BASIC PRINCIPLES

Psychrometrics is the science of the physicalproperties of air and its associated water vapor. AllHVACR professionals must have a good command of the basic principles that support this science. Weselect, install, and service mechanical systems thatare designed to manipulate the physical properties of air for the purpose of controlling its temperature,humidity level, and cleanliness. This is done toprovide improved comfort within a conditioned space,as well as for other reasons—to protect constructionmaterials used in the building, for example, or topreserve food products. The following paragraphsreview eight physical properties of air:

þ dry-bulb temperatureþ wet-bulb temperatureþ relative humidityþ specific humidityþ dew pointþ enthalpyþ vapor pressureþ density.

The physical property of air most commonlydescribed by customers is dry-bulb temperature.The temperature value is measured in degrees, either Celsius or Fahrenheit. Heat intensity, not heat quantity, is measured in units of temperature.Warmer molecules vibrate more rapidly than docooler molecules, and this rate of vibration is calledheat intensity. However, a temperature value does not tell you the quantity of heat present in a system.Heat quantity is measured in British thermal units, or Btu. The temperature value of an air sample isonly one measurement of several that must be takento calculate the heat quantity present in the sample.

Energy used to change air temperature is defined assensible heat.

The next most common physical property of airdescribed by customers is relative humidity.Complaints about humidity range from air being toodamp to air being too dry. Relative humidity (RH) isa value used to communicate the ratio of water vaporin an air sample to the maximum amount of watervapor that the air sample can hold at a given dry-bulb temperature. The vapor pressure of the airsample is compared to the vapor pressure of thatsame air sample when saturated with water vapor.The resulting ratio is converted to a percentage thatwe call relative humidity.

For example, air at a dry-bulb (db) temperature of75°F has a vapor pressure equal to 0.88 inches ofmercury (in. Hg) when saturated with water vapor—that is, when its relative humidity is 100%. The sameair sample would have a relative humidity value of50% when the water vapor content of the air createsvapor pressure equal to 0.44 in. Hg:

Air at 50% RH can accept more water vapor. Anyadditional water vapor added to air that is alreadycompletely saturated will fail to dissolve into the air,and fog will appear.

The relative humidity of air is important not only to comfort, but also to providing a healthy indoorenvironment. Flu viruses, for example, thrive in airthat is too dry. On the other hand, dust mites cannotsurvive in dry air. These tiny creatures lack theability to drink water. They must absorb water from

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PRACTICAL APPLICATIONS OF PSYCHROMETRICS

by Harlan Krepcik, CMSAssociate Professor of HVACR, Tidewater Community College, Portsmouth, VA

© 2010 by the Refrigeration Service Engineers Society, Des Plaines, ILSupplement to the Refrigeration Service Engineers Society.

Service Application ManualSAM Chapter 650-002

Section 3A

the surrounding air to live. Dust mites cannot absorbwater from air when its relative humidity value is lessthan 50%, and as a result they dehydrate and die.The job of HVACR professionals is to provide indoorenvironments that are both comfortable and healthy.To do so, mechanical systems must be able to controlall physical properties of air.

Specific humidity (also referred to as humidity ratio)is the term used to describe a third physical propertyof air. This value is a measure of the actual mass ofwater dissolved into an air sample. It is measured ineither pounds of water or grains of water that havebeen evaporated into one pound of air (7,000 grainsequals 1 lb of water). As a reference of scale, onepint of water has a mass of 1.04 lbm (pounds mass).Energy used to remove water mass from an air sampleor to add water mass to an air sample is measured inunits defined as latent heat. The quantity of heat(Btu) required to change the moisture content of airvaries with the temperature of the air sample. Forexample, at a temperature equal to 212°F, heatquantity equal to 970 Btu is required to change thephase of 1 lb of water from liquid to vapor, or viceversa. However, when the temperature of standardair is cooled to 60°F by an evaporator, heat quantityequal to approximately 1,060 Btu is required tochange the phase of 1 lb of water.

When air is cooled, its specific volume decreases, as does its ability to hold water in solution. This isdue to the increase in the density of air as it cools.Specific volume is the reciprocal of density—in other words, the two properties are inversely related.Consider an air sample at 75°F db and 50% RH. The density of the air equals 0.073126 lb/ft3 and itsspecific volume equals 13.675 ft3/lb (1 ÷ 0.073126).Try to visualize this volume by picturing a ball of air35.6 in. in diameter. At 50% RH, 1 lb of the airsample holds 64.9 grains of water in solution. Whensaturated, 75°F air can hold 131.8 grains. Notice thatrelative humidity is a ratio of the vapor pressures ofthe two air samples, not a ratio of the water massdissolved into each air sample. When this air sample(50% RH at 75°F) is cooled to 55.15°F, its densityincreases to 0.0759 lb/ft3. At this point, its specificvolume decreases to 13.175 ft3/lb. The “ball” of airin question now has a smaller diameter of 35.16 in.

This smaller and denser “container” cannot hold asmuch water mass in solution as the larger “container”at 75°F. At the lower temperature of 55.15°F, theexisting 64.9 grains of water in each pound of air(specific humidity) results in a relative humidityequal to 100%.

The reverse occurs when air is heated—the densityof the air decreases and its specific volumeincreases. As the “container” or volume of airexpands, it can hold more water in solution. Considerthe same air sample at 75°F db and 50% RH. Whenheated to a temperature of 100°F db, the sample canhold 302.5 grains of water in each pound of air atsaturation (100% RH). At this new highertemperature, the specific volume of the sampleincreases to 15.083 ft3/lb of air. The “ball” of airincreases to a diameter equal to 36.79 in. Theexisting 64.9 grains of water create only enoughvapor pressure to result in a relative humidity equalto 22.6%. This physical property explains why winterair that infiltrates into a home and is heated to roomtemperature has a lower relative humidity.

Clearly, as work is done on an air sample by amechanical system, changes in the physicalproperties of that air are significant. An HVACRprofessional must be able to measure and calculatethe impact of work done on an air sample for thepurpose of predicting the acceptability of an airstreamfor a given application. The psychrometric chart is auseful tool that allows you to make those calculationswithout the need for higher-level mathematics.

Another term with which technicians must befamiliar is wet-bulb temperature. The wet-bulb (wb)temperature is measured using a thermometer withits bulb covered with a water-saturated wick. Air is caused to flow over the wick at approximately 900 ft/min, resulting in evaporation of water from the wick. The evaporation process draws heat fromthe bulb of the thermometer, causing a decrease intemperature. When the temperature equalizes, thewet-bulb temperature of the air has been measured.This temperature is subtracted from the dry-bulbtemperature of the air, and the difference is calledthe wet-bulb depression. A large wet-bulb depressionindicates that the air is dry and readily accepts water

2

from the wick, thus accelerating the evaporation rate.A small wet-bulb depression indicates that the air is damp and lacks ability to accept water from thewick, thus reducing the evaporation rate.

Electronic instruments typically are used to measurethe physical properties of an air sample, since theyafford greater accuracy than sling psychrometers. A manually operated sling psychrometer thatincludes a slide rule calculator is shown in Figure 1.An example of a digital psychrometer used morecommonly today is pictured in Figure 2.

The dew point temperature of an air sample is alsoimportant. When air is cooled to its dew pointtemperature, water vapor in the sample cannot beheld in solution and will condense out of the air.An evaporator coil must be at a temperature belowthe dew point of an airstream to reduce its moisturecontent. If the coil temperature is above the air’s dewpoint, the dry-bulb temperature of the airstream willdecrease without a change in its moisture content.This results in poor control of relative humiditywithin the conditioned space. There is another

important issue that relates to dewpoint temperature. The dew point of the outdoor ambient air is oftenhigher than the dry-bulbtemperatures of the indooratmosphere and constructionmaterials used to form the buildingwalls. Water vapor in the infiltratingair can condense inside cool walls,causing damage to constructionmaterials and the formation ofcontaminants that are unhealthy foroccupants. Therefore, a good vaporbarrier and adequate insulation areimportant to the success of aclimate-controlled structure.

The last physical property of air to be consideredhere is enthalpy. Enthalpy is a measure of the totalheat content (measured in Btu/lb) of an air sample.Psychrometric charts do not use enthalpy valuesreferenced to absolute zero (–460°F). Values used onthe chart are referenced to the heat content of liquidwater at 32°F and dry air at 0°F. This method issimilar to that used on pressure gauges. Pressuremeasurements are referenced to atmosphericpressure. A gauge reading of 0 psig does not meanthat the space is under no pressure. Rather, a readingof 0 psig indicates that the space is at atmosphericpressure, which is 14.696 psi at sea level. Therefore,pressures below 0 psig are referred to as vacuumreadings. They are merely measurements of pressurebelow atmospheric pressure. As with pressuregauges, some psychrometric charts indicate negativeenthalpy values for working with frigid air. In thisdiscussion, we will utilize an ASHRAE psychrometricchart formatted for air conditioning using temperaturesabove the 0°F and 32°F reference points. Valuescalculated will be positive—that is, the enthalpyvalues calculated will be in Btu per pound of airabove the enthalpy of air at 0°F and liquid water at 32°F.

The HVACR industry uses standard airequations rather than mass flow ratecalculations to evaluate the work done on an airsample. This practicesimplifies the evaluationprocess. Careful use ofstandard air equationsproduces results that aretypically within 2% ofmass flow rate calculationsfor the same work processperformed on a givenairstream.

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Standard air equations are based on the physicalproperties of air at sea level. “Standard air” isdefined as having a dry-bulb temperature equal to70°F at a pressure equal to 14.696 psi and a densityequal to 0.075 lb/ft3 of air. At these conditions, theenthalpy of dry air is equal to 16.8 Btu/lb and itsspecific heat is 0.24 Btu/lb. The standard airequations are stated as follows:

þ Sensible heat equation: QS = cfm × 1.08 × ∆Tþ Latent heat equation: QL = cfm × 0.68 × ∆WGR

þ Total heat equation: QT = cfm × 4.5 × ∆h

where:Q = Btuh

cfm = cubic feet of air per minuteT = dry-bulb temperature (°F)

WGR = grains of water per pound of airh = enthalpy.

The numeric values shown in the three equationsabove (1.08, 0.68, 4.5) are not constants—they arefactors that have been derived by using the knownphysical properties of standard air and dimensionalanalysis. The factors serve the purpose of convertingall the units of measurement used in the equations toBtuh. The mathematical calculations used to deriveall three factors are shown in the shaded box below.

Standard air equations are useful tools that can beused to solve complex psychrometric problems. Twoof the physical properties of an air sample must beknown in order to establish state points and generatea solution schematic on a psychrometric chart.Relatively common instruments are used to measure

the needed physical properties. From the state pointplotted on the chart, the remaining physicalproperties of the air sample can be determined.

SOLUTION SCHEMATICS

It is important for you to be able to calculate theprojected heat load on a cooling coil accurately, so that the system chosen for an application hasadequate sensible and latent capacities. A systemdesigner can determine these capacity values byusing higher-level mathematics or by utilizing apsychrometric chart. A solution schematic drawn on a psychrometric chart helps a designer visualizeand calculate the work done on an airstream withoutthe need for advanced mathematics.

Typically, a technician will use a sling psychrometeror an electronic psychrometer to measure the dry-bulb and wet-bulb temperatures of the air. These twovalues are used to position a state point on the chart.Figure 3 illustrates how the state point is establishedfor an air sample that measures 80°F db and 67°F wbtemperatures. This state point is the AHRI enteringair test point used to establish cooling performancefor HVAC systems. Notice that dry-bulb temperaturevalues are provided on the horizontal axis at thebottom of the chart. The dry-bulb value lines aredrawn vertically upward, almost perpendicular to thehorizontal axis. Wet-bulb values are provided on thesaturation curve located on the left side of the chart.The wet-bulb value lines are drawn descending fromthe saturation curve downward and to the right.Arrows have been superimposed on the chart toillustrate how these value lines are used to position

the state point for thisair sample.

When sensible work isperformed on an airsample, the state pointshifts horizontally alongthe dew point line onthe psychrometricchart. Dew point andhumidity ratio valuesshare horizontal lineson the chart. The state

4

5

Figure 3

ASHRAE

point will shift to the right when sensible heat isadded to the air sample. It will shift to the left whensensible heat is removed. When latent work isperformed on an air sample, the state point movesvertically. It will shift upward along the dry-bulbtemperature line when water vapor is added anddownward when water vapor is removed from the airsample.

In practical applications, cooling processes mostoften include both sensible heat and latent heattransfer. This shifts the airstream’s state pointdiagonally, depending on whether heat energy isbeing added or removed. Condition lines are drawnbetween the initial and final state points to provide a visual representation of the work done on theairstream. The angle of the condition line indicatesthe ratio of sensible heat transfer to total heattransfer. Steep condition lines depict work processesthat have small sensible heat ratios. Condition lineswith shallower slopes depict work processes thathave larger sensible heat ratios (meaning that moresensible heat than latent heat was transferred).

Figure 4 illustrates an airstream that has been cooledfrom 80°F db, 67°F wb to 60°F db, 56.3°F wb. Thediagonal condition line illustrates that both the airtemperature and the water vapor content have beenchanged. As sensible heat was removed from theairstream, the state point shifted to the left. Thedownward slope of the condition line indicates thatwater vapor was removed during the process. Therelatively shallow angle of the line indicates thatmore sensible work than latent work was done on theairstream. Standard air equations provided earliercan be used to calculate the quantity of sensible andlatent work performed. To utilize these equations, theair flow volume rate in cubic feet of air per minute(cfm) must be known. Methods used to determine air flow are outside the scope of this discussion. For now, assume that blower data provided by themanufacturer were used to verify an air flow volumerate of 8,000 cfm.

The change in dry-bulb temperature is used tocalculate the sensible heat removed from theairstream. In this example, the dry-bulb temperaturedecreased from 80 to 60°F, for a temperature

difference equal to 20°F. The known values are nowinserted into the sensible heat equation, which issolved as follows:

QS = cfm × 1.08 × ∆T= 8,000 cfm × 1.08 × 20°F = 172,800 Btuh

Change in the specific humidity of an airstream mustbe known in order to calculate the quantity of latentheat removed from the air. These values are takenfrom the psychrometric chart, as shown in Figure 5on page 8. Specific humidity values (also known ashumidity ratio values) are shown on the vertical axislocated on the right side of the chart. Trace a linefrom each state point horizontally to the right andread the specific humidity on the humidity ratioscale. Note that these values are provided in poundsof water per pound of air on the ASHRAE chartshown in Figure 5. They must be converted to grainsof water per pound of air for use in the standard airequation. The initial humidity ratio is 0.0112 poundsof water per pound of air and the final value is0.0088 pounds of water per pound of air, whichyields a difference equal to 0.0024 pounds of waterper pound of air. Since one pound of water equals7,000 grains, the change in water vapor content ofthe air is 16.8 grains of water per pound of air(0.0024 × 7,000). This converted value is nowinserted into the latent heat equation, which is solved as follows (the answer is rounded off):

QL = cfm × 0.68 × ∆WGR

= 8,000 cfm × 0.68 × 16.8 WGR = 91,390 Btuh

In order to calculate the amount of water massremoved by the cooling process, you need to utilizethe thermodynamic properties of water at saturation.These values can be found in the ASHRAEFundamentals Handbook (Table 3 of Chapter 1 in the2009 edition). The specific enthalpy for evaporationat the cooling coil temperature in this example is1,060 Btu per pound of water. You can use this valueto show that approximately 86.2 lb (91,390 ÷ 1,060)or about 10.3 gallons (86.2 ÷ 8.33) of water per hourof operation will be condensed from the airstream.

The sum of sensible heat and latent heat valuesequals the total cooling work done on the air, which

6

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Figure 4

ASHRAE

8

Figure 5

ASHRAE

9

Figure 6

ASHRAE

in this example is 264,190 Btuh (172,800 + 91,390).The ratio of sensible heat transferred is calculated by dividing the sensible work performed by the totalwork performed. In this case, the sensible heat ratioof the cooling process is 0.654 (172,800 ÷ 264,190),indicating that more sensible heat than latent heatwas transferred. Recall that we were able to predictthat outcome from the slope of the condition line inFigure 4. By plugging values taken from the solutionschematic into the standard air equations, you havenow determined that 65.4% of the total work done onthe air utilized sensible heat energy, which meansthat 34.6% utilized latent heat energy.

You can check your work by using enthalpy valuestaken from the solution schematic to calculate thetotal heat energy transferred during this coolingprocess. Enthalpy values are plotted on a scale thatin reality wraps around the entire perimeter of thepsychrometric chart. Look at Figure 6 on page 9. In order to keep the drawing from becoming toocongested, only a few of the relevant enthalpy valuelines are continued through the main body of thechart—see the diagonal lines extending below thehorizontal axis (between 20 and 25 Btu/lb) andbeyond the vertical axis on the far right-hand side of the chart (between 30 and 35 Btu/lb). Notice thatthese enthalpy value lines descend from left to righton the chart, as do the wet-bulb temperature lines.The two sets of lines are almost, but not quite,parallel. Some system designers simply extend thewet-bulb temperature line from a state point upwardto the left to intersect the enthalpy scale, and in thisway attempt to determine the total heat content of anair sample. This practice yields an enthalpy valuethat approximates the correct value—however, a slighterror is introduced into the calculation. Therefore, itis a better practice to trace a line through each statepoint connecting equal enthalpy values on the twosides of the scale, as illustrated in Figure 6.

In this example, the initial heat content of theentering air is 31.5 Btu/lb and the heat content of the leaving air is very close to 24 Btu/lb. The changein total heat content caused by the cooling processtherefore equals 7.5 Btu per pound of dry air, and this enthalpy value is entered into the standard airequation for total heat as follows:

QT = cfm × 4.5 × ∆h= 8,000 cfm × 4.5 × 7.5 ∆h = 270,000 Btuh

The sum of the sensible heat and latent heat valuesfrom earlier calculations was 264,190 Btuh. Thisvaries from the solution to the enthalpy equationabove by 5,810 Btuh (270,000 – 264,190), or 2.15%[100 × (5,810 ÷ 270,000)]. This variation is easilyaccounted for by the rounding off that is done inderiving the conversion multipliers used in thestandard air equations. When the total heat valueyielded by the standard air equation for enthalpy iswithin 2 or 3% of the total heat value yielded bysumming sensible and latent heat values, your workis reasonably accurate. The two total heat values willrarely be equal—the comparison merely serves as acheck of the calculations performed. Remember, yourwork must be accurate, which means that it must beerror-free. Measurement instruments used in the fieldand the resolution of a psychrometric chart do notallow your work to be precise, which means exact.

Next, you will apply what you have learned aboutdeveloping a solution schematic to HVAC systemsthat mix outdoor ventilation air with air recirculatedfrom the conditioned space. Developing this skillallows a system designer to calculate accurately theheat loads that will be imposed on a cooling coil andmake better equipment selections.

MIXED-AIR PROCESSES

In many residential applications and in all commercialapplications, outdoor air is mixed with recirculatedair to provide supply air that satisfies ventilationrequirements. The quantity of outside air that mustbe brought into an occupied space varies with theactivity of the occupants. A portion of Table 403.3taken from the 2009 International Mechanical Code®

is reproduced in Figure 7. It lists the outdoor airventilation requirements for various kinds ofcommercial spaces. The requirement of air perperson is added to the requirement of air per squarefoot of floor area to arrive at the total ventilation airrequired. The introduction of outdoor air adds heatload to a cooling coil during warm periods insummer. Additional heating capacity is needed totemper winter ventilation air. In this section, you will

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Figure 7. Excerpt from Table 403.3: Minimum ventilation rates

Occupancy classification

OfficesConference roomsOffice spacesReception areasTelephone/data entryMain entry lobbies

Private dwellings, single and multipleGarages, common for multiple unitsGarages, separate for each dwellingKitchensLiving areas

Toilet rooms and bathrooms

Public spacesCorridorsElevator carShower room (per shower head)Smoking loungesToilet rooms—publicPlaces of religious worshipCourtroomsLegislative chambersLibrariesMuseums (children’s)Museums/galleries

Retail stores, sales floors, and showroomfloors

Sales (except as below)Dressing roomsMall common areasShipping and receivingSmoking loungesStorage roomsWarehouses (see storage)

Specialty shopsAutomotive motor-fuel dispensing stationsBarberBeauty and nail salonsEmbalming roomPet shops (animal areas)Supermarkets

Sports and amusementDisco/dance floorsBowling alleys (seating areas)Game arcadesIce arenas without combustion enginesGym, stadium, arena (play area)Spectator areasSwimming pools (pool and deck area)Health club/aerobics roomHealth club/weight room

People outdoorair flow rate in

breathing zone,cfm/person

55555

———

0.35 ACH butnot less than

15 cfm/person

—

———60—5555

7.57.5

7.5—7.5—60——

—7.520—7.57.5

20107.5——7.5—2020

Area outdoorair flow rate in

breathing zone,cfm/ft2

0.060.060.060.060.06

————

—

0.06————

0.060.060.060.120.120.06

0.12—

0.060.12—

0.12—

—0.060.12—

0.180.06

0.060.120.180.300.300.060.480.060.06

Default occupantdensity, lb/1000 ft2

505306010

———

Based on number ofbedrooms. First

bedroom, 2. Eachadditional bedroom, 1.

—

———70—

1207050104040

15—40—70——

—2525—108

1004020——

150—4010

Exhaust air flowrate, cfm/ft2

—————

0.75100 cfm per car

25/100—

20/50

—1.0

50/20—

50/70——————

—0.25—————

1.50.50.62.00.9—

———0.5—————

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study a draw-through equipment design in which themixed air travels first through the cooling coil beforepassing through the indoor blower. Fan energy willprovide reheat and slightly longer “ON” cycles forimproved humidity control in the space. For the sakeof clarity, conditioned air leaving the cooling coil isreferred to here as primary air, and air entering theconditioned space through registers or grilles isreferred to as supply air. The temperatures of primaryair and supply air differ according to the amount ofheat gained by the airstream from the fan motor andthrough the supply ductwork. This example consistsof packaged rooftop equipment with the supplyductwork installed in the plenum cavity above a dropceiling. The plenum cavity has been sealed for use asa return-air pathway.

A system designer must begin the process bycompleting a heat gain/heat loss load calculation forthe conditioned space. The outcome of this exercisewill describe how the space gains (or loses) heatenergy. In commercial buildings, heat transfer throughexterior walls affects interior spaces that are within 10 ft of those walls. For core areas of the building,heat gain is almost entirely internally generated. In many regions, core areas experience a demand for cooling 12 months out of the year. The magnitudeof internal heat gain can vary widely, depending onlighting, machines, equipment, the number of peopleworking within the space, and the type of work beingdone. Only heat gain appearing within the space isused to determine the volume of supply air needed to cool the conditioned area. This includes heattransmission from outdoors as well as internallygenerated heat. The heat load provided to the coolingcoil by ventilation air taken from outside is notfactored into the equation used to determine theneeded indoor blower air volume rate. Several loadcalculation procedures are available for commercialapplications. The ACCA Manual N procedure and the ASHRAE load calculation procedure are twocommonly used by industry professionals. Equipmentmanufacturers such as Carrier and Trane also haveprocedures available for this purpose. A loadcalculation for an interior space does not reflect thetotal heat load that appears at the cooling coil inmixed-air applications. The heat load imposed by theoutdoor ventilation air must be accounted for also.

A psychrometric chart can be used to calculate theheat loads that must be handled by the cooling coil.The physical properties of the mixed airstream mustbe known to make that calculation. To determinethose properties, a system designer must plot statepoints for both the air recirculated from theconditioned space and the outdoor air used forventilation. In this example, the indoor air will bemaintained at a 75°F db temperature and at 45%relative humidity. On design day, the ventilation airwill be 91°F db and 80°F wb. The state pointsrepresenting these two air samples are plotted asshown in Figure 8.

A room sensible heat ratio (RSHR) line must be tracedon the chart in order to visualize how the conditionedspace gains sensible and latent heat. This ratio isderived from the heat load calculation. For thisexample, let’s say that the room sensible heat gain on design day is 153,600 Btuh and the room latentheat gain is 20,950 Btuh. Total room heat gain equals174,550 Btuh, or 14.54 tons (174,550 ÷ 12,000). The RSHR, therefore, is 0.88 (153,600 ÷ 174,550).This ratio is first traced onto the protractor located inthe upper left corner of the ASHRAE psychrometricchart. Sensible heat ratio (SHR) values are printed on the inside of the arc. A reference line is tracedfrom the origin of the protractor through the arc at0.88 SHR. The slope of this line represents themanner in which the space experiences heat gain.The shallow angle of the SHR line indicates thatmore sensible heat than latent heat is gained ondesign day. Next, an RSHR line parallel to the SHRreference line on the protractor is traced from the roomsetpoint condition downward and to the left until itintersects the saturation curve. In this example, theRSHR line intersects the saturation curve at 49.5°F,as illustrated in Figure 9 on page 14.

Supply air with physical properties that yield a statepoint positioned anywhere on the RSHR line willmaintain room setpoint conditions. However, the dry-bulb temperature of the supply air determinesthe amount of supply air needed to condition a space.The air flow needed decreases as the temperature ofthe supply air decreases. As the temperature of thesupply air increases, the air flow required to maintaindesign conditions also increases.

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Figure 9

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Figure 10

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A designer must consider another factor. Supply airentering the conditioned space acts like a “sponge,”absorbing sensible heat and humidity from the roomair. As it does so, the physical properties of the supplyair shift in a direction parallel to the RSHR line.Supply air represented by a state point located abovethe RSHR line can maintain the room’s dry-bulbtemperature—however, the resulting room relativehumidity will be higher than the specified value of45%. This is illustrated in Figure 10 on page 15, inwhich supply-air state point A is situated above theRSHR line. As the supply air gains temperature andhumidity, it shifts to state point A', yielding a roomtemperature equal to 75°F and a relative humidityabove 45%. Supply air represented by a state pointlocated below the RSHR line can maintain the roomdry-bulb temperature, but the room relative humiditywill be lower than 45%. This is also illustrated inFigure 10, which shows state point B shifting to statepoint B'.

The position of the primary-air state point varies withthe type of cooling equipment used. Direct expansion(DX) cooling coils typically yield primary air withrelative humidity between 75 and 90%. Chilled watercoils can yield primary air with relative humidity ashigh as 100% and dry-bulb temperatures lower thanthose produced by DX equipment. A designer mustbe familiar with the performance of the equipmentbeing installed to make a good judgment call whenplotting the primary-air state point. In this example,we will plan for a DX coil that providesprimary air with a relative humidity of85%. On the solution schematic, theprimary-air state point will be positionedat 85% RH, approximately 2°F to the left of a horizontal intersection with theRSHR line (to allow for the heat riseresulting from fan motor heat and supplyduct gain). This heat gain will cause the primary-air state point to movehorizontally to the right. Supply airentering the room will therefore be about2°F warmer than the primary air leavingthe cooling coil in this example. Statepoints representing the primary air andthe supply air are shown in Figure 11.Note that the supply-air state point now

rests directly on the RSHR line, indicating that thephysical properties of the supply air are sufficient tomaintain room design conditions. As the supply airgains sensible and latent heat from the room air, itsstate point will be altered in a manner that tracks theRSHR line to reflect the ratio of heat sources in thespace. The supply air entering the room will absorbsensible heat and latent heat in a ratio that causes itsstate point to ascend to the right until it reaches theroom’s design setpoint.

The supply air volume needed to maintain roomdesign conditions is calculated from the data plottedon the chart up to this point. The supply air dry-bulbtemperature is 57°F, which is 18°F cooler than the room design temperature. This temperaturedifference, together with the room’s sensible heatgain, will be used in the sensible heat equation todetermine the supply air flow rate needed for thespace. In the equation shown below, “q” equals cubicfeet of air per minute (cfm). Again, the answer hasbeen rounded off:

At this point, the designer must calculate the physicalproperties of the air entering the cooling coil. Thisairstream will be a mixture of air recirculated fromthe space and outdoor air used for ventilation. Forthis example, let’s assume that the design calls for

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Supply air

Primary air

Figure 11

“neutral” space pressure—which means that theexhaust air volume equals the volume of ventilationair being introduced from outdoors. Regulating codesmust be followed when choosing ventilation air for anapplication. Say that ventilation air equal to 1,200 cfmis required. Then 6,700 cfm (7,900 – 1,200) of roomair will be recirculated, while 1,200 cfm of room airwill be exhausted. Air returning from the conditionedspace can increase in temperature while movingthrough the return plenum or ductwork.

Applications with ducted returns that have beencorrectly sealed, insulated, and installed inunconditioned interior spaces are subject to less heatgain than ductwork installed outdoors. Any heat gainthrough the return-air duct will cause an increase inair temperature, which must be accounted for on thesolution schematic that you create on a psychrometricchart.

In a plenum return application, the return-airtemperature will increase due to heat emitted byrecessed ceiling lights, as well as heat gain that takesplace through the roof, exterior walls, and partitions.Partitions are walls separating the plenum cavityfrom unconditioned interior spaces. This rise must be accounted for by moving the room state pointhorizontally to the right by an amount equal to thetemperature gain. In this example, assume that acalculation made to determine gain in the plenumfinds the rise to be equal to 3°F. The room-air statepoint is shifted to the right by that amount. This

yields a return-air dry-bulb temperatureequal to 78°F (75 + 3). Return air withthese new properties is then mixedwith ventilation air inside the HVACequipment. Ventilation air from outsideis supplied through a manual outdoorair damper or an economizer, asillustrated in Figure 12.

Figure 13 on page 18 shows the return-air state point located 3°F to the right of the room-air state point on the RSHR line. Next a line is drawnon the psychrometric chart connectingthe state points of the outdoor air andthe return air. A new state point

representing the air mixture entering the cooling coil then must be positioned somewhere on thisconnecting line. A convenient method for positioningthe mixed-air state point on the chart is first tocalculate the dry-bulb temperature of the mixedairstream. This is done by using the air volumes anddry-bulb temperatures of the returning room air andthe ventilation air. The return air (6,700 cfm) reachesthe HVAC unit at a dry-bulb temperature of 78°F,and the ventilation air (1,200 cfm) has a dry-bulbtemperature of 91°F. These values are used to findthe dry-bulb temperature of the mixed air as follows:

A state point representing the mixed air now can bepositioned on the line that connects the outdoor-airand return-air state points. The mixed-air state point,shown fixed on that line at 80°F db temperature inFigure 13, represents primary air entering the coolingcoil. The curved condition line drawn on the chartillustrates how the physical properties of the primaryair change as work is done on the airstream by thecooling coil. Notice that the initial work done on theairstream is the transfer of sensible heat, causing

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ASHRAE

only a change in dry-bulb temperature. As theairstream approaches its dew point temperature,moisture begins to condense from the air, causing adownward shift in the condition line. The enteringand leaving primary-air state points are now used tocalculate the work that must be done on the airstreamby the cooling coil.

The sensible capacity that must be provided by theDX equipment can be calculated once you know the change in temperature of the primary air from the entering point (80°F db) to the leaving point(55°F db). In this case, a temperature difference of 25°F is plugged into the sensible heat equation,together with the air flow volume (7,900 cfm), toselect the sensible equipment capacity needed:

QS = cfm × 1.08 × ∆T= 7,900 cfm × 1.08 × 25°F = 213,300 Btuh

The needed latent equipment capacity is calculatedby finding the change in the humidity ratio of theprimary air, from the entering point (0.010 pounds of water per pound of air) to the leaving point (0.0077 pounds of water per pound of air), as takenfrom the psychrometric chart shown in Figure 14 onpage 20. The difference (0.0023 pounds of water perpound of air) must be converted to change in grainsof water by multiplying it times the conversion factorof 7,000 grains per pound of water, yielding a changein specific humidity equal to 16.1 grains of moistureper pound of dry air (7,000 × 0.0023). The air flowvolume rate and the change in humidity ratio canthen be used in the latent heat equation as shownbelow:

QL = cfm × 0.68 × ∆WGR

= 7,900 cfm × 0.68 × 16.1 WGR = 86,490 Btuh

Adding the sensible heat and latent heat valuesyields the needed total cooling capacity that must bedelivered by the HVAC equipment selected, which inthis example equals 299,790 Btuh (213,300 + 86,490),or 24.98 tons (299,790 ÷ 12,000). It has been shownthat a nominal 25-ton unit must be selected to handlethe space load (14.54 tons) plus the load imposed bythe ventilation air (10.44 tons). This translates to oneton of load on the cooling coil per 115 cfm of outdoor

air (1,200 cfm ÷ 10.44 tons) in this application.Certainly, the cost of an energy recovery ventilator(ERV) can be justified to reduce the size of the coolingequipment needed to condition the environment inthis space. An ERV that reduces the load imposed bythe ventilation air by 50% (5.2 tons) would reducethe capacity needed for the HVAC cooling equipmentfrom 25 nominal tons to 20 nominal tons, and loweroperating costs significantly.

As a general guide, DX equipment can handleadditional load from outdoor air when the ventilationair makes up no more than 20 to 25% of the primaryair. This value varies with the equipment selected.When more outdoor air is needed, a designer mustshed load from the outdoor air using ERV equipmentor select a chilled water system for the application(or both).

Enthalpy values taken from the solution schematicshown in Figure 14 can be used to check the abovecalculations. The enthalpy of the entering andleaving airstreams are approximately 30.2 Btu/lb and21.7 Btu/lb, respectively, for a total change in heatcontent equal to 8.5 Btu per pound of dry air. Thechange in enthalpy and the air flow volume rate areused as shown below to calculate the total heattransferred in this cooling process:

QT = cfm × 4.5 × ∆h= 7,900 cfm × 4.5 × 8.5 ∆h = 302,175 Btuh

This value is within 2,385 Btuh of the total coolingcapacity calculated earlier (302,175 – 299,790) andindicates that calculations made using the solutionschematic are reasonably accurate. Variation from theenthalpy calculation amounts to a discrepancy of lessthan 1% [100 × (2,385 ÷ 299,790) = 0.79%].

EVAPORATIVE COOLING PROCESSES

In arid regions, it is not necessary to removehumidity from the conditioned space during thecooling cycle. The following paragraphs describeapplications in which water may be used in a directevaporative process to provide comfort cooling insuch regions of the world. These systems do notrecirculate indoor air—instead, they utilize 100%

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Figure 14

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A

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Figure 15

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outdoor air to cool a space. A system designer mustbe aware that the entering airstream’s wet-bulbtemperature limits direct evaporative cooling.However, design wet-bulb temperatures are rarelyhigher than 78°F, making direct evaporative coolingeconomical in arid regions.

Direct evaporative cooling is an adiabatic exchange ofheat. By definition, no heat is added to, or extractedfrom, an adiabatic process. The initial and final airconditions fall on a line of constant total heat(enthalpy), which nearly coincides with a line ofconstant wet-bulb temperature. Heat must be addedto evaporate water. In the case of direct evaporativecooling, the heat is supplied by the air into whichwater is evaporated. The dry-bulb temperature of the air is lowered, and sensible cooling results. The amount of heat removed from the air equals the amount of heat absorbed by the water evaporated.When water is recirculated in the direct evaporativecooling apparatus, the water temperature in thereservoir will approach the wet-bulb temperature of the air entering the process.

The maximum possible reduction in the dry-bulb airtemperature is the difference between the enteringair dry-bulb and wet-bulb temperatures. If the aircould be cooled to the wet-bulb temperature, theprocess would be 100% effective and the leaving airwould be saturated. Effectiveness is defined using amathematical calculation in which the depression ofthe dry-bulb temperature of the leaving air is dividedby the difference between the dry-bulb and wet-bulbtemperatures of the entering air. Theoretically,adiabatic direct evaporative cooling is less than100% effective. However, evaporative coolers can be 95% (or more) effective and are a practical choicefor comfort cooling in arid regions.

By way of example, consider entering air at 96°F dband 63°F wb. The resulting initial difference is 33°F(96 – 63). If the effectiveness of the evaporativecooler is 80%, the depression in air temperature will be approximately 26°F (33 × 0.80). The dry-bulbtemperature of the air leaving the cooler then wouldbe 70°F (96 – 26). This process is illustrated on thepsychrometric chart shown in Figure 15 on page 21.First, a state point is plotted to represent the condition

of the entering air, which is point A on the chart. A line is traced upward to the left along the enthalpyline that intersects the entering-air state point. In thisexample, the enthalpy is shown to be approximatelyequal to 28.3 Btu/lb. (Since wet-bulb lines nearlyparallel enthalpy lines, it is the practice of somesystem designers to trace their line along the wet-bulb line that intersects the entering-air state point,although this introduces a very slight error.) The statepoint representing the physical properties of theleaving air is positioned on the traced line at itsintersection with the 70°F dry-bulb line, which ispoint B on the chart in Figure 15.

In the adiabatic evaporative cooler, only part of therecirculated water evaporates and the water supply is recirculated. The recirculated water will reach anequilibrium temperature approximately equal to thewet-bulb temperature of the entering air.

A heat gain load calculation is consulted todetermine the air flow volume rate needed tomaintain the desired space temperature. For thisexample, assume that the sensible room gain equals42,000 Btuh, and that the room design setpointtemperature is 80°F. The temperature differencebetween the room setpoint and the air leaving thedirect evaporative cooler is 10°F (80 – 70). Thesensible heat equation then can be used to calculatethe air flow volume rate (q) required to maintain thesetpoint, as shown below. The answer is rounded off:

If air is not exhausted freely, the increased staticpressure within the space will reduce air flow throughthe evaporative cooler. The result is an increase inthe moisture and heat absorbed by the air leaving theevaporative cooler. Reduced air flow also reduces theair velocity in the room. These effects combine toreduce the comfort level for the occupants. Properlydesigned systems should have a minimum of 2 ft2

of exhaust area for every 1,000 cfm of supply air. If the exhaust area is not sufficient, a poweredexhaust should be used. The amount of poweredexhaust depends on the total air flow and the amountof free or gravity exhaust. Some applications require

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powered exhaust capacity equal to the cooler cfmoutput.

When a direct evaporative cooling unit alone cannotprovide the desired conditions, several alternativescan be employed to satisfy application requirementsand still be energy-efficient. One possible option is to increase the volume of the recirculating watersupplying the direct evaporative cooling, whilechilling it by mechanical refrigeration. This processprovides lower leaving wet-bulb and dry-bulbtemperatures. Compared to the cost of usingmechanical refrigeration only, this strategy canreduce operating costs by as much as 25 to 40%.

STEAM HUMIDIFICATION

It is a common practice to inject steam or liquidwater into an airstream to raise its humidity. Ifmixing is adiabatic, the final state point of the moistair lies on a straight line in the direction fixed by thespecific enthalpy of the injected water drawn throughthe initial state point of the air. A typical example isthe addition of humidity to dry winter air used forventilation.

For example, consider winter air at 35°F db and24.8°F wb (see point A in Figure 16 on page 24). A horizontal line can be traced to the right todetermine the humidity ratio of the ventilation air.The specific humidity (humidity ratio) of this air is0.0007 pounds of water per pound of air, orapproximately 4.9 grains of water per pound of air(0.0007 × 7,000). The relative humidity of the air is 16.6%. When brought into an indoor space andheated to 70°F db, its resulting relative humidity willbe a mere 4.6%. Humidity must be added to improvethe quality of the air for use at a comfortable indooratmosphere.To achieve a 70°F db temperature with45% relative humidity (see point B in Figure 16), afinal specific humidity of 0.007 pounds of water perpound of air, or 49.0 grains of water per pound of air(0.007 × 7,000) must be attained. This means that44.1 grains (49.0 – 4.9) of moisture must be added to every pound of ventilation air. In this example,humidification will be accomplished by adiabaticinjection of saturated steam at 230°F. It is importantto note that sensible heat must be added to the air

before steam is injected into the airstream. Steamadded to the cold ventilation air will condense insidethe ductwork.

The sensible heat and latent heat required tocondition the ventilation air can be determined bydeveloping a solution schematic on a psychrometricchart. The injected steam will provide a portion ofthe sensible heat that must be added to the airstream.To generate the solution, the system designer mustfirst consult Table 3 of the chapter on psychrometricsin the ASHRAE Fundamentals Handbook to extractthe specific enthalpy value for the saturated steambeing added to the ventilation air. A portion of thattable is reproduced in Figure 17 on page 25. As youcan see, the specific enthalpy of 230°F steam isapproximately 1,157 Btu per pound of water. Nowlook again at Figure 16. On the protractor printed in the upper left corner of the psychrometric chart, a line is traced from the origin of the protractorthrough the point on the curve corresponding to1,157 Btu/lb. Specific enthalpy on the protractor is the change in enthalpy divided by the change inhumidity ratio (∆h/∆W ). Those values are shown onthe outside of the protractor curve.

A state point for the final desired air condition hasalready been plotted on the chart as (point B). Next,a sloped line is drawn parallel to the line traced onthe protractor downward from point B to intersect thehorizontal line drawn from point A. The intersectionpoint is labeled point C on the solution schematic inFigure 16. The slope of this line is used to reveal thetemperature gain from sensible heat added by thesteam. In this example, temperature gain from thesteam is approximately 1.5°F.

With this information, the sensible heat that must beadded to the air can be determined for a known airflow volume rate. In this example, assume that theneeded ventilation air from outside equals 3,000 cfm.Sensible heat must be added from a new energysource to warm the air from 35°F to 68.5°F, for a totaltemperature difference of 33.5°F. The sensible heatequation is used to make the calculation as follows:

QS = cfm × 1.08 × ∆T= 3,000 cfm × 1.08 × 33.5°F= 108,540 Btuh

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Figure 16

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The sensible heat added by the injected steam alsocan be determined using the sensible heat equation,as shown below:

QS = cfm × 1.08 × ∆T= 3,000 cfm × 1.08 × 1.5°F= 4,860 Btuh

Dividing by the conversion factor of 3,413 Btu/kWh,you can determine that 108,540 Btuh = 31.8 kW, and4,860 Btuh = 1.42 kW.

Earlier in this discussion, it was shown that thequantity of water that must be injected into theairstream equals 44.1 grains of moisture per pound

of air. The latent heat added by the steam thereforecan be calculated using the appropriate standard airequation, as shown below. As is customary, the answerhas been rounded off:

QL = cfm × 0.68 × ∆WGR

= 3,000 cfm × 0.68 × 44.1 WGR = 89,960 Btuh

The specific volume of the entering ventilation air is12.48 ft3/lb. This means that approximately 240.38 lbof ventilation air is introduced into the space everyminute, or about 14,423 lb per hour. Dimensionalanalysis is used to make this calculation, as follows:air flow of 3,000 cfm × 1 lb of air per 12.48 ft3 of

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Figure 17. Excerpt from Table 3: Thermodynamic properties of water at saturation

Absolute Specific volume, ft3/lb Specific enthalpy, Btu/lb Specific entropy, Btu/lb/°FTemp, pressure,

°F psia Sat. solid Evap. Sat. vapor Sat. solid Evap. Sat. vapor Sat. solid Evap. Sat. vapor

190 9.3497 0.01657 40.901 40.918 158.05 983.78 1141.83 0.2787 1.5143 1.7930191 9.5515 0.01658 40.092 40.108 159.05 983.17 1142.22 0.2802 1.5110 1.7912192 9.7570 0.01658 39.301 39.317 160.06 982.55 1142.61 0.2818 1.5077 1.7895193 9.9662 0.01659 38.528 38.545 161.06 981.94 1143.00 0.2833 1.5045 1.7878194 10.1791 0.01659 37.773 37.790 162.07 981.32 1143.39 0.2849 1.5012 1.7861195 10.3958 0.01660 37.036 37.053 163.07 980.71 1143.78 0.2864 1.4980 1.7844196 10.6163 0.01661 36.315 36.332 164.08 980.09 1144.17 0.2879 1.4948 1.7827197 10.8407 0.01661 35.611 35.628 165.08 979.47 1144.56 0.2895 1.4916 1.7810198 11.0690 0.01662 34.924 34.940 166.09 978.86 1144.94 0.2910 1.4884 1.7793199 11.3013 0.01663 34.251 34.268 167.09 978.24 1145.33 0.2925 1.4852 1.7777

200 11.5376 0.01663 33.594 33.611 168.10 977.62 1145.71 0.2940 1.4820 1.7760201 11.7781 0.01664 32.952 32.968 169.10 976.99 1146.10 0.2956 1.4788 1.7743202 12.0227 0.01665 32.324 32.341 170.11 976.37 1146.48 0.2971 1.4756 1.7727203 12.2715 0.01665 31.710 31.727 171.12 975.75 1146.87 0.2986 1.4724 1.7710204 12.5246 0.01666 31.110 31.127 172.12 975.13 1147.25 0.3001 1.4693 1.7694205 12.7819 0.01667 30.524 30.540 173.13 974.50 1147.63 0.3016 1.4661 1.7678206 13.0437 0.01667 29.950 29.967 174.14 973.88 1148.01 0.3031 1.4630 1.7661207 13.3099 0.01668 29.389 29.406 175.14 973.25 1148.40 0.3047 1.4599 1.7645208 13.5806 0.01669 28.840 28.857 176.15 972.62 1148.78 0.3062 1.4567 1.7629209 13.8558 0.01669 28.304 28.321 177.16 972.00 1149.15 0.3077 1.4536 1.7613

210 14.1357 0.01670 27.779 27.796 178.17 971.37 1149.53 0.3092 1.4505 1.7597212 14.7094 0.01671 26.764 26.781 180.18 970.11 1150.29 0.3122 1.4443 1.7565214 15.3023 0.01673 25.792 25.809 182.20 968.85 1151.04 0.3152 1.4382 1.7533216 15.9149 0.01674 24.862 24.879 184.21 967.58 1151.79 0.3182 1.4320 1.7502218 16.5475 0.01676 23.971 23.988 186.23 966.31 1152.54 0.3211 1.4259 1.7471220 17.2008 0.01677 23.118 23.135 188.25 965.03 1153.28 0.3241 1.4198 1.7440222 17.8753 0.01679 22.301 22.317 190.27 963.75 1154.02 0.3271 1.4138 1.7409224 18.5714 0.01680 21.517 21.534 192.29 962.47 1154.76 0.3300 1.4078 1.7378226 19.2896 0.01681 20.766 20.783 194.31 961.19 1155.49 0.3330 1.4018 1.7348228 20.0307 0.01683 20.046 20.063 196.33 959.89 1156.22 0.3359 1.3959 1.7318

230 20.7949 0.01684 19.356 19.373 198.35 958.60 1156.95 0.3388 1.3899 1.7288232 21.5830 0.01686 18.693 18.710 200.37 957.30 1157.68 0.3418 1.3840 1.7258234 22.3955 0.01687 18.057 18.074 202.40 956.00 1158.40 0.3447 1.3782 1.7229

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Figure 18

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air × 60 minutes per hour. The math is easier tovisualize if it is written as follows:

Each pound of ventilation air must be injected with 44.1 grains of moisture, or 0.0063 lb of water(44.1 ÷ 7,000) to achieve the needed humidity levelin the conditioned space. This means that a steamhumidifier must deliver 91.2 lb, or 10.9 gallons, ofwater every hour to the airstream in this example.The solution for the needed water is calculated asfollows: 14,423 lb of air per hour × 0,00632 lb ofwater per pound of air × 1 gallon of water per 8.33 lbof water, or:

It is important to note that fog can appear in ductworkwhen two airstreams are mixed. This can result incondensate forming on the duct walls, leading to thegrowth of unhealthy contaminants. The psychrometricchart can be used as a forensic tool to predict whensuch an episode might occur. Consider the mixing ofcold winter ventilation air with warm, moist air froman industrial area. Figure 18 shows the mixing ofventilation air at 35°F db and 90% RH (point A)with indoor air at 85°F db and 70% RH (point B).The mixed air will fall somewhere on the conditionline connecting these two airstreams.

Note that the condition line invades the “region offog” on the chart, which is that area to the left of thesaturation curve. When a solution schematic yields

a condition line that crosses through this region ofthe chart, fog will appear in the ductwork sometimeduring the mixing process.

To ensure that fog does not develop in the ductwork,sensible heat must be added to the ventilation airbefore it is mixed with the recirculated indoor air in this example. The sensible heat added must shiftthe state point for the ventilation air to the right untilthe condition line does not invade the region of fog.In this example, heating the ventilation air to 40°F(point A' in Figure 18) will shift the condition line out of the region of fog with a comfortable margin of safety.

SUMMARY

This discussion is not a definitive exploration of the many uses for a psychrometric chart. All HVACprofessionals must have good working knowledge of psychrometrics and be able to create solutionschematics proficiently. Solution schematics areexcellent visualization tools that communicate theeffect of heat transfer, both sensible and latent,imposed on an airstream. The use of standard airequations simplifies the mathematics needed tocalculate accurately the work done on an air sample.The output from these calculations can be used toindicate the load that will be imposed on a coolingcoil. Evaporative cooling processes can be easilymodeled on a psychrometric chart and used tocalculate the needed air flow volume rate.Humidification calculations can be made quickly and effectively without the need for higher-levelmathematics. Forensic observations can predict the formation of fog in the ductwork of mixed-airsystems, which leads to the growth of unwantedcontaminants on duct walls. For all of these reasons,the psychrometric chart should be a familiar friend to anyone working with HVAC systems.

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