Ventilation - Systems - Page 1 2 3


Stack Driven Ventilation Calculations

Stack ventilation calculations in the simplest form ignore wind effects, although these can be allowed for in a more complex analysis.

The pressures developed in stack systems can be determined from the following formula.

dPs = - rins . g . T ins (h2 h1) (1/Tout - 1/Tins)

where;

dPs = Stack Effect Pressure (Pa)

rins = Air density inside stack (kg/m3)

g = Acceleration due to gravity (9.81 m/s2)

h1 = Height of inlet of stack above datum (m)

h2 = Height of outlet of stack above datum (m)

Tout = Temperature of air outside stack (oK)

Tins = Temperature of air inside stack (oK)

 

The equation below can be used to determine air flow rates in stack driven ventilation or the opening areas required.

Q = Cd . A [ ( 2 / rins) rins . g . (hnpl h ) (Tins - Tout / Tins ) ]

where;

Q = Air flow rate through a large opening (m3/s)

Cd = Discharge coefficient (0.61 for large openings)

A = Opening area (m2)

rins = Air density inside stack (kg/m3)

g = Acceleration due to gravity (9.81 m/s2)

hnpl = Height of neutral pressure level above datum (m)

h = Height of opening above datum (m)

Tout = Temperature of air outside stack (oK)

Tins = Temperature of air inside stack (oK)

 

Neutral Pressure Level

 

This is where the outside pressure equals the internal pressure.

At this level there would be no flow of air in or out of the building.

This is usually high up in a building otherwise the stack effect would not work.

The neutral pressure level for most buildings is about 0.25 metres above the level of the top floor ceiling.

Temperatures

 

The internal room temperatures need to be calculated since in summer heat gains elevate the room temperature.

This can be done using software where summertime temperature can be predicted along with required air flow rates to keep room temperatures to acceptable levels.

The HEVACOMP software package and other programmes may be used.

 

Outside summer temperatures may be obtained from the CIBSE guide A section 2.

It would seem that the outdoor temperature in summer rarely exceeds 27oC, and if the temperature does rise above 27oC it is only for a maximum of 4 days in the south of England and less than one day in the north of the U.K.

 

If a solar chimney is used to assist stack suction pressure then the temperature inside the stack would have to be altered.

 

It is important to obtain accurate inside and outside temperatures since this difference creates the driving force inside the stack or the pressure difference to move air up the stack to outside.

 

 

 

Example 1

 

Calculate the ventilation opening area required in a Stack ventilation system for the building shown below.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


DATA:

The flow rate required each room is 4 air changes per hour.

Each lecture room measures internally 24 m x 10 m x 4m high.

ANSWER:

Air flow rate for each room Q = Room volume x Air change rate / 3600

Q = 24 x 10 x 4 x 4 / 3600

Q = 960 x 4 / 3600 = 1.07 m3/s

 

 

Rearranging above formula for Area (A) gives;

A = Q / Cd [ ( 2 / rins) rins . g . (h npl h) (Tins - Tout / Tins ) ]

For Ground floor room;

 

A = 1.07 / 0.61 [ ( 2 / 1.1656 ) 1.1656 x 9.81 ( 9 1) ( 301 298 / 301 )

A = 1.07 / 0.61 [ 1.716 x 1.1656 x 9.81 x 8 x 0.00997 ]

A = 1.07 / 0.61 [ 0.19557 x 8 ]

A = 1.07 / 0.61 x 1.565

A = 1.121 m2.

 

For First floor room;

 

A = 1.07 / 0.61 [ ( 2 / 1.1656 ) 1.1656 x 9.81 ( 9 5) ( 301 298 / 301 )

A = 1.07 / 0.61 [ 0.19557 x 4 ]

A = 1.07 / 0.61 x 0.782

A = 2.242 m2.

 

Note: The upper floor has less stack suction pressure so openings are larger.

 

Example 2

 

Calculate the ventilation opening area required and the size of fresh air louvre required in a Stack ventilation system for the building shown below.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


DATA:

The flow rate required for the Class room is 10 air changes per hour.

The Class room measures internally 18 m x 10 m x 4m high.

The Fresh air louvre has a 50% free area.

ANSWER:

Air flow rate for each room Q = Room volume x Air change rate / 3600

 

Q = 18 x 10 x 4 x 10 / 3600

Q = 720 x 10 / 3600 = 2.0 m3/s

 

 

Rearranging above formula for Area (A) gives;

A = Q / Cd [ ( 2 / rins) rins . g . (h npl h) (Tins - Tout / Tins ) ]

 

A = 2.0 / 0.61 [ ( 2 / 1.1605 ) 1.1605 x 9.81 ( 8 1) ( 299 296 / 299 )

A = 2.0 / 0.61 [ 1.723 x 1.1605 x 9.81 x 7 x 0.01003 ]

A = 2.0 / 0.61 [ 1.377 ]

A = 2.0 / 0.84

A = 2.38 m2 fresh air area required

 

The fresh air louvre has a 50% free area so the size of louvre is;

 

Louvre area = fresh air area / ( percent free area / 100 )

Louvre area = 2.38 / ( 50 / 100 )

Louvre area = 4.76 m2.

 

 

 

Stack Outlet

 

The opening at the top of a stack can be sized in a similar manner to the fresh air inlets.

The height difference in the formula is between the NPL and the stack outlet.

The flow through the stack outlet is the sum of all the flows through the rooms in a building feeding the stack.

 

A outlet = Q total / Cd [ ( 2 / rins) rins . g . (h h npl) (Tins - Tout / Tins ) ]

where;

Q total = Total air flow rate through stack (m3/s)

Cd = Discharge coefficient (0.61 for large openings)

A outlet = Stack outlet area (m2)

rins = Air density inside stack (kg/m3)

g = Acceleration due to gravity (9.81 m/s2)

hnpl = Height of neutral pressure level above datum (m)

h = Height of stack outlet above datum (m)

Tout = Temperature of air outside stack (oK)

Tins = Temperature of air inside stack (oK)

 

 

Example 3

 

Calculate the stack outlet opening area required in the system given in Example 1.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


DATA:

The flow rate required each room is 4 air changes per hour.

Each lecture room measures internally 24 m x 10 m x 4m high.

 

ANSWER:

Air flow rate for each room Q = 1.07 m3/s (already calculated in Ex.1)

Total air flow rate Q total = 1.07 x 2 = 2.14 m3/s

 

A outlet = Q total / Cd [ ( 2 / rins) rins . g . (h h npl) (Tins - Tout / Tins ) ]

 

A = 2.14 / 0.61 [ ( 2 / 1.1656 ) 1.1656 x 9.81 ( 11 9) ( 301 298 / 301 )

A = 2.14 / 0.61 [ 1.716 x 1.1656 x 9.81 x 2 x 0.00997 ]

A = 2.14 / 0.61 [ 0.19557 x 2 ]

A = 2.14 / 0.61 x 0.391

A = 8.97 m2.

 

 

Fittings Pressure Drop

 

In a rigorous analysis of a stack ventilation system the pressure drop from fittings such as intake and exhaust louvres should not exceed the driving pressure from the stack and the stack pressure drop.

The driving pressure (dPs) can be found from the formula at the beginning of this section and the pressure drop from the stack and fittings can be determined by the normal method for ductwork fittings.

 

Using Curves for Q/A

 

The following curves may be used to calculate air flow rates if the areas of openings are known.

 

 


Ventilation - Systems - Page 1 2 3