Energy Efficient BuildingsPsychrometrics

IntroductionThe air around us is mixture of dry air and water vapor, and can be modeled as a mixture of these two ideal gasses. The study of moist air is called psychrometry. The properties of moist air at a given pressure are displayed on a psychrometric chart. The psychrometric chart contains a rich density of information, and learning to visualize air heating and cooling properties as they occur on the psychrometric chart is very useful.

The two cardinal dimensions on the psychrometric chart are dry bulb temperature, T (F), on the horizontal axis and specific humidity, w (lbw/lba), on the vertical axis. Extensive properties, such as enthalpy and volume are expressed in terms of pounds of dry air. The enthalpy of air increases with both temperature and humidity. The relative humidity of air is the faction of water vapor the air can hold at a given temperature. The relative humidity of air is bounded at 100% along the left side of the psychrometric chart. The specific volume of air is the reciprocal of the density, and increases with temperature. Any two independent properties fix the state.

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For many years, the most common method for measuring humidity was by using a “sling psychrometer”. A sling psychrometer has two thermometers attached to a handle so that the thermometers can rotate around the handle. The bulb of one thermometer is wrapped in wet cloth and is called the wet bulb. The other bulb is called the dry bulb. The temperature of the web bulb thermometer will decrease due to evaporation as the thermometers are swung about. The quantity of evaporation depends on the humidity of the air; thus the difference between wet and dry bulb temperatures is a measure of ambient humidity.

Conceptually, the wet-bulb temperature is the same as the adiabatic saturation temperature, which is the exit temperature of air traveling over water through an infinitely long chamber. Because the evaporation process occurs at constant enthalpy, lines of constant enthalpy on the psychrometric chart are essentially parallel to lines of constant wet-bulb temperature.

CoolingThe total heat removed from air during a cooling process can be calculated from mass and energy balances on a system in which air enters and leaves a cooling chamber and heat is removed from the chamber.

Conservation of mass on a control volume states that the sum of all the mass flows in minus the sum of all mass flows out equals the change in mass stored in the system. At

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steady state conditions, no mass is accumulated in the cooling chamber. Thus, for this system, conservation of mass gives:

Min - Mout = Mstoredma1 – ma2 = 0ma1 = ma2

where ma1 and ma2 are the mass flow rates of moist air entering and leaving the chamber.

Conservation of energy on a control volume states that the sum of all energy flows in minus the sum of all energy flows out equals the change in energy stored in the system. At steady state conditions, no energy is accumulated in the cooling chamber. The energy of non-reacting mass flows is the sum of the kinetic, potential and internal flow energies. In most cooling applications, the change in velocity (V) and elevation (z) between the entrance and outlet of the chamber are negligible. Thus, for this system, conservation of energy gives:

Ein - Eout = Estoredma1 (V1

2/2 + gz1 + h1) – ma2 (V22/2 + gz2 + h2) – Qtot = 0

Qtot = ma1 (h1 – h2)

where Qtot is the total heat removed from the moist air and h1 and h2 are the enthalpies of moist air entering and leaving the chamber.

Sensible CoolingWhen warm air is cooled as it passes over a cooling coil, the temperature begins to decrease. As the temperature decreases, the air can hold less moisture before becoming 100% saturated. If the air leaves the coil before becoming saturated, then no condensation occurs. On a psychrometric chart, the exiting air is at a lower temperature than the incoming air, while the humidity ratio remains constant since no moisture is condensed from the air. Reducing the temperature of air without changing the quantity of water in the air is called sensible cooling, Qsen, and is shown schematically below.

120Δω

2am

1am

Qtot

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Example20,000 cfm of air is sensibly cooled from a temperature of 90 F and a relative humidity of 60% to a temperature of 75 F. Determine the total heat removed from air.

V 1=20,000 cfm from T = 90°F, φ = 60% to T = 75°F

1: T=90 ° F φ=60% h1=41 .74 [Btu

lba ] v1=14 . 26[ft3

lba ] w1=. 01827[ lbwlba ]

2: T=75 ° F w1=w2 h2=38 .02[Btu

lba ]M-balance:

ma1=ma2 and ma1=V 1

v1(Note V 1≠V 2 because v1≠v2)

E-balance:ma1h1−Qtot−ma2h2=0 (SS)

Qtot=ma1(h1−h2)=V 1

v1(h1−h2)=20,000[ft3

min ]×60[minhr ]×1

14 .26 [ lbaft3 ]×( 41.74−38.02 )[Btu

lba ]=313,705[Btu

hr ]

Sensible cooling can also be calculated from the change in air temperature, rather than the change in air enthalpy. Specific heat at constant pressure is defined as:

c p≡(dhdT )

p

Thus the change in enthalpy is then given by the following differential equation.

∫ dh=∫ c pdT

Most cooling operations occur at constant pressure. At atmospheric conditions, the specific heat of water is about 1.0 Btu/lb-F and the specific heat of dry air is about 0.24 Btu/lb-F. In sensible cooling applications, the quantity of water in the air does not change. Further, the change in temperature during the cooling process is small enough that the specific heats are nearly constant. Thus, the specific heat of moist air can be assumed to be constant, and can be taken outside the integral. Thus, in the case of sensible cooling, the change in enthalpy is:

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(h2−h1)=c p (T 2−T 1)

The specific heat, cp, of moist air is approximately 0.26 Btu/lb-F. The density, p, of moist air is approximately 0.075 lb/ft3. Thus, the product of the density and specific heat of moist air, pcp, is approximately 0.018 Btu/ft3-F. Using this approximation, sensible cooling, Qsen, in which the quantity of water in the air is unchanged, can be calculated from the volume flow rate of air and the change in air temperature.

Qsen = V1 pcp (T1 – T2)

Example20,000 cfm of air is sensibly cooled from a temperature of 90 F and a relative humidity of 60% to a temperature of 75 F. Determine the sensible cooling using the approximation that pcp = 0.018 Btu/ft3-F. Also determine the error between the enthalpy-based calculation and this approximate method.

Qsen=ma (h1−h2)=m c p (T 1−T2 )=Vv cp (T 1−T 2)=V ρcp (T1−T 2)

=20,000[ft3

min ]×60 [minhr ]×. 018[Btu

ft3⋅° F ]×(90−75 ) ° F=324,000 [Btuhr ]

The error between the exact and approximate methods=324,000−313,705313,705

≈3 . 3%

Sensible and Latent CoolingThe dew point temperature Tdewpoint is the temperature when air is 100% saturated and water begins to condense out of the air. When air is cooled below the dew point temperature, Tdewpoint, condensation occurs and moisture leaves the air stream. The exiting air stream is at a lower temperature and humidity ratio than the incoming air stream. This process is shown in the figure below. The cooling to reduce the temperature of the air is called sensible cooling. The cooling to condense water from the air is called latent cooling. Thus, this process includes both sensible and latent cooling. The total cooling, Qtot, is the sum of the latent cooling, Qlat, and the sensible cooling, Qsen.

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Qtot = ma (h1 – h2) = ma (h1 – h2p) + ma (h2p – h2) = Qlat + Qsen

The water removed from the air is:

mw = ma (w1 – w2)

ExampleDetermine the total, sensible and latent cooling required to cool 20,000 cfm of air from a temperature of 90 F and a relative humidity of 60% to a temperature of 55 F and 100% relative humidity. Also determine the mass flow rate of water removed from the air.

Cooling 20,000cfm from (90°F, φ=60% ) to (55 °F, φ=100% )

1: T 1=90° F φ1=60% h1=41 .74 [Btu

lba ] v1=14 .26 [ft3

lba ] w1=.01827 [ lbwlba ]

2: T 2=55 ° F φ2=100% h2=23 . 2[Btu

lba ] w2=. 009186[ lbwlba ]

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ma1h1−Qtot−ma h2=0 (SS)

Qtot=ma (h1−h2)=V 1

v1(h1−h2)

Qtot =(20,000×6014 .26 [ft3⋅lba⋅min

min⋅ft3⋅hr ]×( 41.74−23 .20 )[Btulba ])= 1,560 [kBtu

hr ]Qlat=ma (h1−h2p )=

V 1

v1(h1−h2p' )

Qlat =(20,000×6014 .26 [ft3⋅lba⋅min

min⋅ft3⋅hr ]× (41.74−31 .74 )[Btulba ])= 841.3 [kBtu

hr ]Qsen=ma (h2p−h2)=

V 1

v1(h2p−h2' )

Qsen =(20,000×6014 .26 [ft3⋅lba⋅min

min⋅ft3⋅hr ]×(31.74−23 .20 )[Btulba ])= 719 . 2 [kBtu

hr ]mw=ma (w1−w2)

=V 1

v1(w1−w2)=20,000

14 . 26 [ft3⋅lbamin⋅ft3 ]×( . 01827−.009186 ) [ lbw

lba ]×60[minhr ]=764 . 1 [ lbw

hr ]Latent cooling can also be calculated from the volume flow rate of air, V, and the enthalpy of evaporation of water. The density, p, of moist air is approximately 0.075 lb/ft3. The enthalpy of evaporation (condensation) of water, hfg, at atmospheric pressure is approximately 1,075 Btu/lbw. Using these approximations, latent cooling, Qlat, can be calculated as:

Qlat = V p (w1-w2) hfg

ExampleDetermine the latent cooling using the enthalpy of evaporation of water and an approximate value of air density required to cool 20,000 cfm of air from a temperature of 90 F and a relative humidity of 60% to a temperature of 55 F and 100% relative humidity. Also determine the error between the enthalpy-based calculation and this approximate method.

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Qlat=V ρ (w1−w2 ) h fg

=(20,000 [ft3

min ]×60[ minhr ]×. 075[ lba

ft3 ]×( . 01827−. 009186 )[ lbwlba ]×1,075 [Btu

lbw ])=878 .8 [kBtu

hr ]Error=878 . 8−841. 3

841 .3≈4 . 4%

Cooling Done by Cooling CoilIn HVAC systems, air is typically cooled by passing it over a cooling coil. Most cooling coils are finned-tube heat exchangers in which cool water or refrigerant flows through tubes, and the tubes have external fins to increase heat transfer area. If the air is cooled below the dewpoint temperature, water will condense and must be drained from the bottom of the cooling coil. A schematic of a cooling coil is shown below.

In this situation, Qc is that part of the total heat cooling (Qtot) that leaves with coolant in the cooling coil. Qw is that part of the total cooling (Qtot) that leaves with condensate draining from the bottom of the cooling coil. Thus,

Qtot = Qc + Qw

The energy carried away by the condensing water is equal to product of the the mass flow rate of the condensing water, mw, and the enthalpy of the saturated water, hw. The enthalpy of the saturated water is the enthalpy or air at 100% RH at the temperature of the cooling coil. The process is shown schematically below.

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The process can be modeled using mass balances on the air and water, and an energy balance on the entire process. The steady state mass and energy balances are shown below.

M-balance (air): ma1=ma2

M-balance (water):

mw1=mw2+mwmw=mw1−mw2=ma1w1+ma2w2=ma (w1−w2 )

E-balance:

ma1h1−Qc−mah2−mwhw=0 (SS )Qc=ma (h1−h2)−mwhw

ExampleDetermine the energy removed by the cooling coil and condensate when cooling 20,000 cfm of air from a temperature of 90 F and a relative humidity of 60% to a temperature of 55 F and 100% relative humidity. The fluid inside the cooling coil is at 55 F.

Cooling 20,000cfm from (90°F, φ=60% ) to (55 °F, φ=100% )

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1: T 1=90° F φ1=60% h1=41 .74 [Btu

lba ] v1=14 .26 [ft3

lba ] w1=.01827 [ lbwlba ]

2:

T 2=55 ° F φ2=100% h2=23 . 2[Btulba ] w2=. 009186[ lbw

lba ]

w: hw (Tc = 55 F, φ2=100% )=23 .2[Btu

lba ]

M

ass balance:

mw=ma (w1−w2)

=V 1

v1(w1−w2)=20,000

14 . 26 [ft3⋅lbamin⋅ft3 ]×( . 0183− .0092 ) [ lbw

lba ]×60 [minhr ]=764 . 1 [ lbw

hr ]

E

nergy balance:

Qc=ma (h1−h2)−mwhw

Qc =(20,000×6014 .26 [ft3⋅lba⋅min

min⋅ft3⋅hr ]× (41. 74−23 .2 )[Btulba ])−(764 . 1[lbw

hr ]×23 .2[Btulbw ])

Qc=1,560 [kBtuhr ] - 17 .73 [kBtu

hr ] = 1,543 [kBtuhr ]

Note that the energy carried away by the condensate (17.73 kBtu/hr) is much less than the total latent cooling (841.3 kBtu/hr). Most of the latent cooling is done by the cooling coil. Because the energy removed by the condensate is typically very small compared to the energy removed by the cooling coil, the energy transferred to the cooling coil, Qc, is frequently approximated as the total cooling, Qtot.

Bypass FactorIn most cooling applications, the air leaving the cooling coil is not entirely saturated since some air does not come in contact with the cooling coil. The fraction of air that misses the coil is called the bypass factor, BF. The bypass factor can be determined from the temperature of water supplied to the cooling coil and from the states of incoming and exiting air. For example, consider the cooling process shown below.

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The bypass factor, BF, can be determined from the temperatures of water supplied to the cooling coil and the incoming and exiting air temperatures as:

BF=T 2−Tcc

T 1−T cc

Example

Air enters a cooling coil at (90 ° F , φ=60%) , while the coil is at 55°F and the BF = 0.2. What is the leaving air temperature?

BF=T 2−T cc

T 1−T cc

T 2=BF (T 1−T cc)+Tcc=. 2 (90−55 )+55=62 ° F

Adiabatic MixingAir streams at different conditions are frequently mixed together. The condition of the exiting air stream can be determined from the conditions of incoming streams by applying mass and energy balances to the system. In most cases, heat loss from the system is negligible and the system can be modeled at adiabatic. On a psychrometric chart, the condition of the exiting air stream must be on the line connecting the incoming air streams. The distance along that line is determined by the ratio of the mass flow rates of the incoming streams.

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1

32

1

2

3

Air Mass Balancem1+m2=m3V 1 /v1+V 2 /v 2=V 3 /v3

Total Energy Balance:m1h1+ m2 h2=m3 h3

h3=m1h1+m2h2

m1+m2

h3≈V h1+V 2h2

V 1+V 2( if ρ1≈ ρ2)

Sensible Energy Balance:

m1c pT 1+m2c pT 2=m3c pT 3

T 3=m1T 1+m2T 2

m1+m2

T 3≈V 1T1+V 2T 2

V 1+V 2(if ρ1≈ρ2)

Sensible Heat Ratio, Latent Cooling and Air Flow Rate The sensible heat ratio, SHR, is the ratio of sensible cooling to total cooling.

SHR=Qsen

Q tot=

Qsen

Qsen+Qlat .

On a psychrometric chart, the total cooling line can be decomposed into sensible and latent cooling lines, as shown below.

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ExampleDetermine the sensible heat ratio for cooling 20,000 cfm of air from a temperature of 90 F and a relative humidity of 60% to a temperature of 55 F and 100% relative humidity.

Qtot=m¿

a (h1−h2)=ma (h1−h A )+ma (hA−h2)= Qlat + Qsen

SHR=Qsen

Q tot=ma (hA−h2)ma (h1−h2)

=h A−h2

h1−h2=31. 74−23 .2

41 .74−23 . 2=46%

For a given cooling coil, SHR varies with the volume of air flowing across it, V cc . As V cc

increases, the total cooling capacity, Qtot , increases. However, the increased volume flow rate of air also increases the quantity of moist air that does not come in contact with the cooling coil; which increases the bypass factor, BF. The increased bypass

factor, BF, causes latent cooling, Qlat , to increase less than total cooling, Qtot . The net result is that latent cooling decreases with higher air flow rates.

Thus, for dehumidification, it is advantageous to run air conditioners with low air flow rates for longer periods of time. It also follows that oversized air conditioning systems that meet total cooling loads by cycling on for short periods with high air flows and then remaining off for long periods are less able to provide adequate dehumidification than properly sized units which run more continuously.

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ExamplePerformance specifications from an air conditioning system are shown below. Consider the case when the condenser temperature is 120 F and the evaporator air wet bulb temperature is 72 F. Determine the SHR of the evaporator coil when operating with 9,000 cfm and 15,000 cfm of air across the evaporator. Note that the SHR increases, and latent cooling decreases.

72 67 62 72 67 62 72 67 62TC 389.8 354.2 320.6 410.7 374.9 340.1 424.4 388.0 353.1SHC 192.5 236.8 280.0 213.9 270.6 325.9 233.0 301.2 353.1KW 30.3 29.5 28.5 30.7 30.0 29.0 31.0 30.2 29.4THR 477.6 439.7 403.6 499.7 461.7 424.5 514.1 475.7 438.5TC 372.1 338.1 305.9 391.5 356.8 323.6 404.1 368.6 335.5SHC 185.6 229.9 273.0 206.7 263.3 316.2 226.2 293.8 335.5KW 32.9 31.9 30.7 33.3 32.4 31.4 33.6 32.8 31.8THR 467.3 430.6 395.2 488.0 450.8 414.7 501.3 463.6 427.8TC 348.0 315.6 284.7 365.1 332.1 300.7 376.2 342.8 311.2SHC 176.8 220.6 263.1 197.8 253.7 299.5 216.6 283.9 311.2KW 35.5 34.3 32.9 36.1 34.9 33.7 36.5 35.3 34.1THR 448.8 414.6 380.2 468.5 431.6 398.4 481.5 443.0 409.9TC 330.9 300.3 270.8 345.8 315.3 285.4 355.6 324.8 294.9SHC 170.5 214.3 257.0 190.9 246.9 285.4 209.8 277.1 294.9KW 37.9 36.5 35.0 38.5 37.2 35.9 38.9 37.6 36.3THR 440.7 406.3 372.3 457.5 423.0 389.5 468.3 433.7 400.3TC 320.9 290.2 261.2 335.3 304.9 274.8 344.7 314.2 283.8SHC 167.0 210.4 252.7 187.3 243.0 274.8 206.1 273.2 283.8KW 39.0 37.5 35.9 39.7 38.1 36.7 40.1 38.6 37.1THR 433.7 398.7 365.7 450.2 381.2 381.2 460.8 426.0 391.4

TC - Total Cooling (KBtu / hr) SHC - Sensible Cooling (KBtu / hr)KW - Electric Power Input (KW); Electric Power Input = Compressor Power + Evaporator Fan PowerTHR - Total Heat Rejected (KBtu / hr)

120

130

140

145

Cond Temp (F)

Evap Air cfm / BF9,000 / 0.15 12,000 / 0.18 15,000 / 0.21

Evap Air Entering Wet BulbTemp (F)

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V=9,000 cfm V=15,000 cfmQtot 372.1 404.1Qsen 185.6 226.2Qlat 186.5 177.9SHR = Qsen/Qtot 0.50 0.56

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