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Distillation Process Using Passive Solar Still Design
Evan Kontras
University of Missouri
Department of Mechanical and Aerospace Engineering
March 2010
Abstract
As the global population expands and industrializes the production of potable water is becoming
a global issue. Clean water is a valuable resource in many rural areas of developing countries,
and the processes to provide it can be unaffordable and difficult to implement. A passive solar
still uses only the radiant energy from the sun to evaporate brackish water, condense it and
collect it from the still. This report details a design for a solar still to be used in Mumbai, India,
where ambient temperatures average 30 C. Using nothing more than common extruded
polystyrene as the structure and insulation with a polycarbonate glazing, the solar still will cost
no more than $200. The still has a water surface area of 2.25 , and can hold a maximum of 45kg of water. The still produces a minimum of 2 kg and a maximum of 6.68 kg clean water eachday, depending on cloud conditions. This simple design will provide means for rural areas
without accessible drinking water to have convenient clean water each day at a reasonable cost.
Introduction
In the previous report two viable options for solar desalination devices were presented. A
simple passive solar still, and a flow through desalination device consisting of evaporation and
condensation sections were both considered. A very general idea of the capabilities of both
systems was obtained. After considering the output requirement, construction and operationcosts, and operating conditions, the passive solar still was selected for further design. The
passive solar still can provide the required 2 kg of water each day, while having low construction
costs and little required maintenance.
A simple schematic of a passive solar still is shown in Fig. 1 below to provide the reader with
some terms to be used further in this analysis.
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Figure 1. Passive Solar Still Schematic.
The first objective in detailed analysis for a passive solar still is to determine what
parameters are most critical to its operation. Some initial assumptions will be made for the
design. The still is assumed to be a completely sealed system. The glazing is taken to be
infinitely thin, as its thickness will likely be negligible when compared to other dimensions. If
one assumes a control volume surrounding the solar still as shown in Fig. 1, the following modes
of heat transfer inside the system should be considered. Convective heat transfer will occur
between the basin water surface and glazing. Free convection will occur due to buoyant forces
from the variation in density of the air/vapor mixture with temperature difference. The heat
transfer due to convection from the basin water surface to the glazing surface can be written as,
(1)where is the heat transfer coefficient for convection, is the basin water temperature, isthe glazing surface temperature. The convective heat transfer coefficient for a single surface
passive solar still was found from [1] to be
= (2)where is the saturation pressure at basin water temperature, and the vapor pressure atglazing surface temperature. There is also radiative heat transfer between these surfaces that
should be considered. As solar radiation passes through the glazing, some is absorbed by the
basin water, and then radiated from the basin water to the glazing surface. The radiative heat
transfer between these surfaces can be written as
(3)where is the radiation heat transfer coefficient between the surfaces, and is a function of theemissivity of both the basin water and the glazing,
and
respectively. An effective emissivity
value for the surfaces can be found from the following equation.
(4)
It follows that the radiation heat transfer coefficient is
(5)where is the Stefan-Boltzmann constant. Some heat transfer is due to the actual evaporation ofbasin water, and can be written in a similar form to equations 1 and 3. The evaporative heat
transfer is given by
(6)
where is a function of the convective heat transfer coefficient, and was determined by [1] tobe
(7)Using the initial assumptions for the solar still system along with the heat transfer values
for each mode of heat transfer inside the control volume, an energy balance can be written in
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order to determine the amount of water produced each day. The solar still system is broken down
to individual components for simplicity, and an energy balance for each component follows. The
energy balance for the glazing is given by
(9)with
and
being the radiation and convective heat transfer for the glazing. The total heat
transfer to the glazing from solar radiation is . Similarly, the energy balance for thebasin water can be written in terms of the total heat transfer to the water from solar radiation as
(10)where is the mass of basin water, the specific heat of water at the basin water temperature,and the change in water temperature. Some heat will inevitably be lost to the surroundingsthrough the walls of the still and the basin/ground interface, which is accounted for by the
term. The energy balance for the basin in terms of the total heat transfer from solarradiation absorbed by the basin is then given by
(11)where is the heat loss from the basin due to conduction, the heat loss from the walls
due to conduction, the total surface area of the sides of the solar still, and the surfacearea of the basin.
In order to simplify the analysis, the relative importance of the many values iscompared here. By looking at the radiation absorptance and reflectance values for the different
materials of the solar still, a good idea of which terms may be neglected is found.
Solar energy absorbed by glazing: Ordinary window glass is a very selective transmitter of solar radiation. More than 90%
of solar radiation passes directly through most glasses and transparent plastics such as
polycarbonate. For wavelengths longer than 2.7 m however, these materials are effectively
opaque, preventing nearly all energy transmission by infrared radiation [2]. This is ideal for a
solar still, trapping infrared radiation inside the still and increasing internal temperature. It is
therefore assumed that the solar energy absorbed by the glazing is zero.
Solar energy absorbed by basin water: It has been shown by Phadatare and Verma [1] that the thickness of basin water is
directly related to the evaporation output inside a solar still. The thinner the basin water layer, themore easily and effectively evaporation will take place. Although a parametric study of water
depth could be done here, the results presented by Phadatare and Verma [1] will be assumed
correct. A maximum water depth of 2 cm was used for this design, resulting in 20 kg of water
per square meter of surface area in the basin. The total solar energy going into the water is found
using Eq. 10 once the other terms are characterized.
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Energy transfers for the basin water:
Heat transfer from radiation between basin water and glazing: , To get an order of magnitude estimate for
Eq. 3 and 5 were used to find the radiation
heat transfer for a range of different values for and . Figure 2 shows a plot of the radiationheat transfer over a range of for four constant values of , though the effect of changing is apparent.
Figure 2. Radiative Heat Transfer Between Basin Water and Glazing.
It can be seen from Fig. 2 that for the smaller the difference in temperature between thebasin water and glazing the less heat transfer from radiation will occur. Negative values in Fig. 2
represent heat transfer from the glazing to the basin water, that is, . Trapping thisradiation may seem ideal to maintain a high temperature inside the still, but having a glazing
material with a higher temperature than the basin water would make condensation impossible.
For temperature differences of around 15 C or greater, this heat transfer value cannot be
neglected and must be accounted for as one part of the absorbed solar radiation not going intoevaporation. A general order of magnitude estimate for this temperature difference is
approximately 100 W/, but a more accurate value is obtained once the glazing temperature isfound.
-400
-200
0
200
400
600
800
0 50 100 150HeatTransfer(W/m^2)
Basin Water Temperature (C)
Radiation Heat Transfer from Basin Water
Surface to Glazing
Tg=7 degrees
Tg=27 degrees
Tg=47 degrees
Tg=67 degrees
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Heat transfer from convection on basin water surface: Heat transfer due to convection is very important in a passive solar still, as there is
natural fluid flow from buoyancy effects. The heat transfer coefficient for natural convection isdependent on the pressures at both the water surface, and the inner glazing surface. The pressure
difference is the driving force behind the convective flow. The pressure at the basin water
surface is assumed to be the saturation pressure at , the pressure at the glazing surface isassumed as the vapor pressure at the glazing surface temperature. Equations 1 and 2 are used to
find this value, and values over a range of basin water temperatures are plotted in Fig. 3 below
for an estimated glazing temperature of 45 C to observe the trend.
Figure 3. Convective Heat Transfer from Basin Water.
Heat transfer from evaporation of basin water:
Critical to understanding the mass flow rate of condensate from the solar still is
determining the evaporative heat transfer. This is the amount of energy per unit area that actually
goes into turning the basin water into vapor. The rate at which the basin water is evaporated is
also dependent on the convection conditions within the solar still, hence the dependence of theevaporative heat transfer coefficient on the convective heat transfer coefficient. This value is
determined upon further analysis.
-100
-50
0
50
100
150
0 20 40 60 80 100
HeatTransfer(W/m^2)
Water Temperature (C)
Convective Heat Transfer from Water
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Important Heat Loss/Waste Considerations:
Heat transfer from outer glazing surface due to convection: Natural convection has been considered as a mode of external heat loss for the simple
solar still. The heat transferred or lost to the surroundings is given by Eq. 12 and is dependent on
the temperature of the outer glazing surface and the ambient air, as well as the convection heat
transfer coefficient.
(12)Because the glazing is considered infinitely thin, the outer surface temperature is simply . Thenatural convection heat transfer coefficient can be approximated for temperaturedifferences of around 30 C in air as 4.33 W/ [2]. The ambient air temperature in Mumbai,India is fairly stable year round as shown in Table 1 in the preceding preliminary solar
desalination design report. Using an averaged value of 30 C for , the convective heat transferfrom the outer surface of the glazing over a range of glazing surface temperatures was plotted, as
shown in Fig. 4.
Figure 4. Convective Heat Transfer from Glazing Outer Surface.
Heat transfer from basin and sides due to conduction: , Conduction is dependent on the thermal conductivity of the conducting material, the
temperature difference across the material, and the thickness of the material. By insulating the
0
20
40
60
80
100
120
140160
40 50 60 70 80 90 100
HeatTransfer(W/m^2)
Glazing Surface Temperature (C)
Convective Heat Transfer from
Glazing Surface
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solar still walls and basin properly, heat loss through the surfaces from conduction can be
minimized. The heat transfer from conduction for the basin and sides is given by,
, (13), (14)
where and are the thermal conductivities of the basin material and sides, and the thicknesses respectively, and the temperature difference across the material. Evenfor large temperature differences such as 100 C, the heat transfer across a 10 cm thick piece of
Styrofoam is only 30 W/. A temperature difference so large is unlikely, but even if it was tooccur, this loss is small compared to the other modes of heat transfer already mentioned. In order
to simplify analysis, the worst case of 30 W/ is used in this analysis.To sum up the energy balance, now neglecting the particular terms as discussed, the following
energy balance can be written to determine the heat transfer for evaporation.
(15)
where is the amount of solar energy absorbed by the water.
The total amount of solar energy radiating into the solar still is dependent on the position
of the solar still with respect to the sun, the time of day, and the time of year. The climate of
India is fairly stable though, and therefore only the time of day is considered here. The solar still
will be placed facing west for optimum output. At noon the sun will be directly overhead, and
the ambient temperature will be higher for the remainder of the day while the sun sets in the
west. Higher overall system temperatures increase output [1]. The amount of solar radiation as a
function of the angle between the sun and ground is given by,
(16)where represents dawn, exactly noon, dusk,and W/. The hours of sunlight in India fluctuate between 6 and 9 hoursthroughout the year, therefore the worst case of 6 hours is used for this analysis. Fig. 5 shows the
solar radiation throughout the day.
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Figure 5. Solar Radiation at Different Times Through Day.
Knowing the solar radiation on the still, the temperature of the glazing can found using
(17)where is the glazing mass, the specific heat, and , the final and initial temperaturesrespectively. A polypropylene film or 2mm thick polycarbonate sheet will be used to
approximate the infinitely thin glazing surface. Because the glazing will transmit approximately
90% of the solar radiation, the maximum temperature was found using the energy in 10% of the
maximum radiation for 1 second, and an average ambient air temperature of 30 C. The
maximum increment of increase in glazing temperatures are 14.28 C/s and 0.05 C/s for the
polypropylene film and polycarbonate sheet respectively. The steady state temperature will occur
when the glazing is transferring away just as much energy as is incoming. Using the convection
heat transfer equation with
W/
the steady state temperature of the glazing
is found to be 41.34 C. The polypropylene film will obviously reach this temperature much
faster than a polycarbonate sheet, therefore the polycarbonate sheet is selected for this design.
The thickness of the polycarbonate sheet can be manipulated to affect the time it takes for the
glazing to reach steady state temperature. A time greater than the solar cycle would be ideal
because a lower glazing temperature would result from there simply not being enough time to
reach the steady state temperature of 41.34 C, however this is not a realistic option.
0
200
400
600
800
1000
1200
1400
1600
7:12 AM 9:36 AM 12:00 PM 2:24 PM 4:48 PM
SolarRadiation,W
/m^2
Time of Day
Solar Radiation Throughout
Day
Time of day
q_solar
W/m^2
9:00 AM 0
9:20 AM 235.64058
9:40 AM 464.12133
10:00 AM 678.510:20 AM 872.26279
10:40 AM 1039.5223
11:00 AM 1175.1965
11:20 AM 1275.1629
11:40 AM 1336.3841
12:00 PM 1357
12:20 PM 1336.3841
12:40 PM 1275.1629
1:00 PM 1175.1965
1:20 PM 1039.52231:40 PM 872.26279
2:00 PM 678.5
2:20 PM 464.12133
2:40 PM 235.64058
3:00 PM 0
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Figure 6 shows the effect of changing the glazing thickness on the time it takes the glazing to
reach its steady state temperature.
Figure 6. Steady State Temperature of Glazing as Function of Thickness
It is easy to see that a glazing thick enough to prevent it from heating to steady state within the
solar cycle is not plausible. For such a thick piece of glazing the assumption of 90% total
radiation transmitted is likely incorrect. Therefore 41.34 C is used as the glazing temperature.
Next the energy transfers for the water are considered. The incoming solar radiation going into
the basin water can be simplified and written as
(18)where is the heat transfer per unit area which was found earlier to be 30 W/ in a worstcase scenario. The total area was manipulated here to get a range of estimates for . Fig. 7on the following page shows the change in with basin surface area.
0
50
100
150
200
250
0 1 2 3 4 5
Time,
(min)
Thickness of Polycarbonate Glazing, (cm)
Time to Steady State Temperature
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Figure 7. Percentage of Solar Radiation Absorbed by Basin Water vs. Basin Surface Area.
It is easy to see that as the total surface area of the basin increases, more energy is lost to the
surroundings through conduction, as we should expect. Because the sun is not always directly
overhead, values of were found for the entire range of solar radiation throughout the dayand averaged for individual surface area values of interest, shown in Fig. 8.
Figure 8. Energy Absorbed by Basin WaterTo find the proportion of this energy that actually goes into evaporation, Eq. 1-6, Eq. 15, and the
steady state glazing temperature are used. A range of values for is used to show the sensitivityto this parameter. An average value of 700 W/ was used for the amount of solar energyabsorbed by the basin water. Fig. 9 shows the heat of vaporization for different basin water
temperatures.
0.845
0.85
0.855
0.86
0.865
0.87
0.875
0.88
0 0.5 1 1.5 2 2.5 3
SolarEnergyEnergyGoingIntoBasin
Water(%)
Total Basin Surface Area (m^2)
Percentage of Solar Energy Into Basin Water
665
670
675
680
685
690
695
700705
710
0 0.5 1 1.5 2 2.5
HeatTransfer(W/m^2
)
Basin Surface Area, (m^2)
Averaged Solar Radiation Absorbed
by Basin Water
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Figure 9. Heat of Vaporization of Basin Water.
Although it may seem from Fig. 9 that a low basin water temperature is ideal, this is not the case
because the vapor at less than 41.34 C would not condense on the glazing. Increasing the water
temperature will effect evaporation by also increasing the radiation heat transfer between the
water and glazing. To determine the temperature of the basin water due to the solar radiation, Eq.
17 is used and it is found to be 68.42 C. This corresponds to a heat of evaporation of 307.86
W/. Dividing this value by the latent heat of vaporization of water at 70 C(~68.42 C) themass production rate of distilled water obtained in the solar still is found to be 0.496 kg/hr, per
square meter of water surface. There is on average around 6 hours of production time each day,
evaporating 2.97 kg of water each day per square meter of still. On cloudy days the solarradiation can be expected to be only 30% of that on a clear day [3]. Therefore the size of the
solar still will be increased to again accommodate a worst case scenario. Producing only 0.89 kg
with a 1 square meter area after 6 hours on a cloudy day, more surface area is required. A total
water surface area of 2.25 square meters will be used. This design will meet the output
requirement even in the least ideal conditions producing 2 kg water on cloudy days, and on clear
days produce 6.68 kg of clean water.
Cost:
The cost of construction for a passive solar still is considerably cheaper than a more
complex humidification/condensation flow through system. All that is required is a large
insulated box with solar absorbing material in the basin, and a transparent glazing. Because the
box is not under any loading, most insulating foam boards such as expanded polystyrene,
extruded polystyrene, and polyisocyanurate board can provide structural rigidity and no other
materials will be needed. The cost of construction components is listed below.
0
100
200
300
400
500
600
700
800
0 20 40 60 80 100
HeatofEvaporation(W/M^2)
Temperature of Basin Water (C)
q_evap
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Box Structure/Insulation: Extruded polystyrene foam has the best combination of light weight,
rigidity, and low cost. Foam boards of 2 thickness measuring 4x8 can be purchased for
approximately $20 from sources such as Univfoam and Foam-Control. Three boards are required
for the construction a solar still with base dimensions of 1x2.25 m, with a 20 inclined slope
glazing. The maximum side height is 0.50 m, the minimum side height is 0.14 m.
Glazing: One solid piece of polycarbonate measuring 1x2.25m will be required for the glazing.
This can be purchased from sources such as Eplastics and USplastic for around $70 for a 1/16
thick sheet measuring 4x8. The excess from this sheet will be used to construct the catch for the
distilled water.
Solar Radiation Absorber: Another sheet of the same polycarbonate sheet used for the glazing
can be painted black and used as a solar heat absorber.
A picture of the passive solar still is shown below in Fig. 10, and dimensions are shown in Fig.11. The dimensions of the water refill port are arbitrary, or if tube filling is chosen as the filling
mode, it can be omitted. The actual catch for distill water is not shown, but simply consists of a
strip of polycarbonate fixed to the sloped glazing near the bottom, to catch and direct the
condensate out through the drip spout.
Figure 9. Passive Solar Still Design.
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Figure 10. Design Drawing of Solar Still, Dimensions in cm.
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References
1. M.K. Fadatare, S.K. Verma,Influenceof Water Depth on Internal Heat and Mass Transfer ina Plastic Solar Still, Desalination 217, March 2007.
2. John H. Lienhard IV, John H. Lienhard V,A Heat Transfer Textbook3rd Ed., PhlogistonPress, Cambridge, MA; 2008.
3. K. Kimura, D.G. Stephenson, Solar Radiation On Cloudy Days, Trasnactions, AmericanSociety of Heating, Refrigeration, and Air Conditioning Engineers, Vol 75, Part 1, 1969.
4. F.A. Holland, J. Siqueiros, S. Santoyo, C. Heard, E.R. Santoyo, Water Purification Using Heat
Pumps, E&FN Publishing, London, UK; 1999.
5. SolAqua LLC, Solar Still Basics,http://www.solaqua.com/solstilbas.html, SolAqua; 2008
6. Raymond Chang, Chemistry 8th Ed., McGraw Hill Publishing, New York, NY; 2005
7. S.A. Parsons, B. Jefferson, Potable Water Treatment Processes, Blackwell Publishing, Oxford,
UK; 2006.
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