52 North Carolina Industrial Ventilation Conference

BA-1-3 1

BASICS OF VENTILATION BASICS OF VENTILATION IIIIII(BA(BA--11--3)3)

11

Douglas L. Gunnell, PEGunnell Engineering Services

Clemmons, NC

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The objective of this module is to provide a basic understanding and practical application of the science of Psychrometics that deals with the thermodynamic

Module ObjectiveModule Objective

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that deals with the thermodynamic properties of moist air, i.e., dry air and water vapor. [Psychrometric Charts will be introduced and used extensively.]

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Thermodynamic properties are used to analyze conditions and processes involving moist air. Both normal temperature HVAC

Module ObjectiveModule Objective

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systems including replacement air, as well as high temperature industrial ventilation systems, will be introduced.A review of pertinent properties of air, as well as the Perfect Gas Law is included.

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REVIEWREVIEW

SELECTED PROPERTIES OF AIR

THE PERFECT GAS LAW

44BABA--11--33

PROPERTIES OF AIRPROPERTIES OF AIRDENSITY (DENSITY (ρρ))

Density = mass/volume

Units of measure:

55

lbm/ft³Tons/yd³grams/cm³ grains/ft³

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STANDARDSTANDARD DENSITYDENSITY(Sea Level Density)(Sea Level Density)

pstd = 0.075 lbm/ft³

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(At Standard Conditions (STP): 14.7 psi, 70°F & 0% Rh)

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PROBLEMPROBLEM

7,500 cubic feet of air flows through a duct every minute at Standard Conditions. How many pounds per minute?

AnswerAnswer

77

y p p

562.5 lbm/minute

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Actual ConditionsActual Conditions

Any condition that varies from Standard Conditions - 14.7 psi, 70ºF & 0% RH

88

Applied to correctly size the duct, fan, and air control devices

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ACTUAL DENSITYACTUAL DENSITY

ρact = ρstd x df

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ρact - actual densityρstd - standard densitydf - density factor

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Actual DensityActual Density

With a density factor of 0.55, what is the Actual Density of this air sample?

AnswerAnswer

ρ = (ρ ) (df)

1010BABA--11--33

ρact = (ρstd) (df)

(0.55) ft

lbm 075.0

3

ρact =

3ft

lbm 041.0ρact =

Density Factor (Density Factor (dfdf))

The ratio of actual density to standard density

1111BABA--11--33

ProblemProblem

A gas has a density of 0.043 lbm/ft³, what is density factor?

AnswerAnswer

1212BABA--11--33

52 North Carolina Industrial Ventilation Conference

BA-1-3 3

Density FactorDensity Factor

The density of a gas is a function of elevation (barometric pressure), temperature, moisture content, and

i th d t tpressure in the duct system.

The following equations have been developed to calculate density factors (df):

dfe, dft, dfm, dfp

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Density Density Factor:Factor:Elevation (Elevation (dfdfee))

dfe = [1 – (6.73)(10_6)(z)]5.258

“z” is elevation above Sea Level – feet

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Example:Fan located 2400 feet above sea level

dfe = [1 – (6.73)(10_6)(2400)]5.258 = 0.92

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Density Density Factor:Factor:Temperature Temperature ((dfdfTT))

(dfT) = (70 + 460)/(T + 460)Example:Air at 320°F

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dfT = 0.68 460 320

460 70

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Density Density Factor: Factor: Moisture Moisture ((dfdfmm))

dfm = (1 + ω)/(1 + 1.607 ω)

“ω” is humidity ratio (Specific Humidity) - #H20 / #Dry AirThe value of humidity ratio is readily available on a Psychrometric Chart

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Psychrometric Chart.

Example:Specific Humidity of Air ω = 0.15 #H20/#Dry Air

dfm= 0.93 (0.15)] (1.607) 1[

0.15) 1(

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Density Density Factor: Factor: Duct Duct System System Pressure Pressure ((dfdfρρ))

dfρ=

When Air in the Duct System is under extreme pressure conditions (+/- 20“ w.g.), the change in the density is significant (> 5%).

407

SP 407 duct

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Usually only considered at the Fan Inlet or Fan Outlet

Example:Air enters a fan at - 26“ w.g.

dfρ= 0.94 407

26 - 407

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Total Density FactorTotal Density Factordfdf (total)(total)

df (total) = dfe * dfT * dfm * dfρ

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52 North Carolina Industrial Ventilation Conference

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Total Density FactorTotal Density Factordfdf (total)(total)

df (total) = dfe * dfT * dfm * dfρ

Example:Ai t f t 320°F 0 15 LB H 0/LB D

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Air enters a fan at 320°F, 0.15 LB.H20/LB.Dry Air, and - 26“ w.g. pressure. The Fan is located at 2400‘ ASL. What is the Density Factor of the Air at the fan inlet?

df = 0.92 * 0.68 * 0.93 * 0.94 = 0.55

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Specific Volume (S.V.)Specific Volume (S.V.)(volume/mass (volume/mass -- ftft³/³/lbmlbm))

SV = Ft³Mix/LbmDryAir (as defined by ASHRAE, Trane Co, etc.)

SV = Ft³Mix/LbmMix (as defined by AAF)

Humid Volume (H.V.)

2020

Humid Volume (H.V.)(volume/mass - ft³/lbm

HV = Ft³Mix/LbmDryAir (as defined by AAF)

Note:Mix = dry air + water vapor

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Standard Specific VolumeStandard Specific Volume(Sea Level Specific Volume)(Sea Level Specific Volume)

SVstd ≈ 13.35 Ft³Mix/LbmDryAir (as defined by ASHRAE, Trane, etc.)[At Standard Conditions (STP): 14.7psi, 70ºF, & 0% RH]

Standard Humid Volume

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Standard Humid Volume(Sea Level Humid Volume)

HVstd ≈ 13.35 Ft³Mix/LbmDryAir (as defined by AAF)[At Standard Conditions (STP): 14.7psi, 70ºF, & 0% RH]

Note:Similar to Density, the values of Specific Volume and Humid Volume will vary at Actual Conditions – Conditions other than Standard.

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Perfect Gas LawPerfect Gas LawPV = PV = nRTnRT

P = ρRgT(Form most often applied in Industrial Ventilation)

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[Perfect Gas Law Combines the three Laws (Charles, Boyle, and Gay-Lussac) into a single Equation.]

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Pressure (P)Pressure (P)

Force exerted by the gas per unit area –lbf/in², lbf/ft², “ w.g., “ Hg, and Pascals.

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Volume (V)Volume of the gas - Ft³

Moles (n)Number of Moles - Lbm

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Universal Gas Constant (R) Universal Gas Constant (R) R = 1545.4 ftR = 1545.4 ft--lbflbf//lbmlbm°°RR

Absolute Temperature (T)(R °R ki )

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(R - °Rankine)

T = °F + 460

T = 70 + 460 = 530°R(Standard Air)

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52 North Carolina Industrial Ventilation Conference

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Perfect Gas LawPerfect Gas Law

The Perfect Gas Law can be applied to determine the density and specific volume of air at all of the infinite number of ACTUAL conditions.

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If the mass of the gas is known, the density and the volume it occupies is a function of:

the pressure the gas sees (absolute pressure) the temperature of the gas (absolute temperature) the presence of moisture or any other material mixed

with the gas

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Gas Constant for a Particular Gas Gas Constant for a Particular Gas ((RRgg))

(Rg) = M

R

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R – Universal Gas Constant – 1545.4 ft-lbf/lbmºR

M – Molecular Weight of the Gas

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The Molecular Weight of AirThe Molecular Weight of Air

The components of Air:21% 02 (MW = 32)78% N2 (MW = 28)

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1% Argon (MW = 40)

0.21 x 32 = 6.720.78 x 28 = 21.840.01 x 40 = 0.401.0 28.96 lbm

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Gas Constant for Air (Gas Constant for Air (RRgg))

lbff41545

2828

Rg = 53.36 R - lbm96.28

lbf -ft 4.1545

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Calculate the Standard Specific Calculate the Standard Specific Volume of AirVolume of Air

(Air at Standard Conditions: 14.7 psi, 70°F, and 0% Rh)

PV = nRT (Perfect Gas Equation)

AnswerAnswer

2929

144 x 14.7

530 x 53.36

1

V

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PTn

V/R g

dalbm / cf 13.35 V

Calculate the Standard Density Calculate the Standard Density of Air of Air

(Air at Standard Conditions: 14.7 psi, 70°F, and % Rh)

P = ρRg T

P 144714 x

AnswerAnswer

3030

TR

P

g

(530))36.53(

1447.14 x

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ρ = 0.075 lbm/scf

52 North Carolina Industrial Ventilation Conference

BA-1-3 6

Calculate the Volume of a Calculate the Volume of a Pound Pound Molecule (Mole)Molecule (Mole)

Air at Standard Conditions: 14.7 psi, 70°F, and 0% Rh

PV = nRT

V nRT/P

AnswerAnswer

3131

V = nRT/P

144 x 7.14

530 x 53.36 x 28.96 V

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lb.mole / scf 386.9 V

Pound Molecule (Mole) VolumePound Molecule (Mole) Volume

1 lb. mole = 386 scf

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For any gas @ 70°F, 29.92“ w.g. & 0 moisture

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Calculate the Volume of a Calculate the Volume of a Pound Pound Molecule (Mole) of Molecule (Mole) of Std Std Air Air

386 scf/lb.mole

28.96 1lbmda/lb.mole

= 13.33 scf/lbmda

3333BABA--11--33

The Molecular Weight of WaterThe Molecular Weight of Water

Element MWH 2

3434

H2 20 16

18 lbm

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Calculate the Volume of a Pound Calculate the Volume of a Pound Molecule (Mole) of HMolecule (Mole) of H2200

386 scf/lb.mole = 21 scf/lbm H20

3535

2

18 lbm H20/lb.mole

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PSYCHROMETRICSPSYCHROMETRICS

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PSYCHROMETRICSPSYCHROMETRICS

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52 North Carolina Industrial Ventilation Conference

BA-1-3 7

PsychrometricsPsychrometrics

The science that deals with the thermodynamic properties of moist air – dry air/water vapor mixture, and the utilization of these properties to analyze conditions and processes involving moist

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air.

For the accuracy required in the majority of air conditioning/moist air problems, the perfect gas relations can be applied when calculating the thermodynamic properties.

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PsychrometricsPsychrometrics

Atmospheric air consists of a large number of gases (including oxygen, nitrogen, argon, and carbon dioxide) as well as water vapor and various contaminants

Dry air exists when all water vapor and contaminants have

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Dry air exists when all water vapor and contaminants have been removed from atmospheric air

Moist air is a binary (two-component) mixture of dry air and water vapor. The amount of water vapor in moist air can vary from zero (dry air) to a maximum (saturation) which depends on temperature and pressure

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Properties of Moist AirProperties of Moist Air

Psychrometrics deals with the following defined properties of moist air: Dry Bulb Temperature –the temperature of a gas or mixture of

gases indicated by an accurate thermometer after correction for

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g yradiation. (ºF)

Wet Bulb Temperature – thermodynamic wet bulb, temperature is the temperature at which liquid or solid water, by evaporating into the air, can bring the air to saturation adiabatically at the same temperature. (ºF)

Dew Point Temperature – the temperature at which the condensation of water vapor begins for a given state of humidity and pressure as the temperature of the vapor is reduced. (ºF)

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Relative Humidity–the ratio of the mol fraction of water vapor present in the air, to the mol fraction of water vapor present in saturated air at the same temperature and barometric pressure. (%RH)

idi i f h f ( )

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Humidity Ratio– Ratio of the mass of water vapor (steam) associated with one pound mass of dry air. (Lbmw/Lbmda) or (Grainsw/Lbmda)

Enthalpy– thermodynamic property of a substance defined as the sum of its internal energy plus the quantity PV/J, where P=pressure of the substance, V=it’s volume, J=mechanical equivalent of heat. Often called total heat and heat content (Btu/Lbmda)

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Specific Volume–the ratio of the volume of the mixture to one pound mass of dry air (Ft³/Lbmda)

Vapor pressure– the pressure exerted by a vapor. If a i k t i fi t it li id th t th

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vapor is kept in confinement over its liquid so that the vapor can accumulate above the liquid, the temperature being held constant, the vapor pressure approaches a fixed limit called the maximum, or saturated vapor pressure, dependent only on the temperature and the liquid. (In. Hg)

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PSYCHROMETRIC CHARTPSYCHROMETRIC CHART

The Psychrometric Chart is a graphic representation of the thermodynamic properties of moist air. Since it is possible to have an infinite number of air-vapor combinations, to minimize the complexity of the chart, the air component is

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minimize the complexity of the chart, the air component is a fixed value - per pound of dry air.

Dr. Willis Carrier is credited with the development of the psychrometric chart in 1911. There are a number of psychrometric charts, defined by dry-bulb temperature ranges: Normal Temperature, Low Temperature and High Temperature; as well as, a chart for Humid Air

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52 North Carolina Industrial Ventilation Conference

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COMFORT APPLICATIONSCOMFORT APPLICATIONS

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COMFORT APPLICATIONSCOMFORT APPLICATIONS

For Comfort HVAC Systems, it is common practice to use Standard Air (df = 1) for the calculations when the following properties

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fall within the stated ranges:

Dry Bulb Temperature < 100ºFDew Point Temperature < 80ºF Elevation < 1000 Ft. ASL

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5151 51BABA--11--33 5252BABA--11--33

Sling Sling PsychrometerPsychrometer

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Battery Operated Battery Operated PsychrometerPsychrometer

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52 North Carolina Industrial Ventilation Conference

BA-1-3 10

Digital Digital PsychrometerPsychrometer

5555BABA--11--33 5656 56BABA--11--33

5757BABA--11--33 5858BABA--11--33

5959BABA--11--33 6060 60BABA--11--33

52 North Carolina Industrial Ventilation Conference

BA-1-3 11

Psychrometric EquationsPsychrometric EquationsHT = HS + HL

HT = Total Heat, BTU/Hr

HS = Sensible Heat, BTU/Hr

HL = Lateral Heat, BTU/Hr

HS = (1.08)(CFM)(∆t)

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HL = (0.68)(CFM)(∆W)

HT = (4.45)(CFM)(∆h)

HT = (500)(GPM)(∆t)

CFM - Volumetric airflow rate, Ft2/Min

∆t - Temperature differential, ˚F

∆W - Humidity ratio differential, Grains/Lb.

∆h - Enthalpy differential, BTU/Lb.

GPM - Volumetric water-flow rate

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Continuity EquationContinuity Equation

Q = VA

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Q = Volumetric flow rate, Ft3/MinV = Average velocity, Ft/MinA = Cross-sectional Area, Ft2

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PSYCHROMETRIC CLASS PROBLEMPSYCHROMETRIC CLASS PROBLEM

In an office, the following measurements were made with a sling psychrometer, 70ºF dry-bulb, and 58.5ºF wet-bulb. From a normal temperature psychrometric chart, determine the following thermodynamic properties:

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thermodynamic properties:

Dew Point TemperatureHumidity Ratio Relative Humidity EnthalpySpecific Volume

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CLASS PROBLEM SOLUTIONCLASS PROBLEM SOLUTION

Dew Point Temperature:Humidity Ratio: Relative Humidity:E h l

50.6ºF54.6 Grains/Lbm

50% RH

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Enthalpy:Specific Volume:

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25.33 Btu/Lbm13.5 Ft³/Lbm

TOTAL HEAT =TOTAL HEAT =SENSIBLE HEAT + LATENT HEATSENSIBLE HEAT + LATENT HEAT

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SENSIBLE HEAT + LATENT HEATSENSIBLE HEAT + LATENT HEAT

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SENSIBLE HEAT Heat which changes the temperature of a substance

without changing its state

6666

LATENT HEAT Heat which changes the state of a substance without

changing its substance Two familiar examples: latent heat of fusion (changing

ice to water) and latent heat of vaporization (changing water to vapor)

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52 North Carolina Industrial Ventilation Conference

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6767BABA--11--33 6868BABA--11--33

6969BABA--11--33 7070BABA--11--33

7171BABA--11--33 7272BABA--11--33

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BA-1-3 13

PROBLEMPROBLEM

The following two air streams are mixed in an air handling unit mixing box:

52,000 CFM @ 78°F Dry Bulb and 60% Relative Humidity19,000 CFM @ 94°F Dry Bulb and 78°F Wet Bulb

Determine the following thermodynamic properties of the

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Determine the following thermodynamic properties of the mixed air stream utilizing the normal temperature psychrometric chart:

Dry Bulb Temperature Wet Bulb TemperatureDew Point Temperature Humidity RatioRelative Humidity EnthalpySpecific Volume

BABA--11--33 7474BABA--11--33

SOLUTIONSOLUTION

NOTE:When the mixing of two air streams is plotted on the Psychrometric Chart, the mix point will be located on a line segment that connects the two

7575

gpoints.

Calculate the mix dry bulb temperature on both the mass flow basis and the volume basis (the mass flow basis is the more accurate).

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Solution (cont.)

7676BABA--11--33

Solution (cont.)Solution (cont.)

7777BABA--11--33

PROCESS APPLICATIONSPROCESS APPLICATIONS

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52 North Carolina Industrial Ventilation Conference

BA-1-3 14

CLASS PROBLEMCLASS PROBLEM

The following conditions were recorded at the exit of a process: 220°F dry bulb & 120°F dew point

Determine the properties listed below using the appropriate h t i h t

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psychrometric chart:

Wet Bulb Temperature: ___°FDensity Factor: ____Humid Volume: ____Ft³/LbmDA

Moisture: _____LbmW/LbmDA

Enthalphy: ______BTU/LbmDA

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CLASS PROBLEM SOLUTIONCLASS PROBLEM SOLUTION

Wet Bulb Temperature: Density Factor: Humid Volume:

28 °F0.74519.3 Ft³/LbmDA

8080

Moisture: Enthalphy:

BABA--11--33

0.083LbmW/LbmDA

143 BTU/LbmDA

Moist Air is a Mechanical Binary Moist Air is a Mechanical Binary ((TwoTwo--Component)Component)

Mixture of Dry Air and Water VaporMixture of Dry Air and Water Vapor

Moist air is considered a mixture of independent perfect gases (i.e., dry air and water vapor), each is assumed to obey the perfect gas equation of state as follows:

Dry air: pdaV = ndaRT

Water vapor: pwV = nwRTwhere

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pda = partial pressure of dry airpw = partial pressure of water vaporV = total mixture volumenda = number of moles of dry airnw = number of moles of water vaporR = universal gas constant 1545.32 ft.lbf/lb mol. ºRT = absolute temperature, ºR

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Class ProblemClass Problem

Determine the humid volume of one pound of dry air @ 70°F & 29.92“ Hg to which has been added sufficient water vapor to saturate it. One pound of dry air occupies 13.35 Ft³

F th t t bl th f t t d t

8282

From the steam tables, the pressure of saturated steam (water vapor) at 70°F is 0.74“ Hg & the volume occupied by one pound of water vapor is 868 Ft³

PT = PDA + PW

PDA = 29.92“ Hg - 0.74“ Hg = 29.18“ Hg

Note: The volume of the air will increase when it is saturated with water vapor (the pressure exerted by the dry air component has decreased, therefore its volume will increase).

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Humid Volume = (13.35) = 13.69Ft³

Note: The water vapor will occupy the same volume as the dry air portion of the mixture.

PSYCHROMETRIC CHART SOLUTION:

18.29

92.29

8383

The quantity of water vapor required to saturate one pound of dry air @ 70°F & 29.92“ Hg is ~ 110.4 Grains (From Psychrometric Chart).

(0.01577 Lbmw)(868 Ft³/Lbmw) = 13.69 Ft³

Note: The volume of water vapor is equal to the volume occupied by one pound of dry air at its partial pressure & at 70°F.

Lbm 0.01577 Grain/Lbm 7000

Grains 4.110

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MOIST AIR IS A MIXTURE OF DRY AIR MOIST AIR IS A MIXTURE OF DRY AIR AND WATER VAPORAND WATER VAPOR

Dry Air + Water Vapor = Moist Air (Mixture)To one pound of dry air is added 80 grains of water vapor @ 70°F & 29.92" Hg.Determine the humid volume of the partially saturated mixture of moist air.

8484

(383 SCF/Lb mole)(1/18 Lbw/Lbm mole)80 Grains/7000Grains/Lbm

(21.23 SCF/Lbmw)(0.0114Lbmw)Humid Volume = 13.35 Ft³DA + 0.24 Ft³W

BABA--11--33

= 21.28 SCF/Lbw

= 0.0114 Lbmw

= 0.24 SCF= 13.59 Ft³/LbmDA

52 North Carolina Industrial Ventilation Conference

BA-1-3 15

ADIABATIC COOLINGADIABATIC COOLING

In the Adiabatic Cooling Process, water is introduced into the air stream by either a spray system or by passing the air through a media l d ith t Th th d i i

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laden with water. The thermodynamic process is constant enthalpy (h) – no heat is added or rejected. The dry bulb temperature of the air decreases and the humidity ratio of the air increases at a constant enthalpy. The process is normally plotted on the psychrometric chart on the wet bulb line.

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Example:Example:

The moisture removed during a drying process is 120 Lbm/min. The 20,000 SCFM of dry air required for the process is discharged @ 500F.

Determine the following:W t t D Ai R ti

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• Water to Dry Air Ratio• Dew Point Temperature• Wet Bulb Temperature• Humid Volume• Enthalpy• Density Factor Mixture (Utilize the Psychrometric Chart for Humid Air, Fig.9-j IVM as applicable)

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Water to Dry Air Ratio:LbmW/LbmDA = (120 LbmW/Min)(1/20,000Ft³/Min)(1/0.075 LbmDA/Ft³)

= 0.08 LbmW/LbmDA

Dew Point Temperature: Wet Bulb Temperature:

118°F142°F

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Wet Bulb Temperature: Humid Volume: Enthalpy: Density Factor of Mixture:

If the above conditions exist at the inlet of a wet collector having a humidifying efficiency of 83%, what is the air volume (ACFM) and temperature? Assume that the process proceeds adiabatically.

BABA--11--33

142 F27 Ft³/LbmDA

220 BTU/Lbm0.53

HUMIDIFYING EFFICIENCYHUMIDIFYING EFFICIENCY

nn = IVM p. 9 - 32

Where:Nn = Humidifying Efficiency, %Ti = Dry-Bulb temperature at collector inlet F

100 x Ts - Ti

To - Ti

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Ti = Dry Bulb temperature at collector inlet, FTo = Dry-Bulb temperature at collector outlet, FTs = Adiabatic saturation temperature, F

0.83 =

To = 203 FHumid Volume = 20.7 Ft³/LbmDA

ACFM = (20,000Ft³/Min)(0.075 LbmDA/Ft³)(20.7 Ft³/LbmDA)ACFM = 31,050

142) - (500

) o

T - 500(

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CONSERVATION OF MASSCONSERVATION OF MASS

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CONSERVATION OF ENERGY

BABA--11--33

When mixing two air streams using the When mixing two air streams using the poundpound--mass method, the following mass method, the following

relationships are applied:relationships are applied:

Mass (Ideal Gas)ma = pa (Qa) = pstd (dfa)(Qa)

Conservation of Massma + mb = mc

9090

Conservation of Energyma (ha) + mb (hb) = mc (hc)

For Ideal Gasma (Cp)(Ta) + mb (Cp)(Tb) = mc (Cp)(Tc)Cp cancels out of the equation

ma (Ta) + mb (Tb) = mc (Tc)

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BA-1-3 16

EXAMPLE:EXAMPLE:

Calculate the mixture temperature of the following two air streams, (assume sea level, and no moisture).

1 – 5500 CFM @ 175F2 2500 CFM @ 70F

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2 – 2500 CFM @ 70F

)500,5(175 460

70 460)075.0(

1m

)500,3(70 460

70 460)075.0(

2m

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344.3 Lbm/min

187.5 Lbm/min

m1 (T1) + m2 (T2) = m3 (T3)

Note: m3 = m1 + m2

)(Tm)( Tm

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T3 =

T3 =

= 598 – 460 = 138 F

)m (

)(Tm)(

21

2211

m

Tm

R 598 187.5) 3.344(

0)(187.5)(53 (635))3.344(

BABA--11--33