# Design of Ventilation Systems

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*may be suitable^{ o }C (100 - 120^{ o }F) - For cooling where the inlets are near occupied zones ,
*6 - 8*below room temperature may be suitable^{ o }C (10 - 15^{ o }F) - For cooling where high velocity diffusing jets are used,
*17*below room temperature may be suitable^{ o }C (30^{ o }F)

### 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) wheret_{ 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*and the room temperature

^{ o }C*t*

_{ r }*= 22*, the air flow rate can be calculated as:

^{ o }C

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*, the amount of air can expressed as:

_{ 2 }= 0.008 kg/kg

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.