CHAPTER 3

MOIST AIR PROPERTIES AND THE CONDITIONING PROCESS

PSYCHROMETRICS

Psychometrics: Deals with the properties of moist air.

Willis Carrier made a significant contribution to air conditioning field when he published relations of moist air properties and developed the PSCHROMETRIC CHART.

ATMOSPHERIC AIR

• A mixture of multiple gasses plus water vapor and pollutants.

• In 1949 the International Joint Committee on Psychrometric Data published a “standard composition of dry air” Table 3-1

Subtitles can get confusing when dealing with air water vapor mixtures

Try to remember these subtitles:a: dry airv: water vapor in mixture of air

and water vapor.s: saturated vapor (entirely water vapor)

Based on these properties we can find molecular mass Ma of dry air as:

Ma = (32.00 x 0.2095) + (28.016 x 0.7809) + (39.944 x 0.0093) + (44.010 x 0.0003) =

Ma = 28.965

The ideal gas law tells us that:

P v = P/ρ = RaT

We also know that the gas constant (Ra) of an ideal gas can be expressed as:

Ra = Ř/Ma

Where, Ř is the universal gas constant Ř = 1545.32 (ft-lbf)/(lb mole-R)and Ma is the molecular mass of dry air from table 3-1 = 28.965

Ra = 1545.32/28.965 = 53.352 (ft-lbf)/(lb mole-R)

HVAC Processes involve a mixture of dry air and moist air:

The molecular mass of water (Mv) is 18.015

Therefore we can find the gas constant for water vapor as Rv = Ř/ Mv

Rv = 1545.32/18.015= 85.78 (ft-lbf)/(lb mole-R)

For the purpose of HVAC design and analysis ASHRAE developed a definition of the US

standard atmosphere as:

• Acceleration due to gravity (constant) = 32.174 ft/sec2

• Temperature at sea level = 59.0 F (70 F also used)• Pressure at sea level = 29.921 in. Hg (30 in. often used)• The atmosphere consists of dry air which behaves as a

perfect gas• Density of air at standard sea level = 0.0765 lbm/ft3

To deal with the issue of changes in altitude above sea level at different locations throughout the world, ASHRAE developed the following relation:

P= a + bHWhere, P is pressure in inches of mercury (in. Hg)

H is elevation above sea level in Feet (appendix B)a and b are constants given in table 3-2

Let’s look at moist air properties

Table A-1a in Appendix A gives properties of saturated air (steam) at various temperaturesWe can use these properties to find the mass density of saturated air:

ρ = 1/v = pv /RvT

Where, ρ = mass density of saturated air v = vapor volume (ft3/lb)pv = partial vapor pressure (lbf/in2)

T = temperature (R = F + 460)Rv = gas constant for water vapor (85.78 (ft-lbf)/(lb mole-R)

Let’s look at an example

Find the mass density of a saturated mixture of air and water vapor at 80 F. (100% RH)

• Step 1: Look up saturation pressure in Table A-1a for 80 F. we get Ps = 0.507 lbf/in2.

• For saturated air Ps = Pv

• We know that Rv = 85.78 (ft-lbf)/(lb mole-R)

ρ = Pv/ Rv T = (0.507)(144)/(85.78)(459.67+80)

ρ = 0.001577 lbm/ft3

We could also find density as:

ρ = 1/v

where v is the vapor volume of the saturated air at 80 F from table A-1a which equals:

ρ = 1/632.67 = 0.00158 lbm/ft3

we got ρ = 0.001577 lbm/ft3

Some important psychometric terms:

• Humidity Ratio (W) : the ratio of mass of water vapor (mv) to mass of dry air (ma) in a mixture. Defined as:

W= mv / ma

• Relative Humidity (φ) : the ratio of the mole fraction of water vapor (xv ) in a mixture to the mole fraction of water vapor (xs ) in a saturated mix at same temp. and pressure.

φ = [xv / xs ] t,P

Recall that In a mixture of ideal gases, each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature. The total pressure of a gas mixture is the sum of the partial pressures of each individual gas in the mixture such that for a mixture of dry air and water vapor we get;

p = pa + ps

For a mixture of perfect gasses the mole fraction can be expressed as the partial pressure of each constituent:

Xv = pv /P and Xs = ps /P

So now we can reduce the relative humidity (φ ) to:

φ = [xv / xs ] t,P = (pv /p)/ (ps /p)

φ = pv /ps

So far, we’ve derived the relative humidity (φ) to be equal to the partial pressure of the water vapor (pv) in the mixture divided by the partial pressure of the vapor in a saturated mixture (ps) such that φ = pv /ps

we can go another step, we derived earlier that: ρv = pv / Rv T and ρs = p/ Rv T

Since we assume temperature of dry air and vapor in the mixture to be the same, T and R cancel out leaving us with:

φ = [ρv / ρs ] t,P

So we now have:

φ = [ρv / ρs ] t,P

Where the densities ρv and ρs are referred to as the absolute humidity of the water vapor which defines the mass of water per unit volume of a mixture.

Values for ρv and ρs can be found in table A-1a

The humidity ratio (W) can also be stated as:

W = m v p v / m a p a = 18.015 p v /28.965 p a

W = 0.6219 p v / p a

We can derive a relationship between relative humidity (Φ) and humidity ratio (W). We have:

W = 0.6219 p v / p a and φ = Pv /Ps

W = 0.6219 p v / p a and φ = Pv /Ps

φ = pv /ps : pv = Φ ps

W = 0.6219 p v / p a and W = 0.6219 p s / p a

Substituting pv = φ ps yields,

W = 0.6219 φ ps / p a

φ = Wp a / 0.6219 p s

• The degree of saturation (µ) is the ratio of humidity ratio (W) to the humidity ratio (Ws) of a saturated mixture (at same T and P)

µ = [W/Ws] t,p

• The dew point (td): the temperature that moist air first starts to condense when cooled. The dew point is fixed by the humidity ratio (W) or by the partial pressure of the water vapor (pv).

• The ehthalpy (i) of a mixture of perfect gasses is equal to the sum of the ehthalpy of each constituent,

i = ia + W iv

Where, ia is the enthalpy of dry air and

iv is the enthalpy of water vapor

Because the amount of moisture in a air-water vapor mix can vary in some processes we reference the enthalpy to unit mass of dry air.

Therefore, enthalpy values have the units of Btu/Lb

dry air and kJ/kg dry air

If we assume ideal gas behavior and select T=0 as our reference state the enthalpy becomes a function of temperature ONLY since ia = 0 at T=0°F and is = 1061 Btu/lb at T=0F

Using T=0° as a reference point allows use to use the following simple relations:

ia = cpa t

iv = is + cpv t

ia = cpa t

iv = is + cpv t

We know is = 1061Btu/lb at T=0F

We can check these relations by setting t=0We get:

ia = ccp (0) = 0

iv = is + ccv t = 1061 + ccv (0)= 1061 Btu/lb

IT WORKS!!!

If we combine these two relations to reflect a mixture of dry air and water vapor we get:

im = cpa t + is + cpv t

For HVAC work it’s acceptable to assume cpa and cpv remain constant as;

Cpa = 0.240 Btu/(Lbm-F) Cpa = 1 kJ /kg-C

Cpv = 0.444 Btu/(Lbm-F) Cpv = 1.86 kJ /kg-C

If we plug the values for Cpa and Cpv into these relations for I we get:

im = 0.240(t) + W(1061.2 + .444(t)) Btu/Lba

and

im = 1.0(t) + W[2501.3 + 1.86(t)] kJ/kga

Remember these relations!!!

Let’s try an example:

Compute the enthalpy of saturated air at 60°F and standard atmospheric pressure (14.696).

use im = 0.240(t) + W(1061.2 + .444(t)) Btu/Lba

Hint: you first have to find Ws using Ws = 0.6219 p s / p a where p s is found in table A-1a for t = 60°F

OK, one more hint, you can find p a as p (atmospheric) - p s

Ws = 0.6219 (p s / p a )= 0.6219 (p s / p-p s )

We find p s from table A-1a for 60°F

p s = 0.2563 lb/in2

Ws = 0.6219 [0.2563/(14.696-0.2563)]=0.01104 lbmv/lbma

Note that units of pressure lb/in2 or lb/ft2 doesn’t matter because they cancel out anyway!!

im = 0.240(t) + W(1061.2 + .444(t))

im = 0.240(60) + 0.1104(1061.2 + .444(60))

im = 26.41 Btu/Lba

In HVAC Design, we are often given certain parameters such as inlet air temperature, exiting

air temperature, humidity ratio etc.

The concept of Adiabatic saturation yields certain useful relations.

Air exiting the device at point 2 is assumed to be saturated such that the relative (φ) is 100%.

t*2 is adiabatic saturation temperature or thermodynamic wet

bulb temperature.

If we consider that the device is operating at a steady –flow-steady-state an energy balance on the device yields several useful relations.

pv2 = ps2 at t*2

i *fg2 = enthalpy of vaporization at t*

2

i *fg2 = iv1 - i *

w at t*2

iv1 = enthalpy of the vapor at t1

i *w = enthalpy of water at t*

2

All of these values can be found in table A-1a

Now the Adiabatic Saturation Relations

W1 = [((t*2- t1)) + (W*

s2)(i*fg2)] / (iv1 - i*

w)

Let’s try the example in the text together

To solve this problem we’ll use our saturation equation:W1 = [((t*

2 - t1)) + (W*s2)(i*

fg2)] / (iv1 - i*w)

We either are given or can look up in table A-1a except W*

s2 but we now know that:

Let’s find all the values together

We’re given t1= 80°F, t2= 64°F, patm = 14.696 lbf/in2

Use Table A-1a • Find pv2 as pressure at t2= 64°F (interpolate): pv2 = 0.299

• Find i *fg2 as iv1 - i *

w at t*2 = 64°F (interpolate)

60°F : iv1 - i *w = 1087.4 - 28.0 = 1059.4

70°F : iv1 - i *w = 1091.7 - 38.0 = 1053.7

Interpolate for 64°F: [(1059.4 - 1053.7)/10] x 4 = 2.281059.4 - 2.28 = 1057.1

• Find iv1 at t1= 80°F iv1 = 1096

• Find i *w at t*

2 = 64°F (interpolate) i *w = 32

Now it’s all “plug and play”t1= 80°F, t*

2= 64°F, patm = 14.696 lbf/in2, pv2 = 0.299

i*fg2= 1057.1, iv1= 1096, i*

w= 32, = 0.24

First let’s find W*s2 = =

W*s2= 0.0129 lbmv/lbma

Now we can find W1= [((t*2 - t1)) + (W*

s2)(i*fg2)] / (iv1 - i*

w)

W1= [((64 - 80)) + (0.0129)(1057.1)] / (1096 - 32)

W1= 0.0092 lbmv/ lbma

Now we need to find relative humidity φ1

We know that the relative humidity φ1 = pv1 /ps1

We can easily find ps1 for the saturated vapor exiting at 80°F from table A-1a as ps1 = 0.507 so,

So all we need to do now is find pv1 but how?

We can use the relation:W1 = 0.6219 (p v1 / p-p v1 )

We found W1= 0.0092 lbmv/ lbma so then we should be able to solve for p v1

= 0.2142 lb/in2

And now we can find φ1as:

φ1 = pv1 /ps1 = 0.2142/0.507 = 0.423 or 42.3 % RH

YOUR TURN!!!!

Find the humidity ratio (W), enthalpy (i), and specific volume (v) for saturated air at standard atmospheric pressure (p=14.696 lb/in2) at a temperature of 80°F.

Solutiont = 80°F p = 14.696 lbm/in2

Pv =0.507 lbm/in2

=

0.0222 lbv/lba

use im = 0.240(t) + W(1061.2 + .444(t)) Btu/Lba

im = 0.240(80) + 0.0222(1061.2 + .444(80)) = 43.55 Btu/Lba

And,v = Ra T / Pa=[ 53.35(460 + 80)] / [(14.6965-0507)(144)]

v = 13.61 ft3/lbm

Summary of Moist Air Relations

Subtitlesa: dry airv: water vapor in mixture of air and water vapor.s: saturated vapor (entirely water vapor)

Gas Constants

Dry Air: Ra = 53.352 (ft-lbf)/(lb mole-R)Water Vapor: Rv = 85.78 (ft-lbf)/(lb mole-R)

Densityρ = 1/v = pv /RvT

Where, ρ = mass density of saturated air v = vapor volume (ft3/lb)pv = partial vapor pressure (lbf/in2)

T = temperature (R = F + 460)Rv = gas constant for water vapor (85.78 (ft-lbf)/(lb mole-R)

Also remember that,ρ = 1/v

Humidity Ratio

W = and Ws =

W =0.6219 and Ws= 0.6219

Relative Humidity

φ = and φ =

φ =

Also remember that pa = patm - ps

Enthalpy of moist air

ia = cpa t

iv = is + cpv t

imix = ia + W iv and im = cpa t + is + cpv t

Specific heat for HVAV work (assume remains constant)

Cpa = 0.240 Btu/(Lbm-F) Cpa = 1 kJ /kg-C

Cpv = 0.444 Btu/(Lbm-F) Cpv = 1.86 kJ /kg-C

So, entering specific heat values yields:

im = 0.240(t) + W(1061.2 + .444(t)) Btu/Lba

And

im = 1.0(t) + W[2501.3 + 1.86(t)] kJ/kga

Examples

Find density of sat. mix air and water vapor at 80°Fρ =

Find at 80 F in table A-1a as Pv= 0.507 lbf/in2

We know that Rv = 85.78 (ft-lbf)/(lb mole-R)

So then,

ρ= = = ρ = 0.001577 lbm/ft3

Find density of sat. mix air and water vapor at

a) 32°Fb) 45°Fc) 95°Fd) 212°F

ρ =

ρ = = = ρ = 0.0003037 lbm/ft3

ρ = = = ρ = 0.0004986 lbm/ft3

ρ = = = ρ = 0.002494 lbm/ft3

ρ = = = ρ = 0.02932 lbm/ft3

Find the humidity ratio (W) of saturated air at atmospheric pressure and temperature of:

a) 32°Fb) 60°Fc) 95°Fd) 110°F

Ws = 0.6219 and = –

at t= 32°F = 0.089 lbm/in2

Ws 32°F= 0.6219 = 0.6219

Ws32°F = 0.003789 lbwater vapor/lbair

Ws 60°F = 0.6219 = 0.6219

Ws 60°F= 0.01103 lbw v/lbair

Ws95°F= 0.03697 lbw v/lbair

Ws110°F= 0.0591 lbw v/lbair

Find the enthalpy (i) of saturated air at atmospheric pressure and temperature of:

a) 32°Fb) 60°Fc) 95°Fd) 110°F

is = 0.240(t) + W(1061.2 + .444(t)) Btu/Lba

At 32 we found Ws 32°F = 0.003789 lb wv/lb air

So,is 32°F = 0.24(32) + (0.003789)[1061.2 + (.444)(32)]= 15.78 Btu/Lba

is 60°F = 0.24(60) + (0.01103 )[1061.2 + (.444)(60)]= 26.40 Btu/Lba

is 95°F = 0.24(95) + (0.03697)[1061.2 + (.444)(95)]= 63.59 Btu/Lba

is 110°F = 0.24(110) + (0.0591)[1061.2 + (.444)(110)]= 91.74 Btu/Lba

Find the humidity ratio (W), enthalpy (i), and specific volume (v) for saturated air at standard atmospheric pressure (p=14.696 lb/in2) at a temperature of 80°F.

Solutiont = 80°F p = 14.696 lbm/in2

Pv =0.507 lbm/in2

=

0.0222 lbv/lba

use im = 0.240(t) + W(1061.2 + .444(t)) Btu/Lba

im = 0.240(80) + 0.0222(1061.2 + .444(80)) = 43.55 Btu/Lba

And,v = Ra T / Pa=[ 53.35(460 + 80)] / [(14.6965-0507)(144)]

v = 13.61 ft3/lbm

THE PSYCHROMETRIC CHART

30 40 50 60 70 80 90 100 110 115DRY BULB TEMPERATURE - °F

ENTH

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ENTHALPY - Btu per lb. of dry air and associated moisture

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Chart by: HANDS DOWN SOFTWARE, www.handsdownsoftware.com

STANDARD AIR

1.000.95

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0.300.250.20SENSIBLE HEAT RATIO = Qs / Qt

SE

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BAROMETRIC PRESSURE: 29.921 in. HG

Chart by: HANDS DOWN SOFTWARE, www.handsdownsoftware.com

75F db 60F wb

90F db 70F wb

90F db 60% RH

90F db, W=0.012

90F db 60% RH72F db, 50% RH

Try these on your own:

a. 95F db, 70% RH find twb, Dew Point td, enthalpy (i), grains of moisture/lba

b. 70F db, 55% RH find twb, Dew Point td, enthalpy (i), grains of moisture/lba

c. Find the change (∆) in enthalpy for air cooled from 90F, 60%RH to 70F, 50%RH in BTU/lba

d. Find the grains of moisture removed from the air in c. above in lbs moisture/lba

a) 95F db, 70% RH twb = 86F, Dew Point td = 84F, enthalpy (i) = 51 BTU/lba, grains of moisture/lba = 177

b) 70F db, 55% RH twb =60F, td = 53F , enthalpy (i) = 26, grains of moisture/lba = 60

c) 41.8- 25.3 = 16.5 BTU/lba

d) 128.5 – 55 = 73.5 grains of moisture/lba

73.5/7000 = 0.0105 lb v/lba

Heating and cooling of moist air

When working with moist air, the water vapor must always be accounted for:

For dry air we have: q = m a i2 - m a i1

For moist air we account for water vapor using the humidity ratio (W):i1 = ia1 + W1 iv1

and

i2 = ia2 + W2 iv1

Also:

cp = cpa + W cpv

Let’s try one together

Find the heat transfer rate required to heat 1500 cfm (ft3/min) of air at 60 F and 90% RH to 110 F Without the addition of moisture.

To solve this problem we find point 1 on the chart at t=60F and RH =90%. Then we move horizontally along the humidity ratio line (W) until we get to t = 110 F (no moisture added so W remains constant). We then read the enthalpy (i) at each point.

60F db 90% RHTo 110 F W constant

At point 1 (60F90RH) we find i1 = 25.1 BTU/lb

At point 2 (110F Wconstant) we find i2= 37.4 BTU/lb

So we have:q = m(i2 - i1)

All we need now is m:m = = = 6753 lbma/hr

We find v1 from the chart at point 1 as 13.33 ft3/lbma

q = m(i2 - i1) = 6753 lbma/hr(37.4 BTU/lb- 25.1 BTU/lb)

q = 83,050 BTU/hr

COOLING AND DEHUMIDIFYING MOIST AIR

When air is cooled and water vapor condenses we must account for the change in enthalpy and the heat lost in the exiting condensate (iw mw).

Then q becomes:

q = ma (i1 – i2) – ma (W1 – W2) iw

Let’s try a cooling/dehumidification example

Moist air at 80F db and 67F wb is cooled to 58F db and 80% RH. Flow rate is 2000 cfm and condensed water vapor leaves at 60F. Find heat transfer rate.

q = ma (i1 – i2) – ma (W1 – W2) iw

W1 – W2 = water vapor removed from air

We obtain the following values from the chart:v1 = 13.85 ft3/lbmi1= 31.4 Btu/lbm W1= 0.0112 lbmv/lbma

i2= 22.8 Btu/lbmW2= 0.0082 lbmv/lbma

We can find iwfor the condensed water vapor from table A-1a for liquid water at 60F as iw= 28 Btu/lbm

ma = = 8,646 lbm/hr

q = 8,646(31.4– 22.8) – 8,646 (0.0112– 0.0082) 28q = 73,629 Btu/hr

Cooling and Dehumidifying

Involves BOTH sensible AND latent heat transfer

Sensible Heat: Decrease in DRY BULB temperatureLatent Heat: Decrease in humidity ratio (W).

qs = ma Cp(t2 – t1)

ql = ma (W2 – W1)ifg

Where ifg is latent heat of vaporization (1061 Btu/lb)

We can also say that:qs = ma (i3 – i1)

ql = ma (i2 – i3)

Total heat transfer is therefore:qt= qs + ql

Sensible heat factor (SHF) is:

SHF can be positive or negative depending on whether qsand qlinvolve heat into the system (+)

or heat out of the system (-)

Heating and Humidifying Moist Air

Heat Balance of device yields:

ma i1 + q + mwiw = ma i2

Where mw is mass of water added to system to humidify air and iw is enthalpy of added water

A mass balance yields: maW1+ mw= maW2

By combining both equations we get:

∆i/∆W = q/mw) +

q/ [ma( W2-W2] +

Some Psychrometric charts contain SHF curves

60F, 20%RH to 110F, 35% RH [W1 =.002 to W2 =.0195

Given:

Q= 1600 cfmT1 = 60 F T2 = 115FRH1 = 20% RH2 = 30%

Tw = 212F

First find enthalpy of injected water at 212F from table A-1aiw(212F) = 1150.4 Btu/lbm

We know that ∆i/∆W = iw = 1150.4 Btu/lbm

Next we find given state points 1 and 2 on the chart

And locate ∆i/∆W = 1150.4 Btu/lbm on the protractor scale as follows:

From this we can find ix = 29.2 Btu/lbm

Now that we have ix = 29.2 Btu/lbm we can find required heat transfer as:

q = ma (ix – i1)

ma can be found as:

ma = Q(60)/v1

where, v1 = 13.16 from psych. Chart

ma = 1600(60)/13.16 = 7296 lbm/hr

thus, q = ma (ix – i1)= 7296(29.2-16.8) = 90,500 Btu/hr

and, mv = ma (W2 – W1) = 7296(.0193-.0022) = 125 lbv/hr

Of course we could have done this mathematically

q = ma (i2 – i1) – ma (W1 – W2) iw

q = 7296 (49.2 – 16.8) – 7296 (.0193 – .0022) 1150.4

q = 92,864 Btu/hr close enough for HVAC work!!

Adiabatic Mixing of Two Streams of Moist Air

Energy balance yields: ma1 i1 + ma2 i2 = ma3 i3

Mass balance on dry air yields:ma1 + ma2 = ma3

Mass balance on water vapor yields:ma1 W1 + ma2 W2 = ma3 W3

From the combine energy and mass balance we can derive the following useful equations:

ma1 W1+ ma2 W2 = ma3W3

(2 - 3)/( - )= (W2 – W3)/ (W3– W1) = ma1/ma2

We also see that:

W3 = W1+[(ma2 /ma3) ((W2 – W1)]

To solve this problem mathematically we will use:

W3 = W1 + [(ma2 / ma3) ((W2 – W1)]

We must first find ma1 , ma2 , ma3

Use chart 1a to find v1 and v2 at states given in problem.

We find v1 = 13.2 ft3/lbma and v2 = 14.4 ft3/lbma

ma1 = (1000)(60)/13.2 = 4542 lbma/hrma2 = (2000)(60)/14.4 = 8332 lbma/hrma3 = 4542 lbma/hr + 8332 lbma/hr = 12,874 lbma/hr

W3 = W1 + [(ma2 / ma3) ((W2 – W1)]

Now we have:• ma1 = (1000)(60)/13.2 = 4542 lbma/hr

• ma2 = (2000)(60)/14.4 = 8332 lbma/hr

• Ma3 = 4542 lbma/hr + 8332 lbma/hr = 12,874 lbma/hr

so all we need now is W1 and W2 which are found on psych. charts as W1 = 0.0054 and W2 = 0.0103

W3 = 0.0054 + [(4542 / 12,874) (0.0103 – 0.0054)]

W3 =0.0103 lbmv/lbma

W3 can also be found graphically by the ratio of ma1/ ma2 here ma1/ ma2 = 2:1 so move 2/3 of the way along line to find point 3. Read W3

More about the usefulness of the Sensible Heat Factor (SHF)

We found that SHF = qs/(qs + ql) = qs/qt

Let’s see how the SHF can be used in an example:

Find point 1 on chart, then draw a line on SHF chart at .8 and draw a line from center of SHF chart. Now draw a parallel line from point 1 until you hit 75F db. We now have point2.

Now we can find ma for Q=1500 cfm as:

ma1 = (1000)(60)/13.11 = 6865 lbma/hr

We can also find i1= 21.6 Btu/lba and i2 = 27.8 Btu/lba

With this information we can now find qt as qt = ma (i2 – i1)

qt = 6865 lbma/hr(27.8 Btu/lba – 21.6 Btu/lba ) = 42,600 Btu/hr

We can also find q sensible as qs = qt(SHF) = 42,600 Btu/hr(.8)

Since we know that ql as qt = qs + ql

knowing qt and qs we can easily find ql as qt – qs or:

ql = 42,600 Btu/hr - 6865 lbma/hr (.8) = 8,500 Btu/hr

We find v1 = 13.11 ft3/lbma 3/lbma from the chart