When deciding what size dehumidifier to use, remember that controlled space requirements sometimes exceed the anticipated design peak load. Unusual and unforeseen humidity loads – such as from abnormal weather conditions or the introduction of high-moisture content raw materials – can burden the drying equipment. Here we present a number of issues that must be considered in approaching and solving specific drying problems. Six typical humidity control examples are presented:
- Food and drug manufacturing, specifically raw materials and processing equipment
- Storage or equipment areas (Standby warehouse)
- Product drying
- Controlled humidity and temperature areas
- Specific purposes for dry air production
- Prevention of condensation (Water treatment plant)
Example I: Production of Hard Candy
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.
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%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
V
C x ∆G x F1 x F2 x F3 x F4 = Grains per hour.
V = 60 x 42 x 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.
F1 = 1.94 From Table II – Factor for moisture difference of 75 gr/lb – interpolated
F2 = .5 from Table III
F3 = 1.0 From Table IV – Factor for 8″ masonry
F4 = .75 From Table IV – Factor for 2 coats paint
40,320
14 x 75 x 1.94 x .5 x 1.0 x .75 = 157,140 grs/hr
Door Load
Ohr x A
C × ∆G × F1 = grs/hr
Ohr = 6
A = 6 × 7 = 42 sq ft
C = 7
∆G = 75
F1 = 1.94
6 x 42
7 × 75 × 1.94 = 5,238 grs/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 x 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:
Sizing Dehumidifier
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 x grs/hr
60 ÷ (S – G)
Where:
x = cfm Delivery air rate from dryer to space
grs/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 grs/lb
14 x 187,778
60 ÷ (46 – 14) = 1369 cfm
From the above calculation the space moisture load is 187,778 grs/min. 1369 cfm air at 14 grs/ Ib will maintain the space design conditions.
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 grs/lb = 42,350
1650 cfm × 46 grs/lb = 75,900
2000 cfm 118 ,250
Then 118,250
2000 = 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
C × (S – G) × 60 – Total Moisture Pickup
2000
14 × (46 – 23) × 60 = 197,123 grs/hr total removal capacity
The following worksheet is a demonstration of what the calculations will look like.
In the above calculations, moisture gain or air leakage in the process duct-work 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%.
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.
Example II: Standby Warehouse
Moisture damage in a standby or storage warehouse can be avoided by surrounding the machinery, equipment, or material with dry air.
Physical Facts
- Area to be conditioned – 210′ x 176′ X 45′ = 1,663,200 cubic feet
- Outside design condition – 95° F db; 77 F wb
- Controlled space requirement* – 85°F db; 40% RH
- No physical openings nor appreciable amount of door openings or closings specified
- No people working in the area
- Construction – 8″ masonry.
- See Appendix 5 LINK TO APP5
Problem To determine the size of the dehumidifier required to maintain standby conditions.
Assumptions
- All physical cracks are sealed and the floor properly vapor-proofed.
- If the room is completely vapor-proofed, use Table 4 on page 8.
- Two coats of vapor barrier paint have been applied externally for metal clad construction.*
* External application is recommended because: Outside walls are usually easier to access than inside walls for paint application. Coating the outside walls discourages water permeation into the wall and thus minimizes water accumulation in the wall structure itself.
Space Moisture Loads to be Computed
- Permeation load
- Moisture load.
The Permeation Load is the only moisture load involved in this example.
V C | x ∆G x F1 x F2 x F3 x F4 | = grs/hr |
Where:
V | = 210 × 176 × 45 = 1,663,200 ft.^{3} |
C | = 14 = constant |
∆G | = 58 grs/lb. Outside design web bulb of 77°F gives 130 grs/lb from Table I (LINK) Controlled space requirement of 85°F db, 40% rh yields 72 Grs/lb from a standard Psychrometric chart. Therefore, 130 – 72 = 58. |
F_{1} | = 1.54 From Table II – Factor for moisture difference of 63 grs/lb |
F_{2} | = 0.24 from Table III – extrapolated as straight line for a volume of 1,663,200 cubic feet. |
F_{3} | = 1.0 From Table IV – Factor for 8″ masonry |
F_{4} | = .75 From Table IV – Factor for 2 coats paint |
1,663,200 14 | × 58 × 1.54 × 0.24 × 1.0 × .75 | = 1,910,019 grs/hr |
Refer to schematic blow which shows the load requirements and drying system.
X = C × | grs/hr 60 | ÷ (S – G) |
Where:
X | = cfm Delivery air rate from dryer to space |
C | = 14 = constant |
S | = 72 = grs/lb moisture requirement of controlled space. |
G | = grs/lb of air leaving dryer. |
Enter curve at 72° “Inlet Moisture Condition”. Interpolate “Inlet Air Temperature Curve” between 75° and 95° and find “leaving Moisture” at 33 grs/lb.
X = 14 × | 1,910,019 60 | ÷ (72 – 33) |
X = 11,427 cfm |
Therefore, 11,427 cfm of air (33 grs/lb) from the dehumidifier is needed to maintain a grain level of 72 grs/lb.
In a building of this size and shape, air distribution ducts are practical for effectively spreading the air so it can return to a common point and re-enter the dehumidifier.
Recommendation Use one Bry-Air VFB-150 Dehumidifier at 12,500 CFM in this standby warehouse example with a fan sized to handle the necessary static pressure of the duct system.
Bry-Air Dehumidifier Calculation Sheet Project: Example 2 – Standby Warehouse
Moisture damage in a standby or storage warehouse can be avoided by surrounding the machinery, equipment, or material with dry air.
Physical Facts
- Area to be conditioned – 210′ x 176′ X 45′ = 1,663,200 cubic feet
- Outside design condition – 95° F db; 77 F wb
- Controlled space requirement* – 85°F db; 40% RH
- No physical openings nor appreciable amount of door openings or closings specified
- No people working in the area
- Construction – 8″ masonry.
- See Appendix 5 LINK TO APP5
Problem To determine the size of the dehumidifier required to maintain standby conditions.
Assumptions
- All physical cracks are sealed and the floor properly vapor-proofed.
- If the room is completely vapor-proofed, use Table 4 on page 8.
- Two coats of vapor barrier paint have been applied externally for metal clad construction.*
* External application is recommended because: Outside walls are usually easier to access than inside walls for paint application. Coating the outside walls discourages water permeation into the wall and thus minimizes water accumulation in the wall structure itself.
Space Moisture Loads to be Computed
- Permeation load
- Moisture load.
The Permeation Load is the only moisture load involved in this example.
V C | x ∆G x F1 x F2 x F3 x F4 | = grs/hr |
Where:
V | = 210 × 176 × 45 = 1,663,200 ft.^{3} |
C | = 14 = constant |
∆G | = 58 grs/lb. Outside design web bulb of 77°F gives 130 grs/lb from Table I (LINK) Controlled space requirement of 85°F db, 40% rh yields 72 Grs/lb from a standard Psychrometric chart. Therefore, 130 – 72 = 58. |
F_{1} | = 1.54 From Table II – Factor for moisture difference of 63 grs/lb |
F_{2} | = 0.24 from Table III – extrapolated as straight line for a volume of 1,663,200 cubic feet. |
F_{3} | = 1.0 From Table IV – Factor for 8″ masonry |
F_{4} | = .75 From Table IV – Factor for 2 coats paint |
1,663,200 14 | × 58 × 1.54 × 0.24 × 1.0 × .75 | = 1,910,019 grs/hr |
Refer to schematic blow which shows the load requirements and drying system.
X = C × | grs/hr 60 | ÷ (S – G) |
Where:
X | = cfm Delivery air rate from dryer to space |
C | = 14 = constant |
S | = 72 = grs/lb moisture requirement of controlled space. |
G | = grs/lb of air leaving dryer. |
Enter curve at 72° “Inlet Moisture Condition”. Interpolate “Inlet Air Temperature Curve” between 75° and 95° and find “leaving Moisture” at 33 grs/lb.
X = 14 × | 1,910,019 60 | ÷ (72 – 33) |
X = 11,427 cfm |
Therefore, 11,427 cfm of air (33 grs/lb) from the dehumidifier is needed to maintain a grain level of 72 grs/lb.
In a building of this size and shape, air distribution ducts are practical for effectively spreading the air so it can return to a common point and re-enter the dehumidifier.
Recommendation Use one Bry-Air VFB-150 Dehumidifier at 12,500 CFM in this standby warehouse example with a fan sized to handle the necessary static pressure of the duct system.
Example III: Product Drying
Here we have a room used to remove water vapor from such products as cattle feed mixes, nylon or rayon cord for tires, raw plastic material, granular chemicals, raw paper stock, cardboard stock for coatings, or other similar products.
In this example, the room is used for drying cattle feed mixes, which are contained on drying carts that stand in the room until the specified level of dryness is attained.
Space condition requirements and product movement rate are determined by the manufacturer.
Physical Facts
- Drying room size – 40′ x 65′ x 16′
- Outside design condition – 93° F db; 73° F wb
- Controlled space requirement – 95° F; 15% RH (36 grs/lb)
- One double door: (a) 6′ x 7′ (b) Opens at 2/hr
- There are no other openings.
- There are no workers in room except to bring mix in and out
- Product movement rate -1500 Ib/hr (i.e., carts with trays of mix are moved into the drying room at the rate of 1500 Ib/hr)
- Product enters room at 8% moisture and leaves at 4% moisture. 9. Drying room wall construction – 8″ masonry.
- 350 cfm outside air required by manufacturer.
Problem To determine the size of the dehumidifier.
Assumptions
- All physical cracks are sealed.
- The double door is weather stripped.
- Two coats of vapor barrier paint have been applied to the wall and ceiling construction of the drying room; the floor is suitably protected against vapor permeation.
Moisture Loads to be Computed
- Product load
- Permeation load
- Door load.
Product Load Since the product will lose 4%moisture (by weight) and there are 1500 pounds of product each hour:
1500 lb./hr × (8 % – 4%) = 60 lb./hr water removal Since one pound of water equals 7000 grains, then: 60 × 7000 = 420,000 grs/hr product load.
Note that the time needed to reduce the material to a 4%moisture level would have to be given or experimentally determined. These data would determine the amount of material to process and the size of the drying chamber needed.
V C | x ∆G x F1 x F2 x F3 x F4 | = grains per hour |
Where:
V | = 40 × 65 × 16 = 41,600 ft.^{3} | |
C | = 14 = constant | |
∆G | = 77 Outside design web bulb of 73°F gives 113 grs/lb. Drying room space Requirement of 95°F,15% RH yields 36 grs/lb from a standard Psychrometric chart. | |
F_{1} | =1.99 from Table II – Factor for moisture difference of 84 grains. | |
F_{2} | = 0.50 from Table III – Permeation factor | |
F_{3} | = 1.0 from Table IV – Factor for 8” masonry | |
F_{4} | = 0.75 from Table IV – Factor for 2 coats paint |
41,600 14 | × 77 × 1.99 × .50 × 1.0 × .75 | = 170,742 grs/hr |
Door Load
O_{hr} x | A C | × ∆G × F1 = grs/hr |
O_{hr} | = 1 |
A | = 6 × 7 = 42 sq ft |
C | = 7 (constant) |
∆G | = 77 grs/lb. |
F_{1} | = 1.99 from Table II – Factor for moisture difference of 84 grains. |
2 × | 42 7 | × 77 × 1.99 = 1,839 grs/hr |
Total Moisture Load
420,000 grs/hr | Product Load |
170,742 grs/hr | Permeation Load |
1,839 grs/hr | Door Load |
592,581 grs/hr | Total Moisture Load |
The 350 cfm outside air requirement will be considered at a later stage in the calculation.
Proceed with calculation as follows:
X = C × | grs/hr 60 | ÷ (S – G) |
Where:
X | = cfm – air rate from dryer |
C | = 14 = constant |
S | = 36 = grs/lb drying room controlled space requirement. In the absence of an outside air requirement this would also be the inlet condition at dyer. |
G | = 15 grs/lb – equals condition of air leaving dryer. Enter curve at 36 – intersect 95°F curve at 15 grs/lb. |
X = 14 × | 592,321 60 | ÷ (36 – 15) |
X = 6584 cfm |
Recommendation The Bry-Air VFB-75 Dehumidifier, rated at 7500 cfm, should be adequate. However, the first step should be to determine if this Dehumidifier has enough capacity to handle the 350 cfm outside air in addition to the moisture load in the drying room.
If the dryer has a delivery rate of 7500 cfm and 350 cfm of outside air is to be introduced, there remains 7150 cfm from the conditioned space. Tabulate this air mixture as follows:
350 cfm × 113 grs/lb. = | 39,550 |
7150 cfm × 36 grs/lb. = | 257,400 |
7500 | 296,950 |
296,950 7500 | = 39.6 |
Reference to Chart 1, Appendix 4, shows that air entering the dryer at 39.5 grs/lb. would leave at 17 grs/lb.
7500 14 | × (36 – 17) × 60 = 610,714 grs/hr |
The computed moisture load is 592,581 grs/hr. Therefore, the VFB-75 is adequate to handle the moisture load.
Bry-Air Dehumidifier Calculation Sheet Project: Example 3- Product Drying
Example IV: Controlled Temperature & Humidity Areas
Many air conditioned manufacturing areas often have a required air flow to handle a sensible load in that space. This air quantity requirement and the accompanying dehumidifier size are usually greater than those needed to handle a latent load. By designing a system for the sensible load situation and then determining the appropriate dehumidifier to handle the moisture load, the desired conditions for the space can be maintained.
Physical Facts
- Area to be conditioned – 62.5′ x 55′ x 14′
- Outside design conditions- 95°F db; 7]°F wb
- Controlled space requirement – 55°F db; 30% RH; 20 grs/lb.
- Doors – 1 (6′ X 8′), 6 openings/hr; 1 (3′ x 7′), 40penings/hr
- Other (fixed) openings – 2.8 sq ft, w/tunnel 10′ deep
- Number of people working in area – 10
- Air required for sensible temperature control – 24,715 cfm, 42°F
- Construction – Block walls; drywall ceiling with vapor proofing; concrete floor on grade
- Make-up air required – 2400 cfm
- Air available for make-up – 50°F db/49°F wb; 50 grs/lb.
Problem To determine the size dehumidifier needed in a controlled humidity and temperature area.
Moisture Load to be Computed
- Permeation
- Load through doors
- Load through fixed openings
- Population load.
Permeation Load
V C | x ∆G x F1 x F2 x F3 x F4 | = grs/hr |
Where:
V | = 62.5 × 55 × 14 = 48,125 ft.^{3} | |
C | = 14 = constant | |
∆G | = 110 (Ambient 130 grs/lb – room 20 grs/lb) | |
F_{1} | = 2.76 from Table II – Factor for moisture difference of 110 grs/lb. | |
F_{2} | = 0.48 from Table III – for 48,125 cu. Ft. | |
F_{3} | = 1.0 from Table IV – Frame masonry & frame construction | |
F_{4} | = .9 for vapor proof paint on walls & ceilings, untreated concrete floor |
48,125 60 | × 110 × 2.76 × .48 × 1.0 × .9 | = 450,846 grs/hr |
Door Load
O_{hr} x | A C | × ∆G × F_{1} = grs/hr |
O_{hr} | = 6 openings |
A | = 6 × 8 = 48 sq ft |
C | = 7 = constant |
∆G | = 110 grs/lb. |
F_{1} | = 2.76 from Table II – Factor for moisture difference of 84 grains. |
6 × | 48 7 | × 110 × 2.76 = 12,491 grs/hr |
O_{hr} | = 4 openings |
A | = 3 × 7 = 21 sq ft |
C | = 7 = constant |
∆G | = 110 grs/lb. |
F_{1} | 2.76 from Table II – Factor for moisture difference of 84 grains. |
4 × | 21 7 | × 110 × 2.76 = 3,643 grs/hr |
Fixed Openings
A × 300 C × D | × ∆G × F1 = grs/hr |
A | = area, 2.8 sq. ft. |
300 | = Constant (vel. of vapor) |
C | = 14 Constant |
D | = Depth of tunnels |
∆G | = 110 grs/lb |
F^{1} | = 2.76 |
2.8 × 300 14 × 10 | × 110 × 2.76 = 1,822grs/hr |
Population Load At a db of 55° and working at a “light physical exertion” – 1100 grs/hr/person10 people x 1100 gr = 11,000 grs/hr
Total Room Moisture Load 450,846 grs/hr Permeation 12,491 grs/hr Door Load 3,643 grs/hr Door Load 1,822 grs/hr Fixed Opening Load 11,000 grs/hr Population Load 479,802 grs/hr Total Room Load
The total room latent moisture load is 479,802 grs/hr, which is added into the calculation below to find the entering grain condition needed for the space.
Total cfm 14 | × (S – G) × 60 = Room load (grs/hr) |
Total cfm = 24,715 cfm
14 | = constant |
S | = 20 grs/lb. (condition of controlled space) |
G | = Unknown grain level needed entering space |
60 | = min/hr |
24,715 14 | × (20 – G) × 60 = 479,802 |
G = 15.4 grs/lb.
Thus the air to the room must be 15.4 grs/lb. and the air mixture (return from the room plus the dehumidifier discharge) entering the main system fan should be 15 grs/lb. to allow for possible leakage into the system duct work. Here one must resort to trial and error techniques to select the dehumidifier size.
cfm 14 | × (S – G) × 60 = X |
cfm | = 7500 cfm – dehumidifier capacity (trial) |
14 | = constant |
S | = 20 grs/lb. condition in the controlled space |
G | = 4 grs/lb. air leaving dehumidifier with entering air 53°F, 30 grs/lb. |
7500 14 | × (20 – 4) × 60 = 514,285 grs/hr |
Note that the make-up air of 2400 cfm must mix with 5100 cfm of return air before entering the dehumidifier.
Recommendation The VFB-75 Dehumidifier will satisfy the room load conditions when mixed with the remaining 17,215 cfm of return air and delivered into the conditioned space.
Bry-Air Dehumidifier Calculation Sheet Project: Example IV – Controlled Humidity and Temperature Areas
Example V: Production of Dry Air for a Specific Purpose
Many applications require a specific quantity of outside air to be delivered at a given moisture content and temperature. This requirement may be a need to make up air exhausted from a space or to supply air for a process such as a drying oven.
Physical Facts
- Maximum allowable moisture content – 17 grs/lb of dry air
- Maximum allowable temperature – 115°F
- Quantity of air required – 6,000 cfm
- Maximum condition of outside air – 95°F, 130 grs/lb
From the Typical Performance Curves chart it is obvious that 130 grain air cannot be reduced to 17 grains in a single pass through a dehumidifier, without other conditioning. Examination of Chart 1 shows that to produce 17 grains air leaving the dehumidifier, the inlet condition should be 64 grains or less at 60°F or less.
This is accomplished as shown above by installing a cooling coil upstream of the dehumidifier to reduce the temperature and moisture content of the outside air.
Example VI: Water Treatment Plants
In most water pumping stations, filtration plants, and wastewater control plants, control of humidity in the pipe galleries, pump rooms, and control rooms is of prime importance. By reducing the dew-point temperature of the air below the temperature of the piping and walls, sweating and condensation can be eliminated. By circulating warm, dry air through the areas, water accumulation is avoided, maintenance for electrical controls, motors, and instruments is reduced, and paint lasts longer on the pipes, valves, and flanges.
A standard rule-of-thumb is used to approximate this type of application load:
Volume of space to be conditioned 25 |
= CFM dehumidifier |
(For each 25,000 cu. ft. of space, supply 1,000 cfm of dry air.)
The use of an after-cooling coil for the dry air discharge from the dehumidifier can be omitted in most installations since the warm, dry air (low rh) will help heat the space during cool or winter conditions. Heat should not build up to an objectionable level because the large piping and wall areas are at the same temperature as the water in the system. Warm air also has the advantage of reducing the rh and increasing the air’s capacity to carry away moisture.