Refrigeration and Air Conditioning 3 E Part 9 doc

Convective heat loss will depend on the area of skin exposed, theair speed, and the temperature difference between the skin and the Figure 23.5 Basic CIBSE psychrometric chart Courtesy o

234 Refrigeration and Air-Conditioning

A further property which is shown on the psychrometric chart isthe specific volume of the mixture, measured in cubic metres perkilogram This appears as a series of diagonal lines, at intervals of0.01 m3

The human body takes in chemical energy as food and drink, andoxygen, and consumes these to provide the energy of the metabolism.Some mechanical work may be done, but the greater proportion isliberated as heat, at a rate between 90 W when resting and 440 Wwhen doing heavy work

A little of this is lost by radiation if the surrounding surfaces arecold and some as sensible heat, by convection from the skin Theremainder is taken up as latent heat of moisture from the respiratorytissues and perspiration from the skin (see Table 23.2) Radiant losswill be very small if the subject is clothed, and is ignored in thistable

Convective heat loss will depend on the area of skin exposed, theair speed, and the temperature difference between the skin and the

Figure 23.5 Basic CIBSE psychrometric chart (Courtesy of theChartered Institution of Building Services Engineers)

Figure 23.6

CIBSE PSYCHROMETRIC

Based on a barometric pressure of 101.325 kPa

Sensible/total heat ratio for water added at 30

236 Refrigeration and Air-Conditioning

ambient As the dry bulb approaches body temperature (36.9°C)the possible convective loss will diminish to zero At the same time,loss by latent heat must increase to keep the body cooled This, too,must diminish to zero when the wet bulb reaches 36.9°C

In practice, the human body can exist in dry bulb temperatureswell above blood temperature, providing the wet bulb is low enough

to permit evaporation The limiting factor is therefore one of wetbulb rather than dry bulb temperature, and the closer the upperlimits are approached, the less heat can be rejected and so the lesswork can be done

Figure 23.8 shows the maximum climatic conditions in differentareas of the world The humid tropical zones have high humiditiesbut the dry bulb rarely exceeds 35°C The deserts have an arid

Figure 23.7 Reading the CIBSE psychrometric chart

% saturation

238 Refrigeration and Air-Conditioning

climate, with higher dry bulb temperatures Approximate limits forhuman activities are related to the enthalpy lines and indicate theability of the ambient air to carry away the 90–440 W of body heat.The opposite effect will take place at the colder end of the scale.Evaporative and convective loss will take place much more easilyand the loss by radiation may become significant, removing heatfaster than the body can generate it The rate of heat productioncan be increased by greater bodily activity, but this cannot be sustained,

so losses must be prevented by thicker insulation against convectiveloss and reduced skin exposure in the form of more clothing Thebody itself can compensate by closing sweat pores and reducing theskin temperature

A total assessment of bodily comfort must take into account changes

in convective heat transfer arising from air velocity, and the effects

of radiant heat gain or loss These effects have been quantified inseveral objective formulas, to give equivalent, corrected effective,globe, dry resultant and environmental temperatures, all of whichgive fairly close agreement This more complex approach is requiredwhere air speeds may be high, there is exposure to hot or coldsurfaces, or other special conditions call for particular care

0.030

0.028

0.026

0.024 0.023 0.022

0.020 0.019 0.018

0.016

0.014 0.013 0.012 0.011 0.010 0.009

0.007 0.006 0.005

0.003 0.002 0.001 0.000

Approximate lethal limit

Bahrain

90 80 70 60 50 40 30 20 Percentage saturation

Acute distress Hong Kong

Eliat

W o

rk becomes difficultImpaired efficiency

New Yor k

Lisbon

Too w ar

For comfort in normal office or residential occupation, withpercentage saturations between 35 and 70%, control of the drybulb will result in comfortable conditions for most persons Feelings

of personal comfort are as variable as human nature and at any onetime 10% of the occupants of a space may feel too hot and 10% toocold, while the 80% majority are comfortable Such variationsfrequently arise from lack or excess of local air movement, orproximity to cold windows, rather than an extreme of temperature

or moisture content

Occupied spaces need a supply of outside air to provide oxygen,remove respired carbon dioxide, and dilute body odours and tobaccosmoke The quantities are laid down by local regulations andcommonly call for 6–8 litre/s per occupant Such buildings areusually required also to have mechanical extract ventilation fromtoilets and some service areas, so the fresh air supply must make upfor this loss, together with providing a small excess to pressurize thebuilding against ingress of dirt [2]

24 Air treatment cycles

Buildings lose heat in winter by conduction out through the fabric,convection of cold air, and some radiation The air from the con-ditioning system must be blown into the spaces warmer than therequired internal condition, to provide the heat to counteract thisloss

Heating methods are as follows:

1 Hot water or steam coils

2 Direct-fired – gas and sometimes oil

3 Electric resistance elements

4 Refrigerant condenser coils of heat pump or heat reclaim systemsFigure 24.1 shows the sensible heating of air

heated from 16°C to 34°C Calculate the heat input and the watermass flow for an air heater coil having hot water entering at 85°Cand leaving at 74°C

4.187 × (85 – 74) = 27 kg/s

Note: the 1.02 here is a general figure for the specific heat capacity

of indoor air which contains some moisture, and is used in preference

to 1.006, which is for dry air

the heater coil at 19°C at the rate of 68 kg/s What is the air-supplytemperature?

Figure 24.1 Sensible heating of air

5 0

ulb temper ature (

68 × 7.35 ~ 500 kW

Air entering the conditioning plant will probably be a mixture ofreturn air from the conditioned space and outside air Since noheat or moisture is gained or lost in mixing,

Sensible heat before = sensible heat after

and

242 Refrigeration and Air-Conditioning

Latent heat before = latent heat after

The conditions after mixing can be calculated, but can also be

shown graphically by a mix line joining the condition A and B (see Figure 24.2) The position C along the line will be such that

15

10 5 0

Figure 24.2 Mixing of two airstreams

AC × m a = CB × m b

This straight-line proportioning holds good to close limits of accuracy.The horizontal divisions of dry bulb temperature are almost evenlyspaced, so indicating sensible heat The vertical intervals of moisturecontent indicate latent heat

saturation, and a mass flow of 20 kg/s, mixes with outside air at

28°C dry bulb and 20°C wet bulb, flowing at 3 kg/s What is thecondition of the mixture?

Method (a) Construct on the psychrometric chart as shown in Figure

24.2 and measure off:

Answer = 22°C dry bulb, 49% sat

Method (b) By calculation, using dry bulb temperatures along the

horizontal component, and moisture content along thevertical For the dry bulb, using

AC × m a = CB × m b (t c – 21) × 20 = (28 – t c) × 3

giving

t c = 21.9°CThe moisture content figures, from the chart or from tables, are0.0079 and 0.0111 kg/kg at the return and outside conditions, so

to computing

If air at 21°C dry bulb, 50% saturation, is brought into contact with

a surface at 12°C, it will give up some of its heat by convection Thecold surface is warmer than the dew point, so no condensation willtake place, and cooling will be sensible only (Figure 24.3)

This process is shown as a horizontal line on the chart, sincethere is no change in the moisture content The loss of sensibleheat can be read off the chart in terms of enthalpy, or calculatedfrom the dry bulb reduction, considering the drop in the sensibleheat of both the dry air and the water vapour in it

The effect of spraying water into an airstream will be as shown inFigure 23.2, assuming that the air is not already saturated Evaporation

244 Refrigeration and Air-Conditioning

will take place and the water will draw its latent heat from the air,reducing the sensible heat and therefore the dry bulb temperature

of the air (Figure 24.4)

50% saturation What would be the ultimate condition of the mixture?

No heat is being added or removed in this process, so the enthalpymust remain constant, and the process is shown as a movementalong the line of constant enthalpy Latent heat will be taken in bythe water, from the sensible heat of the air, until the mixture reachessaturation, when no more water can be evaporated

Initial enthalpy of air = 41.08 kJ/kg

Final enthalpy of air = 41.08 kJ/kg

Final condition, 14.6°C dry bulb, 14.6°C wet bulb, 14.6°C dewpoint, 100% saturated

It should be noted that this ultimate condition is difficult to reach,and the final condition in a practical process would fall somewhat

Figure 24.3 Sensible cooling of air

ulb temper ature (

short of saturation, possibly to point C in Figure 24.5 The proportion

AC/AB is termed the effectiveness of the spray system.

Figure 24.4 Adiabatic saturation to ultimate condition

Figure 24.5 Adiabatic saturation – process line

The adiabatic (constant enthalpy) line AC is almost parallel to

the line of constant wet bulb Had the latter been used, the finalerror would have been about 0.2 K, and it is sometimes convenientand quicker to calculate on the basis of constant wet bulb (This

14.6 ° C dry bulb 100% sat.

40

80

5 0

246 Refrigeration and Air-Conditioning

correlation applies only to the mixture of dry air and water vapour,and not to other gas mixtures.)

Moisture can be added to air by injecting steam, i.e water which isalready in vapour form and does not require the addition of latentheat (Figure 24.6) Under these conditions, the air will not be cooledand will stay at about the same dry bulb temperature The steamwill be at 100°C when released to the atmosphere (or may be slightlysuperheated), and so raises the final temperature of the mixture

dry bulb, 50% saturation, at the rate of 1 kg steam/150 kg dry air.What is the final condition?

Moisture content of air before = 0.0079 kg/kg

Moisture added, 1 kg/150 kg = 0.0067 kg/kg

Final moisture content = 0.0148 kg/kg

Figure 24.6 Addition of steam to air

ulb temper ature (

An approximate figure for the final dry bulb temperature can beobtained, using the specific heat capacity of the steam through therange 20–100°C, which is about 1.972 kJ/kg This gives

Heat lost by steam = heat gained by air

0.0067 × 1.972(100 – t) = 1.006(t – 21)

giving

t = 22.02°CWhere steam is used to raise the humidity slightly, the increase indry bulb temperature is small

The process of adiabatic saturation in Section 24.4 assumed thatthe spray water temperature had no effect on the final air condition

If, however, a large mass of water is used in comparison with themass of air, the final condition will approach the water temperature

If this water is chilled below the dew point of the entering air,moisture will condense out of the air, and it will leave the washerwith a lower moisture content (see Figure 24.7)

The ultimate condition will be at the initial water temperature B Practical saturation efficiencies (the ratio AC/AB) will be about 50–

80% for air washers having a single bank of sprays and 80–95% fordouble spray banks (see Figure 24.8)

single-bank air washer having a saturation efficiency of 70% and is sprayedwith water at 5°C What is the final condition?

(a) By construction on the chart (Figure 24.7), the final condition

is 10.4°C dry bulb, 82% saturation

water for every 1 kg air What is the water temperature rise?

248 Refrigeration and Air-Conditioning

Figure 24.7 Air washer with chilled water

Figure 24.8 Chilled water spray

Enthalpy of air before = 45.79 kJ/kg

Enthalpy of air after = 26.7 kJ/kg

Heat lost per kilogram air = 19.09 kJ

C 5° C

Coolant

Spray pump

Heat gain per kilogram water = 19.09/4

= 4.77 kJ

Temperature rise of water =

4.774.187

= 1.1 K

In the previous process, air was cooled by close contact with a waterspray No water was evaporated, in fact some was condensed, becausethe water was colder than the dew point of the entering air

A similar effect occurs if the air is brought into contact with asolid surface, maintained at a temperature below its dew point.Sensible heat will be transferred to the surface by convection andcondensation of water vapour will take place at the same time Boththe sensible and latent heats must be conducted through the solidand removed The simplest form is a metal tube, and the heat iscarried away by refrigerant or a chilled fluid within the pipes Thiscoolant must be colder than the tube surface to transfer the heatinwards through the metal

The process is indicated on the chart in Figure 24.9, taking point

B as the tube temperature Since this would be the ultimate dew

point temperature of the air for an infinitely sized coil, the point B

is termed the apparatus dew point (ADP) In practice, the coolingelement will be made of tubes, probably with extended outer surface

in the form of fins (see Figure 7.3) Heat transfer from the air tothe coolant will vary with the fin height from the tube wall, thematerials, and any changes in the coolant temperature which maynot be constant The average coolant temperature will be at some

lower point D, and the temperature difference B – D will be a function of the conductivity of the coil As air at condition A enters

the coil, a thin layer will come into contact with the fin surface and

will be cooled to B It will then mix with the remainder of the air between the fins, so that the line AB is a mix line.

The process line AB is shown here as a straight line for convenience

of working Analysis of the air as it passes through a cooling coilshows the line to be a slight curve

The proportion AC/AB is termed the coil contact factor The proportion CB/AB is sometimes used, and is termed the bypass

factor

250 Refrigeration and Air-Conditioning

a coil having an ADP of 7°C and a contact factor of 78% What is theoff-coil condition?

(a) By construction on the chart (Figure 24.9), 10.7°C dry bulb,85% saturation

(b) By calculation, the dry bulb will drop 78% of 24 to 7°C:

dry bulb, 52% saturation, to 15°C dry bulb, 80% saturation What

is the ADP?

This must be done by construction on the chart, and gives anADP of 9°C The intersection of the process and saturation linescan also be computed Again, it has been assumed that the processline is straight

Figure 24.9 Cooling and dehumidifying coil – process line

24.8 Sensible–latent ratio

In all cases the horizontal component of the process line is thechange of sensible heat, and the vertical component gives the latentheat It follows that the slope of the line shows the ratio betweenthem, and the angle, if measured, can be used to give the ratio ofsensible to latent to total heat On the psychrometric chart in generaluse (Figure 23.5), the ratio of sensible to total heat is indicated asangles in a segment to one side of the chart This can be used as aguide to coil and plant selection

The sensible heat to be removed is 36 kW and the latent 14 kW.What are the ADP and the coil contact factor if air is to leave thecoil at 5°C?

Plotting on the chart (Figure 24.10) from 23°C/40% and usingthe ratio

ulb temper ature (