Air Distribution

1.3.1.19 Introduction

The aim of this section is to provide a basic introduction to the methods by which air may enter a space and be distributed and to consider the govern­ing equation for the determination of the air quantity and temperature. The governing equation relating to airflow patterns in a space is not covered in this section, as it is discussed elsewhere in the guides.

Ventilation

Ventilation is required in buildings for many different reasons. In this sec­tion, the emphasis is on industrial environment; however, the general method of approach is common to all systems for the following reasons:

• To provide adequate oxygen to support life

• To remove all odors by dilution

• To reduce the bacteria count by providing fresh air to the space

• To reduce any toxic gases, vapors, and dusts

• To remove explosive gases and dusts

• To ensure that adequate combustion air is provided to any combustion process

• To lower the moisture content in the air, thus reducing the risk of condensation and mold growth

• To add to or remove heat from the space

All the above must be carried out in the most efficient manner.

The air that enters a space has to be distributed in a manner suitable for the particular application. Regardless of the method selected, the air enter­ing the space must achieve one or more of the above requirements in the fol­lowing ways:

• Efficient distribution with no stagnation of the air

• Air velocities that will not cause thermal discomfort or disturb papers or manufactured goods such as light powders and fibers

• Low noise transmission level

• Low owning and operating costs

• Capable of maintaining the internal design conditions, regardless of the external environmental conditions

The above factors can be grouped under the following headings:

• Health ventilation

• Safety ventilation

• Comfort ventilation

• Structural heating or cooling

The distribution and the associated extract of air from a space may be

• From high level (the roof or ceiling)

• From the walls, at any level

• From the floor

• From benches

• Natural ventilation

• Mechanical extract-induced inlet

• Mechanical input-forced extract

• Mechanical inlet-mechanical extract

The following sections consider each of these methods.

1.3.1.20 Ventilation Methods

Natural Ventilation

This is the usual method of ventilation in domestic dwellings and many small office buildings and workshops. New standards, however, require build­ings to have set ventilation rates, which require mechanical ventilation sys­tems. However, as covered later, use is made of natural ventilation to control the air-change rate, regardless of the external conditions. This approach is not practical for industrial applications.

The term natural ventilation relates to the airflow in a building that is caused by three natural factors:

1. Temperature differences (thermal density) between the inside and the outside of the building

2. Wind forces around the building

3. A combination of (1) and (2)

Natural ventilation has the advantage that no power supply is required; hence there are no fans and maintenance costs. It has the disadvantage that during the summer months, the temperature difference between the inside and out­side is small, and with low wind forces, poor ventilation will result when it is most required. During the winter months, however, the temperature difference between indoors and outdoors is high, with corresponding strong wind forces, resulting in high ventilation and heat-loss rates.

It is not possible to design a system according to a given natural ventilation rate, as this will vary with weather conditions. The cold outdoor air has a greater density for a given volume than the warmer indoor air; it enters through low-level openings, displacing the warmer indoor air and creating air changes. The displaced warmer air escapes through high-level openings (see Fig. 9.20). Ventilation caused by this method is commonly called the stack effect.

The greater the distance between the low-level inlets and the high-level outlets, the greater the resulting air change rate will be. The resulting airflow patterns in this arrangement will not ensure satisfactory air distribution in many industrial environments.

The science of building aerodynamics considers the influence of wind forces over buildings and the associated mechanics of fluids; these are com­plex in nature and are not considered here. It is sufficient to briefly consider Fig. 9.21, which shows how wind passing over a building produces a positive pressure on one side and a negative pressure on the other side. It is this pres­sure difference that produces airflow through openings. The combined wind and stack effects vary with the seasons.

Figure 9.21 considers the air movement in elevation; the aerodynamic flow in plan is shown in Fig. 9.22. It can be appreciated that adjacent

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Air Distribution

Mechanical Input-Forced Extract

If heating is provided, this is known as a plenum system. This is a ducted system, which may provide air to a space in one or more of the fol­lowing conditions:

• Untreated (no filter)

• Tempered by means of a heater battery (heated to near room conditions)

• Warm, at a temperature high enough to take care of the fabric and ventilation losses from the building

With the air being forced into the space, the pressure is positive; hence, all leakages are outward.

More sophisticated applications of a pressurized system combined with a suitable extractor are found in hospital operating theaters. The positive pres­sure produced by sterile air entering the room ensures that all leakages are outward. This outward leakage ensures that contaminated air at a lower pres­sure in the surrounding rooms will not enter the space.

Figure 9.24 shows a typical plenum system of mechanical input and forced extract. The outlets in Fig. 9.24 may be subdivided into two categories:

1. Diffused air distribution

2. Tangential air distribution

In the case of diffused air, the associated characteristics are these:

• The jet has a high degree of entrainment.

• There are no circular air patterns in the occupied zone.

• Air movement is ideal in the working zone, both for thermal comfort and pollution control.

• Temperature gradients (stratification) are at a minimum.

Air Distribution

The above are achieved by the use of

• Linear air outlets with highly inductive jets discharging directly downward or alternate jets discharging in opposite directions

• Twist outlets with adjustable twist vanes

• Specially designed air outlets

• Whirl outlets, either radial or linear, that have geometrically designed aperture chambers

The tangential method of air distribution has the following features:

• The room airflow patterns are circular.

• Stagnant pockets form between the circular room air patterns.

• The jets come into contact with surrounding surfaces.

In the application of this method of distribution, full use is made of the Coanda effect (wall jet), in which the jet adheres to a surface due to the fact that no entrainment can occur from a solid surface.

The outlets may be square, rectangular, or circular without any air twist being produced.

All outlets must be positioned at such a height that they do not create drafts in occupied zones.

Mechanical Input-Mechanical Extract

In this method of ventilation, powered extract and input systems are provided. The ratio of the supply to extract will control the pressure produced within the space.

If the input fans are handling the same amount of air as the extract fans, a neutral condition will result, with the disadvantages of the natural system.

Figure 9.25 shows a typical layout of this type of system. From the forego­ing statements, the advantages of such a system are obvious.

Positive — or ncgative — pressure space

Mechanical

подпись: 
positive- or ncgative- pressure space
 mechanical
This section has considered methods of ventilation. These should not be considered the same as methods of air distribution, which are now covered.

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Air Distribution

Ductwork system or unit heaters

Tans discharging warm or cold air downward

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I

подпись: i Air Distribution

Coanda effect on ceiling

подпись: coanda effect on ceiling<=>

Low-level extracts

подпись: low-level extractsLow-level extracts

In a wide hall the air in this area will be stagnated

Air Distribution

FIGURE 9.28 Downward system.

Downward Systems

Air-conditioning systems make use of this method, as the cooler, dense air supplied from high level will drop to low level, picking up the space heat gains before extraction at low level.

It is essential that the entering air ts no more than, say, 9 °C cooler than the room air, for if it is, the cold air will drop, causing complaints from the occupants. In the case of air cooling of hot environments, the temperature difference can be greater.

Full use is made of ceiling diffusers, which ensure that the cold air spreads out over a wide area before dropping. Perforated ceilings may be used, with the ceiling void being the plenum chamber. A typical downward system is shown in Fig. 9.28.

The arrangement shown in Fig. 9.28 is used in industrial halls or auditori­ums. Clean rooms have complex systems using laminar flow to ensure that the room is fully ventilated.

Figure 9.29 shows a typical arrangement of a laminar-flow clean room. The flow may be vertical through a perforated ceiling and floor.

Airflow enters through a perforated wall and is extracted at the other side

Air Distribution

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>

подпись: > Air DistributionAirflow

Air Distribution

Workbench

Air Distribution

Full use can be made in the industrial environment of long jet throws, which entrain room air.

It is essential that the velocity envelope in the occupied zone be in the range of 0.2-0.25 m s-1 to avoid drafts. However, in hot industrial environments, these velocities are frequently exceeded in order to provide adequate body cooling.

Mixed Upward-Downward System

The working of this system is shown in Fig. 9.30, from which it will be seen that good air mixing takes place. Care has to be taken to ensure that the high-level inlets and extract are positioned so that short circulating of the air does not occur.

1’he low-level extract is normally about 25% of the total extract; the ac­tual value is selected to suit the design conditions.

9.4.3.4 Air-Handling Equations

Air Mixing

As the outside air conditions vary from hour to hour, day to day, and month to month, to economize on the heating and cooling loads, the recircu­lated and fresh airstreams are mixed in varying proportions. If the air contains toxic or inflammable gases, vapors, or dusts, no direct mixing occurs, and a heat-recovery device is used.

In order to appreciate the mixing arrangement, consider the layout shown in Fig. 9.31.

In this instance the following abbreviations are used:

Qm = mass flow rate of the gas

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Input air

подпись: input air

Extract air

подпись: extract air Air Distribution

} Extract air

подпись: } extract air Air Distribution>nw — mass of water vapor present

Extract air

Air Distribution

Flow stream A tft’A*QmA> ^A

Air Distribution

Ch-a>clmB’fiB! mu’A’ha

FIGURE 9.31 Mixing arrangement.

H = specific enthalpy of the gas

0 = dry-bulb temperature

Consider any two gas streams A and B which, when combined, produce a mixed condition C.

For Mass Flow

Assuming no flow losses occur, the total mass of a gas introduced into and out of the system must be constant. Hence,

4mA + QmB = “7 mC.

For Moisture Content

Similarly, the total quantity of moisture in the system will be the same before and after mixing. However, the amount of moisture per unit mass of the gas will change: hence, in this case the equation must include mass.

QmA x mwA + qmB x mwB = qmcmwc

For Enthalpy

Provided that no gain or loss of heat takes place during the mixing pro­cess, the total heat in the two airstreams before mixing must equal that of the combined airstream after mixing.

4mA *bA + qmB xhg = qmC x hc

For Temperature

When the dry-bulb temperature of a gas is altered, a positive or negative sen — sible-heat transfer takes place, in terms of both the dry bulb and its associated moisture content.

Because the specific heat capacity of the water vapor is different from that of the dry air, the true dry-bulb mixed-stream air temperature can be deter­mined only by means of a heat balance.

Heat gained or lost by stream A — Heat gained or lost by stream B

Heat gained or lost in stream A + Heat lost or gained by associated moisture

Heat lost or gained in stream B + i Heat lost or gained by associated moisture]

TfmA Cpgi^A “ 0C) +

_ <JmA X mwA XCpW(0A — 0C)

<7otb x cpg(®c~ ®b) +

“JmA X muiB XCpW(6c — &B)

<1va(^A~ CpwtnwA) qVB(0c ^ L’pwmu>B)

Total Room Air Movement

It is essential to ensure that the air distribution in a space provides satis­factory air movement for the occupants’ thermal comfort.

When a jet is released into a free space, it induces into its body the room air. The air leaving the grille is the primary air, while that entrained from the room into the body of the jet is the secondary air. The degree of entrainment

That takes place is related to the velocity of the leaving jet. The combined air

Stream is known as the total air.

The induction is expressed by the momentum equation as

Qmpvp + qmsvs = (qVP + <?vs)vr ’

Where

Qm = mass flow rate of the primary air (kg s’1) vp = entering velocity of the primary air (m s’1) qms = mass flow rate of the secondary air (kg s-1) vs = velocity of the secondary air (m s-1) qVP = volume flow of the primary air (m3 s_1)

*7Vs = volume flow of the total air (m3 s_1) vT — velocity of the total air

The induction ratio (R) is the ratio of the total air to the primary air.

% _ Total air _ Primary air + Secondary air Primary air Primary air

Since the throw of a jet is a function of the velocity, and since the rate of veloc­ity decrease is dependent on the rate of induction that occurs, the quantity of air induced into the discharge from an outlet is a direct function of the perim­eter of the cross section of the primary airstream.

In the case of two outlets, each of the same area, the outlet having the largest perimeter will induce the greatest amount of secondary air; hence, the throw will only be short. For a given air quantity and pressure discharged into a space, the minimum induction and a single outlet of circular cross section obtains the maximum throw, while the greatest induction and shortest throw occurs with a single, long, narrow slot.

The spread of a jet is the angle of divergence of the airstream after it leaves the outlet. The outlet design may be for horizontal or vertical spread or both.

Example

The following example provides some indication of the effects of induction. Primary and secondary air are mixed respectively at the rates of 0.5 m3 s“1, the primary air velocity at outlet being 5.0 m s-1, with the secondary air veloc­ity assumed to be zero.

Determine the velocity and the area of the total airstream when complete mixing of the primary and secondary airstreams has taken place.

Solution The area of the primary airstream before induction is

M = 0.1 m2 .

VJ 5.0

Substituting in the momentum equation,

(0.5 x 5) + (0.5 x 0) = (0.5 + 0.5)v3.

Hence v3 = 2.5 m s-1.

The area of the total airstream is

‘jml 4 m2 _ 0.5 + 0.5 = o 4 m2 v3 2.5 ‘ ’

In basic ventilation design, if a given air change is required, it is a simple matter to determine the capacity of the fan required from

_ Volume of space(m3) x Number of air changes per hour qv 3600 ‘

Depending on the application, the air-change rate may range from 0.5 to 100 air changes per hour. It must be remembered, however, that adequate provi­sion must be made for the makeup air to enter the space without creating dis­comfort or other problems.

In the case of sensible and latent heat, the cooling procedure becomes a little more complicated.

In the case of summer cooling, the heat balance is

Heat gain to space

= Heat absorbed by chilled air flowing through the room,

Or

Sensible heat gain (kW) = Air mass flow x Gain in specific enthalpy,

Or

4>s (kW) = qm (kgs-1) x (br — hs) (kj kgf1).

The sensible heat gain is determined by taking into account all the heat gains in a space, such as solar, fabric, infiltration, machines, processes, occupancy, lighting, etc.

The specific enthalpy (hs) of the room-air design condition is obtained from tables of the property of air or the psychrometric chart.

For deciding to what dry-bulb temperature (0S) the incoming supply air has to be cooled, the following approach is applied.

Sensible heat <f>s (kW) — Air mass flow rate qm (kg s_1)

X Specific heat capacity (kj kg-1 K_1) x Air temperature rise

4>s(kW) = qmxcpx(6r-0s)

4 m ~~ 4 v X P 05(kW;SH) = qvx pxcpx (6r — 0S)

As the air density depends on both the temperature and moisture content of the air, it is necessary to apply the general gas equation:

PaV — mRT

Where

Pa is the absolute pressure of the gas (Pa)

V is the gas volume (m3) m is the mass of gas (kg)

R is the gas constant (287.1 J kg-1 K~])

T is the absolute temperature of the gas (K)

„= ”2 = JL M V RV

Assuming that the pressure and moisture content are constant in the flow along the air supply duct.

_ 1 Pi ~ ‘j-‘

If the standard air density is pj and the supply air density is p2, then.

1_

Pi ~ T

And

P2~T/

The above gives

Pl = Tl

Pi T’i

And

Il

Pi Pi •

Correcting the air density for other air supply temperatures,

-t a nn s — s, 273 + 201 3

P2 = 1.1906 x 773~ (f kg m

Where 1.1906 is the density of air at 20 °C and 50% saturated (specific vol­ume 0.8399 m3 kg-1).

Substituting this into the heat balance equation for sensible heat only gives 0>s(kW) = qv x p2 x cp x (6r — 8S)

<t>s(kW) = qv x 1.1906 x x 1.0048 x 1.0048 x (6r — 6S)

,C1 Sor-e,

Qvx35 x 273 + 0J

Where 1.0048 is the specific heat capacity of air (kj kg-1 K~!).

On rearranging to obtain q^

_ Fi^kW) x (273 + 6)

351(0r-6s) ‘

Example

The sensible gain in an industrial complex is 1000 kW, and the space tem­perature has to be controlled at 20 °C.

If the supply air leaves the cooling plant at 11 °C and gains 1 °C in the duct work before entering the space, determine the air supply volume that is required.

Solution

= 1000 kW; 0r = 20 °C; 0, = 11 + 1 °C gain = 12 °C

So &0 = 20 — 12 = 8 °C.

Hence,

A = 1000kWx (273 + 12) = 1Q1 , 3 — i

Qv 351 x (20-12) ‘

Answer: 101.5 m3 s_1.

Question

A workshop is 30 m x 15 m x 4 m and has to be maintained at 20 °C with six air changes per hour with a supply air temperature of 14 °C. Determine the maximum cooling load that can be met.

Solution

_ N air changes per hour x Volume 3600

6 x30x 15×4 — "2 n m 3600 "

Ct) = ‘7„x351(0f+ 0,)

5 273 + 0S ‘

Substituting known values,

3.0×351(20-14) = 22 kw 273 + 14

Answer: 22 kW.

In order to determine the summer supply air temperature for a given vol­ume of air necessary to remove a given sensible heat gain, the following equa­tion is used.

_ CP,(kW) x (273 + 0,) qv 351 (0,-0,) qvx 351 (0r — 0S) = $,(273 + 0S) qv x 351 0r — <$s x 3510S — <5t x 273 + x 0s

Qv x 351 0r — x273 = qv x 3510S + $sx0s

Qv x 3510r-$s x 273 = 0s{qvy~ 351 + $s)

Hence,

0.

подпись: 0.= ‘?t/x3510r-<I>sx273 qv x 351 + <п>s

Example

A process plant occupies a space of 30 m x 20 m x 4 m. The area has to be maintained at 21 °C with 10 air changes per hour of chilled air. The

Refrigeration plant is capable of 35 kW of sensible cooling. Determine the required supply air temperature.

Solution

Qv = 30xi0x4x.10 = 6M mJ s-i

_ qvx 3510r-<J>s x273

подпись: _ qvx 3510r-<j>s x2733600 351flr- qv x 351 + cf>s

Substituting the known values.

„ 6.66,, x 351 x21-35×273

= 16.6 °C

5 6.66 x 351 + 35

Answer: 16.6 °C.

The above examples have dealt with cooling. In the case of heating ap­plications, the supply air temperature must be greater than the room air, hence 0S > 6T.

Example

A workshop has a fabric and ventilation sensible heat loss of 30 kW, and the process requires that the room be maintained at 22 °C dry bulb. Deter­mine the supply air volume required at 34 °C in order to maintain the space at design conditions.

Solution

_3>sx (273 + 0,) Hv 351(0,- 6T)

= 30 x (273 + 34) 351(34 — 22)

= 2.187 m3 s“1

3 c-l

подпись: 3 c-lAnswer: 2.187 m3 s

Example

A plenum system provides two air changes per hour of makeup air at 33 °C to a process area of 20 m x 15 m x 3.5 m in order to maintain the space tem­perature at 21 °C.

Determine the maximum heat loss the above parameters are capable of meeting.

Solution

Qu = 2×20^5×3,5 = Q 583 ^ g_, _ 3>, x (273 + 0,)

351 (0S — 0r)

_ qvx35l(6s-6r) s 273 + es

= 0.583×351 x (33 -21) 273 + 33

The supply air temperature in the case of a heating system is determined from

A _ 351 x qt,0r + 273 x <ps s_ 351qv-es ‘■

Question

Determine the air temperature that is necessary to maintain a volume of 1200 m3 with six air changes per hour at 20 °C, if the total heat loss is 45 kW.

12Q° * 6 = 2.0 m s

подпись: 12q° * 6 = 2.0 m sSolution

3 -1

____ — = / 11 m ~

3600

E = 351.2< 2 X 20 + 273x_45 = 4007 oC s 351×2-45

Answer: 40.07 kW.

When calculating for heater and cooler batteries, it is the mass flow rate gmthat is required, not the volume flow rate qvas used above. This is deter­mined by using the appropriate specific volume or density of the air at the ac­tual design temperature and moisture content.

Moisture Content

The water vapor present in the air exists at its own partial pressure. The molecular mass of water vapor is 18.015 kg kmol-1. This is less than dry air, which is 29 kg kmol-1.

From the above it will be appreciated that the greater the concentration in the air, the less the mixture weighs.

As the moisture content in a dry air mixture increases, the specific volume increases and the density reduces. The specific volume is v (m3 kg-1), and the density is p (kg m~3).

The relationship between the two parameters is simply p = /v.

The base for air density is taken as 101.325 kPa at 0 °C.

Using the general gas law, the standard density is

Pav = mRT,

Hence density

P = M/v = pa/R T.

Taking the vapor pressure of the air supply from hygrometric tables, the value of specific density can be found.

287.1 x (273 + 6,) 3l — i v=’ 101:325 ~ps m kg

Example

Air is supplied to a space at the rate of 2.0 m3 s-1 at 12 °C and 70% satu­rated, at which condition the vapor pressure is 0.9852 kPa. Determine the density and the mass flow rate.

_ 287.1 x (273 + 12) 3 , — i _ n Q1 3 , —

" 101325-0.9852×1000 ™ 8 °*8154 m k8

The mass flow is then

A = вi’ = 2.0 = ? 452 ke s“1

Lm v 0.8154 8 ‘

Answer: 2.452 kg s_1.

Posted in INDUSTRIAL VENTILATION DESIGN GUIDEBOOK


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