Appropriate Community Technology  A Training Manual (Peace Corps, 1982, 685 p.) 
Phase V: Solar agricultural dryers 
Total time: 
2 hours 
Objectives: 
* To examine and discuss the technical design procedures for solar dryers 

* To review and discuss the rules of thumb for solar dryer design 
Resources: 
* Attachment V3A, "Technical Design Information for Solar Dryers" 

* Attachment V3B, "The Psychrometric Chart" 

* Attachment V3C, "Design Rules of Thumb for Solar Dryers" 

* ISES, "Sunworld," 1980, Vol. 1/No. 6, pp. l8081 
Materials: 
Thermometer, gauze, rubberbands, string, newsprint and felttip pens 
Procedures:
Step 1. (5 minutes)
Present the objectives and outline the
activities.
Step 2. (1 hour)
Distribute Attachment V3A. Reviewand
discuss the attachment and the key variables in dryer design.
Trainer Notes Post and review the following key variables in solar dryer design: * Vent Area  the area, in square centimeters, of the lower (intake) or upper (exhaust) vent, whichever is smaller * Solar Gain  the amount of solar heat being absorbed by the collector, in Kgcal/hr (found by multiplying the hourly insolation rate, given in Kgcal/m² hr. by the aperture or area of the solar collector, given in m ) * Height  the distance between the top of the lower vent and the bottom of the upper vent * Change in Temperature (At)  the difference, in degrees centigrade, between the exhaust air temperature (or the maximum allowable temperature for the agricultural product) and the ambient, or inlet, air temperature Guide the participants through the formulas in the attachment, encouraging their questions and comments. Ask how each formula is applied to dryer design. Explain to those people who are having trouble with the mathematics that you will be discussing more general rules of thumb for these same mathematical formulas and that it is not necessary to understand mathematics to design successful solar dryers. 
Step 3. (20 minutes)
Distribute Attachment V3B and have the
participants read and discuss it. Fashion a simple sling psychrometer and
demonstrate its use.
Trainer Notes * Explain wet and dry bulb temperatures and the psychrometric chart. * To fashion the sling psychrometer, fasten wet gauze to the bulb of a thermometer, tie it to a cord, and twirl the thermometer at the end of the cord. * Review the psychrometric chart, Attachment V3B. * Explain that the chart can be used anywhere in the world. * Demonstrate how the chart can be used to diagram what happens during the drying process (See Section F. Attachment V3A). 
Step 4. (20 minutes)
Distribute and review Attachment V3C,
"Design Rules of Thumb for Solar Dryers."
Trainer Notes Explain that in order to design a successful solar dryer, you only need to understand the rules of thumb and the interaction of the key variables in dryer design. 
Step 5. (15 minutes)
Review and discuss the session
activities and objectives.
Trainer Notes * Explain that the participants now have the necessary technical information for solar dryer design. * Explain that they will have an opportunity to use this technical information when they design their solar dryers. * Encourage the participants to think how this information might be communicated to people with little or no formal education. 
TECHNICAL DESIGN INFORMATION FOR SOLAR DRYERS
A. How to find Percent moisture content (wet basis):
_{}
Where:
M = percent moisture
w = weight of wet sample
d = weight
of dry sample*
* dry = oven dried, 222°C (450°F) for 48 hours
Example:
10 kg of fresh fruit which weigh 6 kg when dry.
w = 10 kg
d = 6 kg
_{}
B. Energy Balance for Drying
The Energy Balance is an equation which expresses the following idea mathematically:
The energy available from the quantity and temperature of air going through the dryer should be equal to the energy needed to evaporate the amount of water to be removed from the crop.
The formula is: m_{a}c_{p}(T_{i}T_{f})=m_{w}L
Where:
m_{a} = mass (or weigh) of drying air
c_{p} =
specific heat capacity of the air (i.e., how much heat it holds per degree of
temperature rise)
T_{i} = initial temperature
T_{f} =
final temperature
L = Latent heat of vaporization of water from grain (amount
of heat needed to vaporize each unit of water)
m_{w} = mass (or
weight) of water to be removed by evaporation
NOTE: The task in solar dryer design is to figure and then achieve high enough temperatures (T_{f}) and air flow to remove the specified amount of water (m_{w}). 
C. How to figure how much water (m_{w}) must be removed from your crop:
The formula:
_{}
M_{w}= mass (weight of water to be removed
w_{i} = initial mass (weight) of crop to be dried
M_{i} & M_{f} = initial and desired final % moisture of the crop
Example:
How much water must be removed from 100 kg of groundnuts in reducing from initial moisture of 26% to final moisture of 14%?
Substituting:
_{}
D. Two Constants:
Latent heat of vaporization of water (L):
Amount of energy needed to vaporize (evaporate) each unit (gram, pound, etc.) of water from the crop.
For free water (in open pan), it's about 2,400 KJ* /kg For water from crops, it's more
and varies a bit with temperature
and moisture content: 2,800
KJ/kg.
* KJ = kilo joules
1 KJ = 1 BTU or 1/4 Kcal
Specific heat capacity of air (C_{p}):
Amount
of heat air can hold per degree of its temperature rise.
Varies a bit with humidity and temperature.
For this, use:
1.02 KJ/Kg° C
Example: 
How much heat is given up if the temperature of 3 kg of air drops from 40 to 35° C? 

=1.02 KJ x 3 kg x (4035°) 

=1.02 (3x5) 

=15.3 KJ 
E. How to figure volume (V) of air from weight:
Air is usually quantified as volume at atmospheric pressure (P) and temperature (t).
The formula: PV = m_{a}R_{t}
Where:
P = Pressure (in kilopascals  kPa)
V = Volume (m³)
t
= temperature (degrees kilvin)
m_{a} = the mass (weight of air)
R
= A constant factor, it equals about 0.291 kPa m³/kgk under dryer
conditions
The Rule of Thumb is:
1 kg air at 35°C and normal pressure = 0.9 m³ or use psychrometric chart
* * *
Useful Solar Dryer Formulae:
_{}
_{}
_{}
THE PSYCHROMETRIC CHART
The upper curve of the chart is for saturated air and is label led wetbulb and dewpoint temperature. (The word "dewpoint" arose from the observation that dew forms on grass when the grass cools, by radiaiton to the sky, to a temperature equal to or less than the wetbulb temperature of the air above it.)
The other curves on the psychrometric chart that are similar in shape to the wetbulb line are lines of constant relative humidity (in X). By definition, relative humidity is a ratio: the partial pressure of the water vapor at a given temperature  the saturation pressure of the water vapor at the same temperature. The scale at the left side of the chart gives the pressures.
Graphic
The straight lines sloping gently downward to the right are lines of constant wetbulb temperatures. The intersection of a drybulb and a wetbulb line gives the state of the air for a given moisture content and relative humidity. The lines of constant wetbulb temperature also give values of constant enthalpy (total heat content), measured in heat units per unit weight of dry air.
Other lines sloping more steeply to the right give the specific volume of dry air, the volume occupied by one kilogram of dry air under the indicated conditions.
In examining a psychrometric chart, note that:
* Processes in which air is heated or cooled without change in moisture content give horizontal lines. Heating along such lines will decrease the relative humidity, while cooling will increase it.
* The wetbulb temperature lines, sloping downward to the right, are lines of adiabatic cooling (where there is no change in heat content). These lines typify drying processes in which air is passed over the surface of wet material and is cooled by evaporation of water from the material. Lines of constant total heat parallel these wetbulb tines.
* Although no processes follow the lines giving the specific volume of dry air, these lines show that at any given drybulb temperature, the density of air decreases as either the temperature or the relative humidity rises.
DESIGN RULES OF THUMB FOR SOLAR DRYERS
A. Assorted considerations for solar agricultural dryer designs:
* 1 kg of air at 35°C @ 0.9 m³
* For grain drying, make beds no more than l5 cm thick, giving a maximum loading rate of 9Okg/M² (requires stirring).
* Tropicalmonsoon insolation of 525 MJ* /M² per day. Use 15 MJ/M² per day for estimate (approximately 14,000 BTUs or 3,500 kcal).
* Typical conservative day long efficiency of stationary collection: 25%
(That is, the energy delivered as heated air to the drying crop is 25% of the energy in the sunlight striking a horizontal surface of equal area to the dryer's collector.)
* MJ = Mega Joule or 1 million joules.
B. Collector size:
Making the collector equal to three times the tray area gives a high drying rate dryer.
C. Dryer capacity:
In the tropics, figure on about 180 M of air to remove 1 kg of water.
Figure about 3/4 M² of collector area to remove 1 kg of water per day (i.e., dry 1.5 kg fresh fruits or 5.25 kg grain per day).
D. Dryer temperature:
1. Depends upon insolation, collection area and vent size.
2. Is very sensitive to vent size (cutting vent size by one half increases delta t by about three times (up to some limit).
3. Doubling area of collector increases it by about one half.
4. Raising temperature from 20 to 35°C can triple the water capacity of the air.
E. Dryer air flow rate:
1. Doubling vent area doubles the air flow rate (but drops delta
by about 3/4).
2. Doubling height increases air flow by 0.4.
3. Doubling
collector area increases air flow by about 40% (also increases At by 1/2).
F. Required moisture contents of crops/approximate values:

For Storage 
Fresh 
Fruits 
10% 
70  85% 
Vegetables 
18% 
70  85% 
Grains 
14% 
25  35 
G. Figuring how much air you need for drying:
There are two methods: using the psychrometric chart or using the energy balance equation.
Method #1. Using the Psychrometric Chart
Example:
You want to dry 1 kg. of rice from initial moisture of 22X to final moisture of 14%. Assume ambient air temperature is 30°C at 80X humidity and you preheat the air to 45 for drying.
The path AB represents the heating process. Note that in moving the temperature to 8, the humidity drops to 35%.
The path BC represents the change in the air as it passes through the dryer, cooling and picking up moisture from the rice. Initially (because rice is quite wet), air gets to C. At the end of the process, it only reaches D. As the rice gets dryer, you find points C & D from the table* of equiibrium moisture contents (it's similar for all crops). In this case, the air's humidity ratio rose by about 0.005. (That's how much water the air carried away.)
Graphic
Equilibrium moisture content of rough rice, per cent wet basis.*  
Temperature 
Relative humidity of air (%)  
(°C) 
20 
30 
40 
50 
60 
70 
80 
90 
10 
8 
9 
11 
12 
13 
14 
16 
19 
20 
7 
9 
10 
11 
13 
14 
15 
18 
30 
7 
8 
9 
11 
12 
13 
15 
17 
40 
6 
7 
8 
10 
11 
12 
14 
17 
* Moisture level at which rice will stabilize if exposed to the specified temperatures and humidity conditions.
The amount of water to be extracted from 1 kg of rice in this case can be figured using the equation found in Part C of Attachment A.
_{}
From the definition of humidity ratio (weight of water vapor in the air  the weight of dry gases in the same air), it follows that the mass of air needed (ma) in this case, where humidity ratio rose by 0.005, is:
_{}
We can transform this weight to volume with the equation from Part E of Attachment A:
PV = m_{a} RT
When
P = 101.3 (normal sea level)
T = 308 (35°C)
Then _{}
Method #2. Using Energy Balance Equation (See Example)
We have calculated above that the amount of water to be removed (M_{w}) = 0.093 kg.
We know the two constants:
1. Latent heat of vaporization (L) = 2,800 KJ/kg
2. Specific
heat of air (C_{p}) = 1.02 KJ/kg°C
Assuming initial temperature (T_{i}) = 45°C and final temperature is a mean value of 32°C, we can substitute in the energy balance equation to get m_{a}:
_{}
We can transfer that to M³ using our rule of thumb (1 kg @ 0.9 M³) or PV = M_{a}RT and we get about 17.3 M³ of air.
You will notice that this result is not identical to the 16.5 M³ calculated above using the psychrometric chart. However, the result is close enough for design work.
H. To figure air flow rate:
Example:
Say we want to dry 1,000 kg of rice. We've figured it takes 17 M³/kg, so that's 17,000 M total. If we want this to flow in 30 hours (say, four 71/2 hour solar days), that's:
17,000/30 or 5662/3 M³/hr. or 9.44 M³/min.
I. To figure area of solar collector needed:
You must determine:
1. Mass of water to be evaporated (M_{w})
2. Specific
latent heat of vaporization of water from the crops (L) = 2,800 KJ/kg
3. The
quantity (Q) of insolation per unit horizontal area per day
4. The efficiency
at the collector (e)
Example:
For 1,000 kgs. of rice, we calculated that we must remove 93 kg. of water. We know L = 2,800 KJ/kg. So the heat required is 93 x 2,800 = 260,400 KJ (260.4 MJ)*
* MJ = Megajoule (1,000,000 joules or 1,000 kilojoules)
This heat must come from the available solar energy.
Tropical monsoon insolation is highly variable, depending upon cloudiness: from 5 to 25 MJ/M² per day.
Use 15 MJ/M² per day as a conservative average in absence of data.
Assuming 15 MJ/M² per day and 25% efficiency of the collector yields 3.75 MJ/M² per day or 15 MJ/M² in four days.
So, the total area of collector required is:
_{}
J. How to figure vent area. using two methods:
Method #1:
If you have the required flow rate figured (See Section H.), use this formula:
_{}
Example:
Assume air flow calculations showed a flow rate of 9.4 M³/min required to dry our 1,000 kg or rice in four days (review Section H.). Then checking data sources, assume that 'he desired temperature of the drying rice is 62 C and that the ambient temperature is 30°C. So delta temperature (change in air temperature in the dryer) is 62°  30° = 32°. Assume a height Of 4m for the dryer. Substitute in the equation:
_{}
Method #2:
If you have an aperture (collector) area and some idea of solar intensity, use this formula:
_{}
Assume that a maximum of 15% of the total daily radiation falls in the hottest midday hour. This is 0.15 x 25 MJ = 3.75 MJ/hr M² = 896 Kcal/ hr. m² *
* 1 MJ/m² = 239 Kcal/m² = 88 BTu/m²
Using the aperture area found in Section I, it's 17.5 m². Let Dt = 32ºC and h = 4M as above.
Always assume a high insolation rate so your vents will be large enough to prevent overheating, even under the most intense sun conditions. You can always close the vent to some degree, if necessary.
Then, substituting the formula:
_{}
Note: This is the maximum vent area you would ever need. With a lower insolation rate of 15 MJ/day, 2 the vent area could be cut down to about 2,600 CM².