The procedure below can be used to design ventilation systems:
- Calculate heat or cooling load, including sensible and latent heat
- Calculate necessary air shifts according the number of occupants and their activity or any other special process in the rooms
- Calculate air supply temperature
- Calculate circulated mass of air
- Calculate temperature loss in ducts
- Calculate the outputs of components - heaters, coolers, washers, humidifiers
- Calculate boiler or heater size
- Design and calculate the duct system
1. Calculate Heat and Cooling Loads
Calculate heat and cooling loads by
- Calculating indoor heat or cooling loads
- Calculating surrounding heat or cooling loads
2. Calculate Air Shifts according the Occupants or any Processes
Calculate the pollution created by persons and their activity and processes.
3. Calculate Air Supply Temperature
Calculate air supply temperature. Common guidelines:
- For heating, 38 - 50 o C (100 - 120 o F) may be suitable
- For cooling where the inlets are near occupied zones , 6 - 8 o C (10 - 15 o F) below room temperature may be suitable
- For cooling where high velocity diffusing jets are used, 17 o C (30 o F) below room temperature may be suitable
4. Calculate Air Quantity
If air is used for heating, the needed air flow rate may be expressed as
q h = H h / (ρ c p (t s - t r )) (1)
q h = volume of air for heating (m 3 /s)
H h = heat load (W)
c p = specific heat air (J/kg K)
t s = supply temperature ( o C)
t r = room temperature ( o C)
ρ = density of air (kg/m 3 )
If air is used for cooling, the needed air flow rate may be expressed as
q c = H c / (ρ c p (t o - t r )) (2)
q c = volume of air for cooling (m 3 /s)
H c = cooling load (W)
t o = outlet temperature ( o C) where t o = t r if the air in the room is mixed
Example - Heating Load
If the heat load is H h = 400 W , supply temperature t s = 30 o C and the room temperature t r = 22 o C , the air flow rate can be calculated as:
q h = (400 W) / ((1.2 kg/m 3 ) (1005 J/kg K) ((30 o C) - (22 o C)))
= 0.041 m 3 /s
= 149 m 3 /h
If the outside air is more humid than the indoor air - then the indoor air can be humidified by supplying air from the outside. The amount of supply air can be calculated as
q mh = Q h / (ρ (x 1 - x 2 )) (3)
q mh = volume of air for humidifying (m 3 /s)
Q h = moisture to be supplied (kg/s)
ρ = density of air (kg/m 3 )
x 2 = humidity of room air (kg/kg)
x 1 = humidity of supply air (kg/kg)
If the outside air is less humid than the indoor air - then the indoor air can be dehumidified by supplying air from the outside. The amount of supply air can be calculated as
q md = Q d / (ρ (x 2 - x 1 )) (4)
q md = volume of air for dehumidifying (m 3 /s)
Q d = moisture to be dehumidified (kg/s)
Example - Humidifying
If added moisture Q h = 0.003 kg/s , room humidity x 1 = 0.001 kg/kg and supply air humidity x 2 = 0.008 kg/kg , the amount of air can expressed as:
q mh = (0.003 kg/s) / ((1.2 kg/m 3 ) ((0.008 kg/kg)- (0.001 kg/kg)))
= 0.36 m 3 /s
Alternatively the air quantity is determined by the requirements of occupants or processes.
5. Temperature Loss in Ducts
The heat loss from a duct can be calculated as
H = A k ((t 1 + t 2 ) / 2 - t r ) (5)
H = heat loss (W)
A = area of duct walls (m 2 )
t 1 = initial temperature in duct ( o C)
t 2 = final temperature in duct ( o C)
k = heat loss coefficient of duct walls (W/m 2 K) (5.68 W/m 2 K for sheet metal ducts, 2.3 W/m 2 K for insulated ducts)
t r = surrounding room temperature ( o C)
The heat loss in the air flow can be expressed as
H = 1000 q c p (t 1 - t 2 ) (5b)
q = mass of air flowing (kg/s)
c p = specific heat air (kJ/kg K)
(5) and (5b) can be combined to
H = A k ((t 1 + t 2 ) / 2 - t r )) = 1000 q c p (t 1 - t 2 ) (5c)
Note that for larger temperature drops logarithmic mean temperatures should be used.
6. Selecting Heaters, Washers, Humidifiers and Coolers
Units as heaters, filters etc. must on basis of of air quantity and capacity be selected from manufacture catalogs.
Boiler rating can be expressed as
B = H (1 + x) (6)
B = boiler rating (kW)
H = total heat load of all heater units in system (kW)
x = margin for heating up the system, it is common to use values 0.1 to 0.2
Boiler with correct rating must be selected from manufacture catalogs.
8. Sizing Ducts
Air speed in a duct can be expressed as:
v = Q / A (7)
v = air velocity (m/s)
Q = air volume (m 3 /s)
A = cross section of duct (m 2 )
Overall pressure loss in ducts can be calculated as
dp t = dp f + dp s + dp c (8)
dp t = total pressure loss in system (Pa, N/m 2 )
dp f = major pressure loss in ducts due to friction (Pa, N/m 2 )
dp s = minor pressure loss in fittings, bends etc. (Pa, N/m 2 )
dp c = minor pressure loss in components as filters, heaters etc. (Pa, N/m 2 )
Major pressure loss in ducts due to friction can be calculated as
dp f = R l (9)
R = duct friction resistance per unit length (Pa, N/m 2 per m duct)
l = length of duct (m)
Duct friction resistance per unit length can be calculated as
R = λ / d h (ρ v 2 / 2) (10)
R = pressure loss (Pa, N/m 2 )
λ = friction coefficient
d h = hydraulic diameter (m)
Systems for ventilation and air handling - air change rates, ducts and pressure drops, charts and diagrams and more.
Ventilation systems - air intakes and outlets - rules of thumbs.
Ventilation systems can be classified by functions, distribution strategies or by ventilation principles.
The equal friction method for sizing air ducts is easy and straightforward to use.
The velocity reduction method can be used when sizing air ducts.
Basic equations for heat transfer - selecting criteria for heaters and coolers in ventilations systems.
Concentration of a pollution in a limited space as a room depends on the amount of polluted material spread in the room, supply of fresh air, outlets positions and construction, principles used for supply and outlet from the space.
A rough guide to maximum air volume capacity of circular ducts in comfort, industrial and high speed ventilation systems.
Minor pressure or head loss in ventilation ducts vs. air velocity - minor loss coefficient diagram.