## Session 3. Solar agricultural dryer design procedures and rules of thumb
Step 1. (5 minutes) Step 2. (1 hour)
Step 3. (20 minutes)
Step 4. (20 minutes)
Step 5. (15 minutes)
A.
M = percent moisture Example: 10 kg of fresh fruit which weigh 6 kg when dry. w = 10 kg
B. 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
m
C. The formula:
M w M 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.
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 * KJ = kilo joules
Varies a bit with humidity and temperature.
E. Air is usually quantified as volume at atmospheric pressure (P) and temperature (t).
P = Pressure (in kilopascals - kPa)
1 kg air at 35°C and normal pressure = 0.9 m³ or use psychrometric chart * * *
The upper curve of the chart is for saturated air and is label led The other curves on the psychrometric chart that are similar in shape to the wet-bulb line are lines of The straight lines sloping gently downward to the right are lines of constant wet-bulb temperatures. The intersection of a dry-bulb and a wet-bulb line gives the state of the air for a given moisture content and relative humidity. The lines of constant wet-bulb temperature also give values of Other lines sloping more steeply to the right give the 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 wet-bulb 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 wet-bulb tines. * Although no processes follow the lines giving the specific volume of dry air, these lines show that at any given dry-bulb temperature, the density of air decreases as either the temperature or the relative humidity rises.
A. * 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). * Tropical-monsoon insolation of 5-25 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. Making the collector equal to three times the tray area gives a high drying rate dryer. C. 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. 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. 1. Doubling vent area doubles the air flow rate (but drops delta by about 3/4). F.
G. There are two methods: using the psychrometric chart or using the energy balance equation.
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 pre-heat the air to 45 for drying. The path A-B represents the heating process. Note that in moving the temperature to 8, the humidity drops to 35%. The path B-C 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.)
* 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 When P = 101.3 (normal sea level) Then
We have calculated above that the amount of water to be removed (M We know the two constants: 1. Latent heat of vaporization (L) = 2,800 KJ/kg Assuming initial temperature (T
We can transfer that to M³ using our rule of thumb (1 kg @ 0.9 M³) or PV = M 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.
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 7-1/2 hour solar days), that's: 17,000/30 or 566-2/3 M³/hr. or 9.44 M³/min. I. You must determine: 1. Mass of water to be evaporated (M
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.
If you have the required flow rate figured (See Section H.), use this formula:
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:
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 mid-day 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 Δt = 32ºC and h = 4M as above. Always assume a high insolation rate so your vents will be large enough to prevent over-heating, 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. |

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