ERT 318/4
UNIT OPERATIONS
SEMESTER 1 (2013/2014)
WATER COOLING TOWER
School of Bioprocess Engineering
University Malaysia Perlis
EXPERIMENT
Water Cooling Tower
1.0
OBJECTIVES
1.1 To determine energy and mass balances using steady flow equation on selected system.
1.2 To investigate the effect of cooling load on “ Wet Bulb Approach”.
1.3 To investigate the effect of air velocity on: (a) Wet Bulb Approach, (b) pressure drop
through the packing
2.0
CORRESPONDING COURSE OUTCOME (CO)
This experiment will give students an appreciation of the construction, design and operational
characteristics of a modern evaporative cooling system.
3.0
INTRODUCTION
3.1

Cooling Tower Performance
A study on the performance of a cooling tower is to verify the effect of the following factors on the cooling
tower performance:
 water flow rates 




water temperatures 
air flow rate 

inlet air relative humidity 
The effect of these factors will be studied in depth by varying them. In this way, we can gain an overall
view of the operation of the cooling tower.
3.2
Thermodynamic Property
In order to understand the working principle and performance of a cooling tower, a basic knowledge
of thermodynamic is essential. A brief review on some of the thermodynamic properties is
presented below.
3.2.1
Specific Enthalpy
At water triple point (i.e. 0.00602 atm and 0.01°C), the specific enthalpy of saturated water is
assumed to be zero, which is taken as datum. The specific enthalpy of saturated water (hi) at a
range of temperatures above the datum condition can be obtained from thermodynamic tables.
The specific enthalpy of compressed liquid is given by
h = h f + v f ( p − psat )
(1)
The correction for pressure is negligible for the operating conditions of a cooling tower, thus h ≈
hf at a given temperature.
3.2.2
Specific Heat Capacity, Cp
Specific heat capacity, Cp is defined as the rate of change of enthalpy with respect to temperature (often
called the specific heat at constant pressure). For the purpose of experiment, we may use the following
relationship:
∆h = C p ∆T
and
(2)
h = C pT
(3)
where Cp for water is taken as 4.18 kJ/kg⋅°
3.2.3
Dalton' and Gibbs Lawss
It is commonly known that air consists of a mixture of dry air (O2, N2 and other gases) and
water vapour. Dalton and Gibbs Laws describe the behaviour of such a mixture as:
The total pressure of the air is equal to the sum of the pressures at which the dry air and the water
vapour each and alone would exert if they were to occupy the volume of the mixture at the
temperature of the mixture.
The dry air and the water vapour respectively obey their normal property relationships at their partial
pressures.
The enthalpy of the mixture may be found by adding together the enthalpies at which the dry air and
water vapour each would have as the sole occupant of the space occupied by the mixture and at
the same temperature.
3.2.4
Humidity and Saturation
Absolute or Specific Humidity, ω =
Relative Humidity, φ =
Mass of water vapour
Mass of dry air
_____ Partial pressure of water vapour in Air_____________
Saturation pressure of water vapour at the same temperature
(4)
(5)
_____Mass of water vapour in given volume of Air___________
Percentage Saturation =
Mass of same vol. of sat. water vapour at the same temperature
At high humidity conditions, it can be shown that there is not much difference between the
relative humidity and the percentage saturation and thus we shall regard them as the same.
3.2.5
Psychometric Chart
The psychometric chart is very useful in determining the properties of air/water vapour mixture.
Among the properties that can be defined with psychometric chart are Dry Bulb Temperature, Wet
Bulb Temperature, Relative Humidity, Humidity Ratio, Specific Volume and Specific Enthalpy.
Knowing any two of these properties, the other properties can be easily identified from the
chart provided the air pressure is approximately 1 atm.
(6)
3.3
Orifice Calibration
Psychometric chart can be used to determine the value of the specific volume. However, the
values given in the chart are for 1 kg of dry air at the stated total pressure. For every kilogram of
dry air, there is w kg of water vapour, yielding the total mass of (1 + w) kg. Thus, the actual specific
volume of the air/vapour mixture is given by:
vab
va = 1+ ω
(7)
The mass flow rate of air and steam mixture through the orifice is given by
m = 0.0137√
x
va
x(1 + ω )
= 0.0137√ vab
&
(8)
where m = mass flow rate of air/vapour mixture
va = actual specific volume
vab = specific volume of air at the outlet
x
= orifice differential, mm H2O
ω = humidity ratio of mixture
The mass flow rate of dry air is given by
ma =
=
1 × mass flow rate of air/vapour mixture
1+ ω
1
1+ ω
0.0137 √____
× 0.0137 x√ (1 + ω )
vab
x=
vab (1 + ω )
A simplification can be made since in this application, the value of ω is unlikely to exceed 0.025.
As such, neglecting wb would not yield significant error.
(9)
3.4
Application of Steady Flow Energy Equation
Consider System A for a cooling tower as defined in
Figure 2.1. It can be seen that for this system,
indicated by the boundary line,

heat transfer at the load tank and possibly a
small
 quantity to surroundings 
work transfer at the
pump
 low humidity air enters at point A
 high humidity air leaves at point B

make-up enters at point E, the same amount
as
 the moisture increase in the air stream
Figure 2.1: System A
From the steady flow equation,
Q − P = H out − H in
(10)
= ma hda + ms hs − (mg hda + ms hs ) A − mE hE
The pump power, P is a work input, therefore it is negative.
If the enthalpy of the air includes the enthalpy of the steam associated with it, and this quantity is
in terms of per unit mass of dry air, the equation may then be written as:
Q − P = ma (hB − hA ) − mE hE
(11)
The mass flow rate of dry air, ma through a cooling tower is constant, whereas the mass flow rate
of moist air increases as the result of evaporation process. The term mE hE can usually be
neglected since its value is relatively small.
Under steady state conditions, by conservation of mass, the mass flow rate of dry air and of
water (as liquid or vapour) must be the same at inlet and outlet to any system. Therefore,
(ma ) A = (ma )B
and (ma )A + mE = (ma )B or mE = (ma )B − (ma )A
The ratio of steam to air is known for the initial and final state points on the psychrometric
charts. Therefore,
(ms )A = maωA and (ms )B = maωB
mE = ma (ωB − ωA )
(12)
Let re-define the cooling tower system to be as in
Figure 2.2 where the process heat and pump
work does not cross the boundary of the system.
In this case warm water enters the system at point
C and cool water leaves at point D.
Figure 2.2: System B
Again from the steady flow energy equation, Q − P = H out − H in , where P = 0.
Q may have a small value due to heat transfer between the unit and its surroundings: Q
= ma hB + mw hD − (ma hA + mw hC + mE hE )
(18)
Rearranging,
Q = ma (hB − h A ) − m w (hD − hC ) − m E hE
Q =( ma hB − h A )− m w C p (t D − t C ) − m E hE
(19)
Again, the term mE hE can be neglected.
4.0
MATERIALS AND EQUIPMENT
4.1
Apparatus
1
2
3
7
4
8
5
9
6
10
Figure 1: Parts Identification and Equipment Set-up of Water Cooling Tower
Legend:
1
Orifice
6
Air Blow
2
Water Distributor
7
Control Panel
3
Packed Column
8
Make-up Tank
4
Flowmeter
9
Load tank
5
Receiver tank
10
Heater
4.2
Components of the Water Cooling Tower Unit
4.2.1 Load Tank
The stainless steel load tank has a capacity of 9 litres. It is fitted with two cartridge
heaters, 0.5 kW and 1.0 kW each, to provide a total of 1.5 kW of cooling load. A make-
up tank is fixed on top of the load tank. A float type valve at the bottom of the make-up
tank is used to control the amount of water flowing into the load tank.A centrifugal type
pump (work input = 40 W) is supplied for circulating water from the load tank through a
flow meter to the top of the column, into a basin and back to the load tank. A
temperature sensor and temperature controller is fitted at the load tank to prevent
overheating. A level switch is fitted at the load tank so that the heater and pump will be
switched off if a low level condition occurs.
4.2.2 Air Distribution Chamber
The stainless steel air distribution chamber comes with a water collecting basin and a
one-side inlet centrifugal fan. The fan has a capacity of approximately 235 CFM of air
flow. The air flow rate is adjusted by means of an intake damper.
4.2.3 Tower and Packing
2
The tower is made of clear acrylic with a square cross-sectional area of 225 cm (15
2 3
cm x 15 cm) and a height of 60 cm. The packing density of tower is 110 m /m for
2 3
Column A and 77 m /m For Column B. It comes with eight decks of inclined packing.
A top column that fitted on top of the tower comes with a sharp-edged orifice, a droplet
arrester and a water distribution system
.
4.3 Operation Processes
4.3.1 Water Circuit
Water temperature in load tank will be increased before the water is pumped through
the control valve and flow meter to the column cap. Before entering the cap, the water
inlet temperature is measured. The water is then uniformly distributed over the top
packing deck. This creates a large thin film of water, which will be exposed to the air
stream. The water will be cooled as it flows downward through the packing due to
evaporation process. The cooled water falls into the basin below the lowest deck and
return to the load tank where it is heated again before recirculation. The outlet
temperature is measured at a point just before the water flows back into the load tank.
Evaporation causes the water level in the load tank to fall. The amount of water lost by
evaporation will be automatically compensated by equal amount of water from the
make-up tank. At steady state, this compensation rate equals the rate of evaporation
plus any small airborne droplets discharge with the air.
4.3.2 Air Circuit
A one-side inlet centrifugal fan draws the air from the atmosphere into the distribution
chamber. The air flow rate can be varied by means of an intake damper. The air
passes the dry and wet bulb temperature sensors before it enters the bottom of the
tower. As the air stream passes through the packing, its moisture content increases
and the water temperature drops. The air passed another duct detector measuring its
exit temperature and relative humidity, then through a droplet arrester and an orifice
before finally leaving the top of the tower into the atmosphere.
4.4 Temperature Sensors
Temperature measurements are assigned as follows:
T1
Dry Bulb Inlet Air Temperature
T2
Wet Bulb Inlet Air Temperature
T3
Dry Bulb Outlet Air Temperature
T4
Wet Bulb Outlet Air Temperature
T5
Inlet Water Temperature
T6
Outlet Water Temperature
T7
Make-up Tank Temperature
T8
Hot Water Tank Temperature
The pressure drop across the tower is measured by a differential pressure sensor.
The air flow rate is measured by a differential pressure sensor and an orifice.
The water flow rate is measure by a flow mete
5.0
SAFETY & PRECAUTION
5.1
All operating instructions supplied must be read carefully and understood
before attempting to operate the unit.
5.2
Keep hands away from the blower when it is running to avoid serious injury.
5.3
Do not connect the power if the unit is damaged or even partially wet.
5.4
Do not touch the hot component of the unit.
5.5
Do not set the heater temperature above its maximum operating temperature
of 70°C to avoid pump damage.
5.6
Do not use pliers or other tools to tighten the screws. Use only bare hands.
Over tightening could damage the seals and cause leakage.
5.7
Exercise caution when pouring water into the make up tank to ensure no
water is spilt onto the electrical parts.
Safety
Footwear
Eye
Protection
Protective
Jacket
6.0
PROCEDURE
6.1
General Operating Procedure
6.1.1. General Start-up Procedures
1.
Check to ensure that valve V1 and V6 are closed and valve V7 is
partially opened
2.
Fill the load tank wit distilled or deionised water. It is done by first
removing the make-up tank and then pouring the water through the
opening at the top of the load tank. Replacing the make-up tank onto
the load tank and lightly tighten the nuts. Fill the tank with the distilled
or deionised water.
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 controller to 50 C.
Switch on the 1.0 kW water heater and heat up the water until
o
approximately 40 C
7.
Switch on the pump and slowly open the control valve V1 and set the
water flowrate to 2.0 LPM. Obtain a steady operation where the water
is distributed and flowing uniformly through the packing.
8.
Fully open the fan damper and then switch on the fan. Check that the
differential pressure sensor is giving reading when the valve manifold
is switched to measure the orifice differential pressure.
9.
Let the unit run for 20 minutes, for the float valve to correctly adjust
the level in the load tank. Refill the make-up tank as required.
10.
The unit is ready for use.
o
Note:
i.
It is strongly recommended that only distilled or deionised water be
used in this unit. The impurities existing in tap water may cause the
depositing in cover tower.
ii.
Check that the pressure tubing for differential pressure measurement
are connected correctly. (orifice pressure tapping point to V4;
Column’s lower pressure tapping point to V6; Column’s higher
pressure tapping point to V3; V5 leave to atmosphere)
iii.
To measure the differential pressure across the orifice, open valve V4
and V5; close valve V3 and V6.
iv.
To measure the differential pressure across the column, open valve
v3 and valve V6; close valve V4 and V5
v.
Always make sure that no water is in the pressure tubing for accurate
differential pressure measurement.
5.1.2 General Shut Down Procedure
1. Switch off heaters and let the water to circulate through the cooling tower
system for 3-5 minutes until the water cooled down.
2. Switch of the fan and fully close the fan damper
3. Switch off the pump and power supply
4. Retain the water in reservoir tank for the following experiment
5. Completely drain off the water from the unit if it is not in used
6.2
Experiment 1: General Observation of the Forced Draught Cooling Tower
Objective: To observe the processes within a forced draught cooling tower
1. Perform the general start-up procedures and observe the forced draught cooling
tower proves.
2. As the warm water enters the top of the tower, it is fed into channels from which it
flows via water distribution system onto the packing. The channels are designed
to distribute the water uniformly over the packing with minimum splashing.
3. The packing surfaces are easily wetted and the water spread over the surfaces to
expose a large area to the air stream.
4. The cooled water falls from the lowest packing into the basin and then is pumped
to the simulated load in the load tank.
5. During the process, some water is lost due to evaporation. Thus, “make-up” water
must be supplied to keep the amount of water in the cooling system constant.
The make-up is observed flowing past the float-controlled valve in the load tank
6. A “droplet arrester” or “mist eliminator” is fitted at the tower outlet to minimize loss
of water due to escape of droplets of water( resulting from splashing, etc.) which
is entrained in the air stream. This loss does not contribute to the cooling, but
must be made good by “make-up”. The droplet arrester causes droplets to
coalesce, forming drops that are too large to be entrained and these falls back
into the packing.
7. The fan drives the air upward through the wet packing. At air outlet, the air
leaving the cooling tower is almost saturated, i.e. Relative Humidity is ~100%.
The Relative Humidity at the air inlet. The increase in the moisture content of the
air due to the evaporation of water into steam and the “latent heat” for this
account for most of the cooling effect.
8. When the cooling load is switched off and the unit is allowed to stabilize, it is
found that the water leaves the basin at the temperature close to the wet bulb
temperature of the air entering. Wet bulb temperature is lower than the dry bulb
temperature and this varies according to the local atmospheric conditions ( i.e.
pressure and relative humidity).
9. With no load, the water would be cooled to the incoming wet bulb temperature.
However, the condition cannot be achieved science the work done by the pump
transfers about 40W to the water.
6.3
Experiment 2: Investigation of the Effect of Cooling Load on Wet Bulb
Approach
Objective:
To investigate the effect of cooling loading on “Wet bulb Approach”
Procedure:
1. Prepare and start the cooling tower with according to section 5.1.
2. Set up system under the following conditions and allow stabilizing for about 15
minutes.
Water flow rate
:
2.0 LPM
Air Flow
:
Maximum
Cooling load
:
0.0 kW
3. After the system stabilizes, record a few sets of measurements (i.e. air inlet dry
bulb and wet bulb temperature (T1 and T2), water outlet temperature (T6), orifice
differential pressure (DP1), water flowrate (FT1) and Heater Power (Q1)), then
obtain the mean value for calculation and analysis.
4. Without changes in the conditions, increase the cooling load to 0.5 kW. When the
system stabilized, record all data.
5. Similarly, repeat the experiment at 1.0 kW and 1.5 kW.
6. Finally, measure the cross sectional area of the column.
7. The four tests may be repeated at another constant airflow.
8. The observation may also be repeated at different conditions, i.e. at different
water flow rates, or different air flow rates and with different load.
6.4
Experiment 3: Investigation of the Effect of Air Velocity on Wet bulb Approach
and Pressure Drop
Objective:
To investigate the effect of air velocity on:
a) Wet Bulb Approach
b) The pressure drop through the packing
Procedure:
1. Prepare and start the cooling tower with according to section 5.1.1.
2. Set the system under the following conditions and allow stabilizing for about 15
minutes
Water flow rate
:
2.0 LPM
Air flow rate
:
Maximum
Cooling load
:
1.0 kW
3. After the system stabilizes, record a few sets of measurements (i.e. temperature
(T1-T6),orifice differential pressure (DP1), water flowrate (FT1), heater power
(Q1) and pressure drop across packing (DP2)), then obtain the mean value for
calculation and analysis.
4. Repeat the test with 3 different sets of orifice pressure drop values 975%, 50%
and 25% of the maximum value) without changing the water flow rate and cooling
loads.
5. Finally, measure the cross sectional area of the column.
6. The test may repeated:
i.
At another constant load
ii.
At another constant water flow rate
6.5
Experiment 4: Investigation of the Relationship between Cooling Load and
Cooling Range
Objective:
To investigate the relationship between cooling load and cooling range
Procedure:
1. Prepare and start the cooling tower with according to section 5.1.1.
2. Set the system under the following conditions and allow stabilizing for about 15
minutes
Water flow rate
:
2.0 LPM
Air flow rate
:
Maximum
Cooling load
:
0.0 kW
3. After the system stabilizes, record a few sets of measurements (i.e. temperature
(T1-T6),orifice defferential pressure (DP1), water flowrate (FT1), heater power
(Q1) and pressure drop across packing (DP2)), then obtain the mean value for
calculation and analysis.
4. Without changes in the conditions, increase the cooling load to 0.5 kW. When the
system stabilized, record all data.
5. Similarly, repeat the experiment at 1.0 kW and 1.5 kW.
6. The test may repeated:
7.0
iii.
At another water flow rates
iv.
At another air flow rate
PROCESS FLOW DIAGRAM (PFD)
6.1
8.0
Draw a process flow diagram (PFD).
RESULT & Discussion
8.1
Experiment 2: Investigation of the Effect of Cooling Load on Wet Bulb
Approach
Description
Unit
Packing Density
Air Inlet Dry Bulb, T1
-1
m
o
Air Inlet Wet Bulb, T2
Water Outlet temperature, T6
o
Orifice Differential, DP1
Pa
Water Flow Rate, FT1
LPM
Heater Power, Q1
Watt
0.5kW
1.0 kW
1.5 kW
C
C
o
C
Calculation:
i.
Calculate the value air flow per unit area, total cooling load and wet bulb
approach
Discussion:
i.
8.2
Plot the graph on relationship between cooling load and wet bulb approach
Experiment 3: Investigation of the Effect of Air Velocity on Wet bulb Approach
and Pressure Drop
Description
Unit
Air Flow
100%
Packing Density
Air Inlet Dry Bulb, T1
-1
m
o
Air Inlet Wet Bulb, T2
o
Air Inlet Dry Bulb, T3
Air Inlet Wet Bulb, T4
Water Outlet temperature, T5
Water Outlet temperature, T6
Orifice Differential, DP1
C
C
oC
oC
oC
oC
Pa
Water Flow Rate, FT1
LPM
Heater Power, Q1
Watt
Pressure Drop Across Packing, DP2
Pa
Calculation:
75%
50%
25%
i.
Calculate the value nominal velocity of air, wet bulb approach and pressure
drop
Discussion
i.
Plot the graph on relationship between value nominal velocity of air, wet
bulb approach and pressure drop


8.3
Specific volume of air at outlet (by plotting Air Outlet Dry bulb and Air
3
-1 
Outlet Wet bulb on the Psychometric Chart) = 0.8936 m kg
X= 116 Pa x 1mm H2O/ 10.13 Pa = 11.45 mmH2O

Experiment 4: Investigation of the Relationship between Cooling Load and
Cooling Range
Description
Packing Density
Air Inlet Dry Bulb, T1
Air Inlet Wet Bulb, T2
Air Inlet Dry Bulb, T3
Air Inlet Wet Bulb, T4
Water Outlet temperature, T5
Water Outlet temperature, T6
Orifice Differential, DP1
Unit
0.5kW
m-1
oC
oC
oC
oC
oC
oC
Pa
Water Flow Rate, FT1
LPM
Heater Power, Q1
Watt
Discussion:
i.
Plot the graph between cooling load and cooling range
1.0 kW
1.5 kW