Design of Ventilation Systems
Design procedure for ventilation systems  air flow rates, heat and cooling loads, air shifts according occupants, air supply principles.
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
Air Heating
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)
where
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} )
Air Cooling
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)
where
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
Moisture
Humidifying
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)
where
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)
Dehumidifying
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)
where
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)
where
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)
where
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.
7. Boiler
Boiler rating can be expressed as
B = H (1 + x) (6)
where
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)
where
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)
where
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)
where
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)
where
R = pressure loss (Pa, N/m^{2})
λ = friction coefficient
d _{ h } = hydraulic diameter (m)
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