CHAPTER 4 AIR FLOW THROUGH OPENINGS

Learning Objectives

This chapter introduces the equations that describe airflow through openings. After you havestudied this chapter and completed the personal feedback form you should be able to:

1. Identify the parameters that describe airflow;

2. Understand how airflow through openings can be approximated by a general mathematicalequation;

3. Understand what is meant by flow parameters and pressure drop.

The Calculation methods precented in Chapters 4, 5 and 6 are based on material published in the AIVC AirInfiltration Calculation Techniques Guide, the AIVC Guide to Ventilation and the CIBSE Guide A Handbook(Liddament 1986, 1996, CIBSE 2006).

Introduction

Natural ventilation is governed by the amount and distribution of openings in the envelope of the building, theinternal pattern of flow paths and the pressure generated by the natural driving forces. This Section is concernedwith outlining the flow characteristics of openings and explaining how the flow rate can be estimated from thepressure drop across each opening. Because the variability of climate is so high and the mechanics of flow isso complex, natural ventilation is often treated in very simplified or empirical terms. This includes treatingcomponents individually or making generalisations about flow patterns. Often such rule of thumb approachesare all that can be accomplished. At the other end of the spectrum increasing use is being made of computationalfluid dynamics to predict natural ventilation performance. While results can be obtained, there is muchuncertainty about the necessary approximations needed to reflect the driving parameters. This course follows atheoretical approach that can enable good calculations to be performed. It too, however is restricted by thefollowing:

· Wind Induced Flow through Openings: Classical natural ventilation theory is based the assumption thatwind induced pressure is developed from the transfer of kinetic energy of the wind when it strikes a solidbody. However, if an opening is ‘large,’ kinetic energy is not fully converted to induce a surface pressure.Instead the wind simply drives air under its own momentum through the opening. Some research suggeststhat this breakdown occurs immediately an opening is present but a porosity figure of up to 30% of thewall surface area has been suggested as satisfactory for classical theory to be acceptable.

· Discharge Coefficient: A “discharge coefficient” is frequently used to describe the flow characteristics ifan identifiable opening. This coefficient will depend on many factors including the geometry of theopening and the direction of attack of the wind. The value suggested in this the theoretical sections is aguideline figure within the classical range for an orifice opening in which wind is normal to the opening.The literature reveals an increasing amount of discharge coefficients for a wider range of conditions.

As yet no further practical guidance on these issues can be given other than the need to carry out specific testson components, including wind tunnel analysis, for critical design.

Further guidance on ventilation calculations, based on a component basis, is given in CEN Standard /TC 156Ventilation for buildings — Calculation methods for the determination of air flow rates in buildings includinginfiltration.

4.1 FLOW THROUGH PURPOSE DESIGNED OPENINGS

For a given applied pressure, the nature of airflow is dependent on the dimensions and geometry of the openingitself. For well-defined, purpose provided openings such as vents, airflow is usually assumed to be turbulentand is often approximated by the orifice equation given by:

122

dQ C A pr

é ù= Dê ú

ë û (m3/s) (4.1)

Where:Q = air flow rate (m3/s)Cd = discharge coefficient,ρ = air density (kg/m3);Δp = pressure difference across opening (Pa);A = area of opening (m2).

In this instance the area, A, is the net physical openable area. This is usually much less than the dimension ofthe vent itself because a significant part of the vent area is taken up with insect screen or loose filling.

The Discharge CoefficientThe discharge coefficient is dependent on the opening geometry and also on the direction of approaching flow(e.g. wind direction). For a flat plate orifice in which the air stream is directed at right angles to the opening, ittypically has a value of approximately 0.61 – 0.65. Such a range is widely used in preliminary designcalculations. However for practical components the actual value will be dependent on the component itself andon wind direction. For more detailed analysis, therefore, the compentent should be tested to obtain its actualflow characteristics.

4.2 FLOW THROUGH GAPS AND CRACKS – AIR INFILTRATION

A particular type of opening is the adventitious construction crack that appears at each joint in the buildingenvelope or at each service penetration. Formerly such openings represented a significant component of theoverall porosity of the building which resulted in high and uncontrollable air change rates. In modern buildingsmuch effort goes into minimising such openings and airtightness requirements are included in the buildingregulations of a number of countries. Despite increasing levels of airtightness it is still important to incorporatethe impact of air infiltration on natural ventilation design.

For very narrow cracks with relatively long flow paths, such as may be found in mortar and between tight fittingcomponents, the nature of flow at ambient flow is altogether very different and is dominated by the effects ofviscosity. Airflow through such openings is closer to laminar in which the flow rate and becomes proportionalto the applied pressure difference.

Often the flow rate through less well defined openings such as infiltration openings is represented by the PowerLaw Equation:

( )nQ C p= D m3/s (4.2)

Where:C = flow coefficient;n = flow exponent;Δp = pressure difference across the opening.

The Coefficient, C, is related to the size of the opening (i.e. it increases with opening size) while the exponent,n, characterises the flow regime and varies in value between 0.5 (fully turbulent flow) to 1.0 (fully laminarflow).

This power law is an approximation but is an approach that is most widely used in building airflow analysis.This is because laminar, turbulent and intermediate flow characteristics can all be represented by the sameequation thus simplifying calculation techniques.

Sometimes a quadratic equation is recommended in which the laminar and turbulent terms are separated. Thisform of the equation is given by:

2p Q Qa bD = + (4.3)

Where α and β are constants

Unlike the power law, this equation is dimensionally correct and is often thought to represent flow at low drivingpressures much more accurately. However despite these benefits, use of the quadratic equation has not yet beenpopular.

4.3 A GENERAL FLOW EQUATION FOR OPENINGS

In undertaking design calculations it is important to be able to express the flow through openings, regardless oftype, by a general equation. In this case the power law equation (Equation 4.2) represents the most general caseand can be applied to the various types of opening as follows:

Purpose Provided Ventilation Openings:0.5

2 ; 0.5dC C A nr

é ù= =ê ú

ë û(4.4)

Infiltration Openings

; 0.6C C n= » (4.5)

4.4 PARAMETERS NEEDED TO DETERMINE AIRFLOW RATE

Thus to determine the flow rate through an opening it is necessary to know:

· The flow characteristics of the opening (either in relation to its area for well defined orifices or bymeasuring its flow characteristics through the process of laboratory or site testing to obtain its C and nvalue.

· The pressure drop, Δp, across the opening.

Flow Characteristics: In early design sizing calculations, based on defined openings, C and n can be based onthe orifice equation as outlined in Equation 4.4 above While an allowance for infiltration openings can be madeusing Equation 4.2 directly. It is imperative that every opening is accounted for. More detailed analysis shouldthen be based on the measured flow characteristics of the intended devices. These characteristics should berepresented in terms of the C and n values (or of the quadratic exponents if the quadratic equation (much rarer)is used).

Pressure Drop: The pressure drop across openings is developed by the driving forces of wind and temperatureas well as that induced by mechanical fans. The calculation of pressure drop forms the basis of naturalventilation design since this defines the necessary openable area needed to achieve the required airflow rate.Guidance in calculating the pressure distribution is covered in the following sections.

4.5 QUESTIONS

Answers:

1. An air vent has a dimension of 30 cm x 3 cm and is fitted with a gauze insect screen that reducesthe effective area of the opening by 40%. Based on the orifice flow equation, calculate the airflowrate through the opening at a pressure drop of 5 Pa.

Gross area of vent = 0.30 * 0.03 m2 = 0.009 m2

The area is reduced by 40% therefore the net area of opening is 0.0054 m2

Assume an air density of 1.21 kg/m3 (20°C)

Flow Equation:

122

dQ C A pr

é ù= Dê ú

ë û

Thus Q = 0.61* 0.0054 * (2*5/1.21)1/2

= 0.0095 m3/s = 9.5 L/s

Note: answer will vary according to density chosen

2. How could the estimate of flow rate vs pressure drop be improved?

By measuring the actual flow performance of the vent rather than relying on orifice assumptions.

3. For the same pressure drop, what is likely to happen to the flow rate over an extended period ofinstallation (e.g. over several years)?

The flow performance will reduce over time because the insect screen will steadily trap material thusreducing the area of opening.

CHAPTER 5 WIND DRIVEN VENTILATION

Learning Objectives

This chapter outlines the principles of wind driven ventilation. After you have studied thisChapter and completed the personal feedback form you should be able to:

1. Understand how wind creates a driving pressure across a building opening;

2. Understand how the pressure is influenced by shelter;

3. Use equations and data to approximate the value of wind pressure acting at an opening onthe surface of a building.

Introduction

Wind provides one of the key driving forces for natural ventilation. It is harnessed in many ways varying frompurpose designed wind towers or “wind catchers” to reliance on ventilation grilles and openable windows.

To size a system correctly it is important to understand the wind pressure mechanism and to be able to estimatethe pressure induced at an opening by wind. This Section covers the development of wind pressure and outlinessimple calculation techniques that can be used to give an approximate estimate of wind induced pressure. Moredetailed calculation methods may require wind tunnel analysis, computational fluid dynamics and anunderstanding of how the size of the opening itself modifies the flow characteristics.

5.1 WIND INDUCED PRESSURE DISTRIBUTION

Wind within the lower regions of the earth's atmosphere is characterised by random fluctuations in velocitywhich, when averaged over a fixed period of time, possess a mean value of speed and direction. The strengthof the wind is also a function of height above ground, this function being dependent on surface or terrainroughness, and on the thermal nature of the atmosphere (thermal stability).

On impinging the surface of an exposed building, wind induces a pressure distribution on the building as shownin the examples illustrated in Figure 5.1

Wind deflection occurs which induces a positive pressure on the upwind face. The flow separates at the sharpedges of the building, giving rise to negative pressures along the sides. Negative pressures are also experiencedwithin the wake region on the leeward face.

The pressure distribution on the roof varies according to pitch, with negative pressures being experienced onboth faces for roof pitches of less than approximately 30 º. Above this angle, positive pressures are experiencedon the leading face

Wind incident to the corner of the building will induce a net positive pressure on both upwind faces but, as flowveers towards a particular side, the remaining face will undergo a transition from positive to negative pressure.The angle at which this transition occurs is dependent on the side ratio of the building. For buildings withventilated roof spaces, wind will directly affect the ceiling pressure distribution.

Figure 5.1. Wind pressure distribution for various building shapes and orientations.

Other considerations include protrusions from roofs such as chimneys and flues. The pressure due to wind actingon the mouths of such components is a function of position, roof pitch and height of protrusion.

5.2 ESTIMATING WIND INDUCED PRESSURE

In general it is observed that relative to the static pressure of the free wind, the time averaged pressure acting atany point on the surface of a building may be represented by the equation:

2

2w pp C vr= (5.1)

Where:pw = wind induced pressure;ρ = air density;Cp = wind pressure coefficient;v = wind velocity at a datum level (usually building height).

Terrain and ShieldingSince the strength of the wind close to the earth's surface is influenced by the roughness of the underlying terrainand the height above ground, a reference level for wind velocity must be specified for use in the wind pressurecalculation. In calculating the wind impact on ventilation the wind velocity is commonly expressed as themeasured speed at building height. Unfortunately, as a general rule, "on-site" wind data are rarely availableand therefore data taken from the nearest climatalogical station must usually be applied. Before such data canbe used, however, it is essential that such measurements are corrected to account for any difference betweenmeasurement height and building height and to account for intervening terrain roughness. By nature of thesquare term in Equation 5.1, wind pressure is very sensitive to the wind velocity and, as a consequence, thearbitrary use of raw wind data will invariably give rise to misleading results. This is, perhaps, one of the mostcommon causes of error in the calculation of air infiltration rates.

Suitable correction for the effects of these parameters may be achieved by using a power law wind profileequation of the form:

m

V zV

ga= (5.2)

Where:z = datum height (m);v = Mean wind speed at datum height (i.e. height of building) (m/s);vm = mean wind speed at weather station (m/s);α and γ are coefficients according to terrain roughness (see table **).

This equation and the associated coefficients are taken from BS 5925:1990. Example coefficients are given inTable 5.1.

Table 5.1 Terrain coefficients for use with Equation 5.2

Terrain Coefficient a g

Open, flat 0.68 0.17Country with scattered wind breaks 0.52 0.20urban 0.35 0.25city 0.21 0.33

Such an approach is generally acceptable for winds measured between roof height and a recording height of10m. It is inappropriate for the reduction of wind speeds measured in the upper atmosphere. Alternative methodsof wind correction are also used based on a log law profile.

Pressure CoefficientThe pressure coefficient, Cp, is an empirically derived parameter which is a function of the pattern of flowaround the building. It is normally assumed to be independent of wind speed but varies according to wind

direction and position on the building surface. It is also significantly affected by neighbouring obstructions withthe result that similar buildings subjected to different surroundings may be expected to exhibit similar markedlydifferent pressure coefficient patterns.

Accurate evaluation of this parameter is one of the most difficult aspects of natural ventilation and air infiltrationmodelling and, as yet, is not possible by theoretical means alone. Although pressure coefficients can bedetermined by direct measurements of buildings, most information comes from the results of wind loading testsmade on scale models of isolated buildings in wind tunnels.

Purpose designed tests for specific buildings and shielding conditions may be performed, but this is an expensiveexercise and is therefore rarely possible.

For low buildings of up to typically 3 storeys, pressure coefficients may be expressed as an average value foreach face of the building and for each 45 º sector, or even 30º sector in wind direction.

For taller buildings, the spatial distribution of wind pressure takes on much greater significance, since thestrength of the wind can vary considerably over the height range. In these instances spatial dependent data areessential. Solutions include scale wind tunnel modelling and the use of cfd to predict the surrounding pressurefield.

Representative pressure coefficients for open and urban environments are presented in Table 5.2

Table 5.2 sample face averaged pressure coefficients for low rise buildings.(based on AIVC data guide Orme )

Turbulent FluctuationsTurbulent fluctuations can provide background air change even when there is no net driving force from themean wind speed. Various formulas exist but resultant air exchange is low and should not be incorporated intoa design methodology.

5.3 QUESTIONS

1. Consider the wind pressure distribution as illustrated in the Figure below.

(a) If face average wind pressure coefficients were applied to describe the wind pressure in acalculation model draw, with arrows, the flow direction for openings A and B.

(b) If the actual pressure distribution was applied, rather than face averaged values how would theestimate flow pattern change?

Which of the two answers is most likely to represent the true situation?

· Face average assumption: The pressure distribution is assumed to be uniform on each wall andtherefore the wind pressure at A and B are assumed to be identical. If A and B are the only openingsin the building then there will be no wind driven flow when using this assumption.

· Actual pressure distribution: the wind pressure is more negative at A than at B. Air will enter thespace via opening B and will exhaust via opening A.

· The actual pressure distribution answer will be the most accurate.

2. Given an open site wind speed of 10 m/s at a meteorological measurement height of 10 m aboveground calculate the wind speed at a building height of 8 m for:

(i) A building located on an exposed site

(ii) A building located at an urban site

3. Use the pressure coefficient tables to assess the average pressure coefficient values at openingsA1 and A2 indicated in the Figure below for:

(i) An open terrain with no surroundingbuildings

(ii) An urban environment withsurrounding buildings of equal height

A B

Answers:

1. Consider the wind pressure distribution as illustrated in the Figure below.

(a) If face average wind pressure coefficients were applied to describe the wind pressure in acalculation model draw, with arrows, the flow direction for openings A and B.

(b) If the actual pressure distribution was applied, rather than face averaged values how would theestimate flow pattern change?

Which of the two answers is most likely to represent the true situation?

· Face average assumption: The pressure distribution is assumed to be uniform on each wall andtherefore the wind pressure at A and B are assumed to be identical. If A and B are the only openingsin the building then there will be no wind driven flow when using this assumption.

· Actual pressure distribution: the wind pressure is more negative at A than at B. Air will enter thespace via opening B and will exhaust via opening A.

· The actual pressure distribution answer will be the most accurate.

4 mBuilding Height

= 4.5 m

A2

A11 m

Building Height Windspeed = 4 m/s

A B

2. Given an open site wind speed of 10 m/s at a meteorological measurement height of 10 m aboveground calculate the wind speed at a building height of 8 m for:

(i) A building located on an exposed site 9.68 m/s

(ii) A building located at an urban site 5.89 m/s

3. Use the pressure coefficient tables to assess the average pressure coefficient values at openingsA1 and A2 indicated in the Figure below for:

(i) An open terrain with no surroundingbuildings

A1 = .7 A2 = -.2

(ii) An urban environment withsurrounding buildings of equal height

A1 = .2 A2 = -.25

·

4 mBuilding Height

= 4.5 m

A2

A11 m

Building Height Windspeed = 4 m/s

CHAPTER 6 STACK DRIVEN VENTILATION

Learning Objectives

This chapter outlines the principles of stack driven ventilation. After you have studied thischapter and completed the personal feedback form you should be able to:

1. Understand how temperature difference creates a driving pressure across a buildingopening;

2. Understand how the pressure is influenced by the height of openings;

3. Use equations and data to approximate the value of stack pressure acting at an opening onthe surface of a building.

Introduction

The stack effect arises as a result of differences in temperature and hence air density between the interior andexterior of a building. This produces an imbalance in the pressure gradients of the internal and external airmasses, thus creating a vertical pressure difference (Figure 6.1). When the internal air temperature is higherthan that of the outside air mass, air enters through openings in the lower part of the building and escapesthrough openings at a higher level. This flow direction is reversed when the interna1 air temperature is lowerthan that of the air outside.

Figure 6.1. Principles of stack effect showing airflow derived from the imbalances in pressuredistribution caused by internal/external temperature difference.

The level at which the transition between inflow and outflow occurs is defined as the neutral pressure plane. Inpractice, the level of the neutral plane is rarely known although it can be predicted for straightforward leakagedistributions.

More generally, the stack pressure is expressed relative to the level of the lowest opening or to some otherconvenient datum (for example ground level).

For natural ventilation design, the consequences and significance of the stack effect must be considered for anumber of alternative configurations. These include:

· Single zone uniform internal temperature distribution;· Multi-zone buildings with impermeably separated vertical zones;· Multi-zone buildings in which interconnected vertically placed zones are at different temperatures;· Multi-zone buildings in which horizontally placed zones are at different temperatures,· Large single zone structures subjected to thermal stratification;· Ventilation stacks,

Each of these configurations may be readily analysed and combined by consideration of the vertical pressuregradient of the respective air masses.

6.1 STACK EFFECT: SINGLE ZONE - UNIFORM INTERNAL TEMPERATUREDISTRIBUTION

Assuming a uniform air temperature, the pressure of an air mass at any height z above a convenient datumlevel, zo, (for example ground or floor level) is given by:

0zp p gzr= - (Pa) (6.1)

Where:pz = pressure at datum level zo (Pa);g = acceleration due to gravity (m/s2);z = height above dataum (m).

The resultant pressure gradient is therefore:

dp gdz

r= - (6.2)

which becomes, by consideration of the ideal (Gas Laws):

273o

dp gdz

rq

= - (6.3)

Where:ρo = air density at 273K (kg/m3);θ = absolute temperature of the air mass (K).

Thus the pressure gradient is inversely proportional to the absolute temperature of the air mass.

Pressure gradients and the absolute pressure distribution for a building in which two openings, hl and h2, arevertically separated a distance h apart are illustrated in Figure 6.1 above. The level of alignment of the internaland external pressures (neutral pressure plane) is a function of the overall distribution and flow characteristicsof openings, and is fixed such that a mass flow balance is maintained. Knowledge of this level is not a pre-requisite of natural ventilation modelling. The stack induced pressure at h2, with respect to the pressure at hl,is represented in Figure 6.1 by the net horizontal displacement of the pressure curves at these locations (A + B)and is given by:

2 1i

1 1273( )s oe

p g h hrq q

é ù= - - -ê ú

ë û (Pa) (6.4)

Where:

θ e = Absolute temperature of the outdoor air (K);θ i = Absolute temperature of the indoor air (K)

Stack Pressures are often comparable with wind induced pressures and many ventilation designs concentrate oddeveloping the stack pressure to drive ventilation airflow.

6.2 STACK EFFECT ADVANCED

Multizone Buildings with Impermeably Separated Vertical Zones

Figure 6.2 Stack induced pressure distribution in a multi-storey building in which each floor is isolated.

Often it is necessary to consider the design for a multi-storey building, such as an office or an apartment block,in which each floor is essentially isolated (Figure 6.2). This isolation is usually necessary to ensure theseparation of individual units and to prevent fire spread. In this instance each floor displays its own neutralpressure plane and it is possible to calculate the stack pressure distribution without reference to adjacent zones.Thus the stack pressure equation 6.4 above can be applied to each floor to estimate flow rates stack flow ratesfor opening configurations.

Multi-zone buildings in which vertically interconnected zones are at different temperatures

Figure 6.3 Stack pressure distribution for vertically placed zones at different temperatures.

In practice, a uniform temperature distribution may not always be possible or even desired. By design orotherwise, it may be that the internal temperature of each zone will differ. In dwellings, for example, upstairsbedrooms are frequently maintained at a lower temperature to the living area, while a roof space might not beheated at all. The resultant stack pressure may be analysed in exactly the same way as the previous example,except that the pressure gradient of each cell varies according to the zonal air temperature (Figure 6.3). Thestack pressure at any level is again given by the displacement in pressure gradients.

For example, the stack pressure at level h2 with respect to level h1 (horizontal displacement ‘A’ in the diagram)is given by:

1 1 2 11 2

1 1 1 1273 ( ) ( )s oe e

p g L h h Lrq q q q

ì üé ù é ùï ï= - - - + - -í ýê ú ê úï ïë û ë ûî þ

(Pa) (6.5)

Where:L1 = First floor level (m);θ1 = Internal temperature of ground floor zone (K);θ2 = Internal temperature of first floor zone (K).

Multi-Zone Building in which Horizontally Placed Zones are at Different Temperatures

Figure 6.4. Stack pressure for horizontally spaced zones at different temperatures.

This situation may occur in an office environment or multi-storey building in which the stairwell or publicaccess areas are maintained at a lower temperature than the occupied parts of the building. Again, thetemperature gradients in each of the zones are analysed as previously described with the stack pressure beingexpressed relative to the lowest opening. The corresponding stack pressures at other heights are then determinedby the relative displacements (Figure 6.4 above).

Thus the stack pressure at height h5, between zone 1 and 4 (horizontal displacement ‘B’ in the Figure) is givenby:

5 11 2

1 1273( )sp g h hrq q

é ù= - - -ê ú

ë û (Pa) (6.6)

Where:θ1 = Temperature of each of the vertical zones;θ2 = Temperature of Zone 1.

6.3 QUESTIONS

1. Calculate the stack pressure difference between openings A1 and A2 the figure below:

2. For the temperatures given draw, with arrows, the direction of airflow at openings A1 and A2.

3. If the external temperature was increased from 10 ºC to 25 ºC, while the internal temperature iskept maintained at 20 ºC, how will the flow directions at A1 and A2 change.

Answers:

1. Calculate the stack pressure difference between openings A1 and A2 the figure below:

Stack Pressure between the two openings (3 m apart) = - 1.25 Pa

2. For the temperatures given draw, with arrows, the direction of airflow at openings A1 and A2.

Air enters through A1 into the building and leaves through A2

3. If the external temperature was increased from 10 ºC to 25 ºC, while the internal temperature iskept maintained at 20 ºC, how will the flow directions at A1 and A2 change.

They are reversed

4 mBuilding Height

= 4.5 m

θint =20°C

A2

A11 m

θint =10°C

4 mBuilding Height

= 4.5 m

θint =20°C

A2

A11 m

θint =10°C

CHAPTER 7 COMBINING WIND AND STACK DRIVEN VENTILATIONADDING HYBRID FANS

Learning Objectives

This chapter outlines the process of combining the impact of wind and stack driven pressures. Itthen describes how a hybrid fan can be introduced. After you have studied this chapter andcompleted the personal feedback form you should be able to:

1. Describe the total pressure acting on an opening;

2. Understand how a hybrid fan influences the pressure distribution.

Introduction

It is usually impossible to separate stack driving forces from wind driving forces. In other words both stack andwind forces are normally present at the same time. Therefore despite efforts to design a stack or wind drivenconcept allowance must be made for the impact of the alternative force.

Unfortunately it is not possible to calculate the air change due to wind and stack pressure separately then to addthe two results together. It is possible, however to add the contributions of pressure due to stack and wind ateach opening to obtain a pressure acting on each opening. The calculation of air change or flow rate througheach opening can then be calculated.

In the case of hybrid ventilation, fans are used to enhance the driving pressure. Again it is not possible to addthe flow rate developed by the fan to the flow rate calculated by wind and stack effect. Instead the change inpressure caused by the fan must be calculated. The means by which this may be achieved are described in thenext section

This Section describes the combining of pressures and the inclusion of hybrid fans.

7.1 COMBINING WIND WITH STACK PRESSURE

The total pressure, pti, acting at an opening, i, due to the combined impact of wind and stack effect, is given by:

i i it w sp p p= +

It is important to understand that summing the pressures due to stack and wind effect at each opening is not thesame as summing the flow rates determined by calculating the flow rates due to wind and stack pressureseparately. Summing the flow rates would lead to an erroneous result.

7.2 PRESSURE EFFECTS DUE TO HYBRID VENTILATION

Mechanical ventilation may be analysed in terms of the induced flow rate and pressure imbalance created acrossthe fan and associated ducting. The resultant pressure difference becomes another component to the totalpressure equation.

The additional pressure imbalance created by an extract only ventilation system is very straightforward tocalculate by direct application of the Power Law equation (Equation 4.2). In this case the flow coefficient C isgiven by the flow rate through the fan and the flow exponent, n, is set to zero i.e:

0( )nf fQ C p == D (7.1)

Where:Qf = Specified flow rate through the fan;Cf = Corresponding flow coefficient.

This forces a constant flow condition. A check must be performed to ensure that the fan can achieve the givenflow rate at the resultant induced pressure drop across the fan.

In practice the resultant induced pressure created by the fan (See Section 8).

7.3 QUESTIONS

1. For the Figure below calculate the combined stack and wind pressures at each openingassuming that the building is located in an urban environment and surrounded by buildings ofequal height.

2. Sketch the probable directions of airflow through openings A1 and A2. Explain your reasoning.

Building Height Windspeed = 4 m/s

4 mArea of opening, A1= 0.1 m²

Area of opening,A2 = 0.1 m²

Building Volume =250 m³

Building Height = 4.5 m

θint = 20°Cθext = 10°C

Exposure: Urban, surrounded by buildings of equal height.

A2

A11 m

Answers:

1. For the Figure below calculate the combined stack and wind pressures at each openingassuming that the building is located in an urban environment and surrounded by buildings ofequal height.

Wind Pressure:A1 = 2.02A2 = -2.52

Stack PressureA1 = 0.0 or - 0.417A2 = -1.25 or -1.667

Note: answer depends on the datum level (e.g. base level or level of first opening). However the pressuredifference will always be the same.

2. Sketch the probable directions of airflow through openings A1 and A2. Explain your reasoning.

· Flow through A1 into building;· Flow out of building through A2;· The wind pressure reinforces the stack pressure.

Building Height Windspeed = 4 m/s

4 mArea of opening, A1= 0.1 m²

Area of opening,A2 = 0.1 m²

Building Volume =250 m³

Building Height = 4.5 m

θint = 20°Cθext = 10°C

Exposure: Urban, surrounded by buildings of equal height.

A2

A11 m

CHAPTER 8CALCULATING NATURAL VENTILATION RATEUSING THE FLOW EQUATIONS, WIND PRESSUREAND STACK PRESSURE EQUATIONS

Learning Objectives

This chapter combines the flow Equations developed in Chapter 4 with the pressure equationsdeveloped in chapters 5 to 7 in order to estimate the natural and hybrid ventilation flow through abuilding. After you have studied this Chapter and completed the personal feedback form youshould be able to:

1. Understand the meaning of single and multizone models;

2. Undertake natural and hybrid ventilation flow calculations for a simple single zone buildingstructure;

3. Introducing hybrid fans;

4. Treatment of single sided openings.

Introduction

The calculation of ventilation rate involves:

· Identifying the ventilation openings;· Determining the pressures acting on each opening;· Applying the flow equations at each opening;· Obtaining a flow balance so that the air entering the building (and individual zones in a building) is

balanced by the outgoing air.

The first three steps are covered by the preceding Sections. This section is concerned with the flow balancecalculation. Appropriate algorithms are presented in the common resource module.

Such calculations are needed to:

· Size ventilation openings as part of a ventilation design;· Check on the adequacy of a ventilation design;· Determine the need for and develop a hybrid ventilation strategy;

Suitable calculation techniques include:

· Single zone network models;· Multi zone network models;· Computational fluid dynamics (CFD).

This course provides sufficient information to undertake single zone modelling. While basic descriptions arepresented on multi zone and cfd technique. More details are available in the resource module.

8.1 SINGLE ZONE NETWORK MODELS

This approach is used to calculate infiltration into a single enclosed space. Model parameters include:

· Flow path distribution:

· Flow path characteristics (C and n values);· Building height;· Internal/external temperature difference;· Local wind speed (or reduced from remote site);· Local shielding conditions;· Terrain roughness parameters;· Characteristics of hybrid ventilation.

Any number of flow paths, terminating within the internal zone, can be selected to represent leakage openingsin each face of the building (Inset Figure).

A naturally ventilated building consists of a series of openings through which air will pass. These openingsinclude:

· Purpose provided openings (e.g vents and windows as described in chapter 3 );· Adventitious infiltration openings.

In assessing the natural ventilation performance of a building, both types of openings must be included. Evenalthough buildings are less leaky than in the past, gaps and cracks can still adversely add to flow rates, especiallyin winter when driving forces are usually higher.

Single Zone Buildings: In some cases the building itself may be considered as a single open space (e.g. anopen plan office, warehouse or a house in which internal doors are predominantly left open.

Multi Zone Buildings: In other cases a building may be considered as a multizone building in which each roomis separated form its neighbours by partitions and walls.

8.2 CALCULATING THE NATURAL VENTILATION RATE

By referring to Figure 8.1 above, the calculation of naturally induced airflow through the building requires thefollowing steps:

· Apply the general flow equation (chapter 4) to each path;· Calculate wind pressure for each path;· Calculate stack pressure for each path;· Determine the total pressure (add wind +stack) for each path;· Determine an internal pressure for the space such that the total airflow into a space is balanced by the

total airflow out of the space.

This final step is the central component of ventilation calculations. Except for very simple networks, thecalculation is not direct and the process of ‘iteration’ is required. It is this element that makes ventilationcalculation approaches so difficult to follow. This section attempts to present a simple explanation of theprocess.

Taking all the flow paths, the conservation of mass requires a flow balance between the ingoing and outgoingairflow. This is expressed by:

10

j

i ii

Qr=

=å (kg/s) (8.1)

Where:ρi = Density of air flowing through flow path i (kg/m³);Qi = Volume airflow rate through flow path i (m³/s).

The air density, ρi, of incoming air is given by that of the value for the outside air while, for outflowing air, itis given by that of the internal air density. If the density differences between the internal and external air massesare negligible in comparison to the magnitude of the overall density of air then density term may be ignoredand, instead, calculation can be considered in terms of conservation of volume flow rather than mass flow.

Working in terms of volume flow rate rather than mass flow rate helps simplify iterative calculation and presentsresults in the familiar format of volume flow.

Thus Equation (8.1) above is simplified to:

10

j

ii

Q=

=å (8.2)

To determine the flow rate through each flow path, it is necessary to express Qi by the chosen flow equation (inthis module the popular power law equation is used although alternative equations such as the quadraticapproach are equally applicable). The power law equation gives:

( )1

0ij

ni i

iC p

=

D =å (8.3)

Where:

Ci and ni = the flow coefficient and flow exponent of the i’th flow path respectively(Δpi) = the pressure difference acting across the path.

The pressure difference (Δpi) is made up of the wind and stack pressure acting on the outside of the openingand the internal pressure of the space, pint. This internal pressure adjusts itself to preserve flow balance and isthe unkown that must be determined in order calculate the flow through each opening.

For the airflows to add up to zero therefore some flows must be positive (air coming into the building) andothers must be negative (air leaving the building). This means, also, that some Δp’s will be positive and someΔp’s will be negative. Unfortunately it is not possible to raise a negative Δp to the power n and therefore somerearrangement of the Equation 8.3 above is needed. Incorporating this rearrangement and replacing Δpi by pi -pint gives:

intint

1 int

0ij

n ii i

i i

p pC p pp p=

æ ö-- =ç ÷ç ÷-è ø

å (8.4)

(Term 1) (Term 2) (Term 3)

In this rearrangement, Term 2 is always positive (and hence can be raised to the power n) while term threebecomes either +1 or -1 and hence defines the flow direction.

AdvantagesThe single cell network approach offers many advantages. These include:

· Comparative ease of calculation;· The incorporation of any number of flow paths;· The inclusion of any combination of wind, stack and mechanically induced pressures;· The ability to assess the effect of flow path distribution on air change rates);· The ability to identify the flow direction and the magnitude of the flow rate through each of the defined

openings;· The calculation of internal pressure;· The ability to determine the neutral pressure plane using the external and internal pressure data.

DisadvantagesThe principle disadvantage is:

· Only applicable to buildings that can be approximated by a zone of single uniform pressure;

8.3 MULTI-ZONE NETWORK MODELS

A Multizone network model is an approach in which individual rooms or zones within a space are separatelydivided. This approach is needed when internal partitioning presents an impedance to the movement of air.Possible exceptions to using a multizone approach may be permissible when:

· Only the absolute maximum air change rate for the building is required;· When internal doors are very leaky or are generally left open.

There are therefore many instances in which single zone approaches will be of little value and, as a result,consideration must be given to an internal flow structure.

Similar model parameters to the previous network approach apply, including:

· Flow path distribution (internal and external);· Flow path characteristics;· Building height;· Internal/external temperature differences and the temperature of each zone;· Wind speed;· Local shielding conditions;· Terrain roughness parameters;· Details of mechanical systems.

Again, any number of flow paths, terminating within each internal zone, can be selected to represent leakageopenings in the building envelope. Additionally paths are selected to represent leakage openings across internalzones. For the m'th such zone with a total of j. flow paths, the mass flow balance is given by:

1

0m

im

m

jn im m

im im im mi im m

p pk p pp p

r=

æ ö-- =ç ÷ç ÷-è ø

å (8.5)

For q zones the mass balance is applied to each zone to give:

1 1

( ) 0m

im

m

jqn im m

im im im mm i im m

p pk p pp p

r= =

æ ö-- =ç ÷ç ÷-è ø

å å (8.6)

Unlike the 'single zone' approach, where there was only one internal pressure to determine, there are now manyvalues. This adds considerably to the complexity of the numerical solution method. Suitable algorithms areoutlined in the resource module.

8.4 INTRODUCING HYBRID VENTILATION

Hybrid fans are introduced by including in the network a fan equation of the type described in Chapter 7Equation 7.1.

8.4.1 Treatment of Single Sided Openings

There are circumstance where other mechanisms may be relevant, in particular where a space within a buildinghas openings on one wall only and has either no flow paths to other zones or where such flow paths areconsiderably smaller than the area of opening on the single outside wall. This could occur for instance whereoffices or classrooms have one external wall and limited internal flow paths to ensure privacy or reducedisturbance from other activities within the building. Equations have been derived which allow the ventilationrate of such spaces to be determined, based upon adapting the wind and stack equations as outlined below.

Wind effect: Based upon full-scale measurements and wind-tunnel studies, a simple empirical equation hasbeen derived to predict wind driven ventilation:

refQ k.A.U= (8.7)

Where, A is the area of opening and Uref is the speed of the undisturbed wind at the height of the buildingcontaining the space under consideration. In practice the value of k will depend upon a number of factors,including wind direction relative to the façade containing the opening, surrounding structures and any localobstruction at the surface of the building in proximity of the opening. However, a conservative value for designpurposes may be taken as 0.025.

Stack effect: The basic equations for stack effect set out in Section 6.1 can be applied to flow through a singleopening. The neutral plane (see Figure 6.1) occurs within the opening and, if internal air temperature is higherthan outside air, air flow out above the opening and in at the lower part of the opening. The position of theneutral plane is determined by the equality of the mass flow in and mass flow out. For a plane rectangularopening of height, h, and area A, the ventilation rate is given by the following equation;

0.5

dA θ.h.gQ C . .3 θ 273

Dæ ö æ ö= ç ÷ ç ÷+è ø è ø(8.8)

where,Δθ is the difference between internal and external air temperatureθ is the mean of internal and external air temperaturesg is the gravitational constantCd is the discharge coefficient of the opening (see Chapter 4)

Wind and Stack Effect: In the case of both wind and temperature difference being present, the contribution dueto wind and stack effect should be determined separately and the larger of the two values applied.

8.5 QUESTIONS

1. For the figure and conditions illustrated calculate the flow airflow rate through openings A1and A2 assuming wind driven flow only.

2. For the same configuration calculate stack driven flow only.

3. Calculate the airflow rates due to the combined influence of wind and stack pressure.

4. For the combined case what is: (a) The airflow direction through openings A1 and A2?(b) The air change rate in air changes per hour (ac/h)?

Answers:

1. For the figure and conditions illustrated calculate the flow airflow rate through openings A1and A2 assuming wind driven flow only.

60.83 L/s

2. For the same configuration calculate stack driven flow only.

144.5 L/s

3. Calculate the airflow rates due to the combined influence of wind and stack pressure.

156.72 L/s

4. For the combined case what is: (a) The airflow direction through openings A1 and A2?(b) The air change rate in air changes per hour (ac/h)?

(a) Into A1 out of A2, (b) 2.26 ac/h

Building Height Windspeed = 4 m/s

4 mArea of opening, A1= 0.1 m²

Area of opening,A2 = 0.1 m²

Building Volume =250 m³

Building Height = 4.5 m

θin

t

= 20°Cθex

t

= 10°C

Exposure: Urban, surrounded by buildings of equal height.

A2

A1 1 m