Hvac design

AIR CONDITIONING

Introduction to Air Conditioning Air conditioning may be required in buildings which have a high heat gain and as a result a high internal temperature. The heat gain may be from solar radiation and/or internal gains such as people, lights and business machines. The diagram below shows some typical heat gains in a room.

If the inside temperature of a space rises to about 25oC then air conditioning will probably be necessary to maintain comfort levels. This internal temperature (around 25oC) may change depending on some variables such as: • type of building • location of building • duration of high internal temperature • expected comfort conditions. • degree of air movement • percentage saturation In some buildings it may be possible to maintain a comfortable environment with mechanical ventilation but the air change rate will tend to be high (above about 8 air changes per hour) which can in itself cause air distribution problems. Since air conditioning is both expensive to install and maintain, it is best avoided if possible. This may possibly be achieved by careful building design and by utilising methods such as: window blinds or shading methods heat absorbing glass heat reflecting glass openable windows higher ceilings smaller windows on south facing facades

alternative lighting schemes. The diagram below shows some of these methods.

If air conditioning is the only answer to adequate comfort in a building then the main choice of system can be considered.

Full comfort air conditioning can be used in summer to provide cool air (approx. 13oC to 18oC) in summer and warm air (approx. 28oC to 36oC) in winter. Also the air is cleaned by filters, dehumidified to remove moisture or humidified to add moisture. Air conditioning systems fall into three main categories, and are detailed in the following pages; 1. Central plant systems. 2. Room air conditioning units. 3. Fan coil units. Central plant systems have one central source of conditioned air which is distributed in a network of ductwork. Room air conditioning units are self-contained package units which can be positioned in each room to provide cool air in summer or warm air in winter. Fan coil units are room units and incorporate heat exchangers piped with chilled water and a fan to provide cool air.

1.0 Central Plant Systems A typical central plant air conditioning system is shown below. Cooling coil

Heater Battery

Supply fan Fresh air

-

Supply air

+

Room

Recirculated air Exhaust air

Return air

Return air fan SCHEMATIC DIAGRAM OF CENTRAL PLANT AIR CONDITIONING SYSTEM

The system shown above resembles a balanced ventilation system with plenum heating but with the addition of a cooling coil. For information on balanced ventilation see VENTILATION section. In winter the heater battery will be on and the cooling coil will probably be switched off for the majority of buildings. In summer the heater battery will not need to have the same output and the cooling coil will be switched on.

A humidifier may be required to add moisture to the air when it is 'dry'. This is when outdoor air has a low humidity of around 20% to 30%. In the U.K. low humidities are rare and therefore humidification is sometimes not used. In dryer regions humidification is required through most of the year whereas in tropical air conditioning one of the main features of the system is the ability to remove moisture from warm moist air. Dampers are used in air conditioning central plant systems to control the amount of air in each duct. It is common to have 20% fresh air and 80% recirculated air to buildings. In buildings with high occupancy the fresh air quantity should be calculated based on C.I.B.S.E. data., this may require a higher percentage of fresh air (i.e. more than 20%). See Ventilation section for examples of fresh air rates.

Filters are required to remove particles of dust and general outdoor pollution. This filter is sometimes called a coarse filter or pre-filter. A removable fibreglass dust filter is positioned in the fresh air intake duct or in larger installation an oil filled viscous filter may be used.

The secondary filter, after the mix point, is used to remove fine dust particles or other contaminant picked up in the rooms and recirculated back into the plant. A removable bag filter is generally used for this where a series of woven fibre bags are secured to a framework which can be slid out of the ductwork or air handling unit (A.H.U.) for replacement.

Air Handling Units Air handling units (A.H.U.) are widely used as a package unit which incorporates all the main plant items as shown below. Pipework, ductwork and electrical connections are made after the unit is set in place on site. Since air conditioning plant rooms tend to be at roof level, the larger A.H.U.'s are lifted into place by crane before the roof is fixed.

In some cases it is usual to place the fan in front of (that is upstream of) the heater battery and cooling coil. This is because fans operate best if the system resistance is at the outlet rather than the inlet of the impeller. This is shown on the schematic diagrams above. The photograph below shows a typical air handling unit with handles on the doors for access to equipment.

2.0 Room Air Conditioning Units These units use refrigerant to transfer cooling effect into rooms. Room air conditioning units fall into two main categories: 1. Split type 2. Window/wall units. Split Air Conditioners Split air conditioners have two main parts, the outdoor unit is the section which generates the cold refrigerant gas and the indoor unit uses this cold refrigerant to cool the air in a space. The outdoor unit uses a compressor and air cooled condenser to provide cold refrigerant to a cooling coil in the indoor unit. A fan then blows air across the cooling coil and into the room. The indoor unit can either be ceiling mounted (cassette unit), floor mounted or duct type. The drawing below shows a ceiling mounted (cassette unit).

The photographs below show a ceiling mounted cassette and an outdoor unit.

Window / Wall Units Window or wall units are more compact than split units since all the plant items are contained in one box. Window units are installed into an appropriate hole in the window and supported from a metal frame. Wall units like the one shown below are built into an external wall and contain all the necessary items of equipment to provide cool air in summer and some may even provide heating in winter.

A small hermetically sealed compressor is used to provide refrigerant gas at the pressure required to operate the refrigeration cycle. The condenser is used to condense the refrigerant to a liquid which is then reduced in pressure and piped to the cooling coil.

3.0 Fan Coil Units These are room air conditioners but use chilled water instead of refrigerant. Units can be floor or ceiling mounted. The chilled water is piped to a finned heat exchanger as in a fan convector. A fan blows room air across the heat exchanger and cool air is emitted into the room, as shown below. Outlet louvre Finned pipe heat exchanger

Cool Air

Chilled water pipes Drip tray and condensate drain

Centrifugal fans

Dust filter

Cabinet

Thermostat

T

FAN COIL UNIT

Room Air

Fan coil units may be looked upon as being small air handling units located in rooms and they can be piped with chilled water for cooling and low temperature hot water (LTHW) for heating if necessary. The room temperature can be controlled with low, medium and high fan speeds and chilled water flow is varied with a two-port or three-port motorised valve. Two-pipe, three-pipe and four-pipe systems have been used. The four-pipe system has two heating and two cooling pipes and may have a single heat exchanger or two separate heat exchangers for heating and cooling. It is useful to have a summer/winter changeover switch in the main control system to avoid both heat exchangers being on at the same time. A three-pipe system used heating flow, cooling flow and common return pipework.

Choosing an A/C System Generally central plant systems are used in large prestigious buildings where a high quality environment is to be achieved. Cassette units and other split systems can be used together with central plant systems to provide a more flexible design. Each system has its own advantages and the following is a summary of some of the main advantages and disadvantages. Central Plant Systems - Advantages: 1. Noise in rooms is usually reduced if plant room is away from occupied spaces. 2. The whole building can be controlled from a central control station. This means that optimum start and stop can be used and a weather compensator can be utilised. Also time clocks can bring air conditioning on and off at appropriate times. 3. Maintenance is centralised in the plant room. Plant is easier to access. Central Plant Systems - Disadvantages: 1. Expensive to install a complete full comfort air-conditioning system throughout a building. 2. Space is required for plant and to run ductwork both vertically in shafts and horizontally in ceiling spaces. 3. Individual room control is difficult with central plant. Many systems have been tried such as Variable Air Volume (VAV), dual duct systems and zone reheaters. Zone re-heaters are probably more successful than the rest. Room Air Conditioning Units - Advantages: 1. 2. 3. 4. 5.

Cheaper to install. Individual room control. Works well where rooms have individual requirements. No long runs of ductwork. Can be used to heat as well as cool if a reversing valve is fitted.

Room Air Conditioning Units - Disadvantages: 1. 2. 3.

Sometimes the indoor unit fan becomes noisy. Noisy compressor in outdoor unit. Each unit or group of units has a filter, compressor and refrigeration pipework that needs periodic maintenance and possible re-charging. Units have course filters therefore filtration is not as good as with AHU’s. 4. The installation may require long runs of refrigerant pipework which, if it leaks into the building, can be difficult to remedy. 5. Not at robust as central plant. 6. The majority of room air conditioners just recirculate air in a room with no fresh air supply although most manufacturers make units with fresh air capability. 7. Cooling output is limited to about 9 kW maximum per unit; therefore many units would be required to cool rooms with high heat gains.

Fan coil units are similar is some respects to Room Air Conditioners. Fan Coil Units - Advantages: 1. 2. 3. 4. 5.

Cheaper to install than all air central plant system. Individual room control. Works well where rooms have individual requirements. No long runs of ductwork. Can be used to heat as well as cool if 3 or 4-pipe system is used.

Fan Coil Units - Disadvantages: 1. 2. 3. 4.

5. 6.

Sometimes the indoor unit fan becomes noisy, especially when the speed is changing with in-built controls. Each unit requires maintenance. Long runs of pipework are required. A chiller is still required to produce chilled water therefore they do not save as much in plant and plant room space as room air conditioners. Also boilers will be required if heating mode is installed. Fresh air facility may not be installed. Cooling output is limited to about 5 kW.

Air Conditioning Air Flow Rates When determining air flow rates for rooms that are air conditioned, the following procedure should be adopted; Calculate heat gains. Complete psychometric chart. Determine mass flow rate of supply air from the following; m

=

H

/

(Cp x (tr –ts))

H m = tr ts

= Sensible heat gain (kW) = mass flow rate of air (kg/s) Specific heat capacity of air (1.005 kJ/kg K) = room temperature (oC) = supply air temperature (oC) from psychometric chart.

where;

Cp

Convert mass flow rate to a volume flow rate: Volume flow rate (m3/s) = mass flow rate (kg/s) / density of air (kg/m3) Convert this to an Air Change rate for comparison. Volume flow rate (m3/h) = Volume flow rate (m3/s) x 3600 Supply Air Rate (AC/h)

=

Volume Flow Rate (m3/h) / Room Volume (m3)

Check out the recommended air flow rate from CIBSE Guide B2 (Ventilation and Air Conditioning) Section 3 - Requirements. Use the higher value of air change rate for design purposes. Example 1 Determine the supply, fresh air and recirculation air flow rates for the Sports Centre (Fitness Room) shown below. The room is to be fully air conditioned. Air Handling Unit (AHU) in roof space above.

Fitness Room

17 m

Ceiling Height = 5.0 m

44 m DATA Occupancy Room volume Room temperature Sensible heat gain

= = = =

Supply air temp.

=

80 people 3740 m3. 18oC from table 3.19 CIBSE guide. 70 kW from calculations (not shown) See Heat Gains section of these notes for further details. 13oC from psychometric chart (not shown) See Psychometrics section of these notes for further details.

Supply Air Rate m

=

H

/

(Cp x (tr –ts))

m

=

70 / (1.005 x (18 -13))

m

=

70 / 5.025

m

=

13.93 kg/s

Volume flow rate (m3/s) Volume flow rate (m3/h)

= = = = = =

mass flow rate (kg/s) / density of air (kg/m3) 13.93 x 1.2 13.93 m3/s Volume flow rate (m3/s) x 3600 16.72 m3/s x 3600 60,192 m3/h

Supply Air Rate (AC/h)

Volume Flow Rate (m3/h) / Room Volume (m3) 50,148 m3/h / 3740 16.1 AC/h

= = =

Table 3.1 Summary of Recommendations (Guide B2) directs to Table 3.19. From Table 3.19 for Fitness Centre the air flow rate is 10 to 12 AC/h. Use higher value of 16.1 AC/h supply air rate as calculated. Fresh Air Rate Section 3.2.1.3 Body odour (CIBSE guide B2) gives the following information; Therefore in the absence of further information, it is recommended that 8 litre.s-1 per person should be taken as the minimum ventilation rate to control body odour levels in rooms with sedentary occupants. Since the level of activity is higher than sedentary in the Fitness Room, we will adopt a fresh air rate of 24 l/s per person. Fresh air rate =

24 l/s = = = =

x 80 people 1920 l/s 1.92 m3/s 1.28 x 3600 = 6912 m3/h 3 6912 m /h / 3740 = 1.9 AC/h

Percentage fresh air; 1.9 / 16.1 (x 100%)

=

11.8%

Recirculation Air Rate Recirculation Air Rate

= = =

Supply air rate - Fresh air rate 16.1 1.9 14.2 AC/h

Percentage recirculation air; 14.2 / 16.1 (x 100%) Cooling Coil

=

88.2% Heater Battery

Supply fan Pre-filter

Bag filter(s)

Pre-heater

1.9 AC/h Fresh 1.92 m3/s air

+

-

+

Supply air 16.1 AC/h

13.93 m3/s

Dampers

Room

Recirc.14.2 AC/h air 12.01 m3/s

1.9 AC/h Exhaust 1.92 m3/s air

Return air 16.1 AC/h

13.93 m3/s Return air fan

HEATSchematic GAIN Diagram Of Central Plant Air Conditioning System

Introduction Heat gains from the sun can lead to increases in internal temperatures beyond the limits of comfort. This is usually about 24oC. It is therefore necessary to determine the amount of solar radiation that is transmitted into buildings through; windows, walls, roof, floor and by admitting external air into the building. Several measures can be adopted to reduce solar radiation in buildings. These are external and internal shading and by careful building design.

Calculating Heat Gains The load on an air-conditioning system can be divided into the following sections: 1. Sensible Transmission through glass. 2. Solar Gain through glass. Internal Heat gains 4. Heat gain through walls. 5. Heat gain through roof. 6. Ventilation and/ or infiltration gains. The heat gain through the glass windows is divided into two parts since there is a heat gain due to temperature difference between outside and inside and another gain due to solar radiation shining through windows. The method adopted uses the CIBSE guide A (2006) and CIBSE Guide J (2002) . The Tables that are referred to are CIBSE guide A (2006) Solar cooling loads in Tables 5.19 to 5.24. CIBSE Guide J (2002) Air and Sol-air temperatures in Table 5.36 (London), Table 5.37 (Manchester) and Table 5.38 (Edinburgh) If internal gains are to estimated then CIBSE Guide A (2006) Table 6.4 to 6.17 are also required. It would be helpful to have these Tables close by, to complete the calculations. An example of a heat gain claculation is given in CIBSE Guide A (2006) section 5.8.2 example 5.3. Heat gains through solid ground floors are minimal and can be neglected.

1.0 Sensible Transmission Through Glass This is the Solar Gain due to differences between inside and outside temperatures. In very warm countries this can be quite significant. This gain only applies to materials of negligible thermal capacity i.e. glass. Qg

=

Ag . Ug (to- tr)

Qg Ag

=

Sensible heat gain through glass (W) Surface area of glass (m2)

........ eqn. 1

Where; =

Ug = 'U' value for glass (W/m2 oC) to = outside air temperature (oC). Can be obtained from CIBSE guide J Tables 5.36 to 5.38 for various months and times in the day. tr = room air temperature (oC) 2.0 Solar Gain Through Windows This gain is when the sun shines though windows. The cooling loads per metre squared window area have been tabulated in Table A9.15 for various times, orientations, latitutes and building weights. These figures are then multiplied by correction factors for shading and air point control as shown in the equation below. Heat load is found from; Qsg where qsg Fc Fs Ag

= = = =

=

Fc . Fs . qsg . Ag

........ eqn. 2

Qsg = Actual cooling load (W) Tabulated cooling load from CIBSE Guide A (2006) Table 5.19 to 5.24) (W/m2) Air point control factor (Table 5.19 to 5.24) or see Table below) Shading factor (Table 5.19 to 5.24) or see Table below) Area of glass (m2)

The Air point control factors (Fc) and Shading factors (Fs) are given in the Table below for various types of glass, building weights and for open and closed blinds. Air point control factors (Fc) Building Weight Light Heavy

Single Glazing

Double glazing

Horizontal blind 0.91 0.83

Horizontal blind 0.91 0.90

Single Glazing Open Closed horizontal horizontal blind blind

Double glazing Open Closed horizontal horizontal blind blind

Shading factors (Fs) Type of glass Building Weight

Light Heavy Light Bronze tinted 6mm Heavy Bronze tinted Light 10mm Heavy Light Reflecting Heavy

1.00 0.97 0.86 0.85 0.78 0.77 0.64 0.62

Clear 6mm

0.77 0.77 0.77 0.77 0.73 0.73 0.57 0.57

0.95 0.94 0.66 0.66 0.54 0.53 0.48 0.47

0.74 0.76 0.55 0.57 0.47 0.48 0.41 0.41

3.0 Internal Heat Gains - CIBSE Guide A (2006) Internal gains can account for most heat gain in buildings in the U.K. These gains are from occupants, lights, equipment and machinery, as detailed below. OCCUPANTS - Sensible and latent heat gains can be obtained from CIBSE Guide A (2006) - Table 6.3. Typical gains are shown below.

Conditions

Typical building

Sensible Heat Gain (Watts)

Latent Heat Gain (Watts)

Seated very light work

Offices, hotels, apartments

70

45

Moderate office work

Offices, hotels, apartments

75

55

Standing, light work; walking

Department store, retail store

75

55

Walking standing

Bank

75

70

Sedentary work

Restaurant

80

80

Light bench work

Factory

80

140

Athletics

Gymnasium

210

315

LIGHTING – Average power density from Tables 6.4. ELECTRICAL EQUIPMENT - PC’s and Monitors - Tables 6.7 and 6.8. Laser Printers and Photocopiers - Tables 6.9 and 6.10 Electric Motors – Table 6.13 and 6.14. Lift Motors – Table 6.15. Cooking equipment – Table 6.17. Heat load is found from; Q int. = Heat from Occupants + from Cooking

Heat from Lighting + Heat from Electrical Equipment + Heat

4.0 Heat Gain Through Walls This is the unsteady-state heat flow through a wall due to the varying intensity of solar radiation on the outer surface.

4.1 Sol-Air Temperature In the calculation of this heat flow use is made of the concept of sol-air temperature, which is defined as; the value of the outside air temperature which would, in the absence of all radiation exchanges, give the same rate of heat flow into the outer surface of the wall as the actual combination of temperature difference and radiation exchanges. SOL-AIR TEMP, teo

α . I . cos a . cos n +) α Is hso

=

ta + (

........ eqn. 4.1

teo

=

sol-air temperature (oC)

ta α I

= = =

outside air temperature (oC) absorption coefficient of surface intensity of direct solar radiation on a surface at right angles to the rays of the

a n Is

= = =

hso

=

solar altitude (degrees) wall-solar azimuth angle (degrees) 2 intensity of scattered radiation normal to a surface (W/m ) 2 external surface heat transfer coefficient (W/m oC)

where

2

sun. (W/m )

The U.K. values of sol-air temperature are found from CIBSE Guide J (2002) Table 5.36 (London), Table 5.37 (Manchester) and Table 5.38 (Edinburgh). 4.2 Thermal Capacity The heat flow through a wall is complicated by the presence of thermal capacity, so that some of the heat passing through it is stored, being released at a later time. Thick heavy walls with a high thermal capacity will damp temperature swings considerably, whereas thin light walls with a small thermal capacity will have little damping effect, and fluctuations in outside surface temperature will be apparent almost immediately. The thermal capacity will not affect the daily mean solar gain but will affect the solar gain at a particular time. The calculation is, therefore, again split into two components. 1. Mean gain through wall, Qθ where,

=

A . U ( tem - tr)

........ eqn. 4.2

Qθ = heat gain through wall at time θ 2 A = area of wall (m ) 2 U = overall thermal transmittance (W/m oC) (See Thermal Transmission in HEATING section of the notes) tem = 24 hour mean sol-air temperature (oC) (Table 5.36 to 5.38) tr = constant dry resultant temperature (oC). In practice room dry bulb is used.

2. The variation from the mean solar gain is subject to both a decrement factor and time lag.

where

Qφ

=

f ( teo - tem)

........ eqn. 4.3

Qφ φ teo

= = =

tem f

= =

Heat gain through wall at time (θ +φ) time lag in hours (see CIBSE guide A (2006) Table 3.49 to 3.55) sol-air temperature at time specified (oC) (Table 5.36 to 5.38) 24 hour mean sol-air temperature (oC) (Table 5.36 to 5.38) decrement factor (see CIBSE guide A (2006) Table 3.49 to 3.55)

Therefore the Solar Gain through a wall at time ( θ + φ) is;

where,

teo

=

Qθ+φ

=

A . U ( tem - tr)

+ f ( teo - tem)

........ eqn. 4.4

Qθ+φ φ A U tem

= = = = =

heat gain through wall at time θ+φ (Watts) time lag in hours (see CIBSE guide A (2006) Table 3.49 to 3.55) 2 area of wall (m ) 2 overall thermal transmittance (W/m oC) 24 hour mean sol-air temperature (oC) (Table A8.3) constant dry resultant temperature (oC) In practice room dry bulb is used.

tr = f = decrement factor (see CIBSE guide A (2006) Table 3.49 to 3.55) Sol-air temperature when Tabulated cooling load (qsg) is at a maximum (oC) (Table A8.3)

5.0 Heat Gain Through Roof

The heat gain through a roof uses the same equation as for a wall as shown below.

Q θ+φ θ φ Roof

=

A U [( tem - tr) + f ( teo - tem)]

........ eqn. 5

6.0 Ventilation and/or Infiltration Gains Heat load is found from;

where

Qsi

=

n . V (to- tr) / 3

........ eqn. 6

Qsi n V to

=

Sensible heat gain (W)

= = =

tr

=

number of air changes per hour (h-1) (see note below) volume of room (m3) outside air temperature (oC) Can be obtained from CIBSE guide Table A8.3. room air temperature (oC)

Infiltration gains should be added to the room heat gains. Recommended infiltration rates are 1/2 air change per hour for most air-conditioning cases or 1/4 air change per hour for double glazing or if special measures have been taken to prevent infiltration.

Ventilation or fresh air supply loads can be added to either the room or central plant loads but should only be accounted for once. Total Room Load From Heat Gains Q total

=

Q total

Qg

+

Qsg

+

=

Ag . Ug (to - tr)

Qint. +

Qθ+φφWall +

Fc . Fs . qsg . Ag

2. Solar Glass.

+

Qint.

3. Internal

+

+ Qsi

1. Sensible Glass

+

+

Q θ+φφ Roof

A.U [( tem - tr) + f ( teo - tem)] 4. Walls

A.U [( tem - tr) + f ( teo - tem)] 5. Roof +

n . V (to - tr) / 3

6. Ventilation ........ eqn. 7

In the majority of cases, by far the greatest external fluctuating component is the solar heat gain through the windows. Therefore, it will be this gain which determines when the total heat gain to the room is a maximum. Heat gains may be calculated and displayed in table form as shown below. Heat Gain from

Watts

%

1. Sensible transmission through glass 2. Solar gain through glass 3. Internal 4. External walls 5. Roof 6. Ventilation Total

100%

Heat gain per m2 floor area = Heat gain per m3 space =

EXAMPLE 1 The room shown below is to be maintained at a constant environmental temperature of 21oC for a plant operation of 12 hours per day.

The room is on the intermediate floor of an Library located in London latitute 51.7oN. The internal construction is lightweight demountable partitions, lightweight slab floors and suspended acoustically treated ceilings. Calculate the maximum sensible cooling load in the room in July The outside air temperature (to) may be found from CIBSE guide Table A8.3. for July 23rd. The maximum value ocurrs at 15.00 hrs. and is 24.5oC. DATA: Occupants = 100 Infiltration = 0.5 air changes per hour Building classification = lightweight. External wall 'U' value = 0.45 W/m2oC, internal insulation. External wall colour = light. External wall decrement factor f = 0.65 Glass type & 'U' value = clear 6mm, single, U = 2.80 W/m2oC Window blinds = light slatted blind – open. Lighting = 30 Watts / m2 floor area Heat gain from machinery and equipment = 4000 Watts

NOTE: It should be noted that this total heat gain is used to size central plant items such as Chillers, Condensers and Cooling Towers. Cooling coils are sized usually with a pschrometric chart. Answer Areas: Area of window Total area of glass Area of glass facing South

= = =

1.2 x 1.7 = 2.04 m2. 2.04 x 12No. windows 12.24 m2.

Area of wall facing South

=

22.0 m x 4.0 m high = 88 m2 less glass = 88 - 12.24 = 75.76 m2.

=

24.48 m2.

Floor area Room volume

= =

22 x 14 308 x 4

= =

Gains: 1. Sensible transmission through glass Qg Qg Qg

308m2. 1232 m3.

= = =

Ag Ug (to - tr) 24.48 x 2.8 (24.5 – 21) 239.9 Watts

2. Solar Gain through glass Qsg = Fc Fs qsg Ag where: Qsg = Actual cooling load (W) Fc = Air point control factor (Table A9.15 for Clear 6mm, single, lightweight building gives 0.91) Fs = Shading factor (Table A9.15 for Clear 6mm, single, open light slatted blinds,lightweight building gives 1.0 ) qsg = Tabulated cooling load (A9.15 for July 23rd , orientation South, 12.00 hours gives maximum of 252 W/m2) 2 Ag = Area of glass facing South (m )

3.

(Maximum is at 12.00 hrs)

Qsg Qsg

Internal

Qint. Lights (30 W/m2 x 308) + 4000 W + People (100 x 100) 9,240 + 4,000 + 10,000 23,240.0 Watts

4.

Qint. Qint. Qint. Qint.

External wall where, Qθ+φ = A = U = tem = coloured wall facing South) tr = f = teo = 36oC (Table A8.3)

= = = =

Qθ+φWall

=

= =

0.91 x 1.0 x 252 x 12.24 2806.9 Watts

A U [( tem - tr) + f ( teo - tem)]

heat gain through wall at time θ+φ (Watts) 2

area of wall facing South (m ) 2o overall thermal transmittance given in question as 0.45 W/m C. 24 hour mean sol-air temperature (oC) (Table A8.3 gives 22.5oC for light constant dry resultant temperature (oC). Room dry bulb of 21oC is given. decrement factor for wall is given as 0.65. Sol-air temperature when Tabulated cooling load (qsg) is at 12.00 hrs. gives

Qθ+φWall Qθ+φWall Qθ+φWall

= = =

5.

Roof

Qθ+φRoof =

6.

Ventilation

Qsi Qsi Qsi

= = =

75.76 x 0.45 [( 22.5 – 21) + 0.65 ( 36 – 22.5)] 34.092 [ 1.5 + 8.775 ] 350.3 Watts Nil for intermediate floor. n V (to - tr) / 3 0.5 x 1232 (24.5 – 21) / 3 718.0 Watts

7.

Q total Q total Q total

= Qg + Qsg + Qint. + Qθ+φWall + Q θ+φ Roof + Qsi = 239.9 + 2,806.9 + 23,240.0 + 350.3 + 0 + 718.0 = 27,354.1 Watts

The results are shown in the table below. Heat Gain from 1. Sensible transmission through glass

Watts

%

239.9

0.9

2. Solar gain through glass

2,806.9

10.3

3. Internal

23,240.0

84.9

350.3

1.3

0

0

4. External walls 5. Roof 6. Ventilation

718.0 Total

27,354.1

2.6 100%

Heat gain per m2 floor area = 88.8 W/m2 Heat gain per m3 space = 22.2 W/m3 EXAMPLE 2 The room shown below is to be maintained at a constant environmental temperature of 21oC for a plant operation of 12 hours per day. The room is on the intermediate floor of an Office Block located in London. The internal construction is lightweight partitions, concrete hollow slab floors and suspended ceilings. Calculate the maximum sensible cooling load in the room in July. DATA: Occupants = 80 Lighting = 35 Watts / m2 floor area Infiltration = 1.0 air changes per hour Outside air temperature (to) = 26oC. Building classification = lightweight. External wall surface texture = dark. External wall thickness = 300mm, internal insulation, decrement factor is 0.27. Light slatted blinds = Assume open Heat gain from machinery and equipment = 3000 Watts

Answer Areas: Area of window = Total area of glass = Area of glass facing South West Area of wall facing South West 43.8 m2. Floor area Room volume =

1.2 x 1.7 = 2.04 m2. 2.04 x 10No. windows = 10.2 m2. = 18.0 m x 3.0 m high

= 18 x 16 288 x 3 =

=

20.4 m2.

= 54 m2 less glass = 54 - 10.2 =

= 288m2. 3 864 m .

Gains: 1.

Sensible transmission through glass Qg

2.

Solar Gain through glass (Maximum is at 15.00 hrs)

3.

Internal

Qint. Qint. Qint. Qint.

= = = =

Qsg

Qsg = Qsg

= Qg Qg

Ag Ug (to - tr) = 20.4 x 5.8 (26 – 21) = 591.6 Watts

= Fc Fs qsg Ag 0.91 x 1.0 x 278 x 10.2 = 2580.4 Watts

Qint. Lights (35 W/m2 x 288) + 3000 W + People (80 x 100) 10,080 + 3,000 + 8,000 21,080.0 Watts

4.

External wall Qθ+φWall (dark façade) Qθ+φWall = Qθ+φWall Qθ+φWall

=

A U [( tem - tr) + f ( teo - tem)] 43.8 x 0.45 [( 27 – 21) + 0.27 ( 51.5 – 27)] = 19.71 [ 6 + 6.615 ] = 248.6 Watts

5.

Roof

Qθ+φRoof =

6.

Ventilation

Qsi Qsi Qsi

7.

Q total Q total Q total

= = =

Nil for intermediate floor. n V (to - tr) / 3 1.0 x 864 (26 – 21) / 3 1440.0 Watts

= Qg + Qsg + Qint. + Qθ+φWall + Q θ+φ Roof + Qsi = 591.6 + 2,580.4 + 21,080.0 + 248.6 + 0 + 1,440.0 = 25,940.6 Watts

The results are shown in the table below. Heat Gain from

Watts

1. Sensible transmission through glass

%

591.6

2.3

2. Solar gain through glass

2,580.4

9.9

3. Internal

21,080.0

81.3

248.6

1.0

0

0

1,440.0

5.5

25,940.6

100%

4. External walls 5. Roof 6. Ventilation Total Heat gain per m2 floor area = 90.1 W/m2 Heat gain per m3 space = 30.0 W/m3

The heat gain in the previous example was 88.8 W/m2 floor area and the total was 27,354.1 Watts . Therefore there is a slightly lower overall heat gain for this example (25,940.6 Watts) but the gain per m2 floor area is slightly higher (90.1 W/m2) than in the previous example.

EXAMPLE 3 The Restaurant shown below is to be maintained at a constant environmental temperature of 22oC for a plant operation of 12 hours per day. The Restaurant area is on the ground floor of an Single storey building located at 51.7oN. The internal construction is lightweight partitions, concrete hollow slab floors and suspended ceilings. Calculate the maximum sensible cooling load in the Restaurant area in July. DATA: Occupants = 70 Lighting = 22 Watts / m2 floor area Infiltration = 1.0 air changes per hour Outside air temperature (to) = 28oC. Building classification = lightweight. External wall surface texture = dark. External wall 105mm Brick, internal 50mm EPS insulation, 100mm lightweight aggregate concrete block, 13mm dense plaster, cavity wall. Windows Double glazed internal shade, clear 6mm glass, light slatted blinds assume open Roof use information from section 2(d) in Table A3.54. Heat gain from equipment = 2000 Watts

Answer Areas: Area of window Total area of glass Area of glass facing South Area of wall facing South Floor area Ceiling area

= =

= = =

1.4 x 2.0 = 2.8 m2. 2.8 x 10No. windows = 14.0 m2.

= 14.0 m x 3.0 m high

28.0 m2.

= 42 m2 less glass = 42 - 14 = 28.0 m2.

14 x 10 = 140 m2. Length x 2(Rafter length inside) Rafter length inside = 0.5 x room width / cos roof pitch = Rafter length inside = 0.5 x 5 / cos

300. = 5.774 metres Ceiling area Room volume =

= 14 x 2 (5.774) 2 140m x 3 + ( 14 x 5 x 2.9 )

=

Rafter length inside = 0.5 x 5 / 0.866

= =

161.7 m2. 623 m3.

Gains: 1.

Sensible transmission through glass Qg

2.

Solar Gain through glass (Maximum is at 12.00 hrs)

3.

Internal

Qint. Qint. Qint. Qint.

= = = =

Qsg

Qsg = Qsg

= Qg Qg

Ag Ug (to - tr) = 28.0 x 2.8 (28 – 22) 470.4 Watts =

= Fc Fs qsg Ag 0.91 x 0.95 x 252 x 14.0 = 3,050.0 Watts

Qint. Lights (22 W/m2 x 140) + 2000 W + People (70 x 100) 3080 + 2,000 + 7,000 12,080.0 Watts

External wall Find decrement factor (f) from Table A3.54. Using wall data given above the wall is type 8(e) and the decrement factor is 0.43, time lag is 8.6 hours, ‘U’ value 0.43 W/m2oC. (dark façade) Qθ+φWall = Qθ+φWall Qθ+φWall Qθ+φWall

A U [( tem - tr) + f ( teo - tem)] = 28 x 0.22 [( 26 – 22) + 0.43 ( 48 – 26)] = 12.32 [ 4 + 9.46 ] = 165.8 Watts

Roof Find decrement factor (f) from Table 3.54, use information from section 2(d). The decrement factor is 0.89, time lag is 2.9 hours, ‘U’ value 0.22 W/m2oC. Qθ+φRoof Qθ+φRoof Qθ+φRoof Qθ+φRoof 6.

Ventilation

7.

Q total Q total Q total

Qsi Qsi Qsi

= = = =

= = =

A U [( tem - tr) + f ( teo - tem)] 161.7 x 0.22 [(26 – 22) + 0.89 (48 – 26 )] 35.6 [ ( 4 + 19.58 ) ] 838.8 Watts n V (to - tr) / 3 1.0 x 623 (28 – 22) / 3 1246.0 Watts

= Qg + Qsg + Qint. + Qθ+φWall + Q θ+φ Roof + Qsi = 470.4 + 3,050.0 + 12,080.0 + 165.8 + 838.8 + 1,246.0 = 17,851 Watts

The results are shown in the table below. Heat Gain from

Watts

%

470.4

2.6

2. Solar gain through glass

3,050.0

17.1

3. Internal

12,080.0

67.7

4. External walls

165.8

0.9

5. Roof

838.8

4.7

1,246.0

7.0

17,851.0

100%

1. Sensible transmission through glass

6. Ventilation Total Heat gain per m2 floor area = 127.5 W/m2 Heat gain per m3 space = 28.7 W/m3

Types of Heating And Cooling Coils Fin & Tube Heat Exchangers Fin and tube heat exchangers are used extensively for heating and cooling air. They consist of one or more rows of finned tubes connected to headers and mounted within a sheet metal casing with flanged ends suitable for duct mounting. The heating elements are normally manufactured with copper tubes, with the extended surfaces, or fins, being of aluminium or sometimes copper. The most common type of finning arrangements are the spirally wound and the rectangular fin. Heating coils may be used with hot water or steam as the heat transfer media whilst frost pre-heaters usually have electric heating elements. Cooling coils are classified as being either of the water or the direct expansion type depending on the media flowing through the tubes. A heating coil is shown below.

In water coils, hot or chilled water or brine circulates through the tubes of the coil either emitting or absorbing sensible heat as the air flows over the fins attached to the outside surfaces. Usually the flow of water and air are in opposite directions to each other, this being known as counter-flow heat exchanger. This configuration gives maximum heat transfer. Direct Expansion Coils (Evaporator Coils) In the direct expansion coil (DX), or evaporator, a refrigerant evaporates inside the tubes of the coil, as shown below.

Heat Rejection

Expansion valve.

Compressor

Condenser

Air Off Coil Refrigeration pipework Evaporator

Air On

DX coil as evaporator

REFRIGERATION CYCLE WITH DIRECT EXPANSION COIL Latent heat is absorbed by the air stream from the refrigerant as the refrigerant evaporates. With this type of coil, as with steam, there is no distinction made between parallel and counter-flow since the surface temperature is more uniform owing to the refrigerant in the tubes boiling at a constant temperature. When direct expansion coils are used they become the evaporator of the refrigeration cycle, and may be termed either ‘dry’ or ‘flooded’. In the ‘dry’ DX coil only a sufficient quantity of refrigerant is introduced to operate in the predominantly vapour state. In the ‘flooded’ DX coil most of the coil is filled with liquid refrigerant and although this is more efficient, it is not used so much in air-conditioning since the additional refrigerant is expensive.

Evaporator coils come in a variety of shapes and sizes, depending on the type of installation, the amount of cooling capacity needed, and the manufacturer. They are constructed of aluminium finned copper tubing. The copper tubing runs perpendicular to the aluminium fins, making U-turns back and forth until the desired coil size is achieved. Added cooling capacity without an increase in length and width is accomplished by adding more rows of copper tubing. All evaporator coils must have a drain pan to collect the water that condenses as the air flowing across the coil cools. The water can drain away by gravity or be pumped away. The cooling effect that takes place inside the coil requires a pressure drop in the refrigerant. This drop can be accomplished in a number of ways: capillary tube, piston or orifice, or thermostatic expansion valve. A capillary tube is a thin copper tube of predetermined length into which the compressed liquid refrigerant is pumped. The length of the tubing causes the pressure drop and subsequent cooling effect of the refrigerant. A piston or orifice blocks the flow of refrigerant and forces it through a tiny hole, creating the needed pressure drop. A thermostatic expansion valve meters the flow of refrigerant to meet the cooling demand of the coil. It determines this demand by way of a sensing bulb attached to the outlet tube on the coil. Because it can meter the flow to meet demand, the expansion valve can keep the coil at optimum cooling potential. Because the cooling coil is an integral part of the air distribution system, its geometry — size, number of rows, fin spacing, and fin profile — contributes to the airside pressure drop and affects the sound power level of the fans. (Fan power needed to circulate air through the duct system may warrant extra sound attenuation at the air handler.) Water Removal Moisture in air can condense in the air stream or when the air impinges upon a solid cool surface. This can happen at sharp bends where water collects in a puddle at the lower surface on the ductwork. A drain can be fitted on vertical sections of ductwork to remove water that has collected. Ducts can also be insulated in areas where condensing moisture is likely to occur. In hospitals and other situations water in ducts is to be avoided since bacteria can multiply in warm moist conditions. Droplet separators have been developed to remove water droplet carry-over from cooling coils. They consist of a media that can absorb the water droplets and transport them down through the material to the drainage section. The system shown below uses a glass-fibre-based material and the droplet separator works best with air velocities between 0 and 4 m/s. The media is so efficient that it can arrest 100 litres of water per cubic metre of media. Some droplet separators are produced as cassettes to ensure easy handling. several cassettes can be linked together to achieve the desired surface area.

Heat Transfer in Cooling Coils Chilled-water cooling coils are finned-tube heat exchangers consisting of rows of tubes (usually copper) that pass through sheets of formed fins (usually aluminium). As air passes through the coil and contacts the cold fin surfaces, heat transfers from the air to the water flowing through the tubes. The following equation quantifies the heat-transfer process: Q = U × A × LMTD Where, Q = amount of heat transferred, Btu/hr (W) U = heat-transfer coefficient, Btu/hr • ft² • °F (W/m² • °K) A = effective surface area for heat transfer, ft² (m²) LMTD = log-mean temperature difference across the coil surface, °F (°C) Increasing any one of these variables (heat-transfer coefficient, surface area, or log-mean temperature difference) results in more heat transfer. Arguably the most effective way to improve heat-transfer performance is to increase the log-mean temperature difference (LMTD). In the context of a chilled-water cooling coil, LMTD describes the difference between the temperatures of the air passing across the coil fins and the water flowing through the coil tubes: LMTD = (TD2 – TD1) / ln (TD2 / TD1) Where; TD1 = leaving-air and entering-water temperature difference at the coil (°C) TD2 = entering-air and leaving-water temperature difference at the coil (°C) One way to increase LMTD is to supply the coil with colder water. Heat-transfer coefficient, Q = U × A × LMTD Also called U-value or thermal transmittance value, the heat-transfer coefficient describes the overall rate of heat flow through the coil. Three factors determine this rate:

Airside film coefficient describes the “barrier” (resistance to heat transfer) between the passing air stream and the fin surfaces Waterside film coefficient describes a similar “barrier” between the inside surfaces of the copper tubes and the circulating fluid Thermal conductance describes the rate at which heat flows through the aluminium fins and copper tubes of the coil System designers can do little to affect thermal conductance, but they can alter the film coefficients. Increasing the rate of airflow reduces heat-transfer resistance on the air side of the cooling coil. Likewise, increasing the water velocity reduces the waterside resistance to heat transfer. Fin geometry can improve the overall heat-transfer coefficient, too, by lessening the airside film coefficient. Like velocity, fin geometry can be specified as part of the design of the HVAC system. For comfort-cooling applications, coil fins are usually stamped into waveforms resembling corrugated cardboard. These waveforms create turbulence in the passing air stream, which lessens the resistance to heat transfer. More exaggerated waveforms produce more turbulence. Turbulent water flow, like turbulent airflow, also reduces resistance to heat transfer. And, like fin geometry, it can become an important criterion for coil selection. Waterside turbulence can be created by metal ribbons or helical wires inside the tubes. Called turbulators, these devices create eddies as the water flows across them. Both methods of improving the heat-transfer coefficient (increased velocity and turbulence) create higher pressure drops, which can mean additional fan or pump power. Coil surface area, Q = U × A × LMTD

The third determinant of heat transfer is the coil’s surface area. Typically, fin spacing for comfort heating or cooling ranges from 24 to 50 fins per metre. Spacing the fins closer together multiplies the surface area by permitting more fins per linear unit. Although the airside pressure drop may increase, adding fins extends the available surface area without affecting the overall size of the coil. Adding rows of tubes also increases the heat-transfer surface area. Most coils are constructed with same-end connections, so rows are usually added in pairs. The weight and cost of the coil increase accordingly, but the airside pressure drop may not. (Wider fin spacing often accompanies the decision to add rows.) The best way to extend the surface area for heat transfer is to decrease the face velocity of the coil, that is, face area relative to airflow: face velocity = airflow / face area Face velocity can be reduced in one of two ways: by increasing the size of the coil or (paradoxically) by reducing the required airflow. Selecting a physically larger coil increases the initial investment in the coil and the air handler, and may also enlarge the air-handler footprint ... seldom desirable outcomes. So, how can we reduce the required airflow without sacrificing coil capacity? Improving Coil Performance Lowering the supply air temperature reduces the amount of air required for sensible cooling and saves fan energy. From our review of the heat-transfer equation, we know that: less airflow increases airside film resistance, which reduces heat-transfer coefficient U; and requires colder air, which decreases LMTD.

To compensate for the negative effects on coil performance that accompany less airflow, we must find a way to increase U (heat-transfer coefficient) and/or A (surface area). In other words, we must select a cooling coil with better-than-average heat-transfer characteristics. Increase U Recall that turbulent flow reduces the film resistance to heat transfer. Choosing a fin configuration with a more pronounced waveform and/or adding turbulators inside the coil tubes will improve the heat-transfer coefficient. Increase A Any additional increase in heat-transfer capacity must be achieved by physically increasing the available surface area; that is, by: Adding rows Adding fins Increasing the physical size of the coil (which will increase the initial costs of the coil, air handler, and airside accessories).

PSYCHROMETRY CHAT INTRODUCTION

The aim of this section of the notes is to allow students to size air conditioning plant such as cooling coil, heater battery and humidifier. The notes are divided into several sections as follows: PSYCHROMETRY FOR AIR CONDITIONING THE PSYCHOMETRIC CHART EXAMPLES OF PSYCHOMETRIC PROPERTIES AIR CONDITIONING PLANT FOR SUMMER & WINTER BASIC PROCESSES TYPICAL AIR CONDITIONING PROCESSES ANNOTATION AND ROOM RATIO SUMMER AND WINTER CYCLES EXAMPLES

The first section deals with Psychrometry for air conditioning and discusses some properties of moist air. A simplified psychometric chart is shown for familiarisation, and some examples of how to find air properties are provided. A diagram of an air conditioning system is shown in schematic form in the section entitled AIR CONDITIONING PLANT FOR SUMMER & WINTER. Before sizing takes place the student should also understand the processes that take place in air conditioning systems. There are four basic processes for summer and winter air conditioning systems. The following basic processes are explained: 1. Mixing 2. Sensible Cooling and Heating 3. Cooling with Dehumidification 4. Humidification The section on Typical Air Conditioning Processes shows winter and summer schematic diagrams and psychometric charts. There are some more details that may be useful to the designer of air conditioning systems. Further information is as follows: Annotation Room ratio When the processes have been superimposed onto a psychometric chart then calculations may commence. These are as detailed in the following sections of the notes. Summer and Winter Cycles 1. Summer cycle psychometrics 2. Summer cycle calculations 3. Winter cycle psychometrics 4. Winter cycle calculations 5. Duct and Fan gains.. The final section is seven examples of plant sizing using psychometric charts.

Psychrometry for Air Conditioning Psychrometry is the study of air and water vapour mixtures. Air is made up of five main gases i.e. Nitrogen 78.03%, Oxygen 20.99%, Argon 0.94%, Carbon Dioxide 0.03%, and Hydrogen 0.01% by volume. The Ideal Gas Laws are used to determine psychometric data for air so that the engineer can carry out calculations. To make life easier a chart has been compiled with all the relevant psychometric data indicated. This is called the Psychometric Chart. A typical chart is shown below.

Air at any state point can be plotted on the psychometric chart. The information that can be obtained from a Psychometric Chart is as follows: Dry bulb temperature Wet bulb temperature Moisture content Percentage saturation Specific enthalpy Specific volume.

The following is a brief description of each of the properties of air. 1. Dry bulb temperature This is the air temperature measured by a mercury-in-glass thermometer. 2. Wet bulb temperature This is the air temperature measured by a mercury-in-glass thermometer which has the mercury bulb wetted by gauze that is kept moist by a reservoir of water. When exposed to the environment the moisture evaporates from the wetted gauze, which gives a lower reading on the thermometer. This gives an indication of how ‘dry’ or how ‘moist’ the air is, since in ‘dry’ air the water will evaporate quickly from the gauze, which depresses the thermometer reading. 3. Moisture content This is the amount of moisture in air given in kg of moisture per kg of dry air e.g. for room air at 21oC dry bulb and 15oC wet bulb, the moisture content is about 0.008 kg/kg d.a. This is a small mass of moisture ( 0.008 kg = 8 grams) per kg of dry air or 9.5 grams per cubic metre of air. 4. Percentage saturation The Percentage saturation is another indication of the amount of moisture in air. This is the ratio of the moisture content of moist air to the moisture content of saturated air at the same temperature. When air is saturated it is at 100% saturation and cannot hold any more moisture. 5. Specific enthalpy This is the amount of heat energy (kJ) in air per kg. If heat is added to the air at a heater battery for example, then the amount to be added can be determined from Specific enthalpy change. 6. Specific volume This is the volume of moist air (dry air + water vapour) per unit mass. The units of measurement are m3 per kg. Also specific volume = 1 / density.

THE PSYCHOMETRIC CHART The six properties of air previously discussed can be shown on one chart called a Psychometric Chart. One of the purposes of the Psychometric Chart is to size heater batteries, cooling coils and humidifiers. A simplified Psychometric Chart is shown below.

This chart is only for demonstration purposes. If accurate assessments are to be carried out use a C.I.B.S.E. chart. USING THE PSYCHOMETRIC CHART If any two properties of air are known then the other four can be found from the psychometric chart.

Examples of Psychometric Properties EXAMPLE 1 Find the moisture content of air at 25oC dry-bulb temperature and 25oC wet-bulb temperature. Referring to the chart below, a vertical line is drawn upwards from 25oC dry-bulb temperature until it intersects at 25oC wet-bulb temperature. This intersection point happens to be on the 100% saturation line. The intersection point is highlighted and a horizontal line is drawn to the right to find the corresponding moisture content. The moisture content is therefore 0.020 kg/kg dry air.

EXAMPLE 2 Find the specific volume and wet-bulb temperature of air at 20oC dry-bulb temperature and 50% saturation. Referring to the chart below, a vertical line is drawn upwards from 20oC dry-bulb temperature until it intersects with the 50% saturation curve. The intersection point is sometimes referred to as the state point. The specific volume is found to be 0.84 m3/kg and the wet-bulb temperature is 14oC

EXAMPLE 3 Find the specific volume, percentage saturation and moisture content of air at 15oC dry-bulb temperature and 10oC wet-bulb temperature. Referring to the chart below, a vertical line is drawn upwards from 15oC dry-bulb temperature until it intersects with the 10oC wet-bulb temperature line. This intersection is the state point. The specific volume is found to be 0.823 m3/kg, the percentage saturation 52% and the moisture content 0.0054 kg/kg d.a.

EXAMPLE 4 Find the specific volume, wet-bulb temperature, moisture content and specific enthalpy of air at 35oC drybulb temperature and 30% saturation. Referring to the chart below, a vertical line is drawn upwards from 35oC dry-bulb temperature until it intersects with the 30% saturation curve. This intersection is the state point. The specific volume is found to be 0.883 m3/kg, the wet-bulb temperature is 22oC, the moisture content 0.011kg/kg d.a. and the specific enthalpy 65 kJ/kg.

Air Cond. Plant for Summer and Winter In the summer time when cooling is required by the air conditioning plant it will be necessary to operate the cooling coil, reheater and possibly other plant as well. In winter time the preheater and reheater battery will probably be on to provide warm air to overcome heat losses. Other plant may be switched on as well. These plant items are shown in the diagram below.

The photographs below show some plant items.

Basic Air Conditioning Processes 1. MIXING Where two air streams are mixed the psychometric process is shown as a straight line between two air conditions on the psychometric chart, thus points 1 and 2 are joined and the mix point 3 will lie on this line. Two air streams are mixed in air conditioning when fresh air (m1) is brought in from outside and mixed with recirculated air (m2). The resulting air mixture is shown below as (m3). The mixing ratio is fixed by dampers. Sometimes, in more sophisticated plant, modulating dampers are used which are driven by electric motors to control the mixture of air entering the system. The diagrams below show mixing of two air streams.

By the conservation of mass formula: m1 + m2 = m3 By the conservation of energy formula: m1 h1 + m2 h2 = m3 h3 where: m = mass flow rate of air (kg/s) h = specific enthalpy of air (kJ/kg) found from psychometric chart.

Basic Air Conditioning Processes 2. SENSIBLE COOLING AND HEATING When air is heated or cooled sensibly, that is, when no moisture is added or removed, this process is represented by a horizontal line on a psychometric chart.

For sensible heating: The amount of heating input to the air approximates to; H1-2 = m x Cp x (t2 - t1) Or more accurately from psychometric chart: H1-2 = m x (h2 - h1) For sensible cooling: The amount of cooling input to the air approximates to; H2-1 = m x Cp x (t2 - t1) Or more accurately from psychometric chart: H2-1 = m x (h2 - h1) where: H = Heat or cooling energy (kW) m = mass flow rate of air (kg/s) Cp = Specific heat capacity of air, may be taken as 1.01 kJ/kg degC. t = Dry bulb temperature of air (oC) h = specific enthalpy of air (kJ/kg) found from psychometric chart.

Basic Air Conditioning Processes 3. COOLING WITH DEHUMIDIFICATION The most commonly used method of removing water vapour from air (dehumidification) is to cool the air below its dew point. The dew point of air is when it is fully saturated i.e. at 100% saturation. When air is fully saturated it cannot hold any more moisture in the form of water vapour. If the air is cooled to the dew point air and is still further cooled then moisture will drop out of the air in the form of condensate. This can be shown on a psychometric chart as air sensibly cooled until it becomes fully saturated (the dew point is reached) and then the air is cooled latently to a lower temperature. This is apparent on the psychometric chart as a horizontal line for sensible cooling to the 100% saturation curve and then the process follows the 100% saturation curve down to another point at a lower temperature. This lower temperature is sometimes called the Apparatus dew Point (ADP) of the cooling coil.

In reality the ADP of the cooling coil is close to the cooling liquid temperature inside the coil. Chilled water or refrigerant may be the cooling liquid. The psychometric process from state point 1 to 2 to 3 may be shown as a straight line for simplicity as shown above with a yellow line.

The total amount of cooling input to the air approximates to; H1-3 = m x (h1 - h3) The sensible heat removed is: H1-2 = m x (h1 - h2) The latent heat removed is: H2-3 = m x (h2 - h3) where: H = Cooling energy (kW) m = mass flow rate of air (kg/s) h = specific enthalpy of air (kJ/kg) found from psychometric chart. In the absence of a suitable psychometric chart the following formula may be used; The sensible heat removed is: H1-2 = m x Cp x (t1 - t2) The latent heat removed is: H2-3 = m x hfg x (g2 - g3) where: H = Cooling energy (kW) m = mass flow rate of air (kg/s) Cp = Specific heat capacity of air, may be taken as 1.01 kJ/kg degC. t = Dry bulb temperature of air (oC) hfg = latent heat of evaporisation, may be taken as 2454 kJ/kg @20oC. g = moisture content of air from psychometric chart (kg/kg dry air)

Basic Air Conditioning Processes 3.1 COOLING COIL CONTACT FACTOR Some of the air going through a cooling coil does not come into contact with the tubes or fins of the cooling coil and is therefore not cooled to the ADP temperature. A mixing process therefore takes place as two air streams mix downstream of the cooling coil as shown below.

One air stream is cooled down to the ADP and the other air stream by-passes the coil surfaces to give an offcoil air temperature (mixed air stream) a little higher than the ADP. This may be looked upon as an inefficiency of the coil and is usually given as the cooling coil contact factor. The process is shown on the psychometric chart below.

The contact factor of a cooling coil may be found from;

Another expression for contact factor is;

Basic Air Conditioning Processes 4. HUMIDIFICATION If is it necessary to add some moisture to the supply air then this is best done by injecting steam into the air stream. Humidification can be carried out by spraying a fine mist of water droplets into the air but this is not recommended in rooms occupied by people due to the risk of bacteria carry over. Dry steam may be injected from a steam supply pipe or generated in a local packaged unit as shown in the photograph below. A disadvantage of using an existing steam supply is smells may be carried over into the air.

The steam package unit is situated close to the air duct and is sized to meet the maximum requirements; this is usually in winter in the U.K. A steam pipe ( sometimes hoses are used) passes from the packaged unit to the air duct and steam at 100oC is injected into the air stream via. a sparge pipe. The un-used steam is drained from the system via a condensate tundish and drain. It is important to layout the steam pipework so that any condensate will drain back to the unit. The psychometric process is shown below.

See Summer and Winter Cycles section for calculation of amount of moisture added at humidifier.

TYPICAL AIR CONDITIONING PROCESSES The schematic diagram below shows a typical plant system for summer air conditioning.

The psychometric diagram below shows a typical summer cycle.

The schematic diagram below shows a typical plant system for winter air conditioning.

The psychometric diagram below shows a typical winter cycle.

ANNOTATION The state points on a psychometric chart may be given numbers or symbols to identify them. If symbols are used the following system may be adopted:

ROOM RATIO This is the ratio of sensible to total heat in the room for summer or winter. The total heat gain (summer) or loss (winter) will be determined by adding the Latent and Sensible heat in a room or rooms, i.e. (SUMMER) Total heat gain = Sensible heat gain + Latent heat gain (WINTER) Total heat = Sensible heat loss + Latent heat gain The room ratio is used on a psychometric chart to determine the supply air state point. A room ratio line is superimposed from the protractor on the psychometric chart onto the main body of the chart by a line passing through the room state point R. An example calculation is as follows: Sensible heat gain = 9.0 kW Latent heat gain = 2.25 kW Total heat gain = 9.0 kW + 2.25 kW = 11.25 kW. Room ratio = Sensible / Total heat Room ratio = 9 / 11.25 = 0.8 The supply air state point must also be somewhere on this room ratio line to meet the room heat gain requirements i.e. the room ratio line always passes through points R and S.

Examples of Psychometric Calculations for Summer and Winter Example 1. SUMMER CYCLE A room is to be maintained at 22oC dry-bulb temperature, 50% saturation, when the sensible heat gain is 10.8 kW in summer. The latent heat gain is 7.2 kW. Determine the cooling coil and reheater outputs required by using a psychometric chart if the plant schematic is as shown below. DATA: Outdoor condition is 28oC, 80% saturation. The outdoor air and recirculated air ratio is 20%/80%. The Apparatus Dew Point ADP is 8oC Neglect the cooling coil contact factor.

Example 2. WINTER CYCLE A room has a 18.0 kW sensible heat loss in winter and a 4.5 kW latent heat gain from the occupants. Determine the supply air temperature and heater battery load using the following information. DATA: Indoor condition: 21oC dry-bulb temperature, 50% saturation. Outdoor condition: -2oC d.b., 80% saturation. The outdoor air and recirculated air ratio is 20%/80%. No preheating or humidification takes place in this simplified example.

Procedure (Winter Cycle) 1. Draw schematic diagram of air-conditioning plant (see above) 2. Plot room condition R on psychometric chart. 3. Plot outside condition O on psychometric chart. 4. No Preheater condition P 5. Join points O and R

6. Find the mix point M by measuring the length of the line O-R and multiply this by the mixing ratio. On a full size CIBSE psychometric chart this measures 110mm. The ratio of recirculated air is 0.8. 110mm x 0.8 = 88mm Measure up the O-R line from point O by 88mm. This determines point M . If there is more recirculated air than outside air at the mix point, then point M will be closer to point R than point O. 7. Find the room ratio. This is the sensible to total heat ratio. Neglect signs ie. the total heat for the room will be Sensible loss plus Latent gain. Total heat = 18 kW sensible + 4.5 kW latent = 22.5 kW total. Heat ratio = 18 / 22.5 = 0.8 Plot this ratio on the protractor, top segment, on the psychometric chart and transfer this line onto the chart so that it passes through point R.

8. Find the supply air dry bulb temperature by calculation. 9. Plot the supply air condition S on the room ratio line. This is on a horizontal line from point M to the right hand side of the chart, and intersects with the RRL.

The supply air temperature is found to be 32.5oC.

SUPPLY AIR FLOW RATE When the sensible heat loss and supply air temperature in winter are known then the mass flow rate of air is calculated from the following formula: Hs = ma x Cp ( ts - tr ) where: Hs = Sensible heat loss (kW) ma = mass flow rate of air (kg/s) Cp = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature (oC) ..............therefore: ma = Hs / Cp ( ts - tr ) ma = 18 / 1.01 (32.5 - 21) ma = 18 / 11.615 ma = 1.55 kg/s HEATER BATTERY OUTPUT The heater battery output is as follows: H reheater battery = ma ( hS - hM) where: H reheater battery = Reheater battery output (kW) ma = mass flow rate of air (kg/s) hs = specific enthalpy at condition S (kJ/kg) hM = specific enthalpy at condition M (kJ/kg) The specific enthalpies at points S and M are shown on the psychometric chart below.

Hheater battery = ma ( hS - hM) Hheater battery = 1.55 ( 50 - 34) Hheater battery = 24.8 kW Therefore the heater battery load is 24.8 kW.

Example 3. SUMMER CYCLE (Cooling Coil contact factor) An office is to be maintained at 22oC dry-bulb temperature, 50% saturation in summer. The sensible heat gain is 8.0 kW. The latent heat gain is 2.0 kW. Determine the cooling coil and reheater outputs required by using a psychometric chart if the plant schematic is as shown below. DATA: Outdoor condition is 28oC, 80% saturation. The outdoor air and recirculated air ratio is 20%/80%. The Apparatus Dew Point ADP is 8oC

Procedure (Summer Cycle) 1. Draw schematic diagram of air-conditioning plant (see above) 2. Plot room condition R on psychometric chart. 3. Plot outside condition O on psychometric chart. 4. Join points O and R.

5. Find the mix point M by measuring the length of the line O-R and multiply this by the mixing ratio. On a full size CIBSE psychometric chart this measures 85mm. The ratio of recirculated air is 0.8. 85mm x 0.8 = 68mm Measure down the O-R line from point O by 68mm. This determines point M . 6. Find the room ratio. This is the sensible to total heat gain ratio. Total heat = 8 kW sensible + 2 kW latent = 10 kW total. Heat ratio = 8 / 10 = 0.8 Plot this ratio on the protractor, bottom segment, on the psychometric chart and transfer this line onto the chart so that it passes through point R. 7. Plot the Apparatus Dew Point ADP of the cooling coil. This is on the 100% saturation curve. The ADP is 80C.

8. Join points M and ADP.

9. Find the off-coil condition W by measuring the length of the line M-ADP and multiply this by the cooling coil contact factor.. On a full size CIBSE psychometric chart this measures 75mm. The cooling coil contact factor is 0.8. 75mm x 0.8 = 60mm. Measure down along the line M-ADP by 60mm. This determines point W. 10. Plot the supply air condition S. The reheater process will be a horizontal line from point W to point S. Point S is on the room ratio line. The supply air temperature is 17oC.

MASS FLOW RATE When the supply air temperature has been found from the psychometric chart then the mass flow rate of air can be calculated from the following formula: ma = Hs / ( Cp ( tr - ts ) ) where: Hs = Sensible heat gain to room (kW) ma = mass flow rate of air (kg/s) Cp = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature (oC) The supply air temperature is 17oC. ma = 8 / ( 1.01 ( 22 - 17 ) ) ma = 1.584 kg/s COOLING COIL OUTPUT The cooling coil output is as follows: Hcooling coil = ma ( hM - hADP) where: Hcooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychometric chart

The specific enthalpies at points M and ADP are shown on the psychometric Chart below. Hcooling coil = 1.584 ( 50.5 - 25) Hcooling coil = 40.4 kW HEATER BATTERY OUTPUT The heater battery or reheater output is as follows: Hheater battery = ma ( hS - hW) where: Hheater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychometric chart. hW = specific enthalpy at condition W (kJ/kg) determined from psychometric chart. The specific enthalpies at points S and W are shown on the psychometric Chart below.

Hheater battery = 1.584 ( 36.5 - 30.5) Hheater battery = 9.5 kW

Example 4. WINTER CYCLE (with Humidifier) An conference room is to be maintained at 21oC dry-bulb temperature, 50% saturation in winter. The sensible heat loss for the room is 17.0 kW. The latent heat gain is 40 Watts per person (see Air Conditioning section). Determine the preheater and reheater outputs required and the amount of moisture to be added at the humidifier in litre/hour, by using a psychometric chart if the plant schematic is as shown below. DATA: Outdoor condition is -2oC, 80% saturation. The outdoor air and recirculated air ratio is 50%/50%. Maximum occupancy is 250 people. The preheater off coil temperature is 5oC. Supply air quantity is 8 air changes per hour. Room volume is 20 x 12 x 4m high = 960 m3.

LATENT HEAT GAIN The latent heat gain = heat gain per person x number of people The latent heat gain = 40 W/person x 250 = 10,000 Watts The latent heat gain = 10 kW SUPPLY AIR & FRESH AIR QUANTITIES Supply air quantity (m3/h) = air change rate x room volume (m3) Supply air quantity (m3/h) = 8 x 960(m3) Supply air quantity (m3/h) = 7680 (m3/h) Supply air quantity (m3/s) = 7680(m3/h) / 3600 = 2.13 (m3/s) Supply air mass flow rate (kg/s) = Supply air quantity (m3/s) / Specific Volume (m3/kg) Supply air mass flow rate (kg/s) = 2.13 (m3/s) / 0.87 (m3/kg) Supply air mass flow rate (kg/s) = 2.45 kg/s The fresh air flow rate (kg/s) = 2.45 kg/s x 50% = 1.23 kg/s

WINTER CYCLE PSYCHOMETRICS 1. Draw schematic diagram of air-conditioning plant (see above). 2. Plot room condition O, M and R on psychometric chart. 3. Plot the after Preheater condition P. The Preheater process will be a horizontal line from O to P and acts as a frost coil in this case, heating the air to 5oC.

4. Join points P and R. 5. Find the mix point M by measuring the length of the line P-R and multiply this by the mixing ratio. The line measures 82mm long. 82 x 0.5 = 41 mm 6. Find the room ratio. Plot this ratio on the protractor, so that it passes through point R. Total heat is 17 kW sensible + 10 kW latent = 27kW. Ratio is 17/27 = 0.63.

Find the supply air dry bulb temperature by calculation. This is found by calculation because we have already calculated the mass flow rate of supply air from information given in the question. SUPPLY AIR DRY BULB TEMPERATURE The temperature of supply air is calculated from the following formula: Hs = ma x Cp ( ts - tr ) where: Hs = Sensible heat loss from room (kW) ma = mass flow rate of air (kg/s) Cp = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature (oC) ( ts - tr ) = Hs / ma x Cp ( ts - tr ) = 17 / 2.45 x 1.01 ( ts - tr ) = 6.87 deg.C ( ts - 21) = 6.87 deg.C ts = 21 + 6.87 deg.C ts = 27.87 oC say 28 oC. 8. Plot the supply air condition S on the room ratio line. 9. Plot condition H on the psychometric chart. This is vertically down from point S, and horizontally across from point M. This is because M-H is the reheater process and thus a horizontal line and H-S is the humidification process and is close to a vertical line if steam is used.

PREHEATER BATTERY OUTPUT (or frost coil) The preheater battery output is as follows: H preheater battery = maf ( hP - hO) where: H preheater battery = Preheater battery output (kW) m af = mass flow rate of fresh air (kg/s) hP = specific enthalpy at condition P (kJ/kg) hO = specific enthalpy at condition O (kJ/kg) Hpreheater battery = 1.23 ( 12 - 5.5) Hpreheater battery = 8.0 kW REHEATER BATTERY OUTPUT The reheater battery output is as follows: Hreheater battery = ma ( hH - hM) where: H reheater battery = Reheater battery output (kW) ma = mass flow rate of supply air (kg/s) hH = specific enthalpy at condition H (kJ/kg) hM = specific enthalpy at condition M (kJ/kg) Hreheater battery = 2.45 ( 42 - 27) Hreheater battery = 36.8 kW HUMIDIFIER OUTPUT The amount of moisture added to the air may be calculated from the following formula: m moisture added = ma (msS - msH) where: m moisture added = The amount of moisture or added or steam flow rate (kg/s) m a = mass flow rate of air (kg/s) m sS = moisture content at condition S (kg/kg d.a.) m sH = moisture content at condition H (kg/kg d.a.) m moisture added = 2.45 (0.0064 - 0.0054) m moisture added = 2.45 (0.001) m moisture added = 0.00245 kg/s 1 litre of water weights 1 kg, therefore; m moisture added = 0.00245 litre/s m moisture added = 0.00245 litre/s x 3600 = 8.82 litres/hour

Example 5 SUMMER CYCLE (air flows to be calculated) A Lecture Theatre measures 15 m x 10 m x 6 m high. It is to be air conditioned in summer so that the room is maintained at 22oC dry-bulb temperature, 50% saturation. Determine the cooling coil and reheater outputs required by using a psychometric chart if the plant schematic is as shown below. DATA: Outdoor condition is 28oC, 80% saturation. The Apparatus Dew Point ADP is 7.5oC. The latent heat gain is 10.0 kW. The sensible heat gain is 12.0 kW. Maximum occupancy is 200 people. The cooling coil contact factor is unknown at present and should be calculated. Use CIBSE guide Table B2.2 and B2.3 to determine air flow rates and calculate the mass flow rate of fresh air and supply air to the room. The room is a non-smoking area.

MASS FLOW RATES Information from Table B2.2 is as follows (see Ventilation section); The recommended outdoor air rate is 8 l/s/person for non-smoking. Information from Table B2.3 is as follows (see Ventilation section); The recommended total air supply rate is 6 – 10 air changes per hour. FRESH AIR FLOW RATE Fresh air rate = 8 l/s/p x 200 people = 1600 l/s = 1.6 m3/s The specific volume at the outside condition may be determined from a psychometric chart. It is 0.88 m3/kg. Mass flow rate = Volume flow rate / specific volume Mass flow rate (Fresh Air) = 1.6 / 0.88 = 1.82 kg/s . SUPPLY AIR FLOW RATE If the maximum ventilation supply air rate is taken from table B2.3 to be 10.0 air changes per hour, then the mass flow rate can be calculated. Volume flow rate (m3/h) = Volume of room (m3) x air change rate (ac/h) Volume of room (m3) = 15 x 10 x 6 = 900 m3 Volume flow rate (m3/h) = 900 (m3) x 10 (ac/h) Volume flow rate (m3/h) = 9000 m3/h Volume flow rate (m3/s) = 9000 / 3600 = 2.5 m3/s. Mass flow rate = Volume flow rate / specific volume

The specific volume at the supply condition may be approximated at this stage from a psychometric chart. It is 0.834 m3/kg. Mass flow rate (Supply Air) = 2.5 / 0.834 = 3.0 kg/s. FRESH AIR AND RECIRC. RATIO. The ratio by mass is therefore; Fresh air rate = 1.82 kg/s Supply air rate = 3.00 kg/s Recirculation air rate = 3.00 - 1.82 = 1.18 kg/s The ratio of fresh air to total supply air is; 1.82 / 3.00 = 0.6, i.e. 60% fresh air and therefore 40% recirculated air. It is not unusual to have a high percentage of fresh air in a high occupancy room such as a Lecture theatre. The air flows are shown on the schematic diagram below.

SUPPLY AIR TEMPERATURE BY CALCULATION In this example the supply air temperature will be found by rearranging the following formula: Hs = ma x Cp ( tr - ts ) where: Hs = Sensible heat gain to room (kW) ma = mass flow rate of air (kg/s) Cp = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature (oC) Rearranging the above formula gives: ( tr - ts ) = Hs / ( ma x Cp ) ( tr - ts ) = 12 / ( 3.00 x 1.01 ) ( tr - ts ) = 3.96 deg.C since tr = 22oC ts = 22 - 3.96 = 18.04 oC ts = 18 oC approx. The processes can now be plotted on a psychometric chart as shown below.

1. Points O, M and R can be shown on the chart. 2. Point ADP can be indicated and lines drawn between these points as shown. 3. The room ratio line can be drawn. 4. Point S is then shown on the chart, on the room ratio line at 18oC. 5. A horizontal line is then drawn from point S towards the line O – ADP. 6. Point W can then be found where the horizontal line W - S intersects the line O - ADP. From the psychometric chart point W is at approximately 9oC dry bulb. The heat ratio is 12 kW sensible / 22 kW total = 0.545.

The specific enthalpies are shown below. COOLING COIL OUTPUT The cooling coil output is as follows: Hcooling coil = ma ( hM - hADP)

where: Hcooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychometric chart The specific enthalpies at points M and ADP are shown on the psychometric Chart above. Hcooling coil = 3.00 ( 64 - 24) Hcooling coil = 120.0 kW REHEATER BATTERY OUTPUT The heater battery or reheater output is as follows: Hheater battery = ma ( hS - hW) where: Hheater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychometric chart. hW = specific enthalpy at condition W (kJ/kg) determined from psychometric chart. The specific enthalpies at points W and S are shown on the psychometric Chart above. Hheater battery = 3.00 ( 36 - 26.5) Hheater battery = 28.5 kW Example 6 SUMMER CYCLE AND WINTER CYCLE A Concert Hall measures 40 m x 20 m x 8 m high. It is to be air conditioned in summer and winter. Determine the following: Air flow rates Supply air temperature by calculation in summer and winter. Cooling coil contact factor. Cooling coil and reheater outputs in summer Frost coil and heater battery output in winter. Humidifier output in litres/hour. The plant schematic is as shown below.

DATA: Indoor condition all year - 22oC dB temperature, 50% saturation. Outdoor condition summer - 28oC dB temperature, 80% saturation. Outdoor condition winter - -3oC dB temperature, 80% saturation. Maximum occupancy - 1000 people. ADP of the cooling coil - 8oC Fresh air requirement - 12 l/s/person Supply air rate - 8 air changes per hour. Frost off coil temperature - 7oC. Latent heat gain - 40 W/person Sensible heat gain in summer - 100 W/person + 20.5 kW fabric, lights, solar & ventilation gains. Sensible heat loss in winter - 20 W/m3 air volume (estimated) FRESH AIR FLOW RATE Fresh air rate = 12 l/s/p x 1000 people = 12,000 l/s = 12.0 m3/s The specific volume at the outside condition may be determined from a psychometric chart. It is approximately 0.88 m3/kg. Mass flow rate = Volume flow rate / specific volume Mass flow rate (Fresh Air) = 12.0 / 0.88 = 13.64 kg/s. SUPPLY AIR FLOW RATE The ventilation supply air rate is taken from table B2.3 and is given as 8 air changes per hour; the mass flow rate can be calculated. Volume flow rate (m3/h) = Volume of room (m3) x air change rate (ac/h) Volume of room (m3) = 40 x 20 x 8 = 6400 m3 Volume flow rate (m3/h) = 6400 (m3) x 8 (ac/h) Volume flow rate (m3/h) = 51,200 m3/h Volume flow rate (m3/s) = 51,200 / 3600 = 14.22 m3/s. Mass flow rate = Volume flow rate / specific volume The specific volume at the supply condition may be approximated at this stage from a psychometric chart. It can be taken as 0.834 m3/kg. Mass flow rate (Supply Air) = 14.22 / 0.834 = 17.05 kg/s. RECIRCULATION AIR FLOW RATE Recirculation air rate = Supply air rate - fresh air rate Recirculation air rate = 17.05 - 13.64 = 3.41 kg/s FRESH AIR AND RECIRC. RATIO. The ratio by mass is therefore; Fresh air rate = 13.64 kg/s Supply air rate = 17.05 kg/s The ratio of fresh air to total supply air is; 13.64 / 17.05 = 0.8, i.e. 80% fresh air and therefore 20% recirculated air. It is not unusual to have a high percentage of fresh air in a high occupancy room such as a Concert Hall. The air flows are shown on the schematic diagram below.

HEAT GAINS The heat gains are given as: Latent heat gain - 40 W/person Sensible heat gain - 100 W/person + 20.5 kW fabric, lights, solar & ventilation gains. Latent gain = 40 W x 1000 people = 40,000 W = 40 kW Sensible gain = 100 W x 1000 people = 100,000 W = 100 kW + 20.5 kW = 120.5 kW. Total heat gain = 120.5 + 40 = 160.5 kW Summer Heat ratio = sensible / total = 120.5 / 160.5 = 0.75 HEAT LOSS The heat loss in winter is given as 20.0 W/ m3. Volume of room (m3) = 40 x 20 x 8 = 6400 m3 Total heat loss = 20 x 6400 = 128,000 Watts = 128 kW Total heat (winter) = 128 + 40 (Latent) = 168 kW Winter Heat ratio = sensible / total = 128 / 168 = 0.76 SUMMER CALCULATIONS & PROCESS SUPPLY AIR TEMPERATURE BY CALCULATION In this example the supply air temperature will be found by rearranging the following formula: Hs = ma x Cp ( tr - ts ) where: H s = Sensible heat gain to room (kW) ma = mass flow rate of air (kg/s) Cp = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature (oC) Rearranging the above formula gives: ( tr - ts ) = Hs / ( ma x Cp ) ( tr - ts ) = 120.5 / ( 17.05 x 1.01 ) ( tr - ts ) = 7.0 deg.C, since tr = 22oC, ts = 22 - 7.0 = 15.0 oC ts = 15 oC The processes can now be plotted on a psychometric chart as shown below. 1. Points O, M and R can be shown on the chart. 2. Point ADP can be indicated and lines drawn between these points as shown. 3. The room ratio line can be drawn. 4. Point S is then shown on the chart, on the room ratio line at 18oC. 5. A horizontal line is then drawn from point S towards the line O – ADP. 6. Point W can then be found where the horizontal line W - S intersects the line O - ADP.

COOLING COIL CONTACT FACTOR On a full size psychometric chart the length of the line from point M to point ADP is 116mm. The distance from point M to point W is 107mm. The cooling coil contact factor is therefore:

The specific enthalpies are shown below.

COOLING COIL OUTPUT in Summer The cooling coil output is as follows: H cooling coil = ma ( hM - hADP) where: H cooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s)

hM = specific enthalpy at condition M (kJ/kg) determined from psychometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychometric chart The specific enthalpies at points M and ADP are shown on the psychometric Chart above. H cooling coil = 17.05 ( 71 - 25) H cooling coil = 784.3 kW NOTE: The cooling coil output is very high and a lot of energy would be required to provide this amount of cooling. The cooling coil load would probably be spread over several air handling units but the it could be examined with a view to some reduction. The coil output is high because the mass flow rate of supply air is high (17.05 kg/s) and the proportion of fresh air is also high (80%). The mix point M is at approximately 27oC dry-bulb so there is little advantage in recirculation in this instance. It would be advantageous to consider the supply airflow rate to see if a lower rate would be acceptable for this building. If 6 air changes per hour are used as the ventilation rate then this would reduce the mass flow rate of supply air. Also the engineer may consider other methods of air-conditioning a hall with a large volume such as using partial radiant cooling where surfaces are cooled rather than air.

REHEATER BATTERY OUTPUT in Summer The heater battery or reheater output is as follows: H heater battery = ma ( hS - hW) where: H heater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychometric chart. hW = specific enthalpy at condition W (kJ/kg) determined from psychometric chart. The specific enthalpies at points W and S are shown on the psychometric Chart above. H heater battery = 17.05 ( 34 - 28) H heater battery = 102.3 kW WINTER CALCULATIONS & PROCESS WINTER CYCLE PSYCHOMETRICS The processes can now be plotted on a psychometric chart as shown below. 1. Points O, M, P and R can be shown on the chart. 2. Join points O and P and P and R. 3. Find the mix point M The line measures 81mm long. 81 x 0.8 = 65 mm , the distance from point M to R is 65mm. 4. The room ratio is 0.76 from previous calculation. Draw RRL.

5. Find the supply air dry bulb temperature by calculation.

SUPPLY AIR TEMPERATURE BY CALCULATION The temperature of supply air is calculated from the following formula: Hs = ma x Cp ( ts - tr ) where: Hs = Sensible heat loss from room (kW) ma = mass flow rate of air (kg/s) same as summer rate for constant volume systems. Cp = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature(oC) ( ts - tr ) = Hs / ma x Cp ( ts - tr ) = 128 / 17.05 x 1.01 ( ts - tr ) = 7.43 deg.C ts = 22 + 7.43 deg.C ts = 29.43 oC say 29.5 oC. Plot the condition H at 29.5oC dB. on a horizontal line from M. Plot condition S on a vertical line from H on the RRL. Assume the humidity process is vertical.

PREHEATER BATTERY OUTPUT (or frost coil)

The preheater battery output is as follows: H preheater battery = maf ( hP - hO) where: H preheater battery = Preheater battery output (kW) maf = mass flow rate of fresh air (kg/s) hP = specific enthalpy at condition P (kJ/kg) hO = specific enthalpy at condition O (kJ/kg) H preheater battery = 13.64 ( 13 - 3) H preheater battery = 136.4 kW REHEATER BATTERY OUTPUT in Winter The reheater battery output is as follows: H reheater battery = ma ( hH - hM) where: H reheater battery = Reheater battery output (kW) ma = mass flow rate of supply air (kg/s) hH = specific enthalpy at condition H (kJ/kg) hM = specific enthalpy at condition M (kJ/kg) H reheater battery = 17.05 ( 39 - 19) H reheater battery = 341 kW NOTE: The heater battery load may be reduced by using other forms of heating for some of the load, e.g. perimeter convectors or radiators. HUMIDIFIER OUTPUT in Winter The amount of moisture added to the air may be calculated from the following formula: M moisture added = ma (msS - msH) where: m moisture added = The amount of moisture or added or steam flow rate (kg/s) ma = mass flow rate of air (kg/s) msS = moisture content at condition S (kg/kg d.a.) msH = moisture content at condition H (kg/kg d.a.) m moisture added = 17.05 (0.0074 - 0.0035) m moisture added = 17.05 (0.0039) m moisture added = 0.0665 kg/s m moisture added = 0.0665 litre/s m moisture added = 0.0665 litre/s x 3600 = 239.4 litres/hour

Example 7 SUMMER CYCLE (Duct, fan gains) A Computer Suite Theatre measures 24 m x 10 m x 3.5 m high. It is to be air conditioned in summer so that the room is maintained at 22oC dry-bulb temperature, 50% saturation. Determine the cooling coil and reheater outputs required. The air conditioning system is shown schematically below. DATA: Outdoor condition is 27oC, 80% saturation. The Apparatus Dew Point ADP is 8oC The internal latent heat gain is 40 W per person, plus additional gain of 5 kW. The internal sensible heat gain is 200 Watts per computer, 100W per person and 15 W/m2 floor area for lights. The solar gain through windows is 6.0 kW Maximum occupancy is 80 people. Number of computers is 80.

The cooling coil contact factor is 0.8. Duct and fans gains are 2oC. Fresh air, recirculated air ratio is 20%/80%.

HEAT GAINS Sensible - (200 x 80) + (100 x 80) + ( 15 x 24 x 10) = 27,600 Watts = 27.6 kW + solar gain 5.0 kW = 32.6 kW Latent - (40 x 80) = 3200 Watts = 3.2 kW + other 5kW = 8.2 kW Total heat gain = 32.6 + 8.2 = 40.8 kW Room ratio = 32.6 / 40.8 = 0.8 MASS FLOW RATE When the supply air temperature has been found from the psychometric chart then the mass flow rate of air to offset heat gains can be calculated from the following formula: ma = Hs / ( Cp ( tr - ts ) ) where: H s = Sensible heat gain to room (kW) m a = mass flow rate of air (kg/s) C p = Specific heat capacity of humid air (approx.1.01 kJ/kg degC) tr = room temperature (oC) ts = supply air temperature (oC) The supply air temperature is 14.5oC. ma = 32.60 / ( 1.01 ( 22 - 14.5 ) ) ma = 4.3 kg/s The processes can now be plotted on a psychometric chart as shown below. From the psychometric chart point W is at approximately 11oC dry bulb. Point D is 11 oC + 2oC (duct and fan gains given in Data) = 13 oC

COOLING COIL OUTPUT The cooling coil output is as follows: Hcooling coil = ma ( hM - hADP) where: Hcooling coil = Cooling coil output (kW) ma = mass flow rate of air (kg/s) hM = specific enthalpy at condition M (kJ/kg) determined from psychometric chart. hADP = specific enthalpy at condition ADP (kJ/kg) determined from psychometric chart The specific enthalpies at points M and ADP are shown on the psychometric Chart above. Hcooling coil = 4.3 ( 49.5 - 25) Hcooling coil = 105.4 kW

REHEATER BATTERY OUTPUT The heater battery or reheater output is as follows: Hheater battery = ma ( hS - hD) where: Hheater battery = Heater battery output (kW) ma = mass flow rate of air (kg/s) hS = specific enthalpy at condition S (kJ/kg) determined from psychometric chart. hD = specific enthalpy at condition D (kJ/kg) determined from psychometric chart. The specific enthalpies at points S and D are shown on the psychometric Chart above. Hheater battery = 4.3 ( 33.5 - 32) Hheater battery = 6.5 kW