opened

Fully closed

make up the cooling tower assembly. When warm water is

Nominal velocity of air, ms-1

1.4622

1.3786

0

pumped from the load tank to the top of the column before

Approach to wet bulb, K

-2

7.6

10.8

Pressure, mm H2O

6.7127

5.9230

0

being distributed evenly at the top packing, the machine is turned on. The thin layer of water is cooled as it passes downward due to evaporation. The cooled water runs into the basin, then back into the load tank where it is reheated and recirculated. A make-up tank is used to keep the water level in the load tank stable. The machine has a blower that transfers air into the air distribution chamber. The air passes through the wet and dry bulb thermometers before entering the column. The moisture content of the air increases as it ascends the column, and the water cools. The air passes through a mist eliminator at the top of the column before being released into the atmosphere. From the experiment 1, we used different power heater at power 0.5kW, 1.0kW and 1.5kW. At 0.5kW,...