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# 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

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

#### 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)

## Related Topics

• ### Ventilation

Systems for ventilation and air handling - air change rates, ducts and pressure drops, charts and diagrams and more.

## Related Documents

• ### Air Intakes and Outlets

Ventilation systems - air intakes and outlets - rules of thumbs.
• ### Classification of Ventilation Systems

Ventilation systems can be classified by functions, distribution strategies or by ventilation principles.
• ### Duct Sizing - the Equal Friction Method

The equal friction method for sizing air ducts is easy and straightforward to use.
• ### Ducts Sizing - the Velocity Reduction Method

The velocity reduction method can be used when sizing air ducts.
• ### Heaters and Coolers in Ventilation Systems

Basic equations for heat transfer - selecting criteria for heaters and coolers in ventilations systems.
• ### Pollution Concentration in Rooms

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.
• ### Sizing Circular Ducts

A rough guide to maximum air volume capacity of circular ducts in comfort, industrial and high speed ventilation systems.
• ### Ventilation Ducts - Minor Loss Resistance

Minor pressure or head loss in ventilation ducts vs. air velocity - minor loss coefficient diagram.

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## Citation

• The Engineering ToolBox (2003). Design of Ventilation Systems. [online] Available at: https://www.engineeringtoolbox.com/design-ventilation-systems-d_121.html [Accessed Day Month Year].

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12.8.9