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A design procedure of ventilation systems, with air flow rates, heat and cooling loads, air shifts according occupants, air supply principles
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A ventilation system may be designed more or less according the following procedure:
Calculate heat and cooling loads by
Calculate the pollution created by persons and their activity and processes.
Calculate air supply temperature. Common guidelines:
If air is used for heating, the needed air flow rate may be expressed as
qh = Hh / ρ cp (ts - tr) (1)
where
qh = volume of air for heating (m3/s)
Hh =heat load (W)
cp = specific heat capacity of air (J/kgK)
ts = supply temperature (oC)
tr = room temperature (oC)
ρ = density of air (kg/m3)
If air is used for cooling, the needed air flow rate may be expressed as
qc = Hc / ρ cp (to - tr) (2)
where
qc = volume of airfor cooling (m3/s)
Hc =cooling load (W)
to = outlet temperature (oC) where to = tr if the air in the room is mixed
If the heat load is Hh = 0.400 kW, supply temperature ts = 30 oC and the room temperature tr = 22 oC, the air flow rate can be calculated as:
qh = 0.4 (kW) / 1.2 (kg/m3) 1 (kJ/kg) (30 - 22)(oC)
= 0.042 m3/s = 150 m3/h
If it is necessary to humidify the indoor air, the amount of supply air needed may be calculated as:
qmh = Qh / ρ (x2 - x1) (3)
where
qm = volume of air for humidifying (m3/s)
Qh = moisture to be supplied (kg/s)
ρ = density of air (kg/m3)
x2 = humidity of room air (kg/kg)
x1 = humidity of supply air (kg/kg)
If it is necessary to dehumidify the indoor air, the amount of supply air needed may be calculated as:
qmd = Qd / ρ (x1 - x2) (4)
where
qmd = volume of air for dehumidifying (m3/s)
Qd = moisture to be dehumifyied (kg/s)
If added moisture Qh = 0.003 kg/s, room humidity x1 = 0.001 kg/kg and supply air humidity x2 = 0.008 kg/kg, the amount of air can expressed as:
qmh = 0.003 (kg/s) / 1.2 (kg/m3) (0.008 - 0.001) (kg/kg)
= 0.36 m3/s
Alternatively the air quantity is determined by the requirements of occupants or processes.
The heat loss from a duct can be expressed as:
H = A k ((t1 + t2) / 2 - tr)) (5)
where
H = heat loss (W)
A = area of duct walls(m2)
t1 = initial temperature in duct (oC
t2 = final temperature in duct(oC
k = heat loss coefficient of duct walls(kW/m2.K) (5.68 10-3 for sheet metal ducts, 2.3 10-3 for insulated ducts)
tr = surrounding room temperature(oC)
The heat loss in the air flow can be expressed as:
H = q cp (t1 - t2) (5b)
where
q = mass of air flowing (kg/s)
cp = specific heat capacity of air (kJ/kg.K)
(5) and (5b) can be combined to
H = A k ((t1 + t2) / 2 - tr)) = q cp (t1 - t2) (5c)
For large temperature drops should logarithmic mean temperatures be used.
Units as heaters, filters etc. must on basis of of air quantity and capacity be selected from manufactures catalogues.
The 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 manufacturer catalogues.
Air speed in a duct can be expressed as:
v = Q / A (7)
where
v = air velocity (m/s)
Q = air volume (m3/s)
A = cross section of duct (m2)
Overall pressure loss in ducts can be expressed as:
dpt = dpf + dps + dpc (8)
where
dpt = total pressure loss in system (Pa, N/m2)
dpf = major pressure loss in ducts due to friction (Pa, N/m2)
dps = minor pressure loss in fittings, bends etc. (Pa, N/m2)
dpc = minor pressure loss in components as filters, heaters etc. (Pa, N/m2)
Major pressure loss in ducts due to friction can be expressed as
dpf = R l (9)
where
R = duct friction resistance per unit length (Pa, N/m2 per m duct)
l = lengthof duct(m)
Duct friction resistance per unit length can be expressed as
R = λ / dh (ρ v2 / 2) (10)
where
R = pressure loss (Pa, N/m2)
λ = friction coefficient
dh = hydraulic diameter (m)
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