topic 1 pyschrometry notes

7/25/2019 Topic 1 Pyschrometry Notes

1/50

Psychrometric properties of air

Topic 1


2/50

1. Introduction

! In all engineering work, consistent units must be employed. In this subject,

as in the ASHRAE handbooks, two systems of units will be used.

! English Engineering System - most commonly used in the U.S. with

some modification, such as use of inches instead of feet. Also knownas inch-pound or IP system.

! International System or SI - use in engineering practice throughout

most of the world and widely adopted in the U.S.

! Reason: Equipment that was designed using IP units will be operational for

years (or even decades). It will be necessary for engineers to work in either

IP or SI systems of units and to be able to make conversion from one

system to another.


3/50

1. Introduction

! Psychrometryis the study of the properties of mixtures of air and water

vapour.

! Atmospheric airis a mixture of many gases plus water vapour and a number

of pollutants.

! The amount of water vapour and pollutants vary from place to place. The

concentration of water vapour and pollutants decrease with altitude, and

above an altitude of about 10 km, atmospheric air consists of only dry air.

The pollutants have to be filtered out before processing the air.

! Moist air- the processed air which is a mixture of various gases that

constitutes of air and water vapour.


4/50

1. Introduction

! The moist air can be thought of as a mixture of dry air and moisture. For all

practical purposes, the composition of dry air can be considered as constant.

In 1949, a standard composition of dry air was fixed by the International Joint

Committee on Psychrometric data.

! Based on the above composition the molecular weight of dry air is found to

be 28.966 and the gas constant R is 287.035 J/kg.K.


5/50

1. Introduction

! At a given temperature and pressure the dry air can only hold a certain

maximum amount of moisture. When the moisture content is maximum, then

the air is known as saturated air.

! This condition is established by a neutral equilibrium between the moist airand the liquid or solid phases of water.

! For calculation purposes, the molecular weight of water vapour is taken as

18.015 and its gas constant is 461.52 J/kg.K.


6/50

1. Introduction

! Sensible heat

- energy added to a substance that produces a change in its temperature.

! Latent heat

- energy added to a substance that produces a change in its phase.


7/50

2. Properties of moist air

! Mixture of perfect gases according to Gibbs-Dalton law:


8/50


! Dry bulb temperature (DBT)is the temperature of the moist air as measured by a

standard thermometer or other temperature measuring instruments.

! Saturated vapour pressure (psat)is the saturated partial pressure of water vapour at

the dry bulb temperature. It is readily available in thermodynamic tables and charts.

ASHRAE suggests the following regression equation for saturated vapour pressure ofwater, which is valid for 0 to 100oC.


9/50


! Relative humidityis the ratio of the mole fraction of water vapour in moist air

to mole fraction of water vapour in saturated air at the same temperature and

pressure. Using perfect gas equation:

! It is normally expressed as a percentage. When the value is 100 percent, the

air is saturated.


10/50


! Humidity ratio (W)(or specific humidity) is the mass of water associated with each

kilogram of dry air. Assuming both water vapour and dry air to be perfect gases,

the humidity ratio is given by:

! Substituting the values of gas constants of water vapour and air Rvand Rain the

above equation; the humidity ratio is given by:

! For a given barometric pressure pt, given the DBT, we can find the saturated

vapour pressure psatfrom the thermodynamic property tables on steam. Thenusing the above equation, we can find the humidity ratio at saturated conditions,

Wsat.


11/50


! If unsaturated moist air is cooled at constant pressure, then the temperature

at which the moisture in the air begins to condenseis known as dew-point

temperature (DPT)of air. An approximate equation for dew-point

temperature is given by:

! Note: From its definition, the dew point temperature is the saturation

temperature corresponding to the vapour pressure of water vapour, it can

also be obtained from steam tables or using the equation of saturated vapour

pressure.


12/50


! The degree of saturationis the ratio of the humidity ratio W to the humidity

ratio of a saturated mixture Wsat the same temperature and pressure:


13/50


! The enthalpyof moist air is the sum of the enthalpy of the dry air and the

enthalpy of the water vapour. Enthalpy values are always based on some

reference value. For moist air, the enthalpy of dry air is given a zero value at

0oC, and for water vapour the enthalpy of saturated water is taken as zero at

0oC. The enthalpy of moist air is given by:

! The unit of h is kJ/kg of dry air. Substituting the approximate values of cpand

hg, we obtain:


14/50


! From the equation for enthalpy of moist air, the humid specific heatof moist

air can be written as:

! Since (W.cpw) is very small compared to the first term, for all practical

purposes, the humid specific heat of moist air, cpmcan be taken as 1.0216 kJ/

kg dry air.K


15/50


! Specific volumeis defined as the number of cubic meters of moist air per

kilogram of dry air. From perfect gas equation since the volumes occupied by

the individual substances are the same, the specific volume is also equal to

the number of cubic meters of dry air per kilogram of dry air:


16/50

Example 1

On a particular day the weather forecast states that the dry bulb temperature

is 37oC, while the relative humidity is 50% and the barometric pressure is

101.325 kPa. Find the humidity ratio and enthalpy of moist air on this day.


17/50

Example 1 (solution)


18/50

3. Wet-bulb temperature

! The thermodynamic state of moist air is uniquely fixed if the barometric pressure and

two other independent properties are known. As such, at a given barometric pressure,

the state of moist air can be determined by measuring any two independent properties.

! One of them could be the dry-bulb temperature (DBT), as the measurement of this

temperature is fairly simple and accurate. The accurate measurement of otherindependent parameters such as humidity ratio is very di"cult in practice.

! Since the measurement of temperature is easier, it would be convenient if the other

independent parameter is also a temperature. The dew-point temperature (DPT) can be

considered, but it is observed that accurate measurement of DPT is di"cult.

! In this context, a new independent temperature parameter called the wet-bulb

temperature (WBT)is defined. Compared to DPT, it is easier to measure the WBT

of moist air. Thus knowing the DBT and WBT from measurements, it is possible to

find the other properties of moist air.


19/50


! If the percentage saturation of an air sample

is less than 100, i.e. it is less than saturated,

and it comes into contact with water at the

same temperature, there will be a di#erence

in vapour pressures.

! As a result, some of the water will evaporate.

The latent heat required for this change of

state will be drawn from the sensible heat of

the water, which will be slightly cooled.

! This drop in the water temperature provides

a temperature di#erence, and a thermal

balance will be reached where the flow of

sensible heat from the air to the water

provides the latent heat to evaporate a part

of it.


20/50


! The e#ect can be observed and

measured by using two similar

thermometers, one of which has its

bulb enclosed in a wet wick. The

drier the air passing over them, the

greater will be the rate of evaporationfrom the wick and the greater the

di#erence between the two readings

(DBT and WBT). The di#erence

between the two temp is termed wet

bulb depression.


21/50


! In brief, WBT is the temperature measured by ordinary

thermometer when its bulb is covered with wet cloth and

exposed to air

! The sling psychrometer is widely used for measurements

involving room air or other applications where the air

velocity inside the room is small.

! It consists of two thermometers mounted side by side

and fitted in a frame with a handle for whirling the

device through air. The required air circulation (~ 3 to 5

m/s) over the sensing bulbs is obtained by whirling the

psychrometer (~ 300 RPM). Readings are taken when

both the thermometers show steady-state readings.

! In the aspirated psychrometer, the thermometers remain

stationary, and a small fan, blower or syringe moves the air

across the thermometer bulbs.


22/50

4. Adiabatic saturation

! Adiabatic saturation temperature- temperature at which water, by

evaporating into air, can bring the air to saturation at the same temperature

adiabatically.

! An adiabatic saturator is a device using which one can measure theoretically

the adiabatic saturation temperature of air.


23/50


An adiabatic saturation device has

liquid water at the adiabatic

saturation temperature T2is

evaporated at a rate mwinto a moist

air stream flowing at a rate of

mda(1 + W) through the duct. Thusconservation of mass applied

separately to both dry air and water

components yields


24/50


! Above equation maybe used to solve for the upstream (ambient) humidity

ratio.

! Note: da = dry air; wv = water vapour


25/50

5. Psychrometric chart

! A Psychrometric chart graphically represents the thermodynamic properties

of moist air.

! Standard psychrometric charts are bounded by the DBT line (x-axis) and the

vapour pressure or humidity ratio (y-axis). The left hand side of the

psychrometric chart is bounded by the saturation line.

! Psychrometric charts are readily available for standard barometric pressure of

101.325 kPa at sea level and for normal temperatures (0-50oC).


26/50

5. Psychrometric chart

R R

ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE

BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY

Copyright 1992

AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.

SEA LEVEL

10 15 20 25

30

35

40

45

50

55

55

60

60

ENTHALPY - BTU PER POUND OF DRY AIR

15

20

25

30

35

40

45

50

ENTH

ALPY

-BT

UPER

POUND

OFDR

YAIR

SATU

RATIONTE

MPE

RATUR

E-F

35

40

45

50

55

60

65

70

75

80

85

90

95

100

105

110

115

120

DRYBULBTEMPERATURE-F

.002

.004

.006

.008

.010

.012

.014

.016

.018

.020

.022

.024

.026

.028

10%RELATIVEHU

MIDITY

20%

30%

40%

50%

60%

70%

80%

90%

35

3540

4045

45 50

50 55

55 60

60

65

65

70

70

75

75

80

80

85WETBULBTEMPERATURE-F

85

90

12.5

13.0

13.5

14.0

VOLUME-CU.FT.PERLB.DRYAIR

14.5

15.0

HUMIDITYRATIO-POUNDSMOISTUREPERPOUND

DRYAIR

0

1.0 1.0

:-

2.0

4.0

8.0

-8.0-4

.0-2

.0-1

.0

-0.5-0

.4-0.3-0

.2-0.1

0.

10

.2

0.3

0.4

0.5

0.6

0.8

-2000

-1000

0

500

1000

1500

2000

3000

5000

- ::

SENSIBL E HEAT Qs

TOTAL HEAT Qt

ENTHALPY

HUMIDITY RATIO

Dh

DW

R R

ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE

BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY

Copyright 1992

AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.

SEA LEVEL


27/50


28/50


29/50

Example 2

Temperature in a room is 25C. If the wet-bulb temperature is 20C, find

the RH, dew point, humidity ratio, specific volume and specific enthalpy at

that state.


30/50

6. Psychrometric processes

! Sensible heating or cooling

! a psychrometric process that

involves the increase or decrease

of air temperature without

changing its humidity ratio.

! example: Moist air passing over a

room space heater


31/50

Example 3

! Find RH and wet-bulb temp of the post-heated air.


32/50



33/50


! Heating and humidifying


involves the simultaneous

increase of dry-bulb temp and

humidity ratio of the air.


34/50

Example 4

2.5 m3of lumber is being dried at 60C dry-bulb temperature and 52C

wet-bulb temperature. The drying rate of the lumber is 12.5 kg of water per

hour. If outside air is at 27C dry-bulb temperature and 80% relative

humidity, how much outside air is needed per minute to carry away the

evaporated moisture?


35/50



36/50


! Cooling and dehumidifying


involves the removal of water from

the air as the air temp drops

below the dew-point temp.


37/50

Example 5

Moist air at 50C dry-bulb temperature and 32% relative humidity enters

the cooling coil of a dehumidification kiln heat pump system and is cooled

to a temperature of 18C. If the drying rate of 6 m3of lumber is 4 kg/hour,

determine the kW of refrigeration required.


38/50



39/50


! Adiabatic / evaporative cooling


involves the cooling or air without

heat loss or gain.

! sensible heat lost by the air is

converted to latent heat in the

added water vapour.


40/50

Example 6

Referring to the figure, air at state point 1 (65C DBT and 57C WBT)

experiences a temperature drop of 3C as it passes through the 1.2 m

wide stack of lumber. Determine the properties of the air at state point 2

and compare them with those at state point 1. If the air is flowing at a rate

of 2 ms-1, determine the drying rate, assuming that the volume of the stack

of 2.5 cm thick lumber is 2.5 m3

. The stack is 1.2 m wide x 3.6 m long, andthe boards are separated by stickers 3.8 cm wide x 1.9 cm thick that are

spaced 0.6 m apart.


41/50



42/50

6. Pyschrometric processes

! Adiabatic mixing of moist stream air


involves no heat loss or gain

during the mixing of two air

streams.


43/50

Example 7

0.2 kg/s of moist air at 45C (DBT) and 10% RH is mixed with 0.3 kg/s of

moist air at 25C and a humidity ratio of 0.018 kgw/kgda in an adiabatic

mixing chamber. After mixing, the mixed air is heated to a final

temperature of 40C using a heater. Find the temperature and relative

humidity of air after mixing. Find the heat transfer rate in the heater and

relative humidity of air at the exit of heater. Assume the barometricpressure to be 1 atm.


44/50



45/50

Example 8

Hot, humid outdoor air is drawn into the system with a volumetric flow rate V1= 100

cfm (0.0472 m3/s). The temperature of the outdoor air is T1= 95oF (35oC) and it has

relative humidity $1= 0.7. In order to provide a comfortable environment, the air

must be both cooled and dehumidified. The air passes over a chilled water coil inorder to dehumidify it. The air exits the chilled water coil at T2= 45oF (7.2oC). The air

then passes over a steam coil in order to reheat it to a comfortable temperature. The

air exits the steam coil at T3= 65oF (18.3oC).


46/50

Example 8

(a) What is the humidity ratio of the air entering the chilled water coil?

(b) What is the mass flow rate of dry air entering the chilled water coil?

(c) At what temperature does condensation begin as the air flows through chilled

water coil?

(d) What is the humidity ratio of the air leaving the chilled water coil?

(e) What is the volumetric flow rate of condensate that forms on the chilled water

coil?(f) What is the relative humidity of the air leaving the steam coil that is provided to

the conditioned space?

(g) What are the rates of heat transfer to the chilled water in the chilled water coil

and from the steam in the steam coil?


47/50



48/50



49/50

Example 9

Moist air at 60oF (15.6oC) DBT and 20% RH enters a heater and humidifier at

the rate of 1600 cfm (0.755 m3/s), Heating of the air is followed by adiabatic

humidification so that it leaves at 115oF (46.1oC) DBT and a RH of 30%.

Saturated water vapour at 212oF (100oC) is injected. Determined the required

heat transfer rate and mass flow rate of water vapour.


50/50


topic 1 pyschrometry notes

Documents