# 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*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}/ (ρc_{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)