Department of Mechanical Engineering

ME 322 – Mechanical Engineering Thermodynamics

Lecture 33

Psychrometric Properties of Moist Air

Air-Water Vapor Mixtures

• Atmospheric air– A binary mixture of dry air (a) + water vapor

(w)– The air in the mixture is treated as a pure

substance even though it is really a mixture itself

• Applications– Heating, ventilating, and air-conditioning

(HVAC)

• Analysis– HVAC – pressures are always low ~ Patm

• Ideal gas law can be used for both air and water vapor

2

Properties of Moist Air

Both air and water vapor are treated as ideal gases that obey Dalton’s Law of Partial Pressures.

28.97 lbm/lbmol 18.016 lbm/lbmol

0.06855 Btu/lbm-R 0.1102 Btu/lbm-Ra w

a w

M M

R R

Dry Air Water Vapor

Since moist air is a binary mixture,

1a wy y

Btu1.986

lbmol-RR Universal Gas Constant:

3

Properties of Moist Air

The field of psychrometrics (air-water vapor properties) has adopted other properties to represent the composition of the mixture rather than the mole fraction.

w

a

y

y w

a

m

m

These properties are related,

/ 28.971.608

/ 18.016w w a w

a w a a

n n m M m

n n M m m

4

Mole Fraction Ratio Humidity Ratio

Properties of Moist Air

Relative Humidity

,

w w w

w sat sat sat

y P PP

y P P P

T

s

wP

satP

T

T-s diagram of water

State of the water vapor in the mixture

Partial pressure of the water vapor in the mixture

Partial pressure of the water vapor in a saturated mixture

a wP P P

Partial pressure of the dry air

Total pressure of the mixture

dpT

5

Dew Point Temperature

Properties of Moist Air

All of these properties are related. For example,

/

/

/

/

18.016 lbm/lbmol

28.97 lbm/lbmol

w w w

a a a

a w a w

w a w a

w w w

a a a

m P V R T

m P V R T

R P R M P

R P R M P

M P P

M P P

6

0.622 0.622

0.622 0.622

w w

a w

sat sat

a w

P P

P P P

P P

P P P

Example

Given: Moist air at the following state

70 F 14 psia 60% 0.60T P

Find: Various psychrometric properties of the moist air

Solution: Partial pressure of the vapor 0.60 w

sat

P

P

The partial pressure of the water in a saturated mixture can be found from Table C.1a,

0.60 0.3632 psia

0.2179 psia

w sat

w

w

P P

P

P

7

satPT

Example

Dew Point Temperature The dew point temperature is the saturation temperature of the water vapor at its partial pressure. Using Table C.1a,

0.2179 psiawP

Interpolating ...

55.5 FdpT

8

T

s

0.2179 psiawP

0.3632 psiasatP

70 FT

55.5 FdpT

If the mixture drops below this temperature, the water vapor will start condensing.

Example

9

Humidity Ratio

0.2179 psia

0.622 0.62214 0.2179 psia

w w

a a

m P

m P

lbm 7000 grains grains0.009834 68.8

lbm lbm lbmw

a w a

Grains – A new unit! A ‘grain’ is an ancient Egyptian measure of the mass of one grain of barley (7000 grains/lbm). Since the humidity ratios are typically very small, the HVAC industry has adopted the use of ‘grains’ to represent humidity ratio ...

lbm0.009834 0.009834

lbmw

a

Example

Mole Fraction Ratio

1.608 1.608 0.009834 0.015813w

a

y

y

Mole Fraction of each Component

0.0158130.016

1 1 1 0.015813

1 1 10.984

1 1 0.015813

w ww

a w

w aa

a a

y yy

y y

y yy

y y

Notice that: 1v ay y

10

Intensive Moist Air Properties

Consider the enthalpy of the mixture …

a wH H H

Question: How can the specific enthalpy of the air-water vapor mixture be specified?

Answer: The total enthalpy must be divided by a mass value. Which mass value should be used?

11

a a w wH m h m h

a a w wH m h m h

Intensive Moist Air Properties

In air conditioning applications, the water vapor mass can vary due to condensation or evaporation (dehumidification or humidification). Thus, specific properties of the mixture are based on the dry air,

Units: Btu/lbma or kJ/kga

12

a wh h h

wa w

a a

mHh h h

m m

Intensive Moist Air Properties

, ,

a w

a w

p pa pw

v va vw

a a w w

u u T u T

h h T h T

c c T c T

c c T c T

s s T P s T P

Using ideal gas mixing for the components of moist air, the internal energy, enthalpy, heat capacities, and entropy of the mixture can be calculated by,

13

Department of Mechanical Engineering

ME 322 – Mechanical Engineering Thermodynamics

Example

Heating of a Moist Air Stream

14

Example

15

Given: Moist air flowing at 300 cfm enters a heating unit at 65°F, 14 psia with a relative humidity of 50%. The moist air leaves the heating unit at 110°F, 14 psia.

Find: (a) The heat transfer rate required (Btu/hr)(b) The relative humidity of the air leaving the heater

Q

1

1

1

14 psia65 F0.50300 cfm

PT

V

2

2

14 psia110 F

PT

Example

16

Q

1

1

1

14 psia65 F0.50300 cfm

PT

V

2

2

14 psia110 F

PT

The First Law applied to the heating system is,

2 2 2 2 1 1 2 1a a w w a a w wQ m h m h m h m h

The mass flow rate of the dry air does not change (in this case the water vapor mass flow does not change either … why?). Therefore,

1 2a a am m m

Example

17

Q

1

1

1

14 psia65 F0.50300 cfm

PT

V

2

2

14 psia110 F

PT

2 12 2 1 1

w wa w a w

a a a

m mQh h h h

m m m

2 2 2 2 1 1 2 1a a w w a a w wQ m h m h m h m h

Rearranging the First Law,

The mass flow rate of the dry air is,

1

1a

a

Vm

v

2 2 2 1 1 1a w a wh h h h

Example

18

Q

1

1

1

14 psia65 F0.50300 cfm

PT

V

2

2

14 psia110 F

PT

The specific volume of the dry air at state 1 is found using the ideal gas EOS with the partial pressure of the dry air,

11

1

aa

a

R Tv

P

The partial pressure of the dry air at state 1 is found knowing the relative humidity,

11

1

w

sat

P

P

1 1sat satP P T

1 1 1a wP P P

Example

19

Q

1

1

1

14 psia65 F0.50300 cfm

PT

V

2

2

14 psia110 F

PT

The component enthalpy values can be found using the ideal gas model for each component,

The humidity ratio at state 1 can be found,

11

1

0.622 w

a

P

P

Example

20

Q

1

1

1

14 psia65 F0.50300 cfm

PT

V

2

2

14 psia110 F

PT

No water vapor is added to or taken from the moist air from state 1 to 2. Therefore,

1 2

At this point, the problem can be solved for the heat transfer rate. We are also interested in the relative humidity at the exit of the heater. This can be found from the humidity ratio at 2,

22

2

w

sat

P

P

2 2sat satP P T 2 2 2a wP P P

22

2

0.622 w

a

P

P

Example

21

Q

1

1

1

14 psia65 F0.50300 cfm

PT

V

2

2

14 psia110 F

PT

Solution (Key Variables):

Even though the humidity ratio stays constant in this process, the moist air leaving the heater will feel uncomfortably ‘dry’. This is a common problem encountered in heating processes. The ‘dryness’ can be alleviated by injecting water vapor into the moist air stream leaving the heater (humidification).

Example

22

Q

1

1

1

14 psia65 F0.50300 cfm

PT

V

2

2

14 psia110 F

PT

What would happen if the moisture content is neglected and the mixture is treated as dry air?

Since no water vapor is added or removed from the moist air in this process, neglecting the moisture results in a small error. However, neglecting the moisture does not reveal the relative humidity at the exit!