Psychometric

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PSYCHROMETRICS

A. Psychrometric

Information

1. Psychrometricpsychro – meaning ‘cold’

metrics – meaning ‘measure of’

Actually, psychrometrics is

more than the measurement

of cold. It is a study of all

the properties of moist air.

AIRDRY

78% Nitrogen20.9% Oxygen1% Argon.1% Other Gases

WET78% Nitrogen20.9% Oxygen1% Argon.1% Other Gases

PLUSWater Vapor

2. Atmospheric Air

The weight of air pushing down on the

earth is referred to as atmospheric

pressure.

At sea level, the pressure of 70o dry air

is 14.696 lbs/in2(psi).

3. Standard AirAt a barometric pressure of 29.921

inches of mercury (14.696 psi),

one pound of 70o dry air will

occupy 13.33 cubic feet. Air at

these conditions is known as

standard air.

1Specific Density = ------------------------

Specific Volume

1= --------- = .075 lbs/ft3

13.33

4. Specific Density

5. Relationship between Specific Volume and Specific Density

Specific

Volume

Specific Density

5. Relationship between Specific Volume and Specific Density

6. Sensible Heat

If we wish to calculate the Btu’s

needed to raise the temperature of

dry air, we would use the sensible

heat formula.

Sensible Heat Formula

Btu/hr = Sp. Heat x Sp. Density x 60 min/hr x cfm x ΔT

Btu/hr = .24 x .075 x cfm x ΔT

Btu/hr = 1.08 x cfm x ΔT

7. BUT

Moisture is almost always present

in air and has a heat content of its

own.

This is known as latent heat.

8. Total Heat

is

Sensible Heat + Latent Heat

9. Enthalpyis

the term used to indicate the

total heat content

of one pound of air.

Enthalpyis

measured with a

wet bulb

thermometer.

10. Total Heat Formula

We use the total heat formula for

changes in BOTH sensible and

latent heat AND it is useful to

determine the capacity of an air

conditioning system.

Total Heat Formula

Btu/hr = Sp. Density x 60 min/hr x cfm x ΔH

Btu/hr = .075 x 60 x cfm x ΔH

Btu/hr = 4.5 x cfm x ΔH

11. Relative Humidity

is a ratio of the amount of

moisture present in the air to

the amount it can hold at

saturation.

12. Specific Humidity

The amount of moisture present

in the air expressed in grains of

moisture per pound of dry air.

7,000 grains of moisture in

one pound of water.

13. Dew Point

The temperature at which the

water vapor in the air becomes

saturated and starts to

condense into water droplets.

In Summary:

air has the following properties:

• Density(dry or wet)

• Volume

• Sensible Heat

• Latent Heat

and the following measurements can be found:

• Density

• Volume

• Temperature

• Dry bulb

• Wet bulb

• Dew Point

• Relative Humidity

Psychrometric

Chart

And now to the

The psychrometric chart is

simply a tool that can be used to

determine the properties of moist

air.

Construction

of the

Chart

C. PROCESSES• Sensible Heat

• Sensible Heat plus Humidification

• Chemical Dehydration

• Sensible Cooling

• Cooling and Dehumidification

• Evaporative Cooling

1. SENSIBLE HEAT PROCESSEntering Conditions: 69oF dry bulb

(return air) 55oF wet bulb

Determine: grains ________(specific humidity)

SENSIBLE HEAT PROCESSEntering Conditions: 69oF dry bulb


Determine: grains ___42___(specific humidity)




dew point ________oF




dew point ___44___oF





enthalpy ________Btu/lb





enthalpy __23.22__Btu/lb






relative humidity ______%






relative humidity __40__%







specific volume ________Ft3/lb.







specific volume __13.45__Ft3/lb.








specific density _______lbs/ft3








specific density __.074__lbs/ft3

SENSIBLE HEAT PROCESSLeaving Conditions: 95oF dry bulb

(supply air) 64.5oF wet bulb

Determine: grains _______(specific humidity)



Determine: grains __42__(specific humidity)




dew point ______oF




dew point __44__oF




dew point __44__oF

enthalpy ______Btu/lb




dew point __44__oF





dew point __44__oF






dew point __44__oF






dew point __44__oF



specific volume ______Ft3/lb.




dew point __44__oF







dew point __44__oF




specific density ______lbs/ft3




dew point __44__oF








dew point __44__oF





sensible heat factor ______




dew point __44__oF





sensible heat factor __1.00__

Sensible Heat Added


Btu/hr = 1.08 x 1000 x (95 – 69)

Btu/hr = 1.08 x 1000 x 26

Btu/hr = 28,080

2. COOLING and DEHUMIDIFICATION

Entering Conditions: 80oF dry bulb

(return air) 63.5oF wet bulb


COOLING and DEHUMIDIFICATION



Determine: grains __61___(specific humidity)





dew point ______oF





dew point __53.5__oF


































specific volume ______ Ft3/lb.








specific volume __13.78__ Ft3/lb.









specific density ______ lbs/ft3









specific density __.073__ lbs/ft3


Leaving Conditions: 60oF dry bulb











dew point ______oF





dew point __48__oF





dew point __48__oF






dew point __48__oF






dew point __48__oF







dew point __48__oF







dew point __48__oF



specific volume ______Ft3/lb.





dew point __48__oF








dew point __48__oF




specific density ______lbs/ft3





dew point __48__oF









dew point __48__oF





sensible heat factor ______





dew point __48__oF





sensible heat factor __.75__

Sensible Heat Removed


Btu/hr = 1.08 x 1600 x (80 – 60)

Btu/hr = 1.08 x 1600 x 20

Btu/hr = 34,560 Btu/hr

TOTAL Heat RemovedBtu/hr = 4.5 x cfm x ΔH

Btu/hr = 4.5 x 1600 x (28.94 – 22.30)

Btu/hr = 4.5 x 1600 x 6.64

Btu/hr = 47,808

qsSHR = --------

QT

34,560 SHR = -----------

47,808

•SHR = .73

IF1. You extend the process line through

the 100% saturation line, then

2. the air would be completely saturated

as it leaves the coil, thus

3. the air temperature would be equal to

the coil temperature.

What is that temperature?

43oF

a. APPARATUS DEW POINT

The temperature of the air at

which it leaves the coil saturated.

100% RH

What is the temperature at which

the air left our coil?

60oF

WHY?

b. BYPASS AIR

Because some of the air was

bypassed and unaffected by the

coil temperature.

This is known as

Bypass Factor

c. Conditions which affect the

BYPASS FACTOR

1. Fin Spacing2. Number of Rows & Depth of

Coil3. Type of Fin4. Velocity of Air5. If Coil is Wet or Dry6. Conditions of System

D. NOW

application of

PSYCHROMETRICS

APPLICATION 1

new unit installed in

existing building

3 TON

Day 1 Conditions

• Entering Air - 80oDB, 73oWB, 72%RH

• Leaving Air - 68oDB, 65oWB, 85%RH

• Determine:

• Sensible heat

• Latent heat

• Sensible Heat Ratio

Locate these two conditions

on the Psychrometric Chart

Entering Air - 80oDB, 73oWB

Leaving Air - 68oDB, 65oWB

Draw a line connecting the two

points.

Draw a vertical line down from the

entering conditions.

Draw a line horizontally to the

right from the leaving conditions.

At the intersection of these two

lines, draw a line upwards

following the wet bulb line until it

crosses the line connecting the

two points.

• NOTE that the VERTICAL line

represents the latent load, and

• NOTE that the HORIZONTAL line

represents the sensible load.

Where does this point of crossing

occur, in terms of distance from

either point?

SO

Use the SWAG method to determine

the approximate amount of sensible

load and latent load you have.

Another method to determine

the amount of sensible heat to

latent heat is:

1. Locate the 80DB, 67WB reference dot.

2. Place your pencil point on the dot.

3. Lay a straight edge against the pencil

point and use the dot as a pivot point.

4. Rotate the straight edge until it is

parallel to your original line.

5. Read the sensible heat percentage on

the far right of the chart.

NOW

Let us do it again for the ‘Day 2’

conditions.

Day 2 Conditions

• Entering Air - 80oDB, 67oWB

• Leaving Air - 63oDB, 58oWB

• Determine:

• Sensible heat

• Latent heat







points.









two points.







either point?

SO






latent heat is:



3. Lay a straight edge against the pencil

point and use the dot as a pivot point.

4. Rotate the straight edge until it is

parallel to your original line.

5. Read the sensible heat percentage on

the far right of the chart.

NOW

Let us do it again for the ‘One

Week Later’ conditions.

One Week Later Conditions

• Entering Air - 78oDB, 62oWB

• Leaving Air - 53oDB, 51oWB

• Determine:

• Sensible heat

• Latent heat







points.









two points.







either point?

SO






latent heat is:



3. Lay a straight edge against the pencil point

and use the dot as a pivot point.

4. Rotate the straight edge until it is parallel

to your original line.

5. Read the sensible heat percentage on the

far right of the chart.

Comparison of Critical Data

DAY 1

Temperature

Difference

DAY 2

Temperature

Difference

ONE WEEK LATER

Temperature

Difference

12 17 25


DAY 1

Sensible

Heat Ratio

DAY 2

Sensible

Heat Ratio

ONE WEEK LATER

Sensible

Heat Ratio

.43 .62 .86


DAY 1

Temperature

Difference

DAY 2

Temperature

Difference

ONE WEEK LATER

Temperature

Difference

12 17 25

CFM REQUIREMENTS

Temperature DROPFor

COOLING

1. Temperature DROP

18o – 22o

Minimum = 15o

Maximum = 25o

2. Application

Building with Sensible Load

HIGH

Should the temperature drop be closer

to 15 or 25?

2. Application

Building with Latent Load

HIGH

Should the temperature drop be closer

to 15 or 25?

3. Cooling

Temperature

Splits

(temperature drops)

Outdoor DB

Indoor

WB oF

Indoor DB 75oF

Indoor DB 78oF

Indoor DB 80oF

85oF

59

63

67

22

19

15

24

21

17

25

23

19

95oF

59

63

67

21

18

15

23

20

17

24

22

19

105oF

63

67

71

17

14

11

20

17

13

21

18

15

115oF

63

67

71

17

13

10

19

16

13

21

17

14

A17

APPLICATION 2

MIXTURE

TEMPERATURES

Mix outdoor air (OA)

with Return Air (RA)

THEN

The Mixture Air (MA)

passes over the coil

QUESTION:

What should be the

temperature of the mixed air?

Problem

Outdoor Ambient Temperature = 95 DB

Return Air Temperature = 78 DB

Required to have:

25% OA

75% RA

Two Methods

1. Formula

2. Psychrometric Chart

Formula

TEMPMA = (%OA x TEMPOA) + (%RA x TEMPRA)

TEMPMA = (.25 x 95) + (.75 x 78)

TEMPMA = 23.75 + 58.5

TEMPMA = 82.25oF

Psychrometric

Chart

1. Plot the following two points on the chart.

OUTDOOR AIR

95DB, 83WB

RETURN AIR

78DB, 65WB

2. Draw a line between the two points.

3. Locate a point approximately 25%

from the condition which has the

MOST air. Use the SWAG method.

4. That will be the mixed air

temperature.

Question:

What if you want to check the

%OA on an existing job?

Problem

Outdoor Ambient Temperature = 95 DB

Return Air Temperature = 78 DB

Mixed Air Temperature = 82.25 DB

Formula

TMA - TRA

%OA = -----------------

TOA - TRA

TMA - TRA %OA = -----------------

TOA - TRA

82.25 - 78 = ----------------- = .25 or 25%

95 - 78

Psychometric

Documents

5 fenthalpy

5of wet bulb

dew point

dew point

dew point

heat percentage

specific volume

wet bulb line