CHAPTER 2.

HVAC FUNDAMENTALS

2.1 Human Thermal Comfort

2.2 Basic Physics

2.3 Properties of Moist Air

2.4 Energy Transport in HVAC Systems

2.5 HVAC Load Estimation

2.1 HUMAN THERMAL COMFORT

Personal Variables

• Metabolic Rate (type of activity, age, sex)

• Clothing (clo)

Environmental Variables

• Air Temperature (K, ℃)

• Humidity (%, kg/kgda)

• Air Flow Velocity (m/s)

• Mean Radiant Temperature (K, ℃)

2.1.1 Variables Affecting Thermal Comfort Level

1) Metabolic Rate (activity, age, sex)

Thermal Interaction of Human Body and Environment

Energy Balance Equation

The terms in the energy balance equation have units of power per unit area and refer to the surface area of the nude body. The most useful measure of nude body surface area, originally proposed by DuBois and DuBois (1916), is described by:

2) Clothing (clo)

Clothing insulation value may be expressed in clounits.

1.0 clo is equivalent to 0.155 (m2K/W)orR = 0.155 clo (m2K/W)

Because clothing insulation cannot be measured for most routine engineering applications, tables of measured values for various clothing ensembles can be used to select an ensemble comparable to the one(s) in question.

3) Temperature and Relative Humidity

Because people typically change clothing styles to suit seasonal weather, ASHRAE Standard 55 specifies summer and winter comfort zones appropriate for clothing insulation levels of 0.5 and 0.9 clo [0.078 and 0.14 (m2K/W), respectively.

Summer comfort operative temperature:23 – 27 oC

Winter comfort operative temperature:21 – 25 oC

Comfort relative humidity:30 – 60%

Operative temperature (to) is defined as a uniform temperature of a radiantly black enclosure in which an occupant would exchange the same amount of heat by radiation plus convection as in the actual nonuniform environment.

Where,

hc = convective heat transfer coefficient (W/m2K) (kcal/m2h oC)hr = radiative heat transfer coefficient (W/m2K) (kcal/m2hoC) ta = air temperature (K) (oC)tmr = mean radiant temperature (K) (oC)

4) Air Flow Velocity

Air flow speed affects the rates of convective heat loss and evaporation from human body skin.

A study of the effect of air velocity over the whole body found thermal comfort level unaffected in neutral environment by air speed of 0.25 m/s or less (Berglund and Fobelets 1987).

Draft is an undesired local cooling of the human body caused by high speed air movement. Draft has been identified as one of the most annoying factors in offices. When people sense draft, they often demand higher air temperatures in the room or that ventilation systems be stopped.

5) Mean Radiant Temperature

6) ASHRAE Thermal Sensation Scale

Mass/Weight

Density/Specific Gravity/Specific Weight

Power/Force

Work/Energy/Heat

Specific Heat

Sensible/Latent Heat/Enthalpy

Pressure

Fluid Flow

2.2 BASIC PHYSICS

Components of a large HVAC system (Based on hot-chilled water system)

2.2.1 Mass versus Weight (질량:중량)

The mass, m, of an object is a fundamental property of the object; a numerical measure of its inertia; a fundamental measure of the amount of matter in the object. Its SI unit is kilogram(kg, or kgm).

The weight of an object is the force of gravity on the object and may be defined as the mass times the acceleration of gravity, G = mg. Since the weight is a force, its SI unit is the newton(N) or kgf.

G = m(kg) × g (m/s2) = N (newton)

Note the picture at right. All the girl’s weight (force due to gravity) is being supported by the swing set. If one were to stand behind her at the bottom of the arc and try to stop her, one would be acting against her inertia, which arises purely from mass, not weight

Technically, whenever someone stands on a balance-beam-type scale at a doctor’s office, they are truly having their mass measured. This is because balances (“dual-pan” mass comparators) compare the weight of the mass on the platform with that of the sliding counterweights on the beams

Conversely, whenever someone steps onto spring-based or digital load cell-based scales (single-pan devices), they are technically having their weight (force due to strength of gravity) measured.

e.g.) Find the mass and weight of the air in a room at 20 oC with a 4.0m x 5.0m floor and a ceiling 3.0m high.

mair = ρair V = (1.2 kg/m3) x (60 m3) = 72 kgm

Gair = mairg = (72kgm) x (9.8 m/s2) = 700 N or 700 kgf

2.2.2 Density, Specific Gravity, Specific Weight (밀도:비중:비중량)

Mathematically, density is defined as mass divided by volume

Density(ρ) is defined as mass divided by volume:

)/mkg( 3mV

m where, m = mass (kgm), V = volume (m3)

Specific gravity (or relative density) of a material is the ratio of its density to the density of water at 4.0 oC, 1000 kg/m3.

water

materialSG

Specific Weight (γ ) is defined as weight per unit volume. Weight is a force. It is expressed as:

where, γ = specific weight (N/m3)ρ =density (kgm/m3)g = acceleration of gravity (m/s2)

g

Density of air will vary as the temperature and moisture content in the air varies. When the temperature increases, the higher molecular motion results in an expansion of volume and thus a decrease in density.

The amount of water vapor in the air also effects the density. Water vapor is a relatively light gas when compared to diatomic Oxygen and diatomic Nitrogen. Thus, when water vapor increases, the amount of Oxygen and Nitrogen decrease per unit volume and thus density decreases because mass is decreasing.

The two most abundant elements in the troposphere are Oxygen and Nitrogen. Oxygen has an 16 atomic unit mass while Nitrogen has a 14 atomic units mass. Since both these elements are diatomic in the troposphere (O2 and N2), the atomic mass of diatomic Oxygen is 32 and the diatomic mass of Nitrogen is 28.

Water vapor (H2O) is composed of one Oxygen atom and two Hydrogen atoms. Hydrogen is the lightest element at 1 atomic unit while Oxygen is 16 atomic units. Thus the water vapor atom has an atomic mass of 1 + 1 + 16 = 18 atomic units. At 18 atomic units, water vapor is lighter than diatomic Oxygen (32 units) and diatomic Nitrogen (28 units). Thus at a constant temperature, the more water vapor that displaces the other gases, the less dense that air will become.

You may be familiar with the concept that moist air is less dense than dry air. This is true when both have the same temperature or when the moist air is warmer. Said in another way, air with a greater percentage of water vapor will be less dense than air with a lesser percentage of water vapor at the same temperature.

Density of Air (dry air versus moist air)

Often people erroneously believe that moist air is denser than dry air because very moist air is more difficult to breathe than dry air.

2.2.3 Power versus Force (능력 : 가해진 힘)

Power is the rate at which work is performed or energy is converted.Its unit is W (watt). 1W = 1 J/s = 0.86 kcal/h.

Force is a push or pull that can cause an object to accelerate. Force has both magnitude and direction, making it a vector quantity. Its unit is N (newton).

F (N)= m(kgm) x a(m/s2). 1 N= 1 kg m/s2

Pulling force

Pulling force due to mass and gravity

Pushing force due to magnetism

A lineman is both

STRONG(applies a big force)

and FAST(displaces objects in

small times, powerful)

Linemen of Steelers, Pittsburgh, PA.

James Watt(1736-1819), Scottish inventor and mechanical engineer

2.2.4 Work

Work (mechanical): The amount of energy transferred by a force and its SI unit is J (joule).

W (J)= F(N) x d (m), 1 J = 1 Nm

Work (thermodynamic): The quantity of energy (or heat) transferred from one system to another. Its SI unit is J (joule).

1 J = 0.24 cal, 1 cal = 4.186 J

Ws1mPas

mkgm

s

mkgmN1J1 3

2

2

2

James Prescott Joule (1818-1889), British physicist

Joule’s Heat Apparatus 1845, Science Museum, London

Joule's apparatus for measuring the mechanical equivalent of heat

1 J = 1 N mpaddle

weight

2.2.5 Energy

kinetic, potential, thermal, gravitational, sound, elastic, electromagnetic energy

• Kinetic Energy

KE = ½ mv2 Unit: (kg)(m/s)2 = J

• Potential Energy (gravitational)

PEgrav = mgh Unit: (kg)(m/s2)m = (kg)(m/s) 2=J

• Thermal Energy ≈ Heat

2.2.6 Heat

In physics and thermodynamics, heat is energy transferred from one body or thermodynamic system to another due to thermal contact when the systems are at different temperatures. Therefore, heat is thermal energy in transit.

• Specific heat capacity of dry air:

Cpda = 1.006 kJ/kg K = 0.24 kcal/kg oC

• Specific heat of water:

Cpwater = 4.186 kJ/kg K = 1 kcal/kg oC

• Specific heat of water vapor at constant pressure:

Cpwv = 1.84 kJ/kg K = 0.44 kcal/kg oC

2.2.7 Specific Heat

Specific heat is defined as the amount of energy that has to be transferred to or from one unit of mass or mole of a substance to change its temperature by one degree.

1 cal = 4.186 J

Specific heat of water as a function of temperature. The value of c varies by less than 1% between 0 oC and 100 oC.

2.2.8 Sensible Heat / Latent Heat / Enthalpy

Sensible heat is the amount of energy released or absorbed by a chemical substance during a change of temperature.

Qs = m cp ∆T

where,m = mass of the substance (kg)cp = specific heat capacity of the substance (kJ/kgK, kcal/kg oC)∆T = the change in temperature of the substance (K, oC)

Latent heat is the amount of energy released or absorbed during a phase change, such as the condensation of water vapor.

QL = m L

where, m = mass of the substance (kgm)L = specific latent heat for the substance (kJ/kgm)

• Latent heat of fusion (ice to water):

QLf = 333.55 kJ/kg = 79.68 kcal/kg

• Latent heat of vaporization (100 oC water to steam):

QLv = 2256 kJ/kg (at 100 oC ) = 539 kcal/kg

• Evaporation heat of water at 0 oC:

Qwe = 2501 kJ/kg = 597 kcal/kg

1 cal = 4.186 J

Enthalpy is the sum of sensible heat and latent heat. The unit is kJ/kg, or kcal/kg).

The surrounding air is at room temperature, but this ice-water mixture remains at 0 °C until all of the ice has melted and the phase change is complete.

Sensible and latent heats of water at different phases

A = Latent heat of fusion of ice

B = Sensible heat of water

C = Latent heat of vaporization

of water at 100 ° C

2.2.9 Pressure

Pressure is an effect which occurs when a force is applied on a surface. Pressure is the amount of force acting on a unit area. The symbol of pressure is P.

The SI unit for pressure is the Pa (pascal). 1 Pa = 1 N/m2

A

FP

where,F = normal force (N)A = area (m2)

The mercury barometer was first discovered by the Italian Evangelist Torricelli in 1643 and has been since then known as the Torricelli barometer.

1) Atmospheric Pressure

Atmospheric pressure is usually measured with a barometer ("bar" o "meter" - an instrument that measures "bars“).

Standard atmospheric pressure at average sea level:

760 mmHg = 1013.25 mb = 1013.25 hPa(1 hPa=100 Pa)

2) Pressure of Water Head (Column)

In fluid dynamics, head is a concept that relates the energy in an incompressible fluid to the height of an equivalent static column of that fluid. Head is expressed in units of height such as meters or feet.

Water head (mAq, “Aq =Aqua”)

1 mmAq = 1 kg/m2

1 mAq = 1000 kg/m2

2.2.10 Fluid Flow

The mass of a moving fluid doesn’t change as it flows.

Consider a portion of a flow tube between two stationary cross sections with areas A1 and A2. The fluid speeds at these sections are v1 and v2, respectively.

During a small time interval dt, the fluid at A1 moves a distance v1dt, so a cylinder of fluid with height v1dt and volume dV1=A1v1dt flows into the tube across A1. During this same interval, a cylinder of volume dV2=A2v2dt flows out of the tube across A2.

If the fluid is incompressible, so that the density ρ has the same value at all points, the mass dm1 that flows into tube across A1 in time dt is dm1= ρ A1v1dt. Similarly, the mass dm2

that flows out across A2 in the same time is dm2 = ρ A2v2dt. In steady flow the total mass in the tube is constant, so dm1 =dm2

and : ρ A1v1dt = ρ A2v2dt or

1) The Continuity Equation (Conservation of Mass)

A1v1 = A2v2 (continuity equation for incompressible fluid)

The product Av is the volume flow rate dV/dt, the rate at which volume crosses a section of the tube:

dV/ dt = Av (volume flow rate)

Previous equation (A1v1 = A2v2 ) can be generalized for the case in which the fluid is not incompressible (i.e., different density).

If ρ1 and ρ2 are the densities at sections 1 and 2, then:

In HVAC engineering, water is usually treated as incompressible fluid, while air is treated as compressible fluid.

ρ1 A1v1 = ρ2 A2v2 (continuity eq. for compressible fluid)

The mass flow rate (dm/dt) is the mass flow per unit time through a cross section. This is equal to the density ρ times the volume flow rate dV/dt.

dm/dt = ρ(dV/dt) = ρAv (mass flow rate)

EXAMPLE:

As part of a lubricating system for heavy machinery, oil of density 850 kg/m3 is pumped through a cylindrical pipe of diameter 8.0 cm at a rate of 9.5 liters per second.

(1 L = 0.001 m3 or 1000L = 1 m3)

(a) What is the speed of the oil? What is the mass flow rate?(b) If the pipe diameter is reduced to 4.0 cm, what are the new values of the speed and

volume flow rate? Assume that the oil is incompressible.

ANSWERS:

(a) v1 = (dV/dt)/A1 = 1.9 m/s, The mass flow rate is ρ(dV/dt) = 8.1 kg/s(b) v2= (A1/A2)v1 = 7.6 m/s, The volume flow rate has the same value (9.5 L/s) in both

sections of pipe.

2) Bernoulli’s Equation (Conservation of Energy)

Bernoulli's principle can be derived from the principle of conservation of energy. This states that in a steady flow of an incompressible fluid the sum of all forms of mechanical energy in the fluid along a streamline is the same at all points on that streamline.

The total energy at a given point in a fluid is the energy associated with energy from pressure in the fluid, plus the movement of the fluid, plus energy from the height of the fluid relative to an arbitrary elevation.

Deriving Bernoulli’s Equation

1) The net work dW done on the fluid element during dt, due to the pressure of the surrounding fluid:

The pressure at the two ends are p1 and p2; the force on the cross section at a is p1A1, and the force at d is p2A2.

The net work dW done on the element by the surrounding fluid during this displacement is

dW = p1A1ds1 – p2A2ds2

Where, dV=A1ds1=A2ds2

Remember,Pressure = Force / AreaForce = Pressure * AreaWork = Force * displacement

dW = (p1 - p2)dV

2) The net change in kinetic energy during dt:

At the beginning of dt, the fluid between a and b has volume A1ds1, mass ρA1ds1, and kinetic energy:

At the end of dt, the fluid between c and d has volume A2ds2, mass ρA2ds2, and kinetic energy:

The net change in kinetic energy dK during time dt is:

21

22 )(

2

1)(

2

1vdVvdVdK

21

2111 )(

2

1)(

2

1vdVvdsA

)2

1( 2mvK

22

2222 )(

2

1)(

2

1vdVvdsA

))((2

1 21

22 vvdVdK

3) The net change in gravitational potential energy:

At the beginning of dt, the potential energy for the mass between a and b is dm gy1 = ρ dV gy1.

At the end of dt, the potential energy for the mass between c and d is dm gy2 = ρ dV gy2

The net change in potential energy dU during dt is

)( mghU

)()( 12 yygdVdU

4) Combining dW, dK and dU

dW = dK + dU

By dividing above equation by dV we obtain:

Above equation states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the changes in kinetic and potential energies per unit volume that occur during the flow.

Then, above equation can be changed to a more convenient form as:

)()()(2

1)( 12

21

2221 yygdVvvdVdVpp

)()(2

112

21

2221 yygvvpp

22221

211 2

1

2

1gyvpgyvp

The subscripts 1 and 2 in the previous equation refer to any two points along the flow tube, so we can also write the following equation which states that the total energy per unit volume at a point in the fluid flow is:

By dividing above equation by ρg (or specific weight γ) we obtain:

where, 1st term = Pressure head (due to static pressure)2nd term = Velocity head (due to bulk motion of a fluid, kinetic energy)3rd term = Elevation head (due to the fluid weight and height)

)Pa(N/mconstant 2

1 22 orygvp

(m)constant 2

2

yg

v

g

p

In fluid dynamics, head is a concept that relates the energy in an incompressible fluid to the height of an equivalent static column of that fluid.

The venturi meter or manometer (shown on the right) is a common type of flow meter which can be used in many liquid applications to convert differential pressure heads into volumetric flow rate, linear fluid speed, or mass flow rate using Bernoulli's principle.

The reading of these meters (in inches of water, for example) can be converted into a differential, or gauge pressure.

Venturi Meter

The venturi meter in the diagram on the right shows two columns of a measurement fluid at different heights. The height of each column of fluid is proportional to the pressure of the fluid.

From continuity equation: v1A1=v2A2 and

Therefore, Also,

21

2221

222

211

2

1

2

12

1

2

1

vvpp

vpvp

12

12 v

A

Av

}1){(2

1

2

1)(

2

1

2

2

121

21

21

2

121

A

Av

vvA

App

1)(

2

}1){(2

1

2

2

11

2

2

121

21

AA

ghv

A

Avgh

ghpp

Daniel Bernoulli

(Groningen, 8 February 1700 – Basel, 8 March 1782)

He was a Dutch-Swiss mathematician and was one of the many prominent mathematicians in the Bernoulli family.

He is particularly remembered for his applications of mathematics to mechanics, especially fluid mechanics, and for his pioneering work in probability and statistics.

Bernoulli's work is still studied at length by many schools of science throughout the world.

2.3 PROPERTIES OF MOIST AIR

2.3.1 Psychrometry

Moist Air = (dry air molecules) + (water vapor)

Psychrometry or psychrometrics are terms used to describe the field of engineeringconcerned with the determination of physical and thermodynamic properties ofgas-vapor mixtures. The term derives from the Greek psuchron (ψυχρόν) meaning"cold"[1] and metron (μέτρον) meaning "means of measurement“.

A psychrometric chart is a graph of the thermodynamic properties of moist air at aconstant pressure (often equated to an elevation relative to sea level). TheASHRAE-style psychrometric chart was pioneered by Willis Carrier in 1904.

It depicts these properties and is thus a graphical equation of state. The propertiesare:

Dry-bulb temperature (DBT) is that of an air sample, as determined by an ordinary thermometer, the thermometer'sbulb being dry. The SI units for temperature are kelvins(K) or degrees Celsius(°C).

Wet-bulb temperature (WBT) is the reading of a thermometer whose sensing bulb is covered with a wet sockevaporating into a rapid stream of the sample air (see Hygrometer). When the air sample is saturated with water, theWBT will read the same as the DBT.

Dew point temperature (DPT) is that temperature at which a moist air sample at the same pressure would reach watervapor “saturation.” At this point further removal of heat would result in water vapor condensing into liquid water fog or(if below freezing) solid hoarfrost

Humidity ratio (also known as moisture content or mixing ratio) is the proportion of mass of water vapor per unit massof dry air at the given conditions (DBT, WBT, DPT, RH, etc.). Humidity ratio is expressed as grams (or kilograms) ofwater per kilogram of dry air.

Relative humidity (RH) is the ratio of the mole fraction of water vapor to the mole fraction of saturated moist air at thesame temperature and pressure. RH is dimensionless, and is usually expressed as a percentage. Note: the notion that air"holds" moisture, or that moisture “dissolves” in dry air and saturates the solution at some proportion, is an erroneousconcept (see relative humidity for further details).

Specific enthalpy symbolized by h, also called heat content per unit mass, is the sum of the internal (heat) energy of themoist air in question, including the heat of the air and water vapor within. The SI unit of enthalpy is given in joules perkilogram of air.

Specific volume, also called inverse density, is the volume per unit mass of the air sample. The SI unit is cubic metersper kilogram of dry air.

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1) Effect of Temperature on Humidity

The moisture-holding capacity of the air depends on the air temperature.

Warm air can hold more moisture than cold air.

The same humidity ratio results in different relative humidities at differenttemperatures.

Hum

idity

rat

io

Dry bulb temperature

Horizontal movement on the psychrometric chart (no change in absolute humidity)

2) Sensible Heating and Cooling

3) Humidification

Humidification with steam or water spray will increase the humidity ratio and relative

humidity.

Cooling is the most common method for dehumidifying moist air.

If moist air is cooled to the saturation curve, further cooling will reduce temperature and remove moisture

A portable dehumidifier

4) Dehumidification

Evaporative cooling can be accompanied by humidification of warm dry air.

Sensible heat loss = Latent heat gain

(no change in enthalpy)

Upward movement along a line of constant enthalpy

Energy saving cooling device in hot-dry climatic regions.

5) Evaporative Cooling

2.3.2 Specific Enthalpy of Moist Air

wvda hxhh

enthalpy = sensible heat + latent heat

Where,h = specific enthalpy of moist air (kJ/kg, kcal/kg)hda = specific enthalpy of dry air (kJ/kg, kcal/kg)x = humidity ratio (kg/kgda)hwv = specific enthalpy of water vapor (kJ/kg, kcal/kg)

1) Specific heat of dry air (Sensible heat)

Where, Cpda = specific heat capacity of dry air at constant pressure

= 1.006 kJ/kg K = 0.24 kcal/kg °C

t = dry bulb temperature (K, °C)

Therefore,

tCph dada

)kcal/kg(24.0)kJ/kg(006.1 tthda

1 cal = 4.186 J

2) Specific enthalpy of water vapor (Latent heat)

Where, Cpwv = specific heat capacity of water vapor at constant pressure

= 1.84 kJ/kg K = 0.44 kcal/kg °C

t = temperature of water vapor (K, °C)

hwe = evaporation heat of water at 0 °C= 2501 kJ/kg = 597 kcal/kg

Therefore,

wewvwv htCph

1 cal = 4.186 J

)kcal/kg(59744.0)kJ/kg(250184.1 tthwv

Finally, the specific enthalpy of MOIST AIR (Sensible heat + Latent heat) is:

By ignoring the temperature of water vapor:

Total heat of moist air m kg at temperature t °C and humidity ratio x kg/kgda is:

kcal/kg)59744.0(24.0

kJ/kg)250184.1(006.1

txth

txth

kcal/kg59724.0

kJ/kg2501006.1

xth

xth

kcal)59724.0(

kJ)2501006.1(

xtmQ

xtmQ

Example:

What is the enthalpy of moist air at 25 °C with humidity ratio x=0.0203 kg/kgda.(x=20.3g/kgda)

1) By including the temperature of water vapor:

h = (1.006 × 25) + 0.0203(1.84 × 25+ 2501) = 76.85 kJ/kgh = (0.24 × 25) + 0.0203(0.44 × 25 +597) = 18.34 kcal/kg

2) By ignoring the temperature of water vapor:

h = (1.006 × 25) + (2501 × 0.0203) = 75.92 kJ/kgh = (0.24 × 25) + (597 × 0.0203) = 18.12 kcal/kg

Also, try to find the answer from the psychrometric chart.

Energy is obtained from an outdoor bonfire (energy source) using a fire shovel

2.4 ENERGY TRANSPORT IN HVAC SYSTEMS

The energy is transported to an indoor fire pot (destination)

Manual Energy Transport

Mechanical Energy Transport using Fluid (Water, Air, and Steam)

Duct and diffuser to transport energy by air

Hot water coil installed on floor

Hot water or steam radiator

2.4.1 Heat Transport by Fluid Flow (Sensible Heating and Cooling)

HVAC equipment loads, equipment capacity, and output are expressed as quantities per unit time (kcal/h), or rates:

Where, Q = heat flow (kcal/h)G = mass flow (kg/h)Cp = specific heat capacity of fluid (kcal/kg ℃)∆t = temperature difference

tCpGQ

The heat liberated from a quantity of fluid is:

Where, q = liberated heat (kcal)m = mass (kg)Cp = specific heat capacity of fluid (kcal/kg℃)∆t = temperature difference (final – initial temperature) (℃)

tCpmq

1) Heat Transport by Water

Where, Q = heat transported by water (kcal/h)G = mass flow of water (kg/h)1.0 = specific heat capacity of water (1.0 kcal/kg℃)∆t = temperature difference (t2-t1) (℃)

ΔtGQ 0.1

Hot water radiator

2) Heat Transport by Air

Where, Q = heat transported by air (kcal/h)G = mass flow of air (kg/h)0.24 = specific heat capacity of air mass (kcal/kg℃)0.29= specific heat capacity of air volume (kcal/m3 ℃)∆t = temperature difference (t2-t1) (℃)

ΔtVQΔtGQ 29.0or24.0

t1 t2

Specific volume of air = 0.83 m3/kg

ΔtVΔtV

ΔtGQ 29.083.0

24.024.0

where, Q = heat transported by steam (kcal/h)G = mass flow of steam (kg/h)539 = latent heat of vaporization of water at 100 ℃ (kcal/kg)

GQ 539

Steam radiator

2.4.2 Heat Transport by Fluid Phase Change

Water Flow

Airflow

Steam Flow

2.4.3 Determining Fluid Flow Rates for HVAC Systems

)kg/h()(0.1 12 tt

QGwater

)kg/h()(24.0 12 tt

QGair

)kg/h(539

QGsteam

2.3.5 Sensible Heat and Humidity Controls in AHU

)kcal/h()(597)(24.0 1212 xxGttGQ

Boiler

t1, x1

)(597)(24.0 1212 xxttG

Q

)59724.0()59724.0( 1122 xtxt

12 hh

)kcal/kg(12 hhh

1) Heating and Humidification

t2, x2

2) Cooling and Dehumidification

)kcal/h()(597)(24.0 1212 xxGttGQ

)(597)(24.0 1212 xxttG

Q

)59724.0()59724.0( 1122 xtxt

12 hh

)kcal/kg(12 hhh

Chiller

t1, x1

t2, x2

3) Adiabatic Mixing of Two Moist Airstreams

air 1 (RA)

air 2 (OA) air 3

A common process in air-conditioning systems is the adiabatic mixing of two moist airstreams.

The RA(Return Air) and OA(Outdoor Air) are mixed and filtered before heating, cooling, and humidification/dehumidification.

The mass flows of RA and OA are controlled by dampers.

Adiabatic mixing is governed by three equations:

Air Mass: G1 + G2 = G3 (Mass conservation)

Enthalpy: G1h1 + G2h2 = G3h3 (Energy conservation)

Humidity Ratio: G1x1 + G2x2 = G3x3 (Mass conservation)

Eliminating G3 gives:

2

1

13

32

13

32

G

G

xx

xx

hh

hh

The state point of the resulting mixture lies on the straight line connecting the state points of the two streams being mixed, and divides the line into two segments, in the same ratio as the masses of the air in the two streams.

2

1

13

32

13

32

G

G

xx

xx

hh

hh

Therefore, when two airstreams are adiabatically mixed, first draw a straight line connecting the two state points of streams being mixed and use the equation for total mass and one of the two governing equations for enthalpy and humidity ratio.

If the enthalpy equation and mass is used:

G1h1 + G2h2 = G3h3

G1 + G2 = G3

321

2211 hGG

hGhG

EXAMPLE

A stream of 5,000 kg/h of outdoor air at 10 ℃ dry-bulb temperature and 30% relative humidity is adiabatically mixed with 15,000 kg/h of return air at 25 ℃ dry-bulb temperature and 50% relative humidity. Find the dry-bulb temperature, enthalpy and humidity ratio of the resulting mixture.

SOLUTION

1) Mark the states of the two airstreams on the psychrometric chart.2) Find the enthalpy values, humidity ratios of the two streams.3) Use the governing equations to calculate the enthalpy of the resulting mixture.4) Mark the state of the mixture on the psychrometric chart.5) Then, find other psychrometric values such as dry-bulb temperature, wet-bulb

temperature, relative humidity, humidity ratio, etc. from the psychrometric chart.

h2=50.1

h3=?

h1=16

t3=?t1=10 t2=25

x3=?

x1=.0025

x2=0.01

h3 = 41.6 kJ/kg(da)

h2=50.1

h3=41.6

h1=16

t3=?t1=10 t2=25

x3=?

x1=0.0025

x2=0.01

Verification of Enthalpy Values by Calculation

The dry-bulb temperature of the mixture also can be calculated by using the extended form of the governing equation.