T9.2 (Reference)

T9.2: INVESTIGATION OF THE EFFECT OF AIR VELOCITY ON WET BULB

APPROACH AND PRESSURE DROP THROUGH THE PACKING

THEORY :

Basic Principle

First consider an air stream passing over the surface of a warm water droplet or film. If we

assume that the water is hotter than the air, then the water temperature will be cooled down

by radiation, conduction and convection, and evaporation. The radiation effect is normally

very small and may be neglected. Conduction and convection depend on the temperature

difference, the surface area, air velocity, etc. The effect of evaporation is the most

significant where cooling takes place as water molecules diffuse from the surface into the

surrounding air. During the evaporation process, the water molecules are replaced by others

in the liquid from which the required energy is taken.

Evaporation from a Wet Surface

When considering evaporation from a wet surface into the surrounding air, the rate is

determined by the difference between the vapor pressure at the liquid surface and the

vapour pressure in the surrounding air. The vapour pressure at the liquid surface is basically

the saturation pressure corresponding with the surface temperature, whereas the total

pressure of the air and its absolute humidity determines the vapour pressure in the

surrounding air. Such evaporation process in an enclosed space shall continue until the two

vapour pressures are equal. In other words, until the air is saturated and its temperature

equals the surface.

However, if unsaturated air is constantly supplied, the wet surface will reach an equilibrium

temperature at which the cooling effect due to the evaporation equals the heat transfer to

the liquid by conduction and convection from the air, which under these conditions; will

be at a higher temperature. Under adiabatic conditions, this equilibrium temperature is the

"wet bulb temperature".

For a cooling tower of infinite size and with an adequate air flow, the water leaving will be

at the wet bulb temperature of the incoming air. Therefore, the difference between the

temperature of the water leaving a cooling tower and the local wet bulb temperature is an

indication of the effectiveness of the cooling tower. Thus, "Approach to Wet Bulb", an

important parameter of cooling towers, is the difference between the temperature of the

water leaving the tower and the wet bulb temperature of the entering air.

Cooling Tower Performance

A study on the performance of a cooling tower can be done with the help of a bench top

unit. Students shall be able to verify the effect of these factors on the cooling tower

performance:

(i) Water flow rates

(ii) Water temperatures

(iii) Airflow rate

(iv) Inlet Air Relative Humidity

The effect of these factors will be studied in depth by varying it. In this way, students will

gain an overall view of the operation of cooling tower.

Thermodynamic Property

In order to understand the working principle and performance of a cooling tower, a basic

knowledge of thermodynamic is essential to all students. A brief review on some of the

thermodynamic properties is presented below.

At the 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

(hf) 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 = hf + v f (p - p sat ) (1)

The correction for pressure is negligible for the operating condition of the cooling tower;

therefore we can see that h ≈ hf at a given temperature.

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 using bench top cooling tower, we may use the following relationship:

Dh = C p DT (2)

and

h = C pT (3)

Where Cp = 4.18 kJ.kg-1

Dalton’s and Gibbs Laws

It is commonly known that air consists of a mixture of "dry air" (O 2, N2 and other gases)

and water vapour. Dalton and Gibbs law describes the behaviour of such a mixture as:

a) 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

b) The dry air and the water vapour respectively obey their normal property relationships

at their partial pressures.

c) 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.

b) The dry air and the water vapour respectively obey their normal property relationships

at their partial pressures.

c) 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.

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 as the same.

To measure the moisture content of the atmosphere, this bench top cooling tower unit is

supplied with electronic dry bulb and wet bulb temperature sensors. The temperature

readings shall be used in conjunction with a psychometric chart.

Psychometric ChartThe psychometric chart is very useful in determining the properties of air/water vapourmixture. Among the properties that can be defined with psychometric chart are Dry BulbTemperature, Wet Bulb Temperature, Relative Humidity, Humidity Ratio, SpecificVolume, and Specific Enthalpy. Knowing two of these properties, any other property canbe easily identified from the chart provided the air pressure is approximately atmospheric.

In the Bench Top Cooling Tower application, the air inlet and outlet sensor show the drybulb temperature and wet bulb temperature. Therefore, the specific enthalpy, specificvolume, humidity ratio and relative humidity can be readily read from the psychometricchart. The psychometric chart provided with this manual is only applicable for atmosphericpressure operating condition (1.013 bar). However, the error resulting from variation oflocal atmospheric pressure normally is negligible up to altitudes 500m above sea level.

Orifice CalibrationAs mentioned above, the psychometric chart can be used to determine the value of thespecific volume. However, the values given in the chart are for 1 kg of dry air at the statedtotal pressure.However, for every 1 kg of dry air, there is w kg of water vapour, yielding the total massof 1 + w kg. Therefore, the actual specific volume of the air/vapor mixture is given by:

The mass flow rate of air and steam mixture through the orifice is given by

Where,

Thus,

The mass flow rate of dry air

A simplification can be made since in this application, the value of v is unlikely to exceed0.025. As such, neglecting wb would not yield significant error.

Application of Steady Flow Energy EquationConsider System A for the cooling tower defined as in Figure 9.1. It can be seen that forthis system, indicated by the dotted line,

a) Heat transfer at the load tank and possibly a small quantity to surroundingsb) Work transfer at the pumpc) Low humidity air enters at point Ad) High humidity air leaves at point Be) Make-up enters at point E, the same amount as the moisture increase in the air stream

From the steady flow equation,

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:

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,

The ratio of steam to air ( v ) is known for the initial and final state points on thepsychrometric charts. Therefore,

Therefore,

Say, we re-define the cooling tower system to be as in Figure 9.2 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.

Therefore,

Therefore,

Therefore,

Again from the steady flow energy equation,

Q may have a small value due to heat transfer between the unit and its surroundings.

Rearranging

The term can be neglected

Characteristics Column Study

In order to study the packing characteristics, we define a finite element of the tower (dz)

as shown in Figure 9.3, the energy balances of the water and air streams in the tower are

related to the mass transfer by the following equation:

where

In this equation, we assume that the boundary layer temperature is equal to the water

temperature T and the small change in the mass of water is neglected.

Thus, from Equation 20,

Integrating Equation (21),

The numerical solution to the integral expression Equation 22 using Chebyshev

numerical method gives,

Where;

Thermodynamics state that the heat removed from the water must be equal to the heat

absorbed by the surrounding air. Therefore, the following equation is derived:

Where,

The following represents a key to Figure 9.4:

BA = Initial enthalpy driving force

AD = Air operating line with slope L/G

Referring to Equation (22), the tower characteristics could be found by finding the area

between ABCD in Figure 4. Increasing heat load would have the following effects on

the diagram in Figure 4:

1. Increase in the length of line CD, and a CD line shift to the right

2. Increase in hot and cold water temperatures

3. Increase in range and approach areas

The increased heat load causes the hot water temperature to increase considerably faster

than does the cold water temperature. Although the area ABCD should remain constant,

it actually decreases about 2% for every 10 JF increase in hot water temperature above 100

0F. To account for this decrease, an "adjusted hot water temperature" is used in cooling

tower design.

Useful Information

1. Orifice Calibration Formula:

Mass flow rate of air and vapor mixture,

The mass flow rate of dry air,

Where,

2. Pump Work Input = 80W (0.08kW)

3. Column Inner Dimension = 150 mm x 150 mm x 600 mm

APPARATUS:

Bench Top Cooling Apparatus

OBJECTIVE:

To investigate the effect of air velocity on:

(a) Wet Bulb Approach

(b) The pressure drop through the packing

PROCEDURES:

1. Allowing the system for about 15 minutes after set the system under the following conditions.

Water flow rate: 2.0 LPM

Air flow rate: Maximum

Cooling load: 1.0 kW

2. A few sets of measurements are recorded after the system stabilizes. (I.e. temperature

(T1-T6), orifice differential pressure (DP1), water flow rate (FT1), heater power

(Q1) and pressure drop across packing (DP2)), then obtain the mean value for

calculation and analysis.

3. The test with 3 different sets of orifice pressure drop values are repeated (75%, 50% and

25% of the maximum value) without changing the water flow rate and cooling

loads.

4. The cross sectional areas of the column are measured.

5. The test is repeated:

a. At another constant load

b. At another constant water flow rate

RESULT

Description Unit Air Flow100% 75% 50% 25%

Packing Density m-1 110 110 110 110

Air Inlet Dry Bulb, T1 ˚C 23.9 24.3 24.5 24.6

Air Inlet Wet Bulb, T2 ˚C 23.8 23.7 23.9 23.9

Air Outlet Dry Bulb, T3 ˚C 25.4 25.7 25.9 26.4

Air Outlet Wet Bulb, T4 ˚C 25.1 25.4 25.4 25.7

Water Inlet Temperature, T5 ˚C 34.8 35.0 35.3 35.6

Water Outlet Temperature, T6 ˚C 24.3 24.4 24.5 25.1

Orifice Differential, DP1 Pa 130 97.5 65 32.5

Water Flow Rate, FT1 LPM 1.2 1.2 1.2 1.2

Heater Power, Q1 Watt 1.0k 1.0k 1.0k 1.0k

Pressure Drop Across Packing, DP2

Pa 50 40 30 20

Table 2.

Description/Unit

Air Flow100% 75% 50% 25%

Nominal Velocity of Air m/s 1.0541 0.9129 0.7454 0.5269

Wet Bulb Approach K 0.5 0.7 0.6 1.2

Pressure mm H2O 5.0986 4.0789 3.0591 2.0394

CALCULATION.

1. APPROACH WET BULB

a. 100%

Inlet wet bulb temperature (T2) = 23.8 ˚C

Outlet water temperature (T6) = 24.3 ˚C

“ Approach to wet bulb” = 24.3 – 23.8 K

= 0.5 K

b. 75%




= 0.7 K

c. 50%




= 0.6 K.

d. 25%




= 1.2 K

2.NOMINAL AIR VELOCITY

Specific Volume of air at outlet

(by plotting T2 and T6 on the psychrometric chart) = 0.86 m³/kg

Air mass flow rate, ṁ = 0.137 √h/ v

h = orifice differential , DP1

v = specific volume of air

Packing area =15 cm (w) x 15 cm (d) x 60 cm (h) =13500m ~ 0.135mm

a. 100%

Mass flow rate, ṁ

ṁ = 0.0137 √130/0.86 = 0.1685 kg/s

Volume flow rate, v (m³/s)

v= ṁV v= 0.1684(086) =0.1449 m³/s

Nominal air Velocity, V (m/s)

V=v/A

V=0.1449/ 0.0225

= 6.44 m/s

b.75%

Mass flow rate, ṁ

ṁ = 0.0137 √(97.5/0.86) = 0.1459 kg/s


v= ṁV

v= 0.1459(0.86)

=0.11255 m³/s


V=v/A

V=0.1255/ 0.0225

= 5.58 m/s

c) 50%

Mass flow rate, ṁ

ṁ = 0.0137 √(65/0.86) = 0.1191 kg/s


v= ṁV

v= 0.1191(0.86)

=0.1024m³/s


V=v/A

V=0.1024/ 0.0225

= 4.55m/s

d) 25%

Mass flow rate, ṁ

ṁ = 0.0137 √(32.5/0.86) = 0.0842 kg/s


v= ṁV

v= 0.0842(0.86)

=0.0724 m³/s


V=v/A

V=0.0724/ 0.0225

= 3.22m/s

RELATIONSHIP BETWEENN NOMINAL AIR VELOCITY AND

i) WET BULB APPROACHii) PACKING PRESSURE DROP

DATA: COOLING LOAD 1.0Kw WATER FLOW RATE 2.0 LPM

3 3.5 4 4.5 5 5.5 6 6.5 70

1

2

3

4

5

6

5.0986

4.0789

3.0591

2.0394

0.50.70.6

1.2

Wet Bulb Approach KLinear (Wet Bulb Approach K)Pressure mm H2OLinear (Pressure mm H2O)

NOMINAL AIR VELOCITY, (m/s)

WE

T B

UL

B A

PP

RO

AC

H, K

PA

CK

ING

PR

ESS

UR

E D

RO

P, m

m H

2O

DISCUSSION:

In this experiment we are trying to investigate the effect of air velocity on wet bulb approach and pressure drop through the packing. Based on our result and calculation, the nominal velocity for air flow of 100%,75% ,50% and 25% are 6.44 m/s, 5.58 m/s, 4.55 m/s and 3.22 m/s respectively. The wet bulb approach for air flow of 100%,75%,50% and 25% are 0.5 K, 0.7 K, 0.6 K and 1.2 K respectively.

Discussion #1 : ( Abang Amiruddin)

Based on the graph that has been obtained it is found that the packing pressure drop increased uniformly but the wet bulb approach decreased rapidly. Referring from the theory, we find that the reading of wet bulb approach should decrease uniformly but due to error our data are not the same as the theory. We have some error because we have encountered is the sensor is placed in a stream having a lower velocity but at low relative humidities. In order to overcome the error, the sensors must be placed in air with a very high relative humidity and where the air velocity is high.

Discussion #1 : ( Aznol Sanjan)

After we conducted this experiment, based on our result and calculation, the nominal velocity for air flow of 25%,50%,75% and 100% are 3.22m/s, 4.55 m/s, 5.58m/s, and 6.44 m/s. The wetbulb approach for air flow of 25%,50%,75% and 100% are 1.2K,0.6K,0.7K and 0.5K respectively. From the graph we plot,we realize that at high relative humidities, there is little error if the sensor is placed in a stream having a lower velocity but at low relative humidities an appreciable error may occur.

To avoid these things to happened, we had to ease the bung securing the wet bulb sensor from the top of the air chamber. Then, we should draw the sensor upward until the air escapes between the socket and the sleeve. The air velocity over the sleeve will now and the temperature will quickly be indicated by the sensor.

Discussion #1 : ( Alif Falatin bin Abdul Latif / 40493)

From our observations, there are some difficulties for the beginning of the experiment due to a little misunderstanding how does the apparatus work during the experiment. This could be happen due to lack of SOP ( Manual Prosedur Kerja) for the apparatus as specially by the student. As a suggestions to improve the stability for the readings or the data taken, would it be possible if the apparatus being used for this experiment upgraded or more efficient. For example, how do we control the pressure from the blower, where it was control just by closing some of the intake from

the blower. There is no specific measurement or how does it suppose to be control accordingly to the exact amount needed.

In our opinion, we could design some control device or intakes that can help us to gain more accurate readings.

Discussion #1 : ( Nadia)

From the experiment we conducted, there is still a little error if the sensor placed in the stream having a low velocity, but at lower relative humidity an appreciable error may occur.

To overcome this problem, we could draw the sensors upward until the air escapes between the socket and the sleeve. The air velocity over the sleeve will now be about 10m/s and the “sling” tempreture will quickly be indicated by the sensors

CONCLUSION

From this experiment, we are able to investigate the effect of air velocity on the web bulb

approach and the pressure drop through the packing.

T9.2: INVESTIGATION OF THE EFFECT OF AIR VELOCITY ON WET BULB

APPROACH AND PRESSURE DROP THROUGH THE PACKING

THEORY :

Basic Principle

First consider an air stream passing over the surface of a warm water droplet or film. If we

assume that the water is hotter than the air, then the water temperature will be cooled down

by radiation, conduction and convection, and evaporation. The radiation effect is normally

very small and may be neglected. Conduction and convection depend on the temperature

difference, the surface area, air velocity, etc. The effect of evaporation is the most

significant where cooling takes place as water molecules diffuse from the surface into the

surrounding air. During the evaporation process, the water molecules are replaced by others

in the liquid from which the required energy is taken.

Evaporation from a Wet Surface

When considering evaporation from a wet surface into the surrounding air, the rate is

determined by the difference between the vapor pressure at the liquid surface and the

vapour pressure in the surrounding air. The vapour pressure at the liquid surface is basically

the saturation pressure corresponding with the surface temperature, whereas the total

pressure of the air and its absolute humidity determines the vapour pressure in the

surrounding air. Such evaporation process in an enclosed space shall continue until the two

vapour pressures are equal. In other words, until the air is saturated and its temperature

equals the surface.

However, if unsaturated air is constantly supplied, the wet surface will reach an equilibrium

temperature at which the cooling effect due to the evaporation equals the heat transfer to

the liquid by conduction and convection from the air, which under these conditions; will

Documents

T9.2 (Reference)