ventilation system

A ventilation system may be designed more or less according the following procedure:

?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 - 50o C (100 - 120o F) may be suitable

?For cooling where the inlets are near occupied zones , 6 - 8o C (10 - 15o F) below room temperature may be suitable

?For cooling where high velocity diffusing jets are used, 17o C (30o 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 (m3/s)

H h =heat load (W)

c p = specific heat capacity of air (J/kg K)

t s = supply temperature (o C)

t r = room temperature (o C)

ρ = density of air (kg/m3)

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 airfor cooling (m3/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/m3) 1005 (J/kg K) (30 (o C) - 22 (o C))

= 0.041 m3/s

= 149 m3/h

Moisture

If it is necessary to humidify the indoor air, the amount of supply air needed may be calculated as:

q mh = Q h/ ρ (x2 - x1) (3)

where

q m = volume of air for humidifying (m3/s)

Q h = 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)

Dehumidifying

If it is necessary to dehumidify the indoor air, the amount of supply air needed may be calculated as:

q md = Q d/ ρ (x1 - x2) (4)

where

q md = volume of air for dehumidifying (m3/s)

Q d = moisture to be dehumified (kg/s)

Example - humidifying

If added moisture Q h = 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:

q mh = 0.003 (kg/s) / 1.2 (kg/m3) (0.008 (kg/kg)- 0.001 (kg/kg))

= 0.36 m3/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 expressed as:

H = A k ((t1 + t2) / 2 - t r ) (5)

where

H = heat loss (W)

A = area of duct walls(m2)

t1 = initial temperature in duct (o C)

t2 = final temperature in duct (o C)

k = heat loss coefficient of duct walls (W/m2 K) (5.68 W/m2 K for sheet metal ducts, 2.3 W/m2 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 (t1 - t2) (5b)

where

q = mass of air flowing (kg/s)

c p = specific heat capacity of air (kJ/kg K)

(5) and (5b) can be combined to

H = A k ((t1 + t2) / 2 - t r)) = 1000 q c p (t1 - t2) (5c)

For large temperature drops should logarithmic mean temperatures 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 manufactures catalogues.

7. Boiler

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.

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 (m3/s)

A = cross section of duct (m2)

Overall pressure loss in ducts can be expressed as:

dp t = dp f + dp s + dp c(8)

where

dp t = total pressure loss in system (Pa, N/m2)

dp f = major pressure loss in ducts due to friction (Pa, N/m2)

dp s = minor pressure loss in fittings, bends etc. (Pa, N/m2)

dp c = minor pressure loss in components as filters, heaters etc. (Pa, N/m2) Major pressure loss in ducts due to friction can be expressed as

dp f = R l (9)

where

R = duct friction resistance per unit length (Pa, N/m2 per m duct)

l = length of duct (m)

Duct friction resistance per unit length can be expressed as

R = λ / d h(ρ v2 / 2) (10)

where

R = pressure loss (Pa, N/m2)

λ = friction coefficient

d h = hydraulic diameter (m)