THEORY
The practical application of this mass transfer theories are deployed in the industry when high rates of heat transfers are necessary. From Reynolds number,
(Re) = 4 ρ Q / π µ d.
A similar correlation to that shown below could be used to study the relationship between the air flow rate, Re and mass transfer (coefficient). From Gilliland and Sherwood, Kg.d .PBM = 0.023Re0.83Nsc0.44
Dg. π
A mass balance of the water vapor transferred from liquid film to air along the full height of the column is shown by the formula below3.
ρ x Q (H2 - H1) = kg x a x HM ……………… (i)
Re- arranging equation (i) gives
kg = ρ x Q (H2 – H1) ………………….......... (ii)
a x HM
The relationship between mass transfer coefficient and Reynolds number can be described by a mathematical correlation which is similar to the one predicted by Gilliland and Sherwood For constant temperature applications or where temperature variations are negligible, the correlation is of the general form:
kg x d = B x ReN ………………………. (iii)
Where B and N are constants determined experimentally. For turbulent flow, N is approximately equal to 0.83 (according to the Gilliland and Sherwood correlation). The graph of log (kg.d) versus log (Re) should give a straight line with a slope equal to N 3.
Objectives:
In this experiment we aim
- To establish the variation of mass transfer coefficient for humidification of air with air flow rate: increasing the air flow rate in a wetted-wall column, the mass of the water vapor transferred from the liquid bulk to the air bulk will increase. Therefore, at constant temperature and standard atmospheric pressure, the air flow rate is directly proportional to the mass-transfer coefficient in a humidification process.
- To determine the correlation between mass transfer coefficient and Reynolds Number for :-
- Laminar flow.
- Turbulent flow.
At low Reynolds number, the mass transfer coefficient will be decreased and at high Reynolds number the mass transfer coefficient will be increased.
EXPERIMENTAL EQUIPMENT
To carry out this experiment we used the following equipment
- Three vertical glass tubes measuring:
- 4.1cm internal diameter, 74cm high
- 1.4cm internal diameter , 64cm
- Rota meters for measuring air flow
- Wet and dry bulb thermometers for measuring temperatures
- Psychometric chart , for measuring humidity
EXPERIMENTAL LAYOUT
Water enters from the top of the columns and flows downwards, exiting through the drainage pipes on the base of each column and into the drainage tank. The water is taken by the centrifugal pump and pumped back around the system, the flowrate of the water can be adjusted by the rotameter, as can the air flowrate which is adjusted by the air flow rotameter. The air is blown through the columns and its humidity on inlet and outlet is measured by wet and dry-bulb thermometers, as is the inlet and outlet temperatures of the water.
EXPERIMENTAL PROCEDURE
-
There are three columns in this experiment layout but the largest column (column3) was selected and the air valve was turned on while the water was turned off. The air was allowed to flow for 15minutes so as to get the temperature at a steady state. The inlet air dry bulb temperature was measured and recorded after 15 minutes for both the inlet air and the outlet air were taking respectively.
- The water valve was then turned on and set to a constant flow rate of 20litres/hr. It was allowed to circulate through the glass tube in a thin film ensuring that the surface was completely wetted. The water inlet temperature and the water outlet temperature were recorded.
-
The air flow rate was selected to be 4litres/min for the first run on the Rota meter air flew upwards while the thin film of water fell downwards along the internal surface area of the glass tube and the following were measured – water inlet temperature, water outlet temperature, air outlet dry bulb temperature and air outlet wet bulb temperature.
- The stop watch was set and readings were observed every two minutes until it was noticed that the inlet and outlet temperature readings were stable (in steady-state).
- Using the psychometric chart, the temperature readings were converted to air humidity at inlet and outlet temperatures.
- Steps (iv) and (v) were repeated for three lower flow rates of 10litres/min, 40litres/min ,10litres/min and 70litres/min.
- The above procedures were repeated for column 1 at the same air flow rate.
- The results from both experiments were tabulated.
OBSERVATIONS
- Careful consideration was given to the wet bulb in both the inlet and the outlet air hydrometers to make that it’s wet at all times.
- The water flow rate did fluctuate at 20liters/hr and it was monitored at all times to keep it stable.
- At the start of the experiment the wet and dry temperature for the inlet air was recorded.
- The temperatures were allowed to reach a steady state .
- It took a longer time for the air flow rate at 4litres/min to reach steady state in column 1 of internal diameter of 4.1cm and height of 74cm than in column1 of internal diameter of 1.4cm and height of 64cm. This could be difference in the internal diameters.
- Thermometer inconsistence reading took could affect the total result.
EXPERIMENT RESULTS AND CALCULATIONS
Water flow rate = 3.3 10 -4 m/sec
Column 1: with 4.1cm internal diameter, 74cm high.
Table 2
Water flow rate = = 3.3 10 -4 m/sec
Column 1: 1.4cm internal diameter, 64cm high
Table 3
Derived Result for column 1 laminar flow (RE<2000)
Derived Result for column 1 Turbulent flow (RE>2000)
The results obtained for Log Re and Log Kg.d for laminar flow (Re<2000)
The results obtained for Log Re and Log Kg.d for turbulent flow (Re>2000).
SAMPLE CALCULATIONS
For column 3
- Internal diameter 4.1cm to m
= 4.1 x 10-2
= 0.041m
= 74 x10-2
= 0.74m
-
To convert air flow rate from litresmin-1 to m3sec-1
= 4litresmin-1 ÷ 60000
= 6.67 x 10-5 m3sec-1
4 L/min = 4x 10-3m3 / 60s = 6.67 x10-5 m3s-1,
10 L/min = 10 x 10-3m3 / 60s = 1.67 x10-4 m3s-1
40 L/min = 40 x 10-3m3 / 60s = 6.67 x10-4 m3s-1
70 L/min = 70 x 10-3m3 / 60s = 1.17 x10-3 m3s-1
For column 3 at 6.67 x10-5 m3sec-1
Re = 4ρQ
π μ d
Re = 4 x 1.293 x 6.67 x10-5
3.141 x 1.80x10-5 x 0.041
Re = 148.79
Where ρ is the density of air =1.293kg/m3, μ is the viscosity of air =1.80 x 10-5 Pas, internal diameter of column 3 is 4.1cm = 0.041m. Q = 6.77 x10-5m3s-1 π = 3.141
-
Arithmetic mean humidity difference HM,
HM = (H4 – H1) + (H3 - H2) / 2
H1 = 0.004, H2 = 0.0120, H3 = 0.0110 and H4 = 0.0130
HM = (0.0130 – 0.004) + (0.0110 -0.0120) / 2 =
HM = 0.004
-
Mass transfer coefficient (kg)
kg = ρ x Q x (H2 – H1)
a x HM
kg = 1.293 x 6.67 x10-5 x(0.0120 – 0.004)/0.10 x 0.0041
kg = 1.7 x10-3 kg s-1m-2
Where ‘a’ is the wetted surface area of the column, d = 0.041, h = 0.74.
a = πdh = π * 0.041 * 0.74
a = 0.10m2
-
kg x d = 1.7 x10-3 x 0.041m = 6.97x 10-5 kg s-1m-1
Log kg* d = log 6.97x 10-5 = --4.16
Log Re =log(148.79)= 2.17
- gradient of graph, Obtained using excel
For column 1 it was 0.41 and clounm 2 was linear
DISCUSSION
The aim of the above experiment was to establish that mass transfer coefficient for air humidification varies with the inlet air flow rate. This objective is necessary because it enables better understanding of the way mass transfer occurs in process systems. Based on the hypothesis stated above which state that, ‘the rate of mass transfer depends on the condition of the interface which are enhanced or affected by the flow rate of both fluids. Hence if there is low flow rate, there would be a corresponding low mass transfer but if there is a high flow rate more mass transfer would occur’ the calculated results show that this hypothesis is accurate for laminar flow and series of inconsistence result got for turbulent flow.
From the results obtained during the experiment, it can be shown that as the air flow rate increases in the mass transfer coefficient also increases for the laminar flow and the turbulent flow was not that accurate in term of mass transfer. The calculated results also show that as the airflow rate increases, the Reynolds number. This may be the effect of the air flow against the wall of the vertical tubes. In column 3, when the air flow rate was at 4litres/min, 10litres/min and 40litres/min 70litres/min the flow was seen to be laminar as their Reynolds number values were 148.79, 370.54 and 1479.91, 2596.66 respectively. Meanwhile, when the air flow rate was at 70litres/min the flow was turbulent while the Reynolds number was 7438.35 In column 1, when the air flow rate was at 4litres/min and 10litres/min the flow was seen to be laminar. Whereas, when the air flow rate was at 40litres/min and 70litres/min the flow was turbulent while their Reynolds number values were 4240.49 and 7438.35 respectively.
At low flowrates, there is hardly any disturbance at the surface of the liquid, therefore there are no ripples seen. However at greater flow rates, the surface of the film of water is disturbed by ripples traveling down the film therefore making the flow turbulent and increasing the mass transfer coefficients. This shows that the results are concordant with the know theory as the rate of mass transfer (and mass transfer coefficient) will depend on whether the flow is laminar of turbulent.
It was also noticed that from the experimental results that the higher the height and diameter of the column, (column 3) the lower the Reynolds number. This can be shown by the results obtained for column 1 which had a diameter of 0.014m. Column 1 would be preferred because it will permit operation in the turbulent range with Reynolds number between 2000 and 10000 at air flowrate of 40-70 litres/min, using high gas velocities. The mass transfer rate between two fluid phases will depend on the physical properties of the two phases, the concentration difference, the interfacial area, and the degree of turbulence. Mass transfer equipment is therefore designed to give a large area of contact between the phases and also to promote turbulence in each of the fluids 4.
For slope of the graph which is equivalent to exponent N in the laminar flow was 0.41 and for the turbulent flow, unable to obtained the value because linear graph obtained, According to the theory, Gilliland and Sherwood obtained a value of 0.83. This is double the theoretical value and this difference could be a result of possible errors that might have occurred during the experiment. Opening the air pump in one column before shutting off in another column-in which cases some of the air might have gone from one column into the other, it could be reading in fact it could be calculation or conversion.
Also, it could be when the water flow rate was fluctuating and the columns having different heights and diameters.
The graphs plotted for the turbulent flow agreed with the theory because it was a straight line graph with slope N while the graph of the laminar flow did not agree with the theory. This could be a result of experimental error, human error or not taking the readings of the laminar and turbulent flow properly.
The wetted wall column that was experimented is very important in the world of chemical engineering because it is used mainly for the theoretical study of mass transfer …………and mass transfer coefficients. Wetted wall columns are used industrially for mass transfer applications where high rates of heat transfer are required.
As a result of carrying out the experiment, I have learnt how the wetted wall column works and how it is used. I have also learnt how to vary the air flow rates, how to use the psychometric chart to convert the temperature reading to air humidity.
CONCLUSION
From the experimental results and calculations, the following conclusions can be made:
- In column 3, as the air flow rate increases, the mass transfer coefficient also increases while in column 1,
- For a given air flow in a wetted wall column with a smaller diameter, more turbulent flow occur compared to a column with a bigger diameter. The smaller the diameter and surface area of the column, the greater the Reynolds number and subsequently the increase in the mass transfer coefficient.
- The higher the air flow rate, the higher the Reynolds number (i.e. the greater the amount of turbulence) and the increase in the mass transfer coefficient.
-
Based on the hypothesis which that, ‘the rate of mass transfer depends on the condition of the interface which are enhanced or affected by the flow rate of both fluids. Hence if there is low flow rate, there would be a corresponding low mass transfer but if there is a high flow rate more mass transfer would occur’ my calculated results show that this hypothesis is accurate.
NOMENCLATURE
REFERENCE
-
Robert E Treybal, Mass transfer operations. Page 187-188 Third edition McGraw-Hill Book Company.
-
www.armfield.co.uk
- Ref 2: Principles of Separation & Reaction laboratory booklet, Wetted-wall column: Mass transfer coefficients, Pages 1-6.
- Ref 3: J.M. Coulson, J. F. Richardson, J.R Backhurst & J.R. Harker (2002). Coulson & Richardson’s Chemical Engineering, Vol. 1, sixth edition: Fluid flow, Heat transfer and Mass transfer, Butterworth-Heinemann, Oxford.
Appendix 11
SAMPLE CALCULATIONS
For column 3
- Internal diameter 4.1cm to m
= 4.1 x 10-2
= 0.041m
= 74 x10-2
= 0.74m
-
To convert air flow rate from litresmin-1 to m3sec-1
= 4litresmin-1 ÷ 60000
= 6.67 x 10-5 m3sec-1
4 L/min = 4x 10-3m3 / 60s = 6.67 x10-5 m3s-1,
10 L/min = 10 x 10-3m3 / 60s = 1.67 x10-4 m3s-1
40 L/min = 40 x 10-3m3 / 60s = 6.67 x10-4 m3s-1
70 L/min = 70 x 10-3m3 / 60s = 1.17 x10-3 m3s-1
For column 3 at 6.67 x10-5 m3sec-1
Re = 4ρQ
π μ d
Re = 4 x 1.293 x 6.67 x10-5
3.141 x 1.80x10-5 x 0.041
Re = 148.79
Where ρ is the density of air =1.293kg/m3, μ is the viscosity of air =1.80 x 10-5 Pas, internal diameter of column 3 is 4.1cm = 0.041m. Q = 6.77 x10-5m3s-1 π = 3.141
-
Arithmetic mean humidity difference HM,
HM = (H4 – H1) + (H3 - H2) / 2
H1 = 0.004, H2 = 0.0120, H3 = 0.0110 and H4 = 0.0130
HM = (0.0130 – 0.004) + (0.0110 -0.0120) / 2 =
HM = 0.004
-
Mass transfer coefficient (kg)
kg = ρ x Q x (H2 – H1)
a x HM
kg = 1.293 x 6.67 x10-5 x(0.0120 – 0.004)/0.10 x 0.0041
kg = 1.7 x10-3 kg s-1m-2
Where ‘a’ is the wetted surface area of the column, d = 0.041, h = 0.74.
a = πdh = π * 0.041 * 0.74
a = 0.10m2
-
kg x d = 1.7 x10-3 x 0.041m = 6.97x 10-5 kg s-1m-1
Log kg* d = log 6.97x 10-5 = --4.16
Log Re =log(148.79)= 2.17
- gradient of graph, Obtained using excel
For column 1 it was 0.41 and clounm 2 was linear
.
SEPARATION AND REACTION
LIQUID PHASE CHEMICAL
MARKER MS RUKHSANA FAIZ
Contents
- Summary
- Introduction
- Theory
- Experimental layout
- Experiment
- Experiment Results and calculation
- Discussion
- Nomenclature
- Appendaix