Example 1 : Hardy Candy Production
In This Section
During candy and cough drop production, the material is in a plastic state. It must flow and be shaped by stamping machines. If the presence of excess moisture causes the material to become sticky, it will not flow freely and it will adhere to the stamping machine.
To eliminate this material and equipment problem, dry the surrounding air.
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Physical Facts
- Area to be conditioned - 60' x 42' x 16
- Outside design condition - 95 F db*; 75 F wb*
- Controlled space requirement - 75 F db; 35 percent rh
- Physical openings - 1 door (6' x 7'); opened 6/hr
- Number of people working in area - 10
- Construction - 8" masonry
- Make-up air specified by owner - 350 cfm.
* db ~ dry bulb value; wb ~ wet bulb value
Problem
To determine the size of dehumidifier necessary to maintain the desired controlled space conditions.
Assumptions
- The door is adequately weather stripped and is of standard construction.
- Ten workers in the area maintain a moderate pace; each requires ventilating air.
- The interior of the control space is constructed with two coats of vapor barrier paint.
- There are no other openings in and out of the controlled space.
- All physical cracks are sealed.
- A vapor barrier is provided in or under the concrete floor.
Space Moisture Loads to be Computed
- Permeation load
- Load through the door
- Population load.
Permeation Load
| = Grains per hour. |
| V | = 60 × 42 × 16 = 40,320 ft³ | |
| C | = 14 (Specific volume of dry air @ 95°F) | |
| G | = 75 outside design wet bulb of 75°F gives 121 gr/lb from Table I Controlled space requirement of 75°Fdb, 35% rh yields 46 grains per pound from standard Psychrometric chart. Therefore, 121 – 46 = 75 grs/lb. |
|
| F 1 | = 1.94 From Table II – Factor for moisture difference of 75 gr/lb - interpolated | |
| F 2 | = .5 from Table III | |
| F 3 | = 1.0 From Table IV – Factor for 8” masonry | |
| F 4 | = .75 From Table IV – Factor for 2 coats paint | |
40,320 |
× 75 × 1.94 × .5 × 1.0 × .75 = 157,140 grs/hr |
Door Load
| Ohr × | A C |
× G × F1 = grains per hour of additional load |
| Ohr | = 6 |
| A | = 6 x 7 = 42 sq. feet |
| C | = 7 |
| G | = 75 |
| F1 | = 1.94 |
6 × |
42 |
× 75 × 1.94 = 5,238 gr/hr |
Population Load
At a db of 75° F and working at a moderate rate, a person will expel 2,540 grains each hour. Therefore, ten people will add: 10 × 2,540 = 25,400 grains each hour
Total Load
157,140 grs/hr Permeation
5,238 grs/hr Through door
25,400 grs/hr Population load
187,778 grs/hr Total
The drying system and load requirement are shown in the schematic below:
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Note that 350 cubic feet per minute (cfm) outside air is based on a requirement of 30 cfm for each of 10 workers is introduced at the dehumidifier. The effect of this air on the ultimate dehumidifier size will be handled below.
Proceed with the following calculation:
X = C × |
gr/hr |
÷ (S – G) |
| Where: | ||
| X | = cfm Delivery air rate from dryer to space | |
gr/hr |
= total grains per hour in space | |
| C | = 14 = constant | |
| S | = 46 = Grs/lb moisture requirement of controlled space. In the absence of a ventilation requirement this would be the inlet condition at dryer. | |
| G | = Grs/lb of air leaving dryer. Enter curve at 46 grain "Inlet moisture condition." Intersect 75° Inlet air temp curve at 14 gr/lb |
14 × |
187,778 |
÷ (46 – 14) = 1369 cfm |
At this stage in the procedure, it is necessary to resort to the method of approximation to select the correct dryer.
In addition to handling the space load the dryer must handle the moisture load contributed by the 350 cfm outside air requirement. So use a 2000 cfm Bry-Air Dryer (MVB-20-C).
If the dryer has a delivery rate of 2000 cfm, and if 350 cfm of outside air is to be introduced, there remains 1650 cfm of air from the conditioned space. Tabulate this air mixture
350 cfm × 121 gr/lb = 42,350
1650 cfm × 46 gr/lb = 75,900
2000 cfm 118 ,250
Then |
118,250 |
= 59.1 grs/lb |
Referencing the Typical Performance Curves Chart 1, shows that air entering the dryer at 59.1 grs/lb would leave the dryer at approximately 23 grs/lb. (NOTE: Interpolate between the 75° and 85° curves since the air is a mixture of 75° and 95° = 79° F.)
Total moisture pick up X = C |
X |
× (S – G) × 60 – Total Moisture Pickup |
2000 |
× (46 - 23) × 60 = 197,123 grs/hr total removal capacity |
The following work sheet is a demonstration of what the calculations will look like:
Bry-Air Dehumidifier Calculation Sheet
Project: Example I – Production of Hard Candy
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In the above calculations, moisture gain or air leakage in the process ductwork was not considered. If, however, the process and return ductwork did contribute to the moisture load, the total duct volume would be an additional space. Then the permeation calculation in Part Four would be used: V = duct volume; C = 14; F, from Table II, with moisture difference ~G measured from inside process air duct to surround ambient; F3 for tight, good commercial ductwork = 0.6. Add the resultant moisture gain to the room total load. A nominal allowance for process air lost due to duct leakage = 5 percent.
Recommendation
Selecting a VFB-24- at 2000 CFM is the best choice for the hard candy manufacturing example. While it may seem to be an oversized selection, consider that all desiccants in all manufacturers' desiccant dryers will age, will possibly become physically and chemically contaminated with dirt, dust, or chemicals, and will gradually lose their effectiveness. Fortunately, with the VFB-24, higher levels of moisture in the leaving air-up to 24 grs/lb dry air-could be tolerated prior to a desiccant change. So what appears to be an oversized selection would actually allow much longer use of a desiccant charge and provide the economies of longer use.
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