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This is part 19th of the full HVAC course. If you wish to access the previous part, you can get it by the end of this post. In this part, we will see about toilet exhaust in ventilation.
Toilets are ventilated by air passing through them.
Air enters at the bottom of the tank and travels upwards through the toilet bowl. This creates a current that carries away odors from the bowl and freshens the air inside the room where the toilet is located.
The best way to control odor in the bathroom is to have good ventilation. A properly installed exhaust fan is a great way to do this.
Ventilation can be provided by natural means (i.e., open windows and doors) or mechanical (vent fans). Mechanical ventilation is often preferred for bathrooms because it is quieter than other types of exhaust systems, such as ducts. However, if a system is not well designed and maintained, it may create problems by discharging stale air into the room where the toilet resides.
If a space does not have adequate ventilation, it could lead to unpleasant smells and increased humidity levels. Toilets should be vented outside to minimize this problem.
Total Exhaust Ventilation system
This system provides fresh air flow through your entire ventilation system. Fresh air intake can be located at any point within your room, including the ceiling. This allows for maximum fresh air intake without installing fans or other equipment throughout your room. Once the air has been filtered, it then exits out through exhaust vents located on the floor. A filter is included in this unit that will increase airflow and remove harmful fumes from the air before they reach your lungs. Use this total exhaust venting system if you are looking for a solution that won’t add extra costs and maintenance.
While designing for toilets, this can be divided into two types:
* Public Toilets or Commercial Toilets --- 12 - 30 ACH
* PrivateToilets or Residential Toilets --- 6 - 10 ACH
* PrivateToilets/Residential Toilets
According to AHRAE - cfm = 25 to 50
therefore, fresh air cfm = 75% to 85% of exhaust cfm
Fresh air cfm = 0.8 * 50
= 40 cfm
Toilets are ventilation systems that move air across bathroom surfaces. They are not ventilators but instead create airflow that moves out from a central location and over your entire bathroom area, including your toilet. This action pushes stale air away from your home and replaces it with fresh air. A properly designed HVAC system can ensure proper ventilation while reducing energy costs. For example, an under-sink exhaust fan may be installed beneath your sink and will pull hot air away from the pipes, ducts, and registers and release this hot air outside.
Door undercut / Grill sizing
By Continuity equation,
Q = A x V
here, v = 300-700fpm
therefore, Q = AV
40 = A x 500
A = 40/500
A = 0.00744 sq m.
Area A = W x H
0.0074 = 0.8m x H
H = 0.0074 / 0.8
H = 0.0093m
H = 9.3mm
Here W, the width of the door = 0.8m, we can get it from architectural drawings.
Now, we will see how to design the toilets in ventilation, by understanding a solved example.
Example;
Here in public toilets, we have to design fan capacity or we have to do an appropriate fan selection. Fan selection depends upon 2 factors:
* Volume of air - Cfm/CMH
* Static pressure - 10mm wg
- For commercial/public toilets we take 12-30 ACH
- In this problem, the dimensions of the toilets 1, 4, 5, and 8 are the same. They are all typical.
- And the dimensions for the toilets 2, 3, 6 and 7 are the same.
- And the dimensions for passage 9, is different from all others.
* Volume of air [ For toilets 1, 5, 4 and 8]
cfm = (m³ x ACH) / 1.7
= (l x b x h x ACH) / 1.7
= (l x w x h x ACH) / 1.7 (ACH = 12 to 30)
1. For first.
cfm = (l x b x h x ACH) / 1.7
= (3 x 3 x 2.8 x 18) / 1.7
therefore,
cfm = 266.82 ( This will be same for 1, 4, 5 and 8)
hence, cfm = 266.82 x 4
cfm = 1067.28 ----------- (eq 1)
2. cfm = (l x b x h x ACH) / 1.7
= (3.5 x 3 x 2.8 x 18) / 1.7
cfm = 311.25 ( here, for 2 and 7 which are same)
therefore,
cfm = 311.25 x 2
cfm = 622.58 -------------(eq 2)
3.
cfm = (l x b x h x ACH) / 1.7
= (2.5 x 3 x 2.8 x 18) / 1.7
cfm = 222.35 ( Here, 3 and 6 are having same dimensions, they are typical)
therefore,
cfm = 222.35 x 2
cfm = 444.7 ---------------(eq 3)
4. For passage 9,
cfm = (l x b x h x ACH) / 1.7
= (12 x 2 x 2.8 x 18) / 1.7 (l = 3.5+3+2.5+3)
cfm = 711.52 --------------(eq 4)
therefore, Total cfm = eq 1 + eq 2 + eq 3 + eq 4
= 1067.28 + 622.58 + 444.7 + 711.52
Total cfm = 2846.08 cfm
CMH = 1.7 x cfm
= 1.7 x 2846.08
= 4838.33 CMH
Volume of air is 4838.33 CMH
* Static Pressure:
For 1m ---- Straight Duct = 0.004" wg.
90° Elbow = 0.2" wg.
45° Elbow = 0.1" wg.
The above static pressure values are taken from the ASHRAE standards database for duct fittings.
Note :
In static pressure, for designing fan capacity we have to take the longest path (index run) or critical path.
For 1m static duct = 0.004"
therefore, for 17m = ?
x = 17 * 0.004
For 1m straight duct = 0.068" wg
For 1, 90° Elbow = 0.2" wg
1, 45° Elbow = 0.1" wg
Therefore, Total Static pressure = 0.068 + 0.2 + 0.1
= 0.368
= 0.368 x 25.4
= 9.3 (approximately 10mm of wg straight duct)
Access Previous parts of this course, if you have missed them, by clicking Below...
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