Chapter 1 Student

Heating, Ventilating, and Air

Conditioning

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Department of Mechanical Engineering Technology

Yanbu Industrial College MET

D

EP A R T M

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MEC

HAN

ICAL

ENGINEERING TECHNO

LOGY

MET 412 Heating, Ventilating, and Air Conditioning

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Chapter 1

Psychrometric Properties of Moist Air and Conditioning processes.

1.1- Introduction

Air Conditioning can be defined as the process of treating air in an internal environment to establish and maintain required standards of temperature, humidity, cleanliness, and motion.

There are numerous applications for Air Conditioning in which human comfort is the prime consideration such as, shops, offices, factories, transport, --------etc, but there are also many applications human comfort is not the prime consideration. These include textile and printing industries, mainframe, servers, computers -----------etc. Therefore understanding of the properties of moist air and the ability to analyze the various processes involving air and water vapour is basic to the refrigeration and HVAC (Heating Ventilation and Air Conditioning) Engineers.

1.2- Composition of Air

The ASHRAE handbook of fundamentals gives the following approximate composition of dry air by volume fraction.

Constituent Molecular weight

kg/kmol Volume fraction

Nitrogen 28 0.7809

Oxygen 32 0.2095

Argon 39.94 0.0093

Carbon Dioxide 44 0.0003

Table (1)

Based on the composition of dry air in Table (1), the molecular weight Ma of dry air is 28.965, then the gas constant Ra is

) K kg kJ/ ( 0.287 .

.

M

R R

a

a 96528

31458

The basic medium in air conditioning practice is a mixture of dry air and water vapor. The water vapor constant Rv is

) K kg kJ/ ( 0.462 .

M

R R

v

v 18

31458


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1.3- Maximum Water Vapour Content

The amount of water vapour in air varies greatly with the particular locality and the weather conditions. Air will contain maximum amount of water vapour when the water vapour in the air exerts the maximum possible pressure. Since the maximum pressure that can be exerted by any vapour is the saturation pressure corresponding to its temperature. Then the air will contain the maximum amount of water vapour when the pressure exerted by water vapour is equal to the saturation pressure corresponding to the temperature of the air.

1.4- Daltons Law

The Gibbs Dalton Law for a mixture of ideal gases sates that the mixture pressure is equal to the sum of the partial pressure of the constituents.

n

i

iP P 1

where: n number of constituents in a gaseous mixture.

Because the various constituents of the dry air may be considered to one gas. Then the total pressure of moist air is,

P = Pa + Pv

where Pa Partial pressure of dry air = Patm - Pv, Patm =101.325 kPa

Pv Partial pressure of water vapour.

1.5- The Psychrometric Chart

Data relating to absolute humidity, air stream temps (Dry Bulb and Wet Bulb) and relative humidity are plotted on Psychrometric charts which have the following form (see the attached chart).

Tdb

W


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1.6- Some Basic Definitions

Dew Point Temperature (DP)

The temperature at which the water vapour in the air is saturated or the minimum temperature any surface exposed to airflow can have without visible condensation occurring on it. Obviously then the DP temperature of the air is always the saturation temperature corresponding to partial pressure exerting by the water vapour. Consequently when the partial pressure exerted by the water vapour is known, the DP temperature of the air can be determined from the saturated water tables. Likewise when DP temperature of the air is known, the partial pressure exerted by the water vapour can be determined from the saturated water tables.

Example 1.1

Air has a temperature of 26 oC and partial pressure exerted by water vapour in the air is 1.227 kPa. Determine the DP temperature of the air.

Solution

From the saturated water tables at P =1.227 kPa. DP = 10 oC.

Example 1.2

Air has a temperature of 26 oC and has a DP temperature of 16 oC. Determine the partial pressure exerted by the water vapour.

Solution

From saturated water tables at T=16 oC. Partial pressure of water vapour = 1.8168 kPa.

Relative Humidity (RH)

Relative humidity can be defined as, the ratio of the actual partial pressure exerted by the water vapour in any volume of the air to the partial pressure that would be exerted by the water vapour if the water vapour in the air is saturated at the temperature of that air,

100100100 x

x

TR/P

TR/P x

P

P

100 x saturationatpressurePartial

pressurepartialActual RH

p,ts

v

vs

vv

s

v

Example 1.3

Air has a temperature of 26 oC and DP temperature of 10 oC. Determine the RH.


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Solution

36.5% 100x 3.36

1.227

C 26 at P

C 10 at P RH

o

s

o

v

Humidity Ratio (Moisture content)(W)

Sometimes called specific humidity, is the ratio of the mass of the water vapour mv to the mass of dry air ma.

airdry of kg

kg

m

m W

a

v

Using the ideal gas law we can drive a relation between the humidity ratio and relative humidity.

TR

VP m

TR

VP m

a

a

a

v

vv

Then

vatm

v

atmvatma

a

v

a

v

av

va

PP

P. W

kPa.P,PPP

P

P.

P

P

.

.

PR

PR W

6220

325101

62204620

2870

and

s

a

P.

W PRH

6220


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Example 1.4

Determine the W and RH for air at saturated atmosphere having a temperature of 26 oC and DP temperature of 10 oC.

Solution

vatm

v

PP

P. W

6220

r0076202271325101

22716220 ai kg/kg dry.

..

.. W

Saturation Ratio (Degree of saturation)

Is the ratio of the mass of water vapour in the air per unit mass of dry air to the mass of water vapour required for saturation of the same air sample.

100xW

W

s

Example 1.5

Air at normal atmosphere pressure has a temperature of 26oC and DP temperature of 10 oC. Determine the saturation ratio of the air.

Solution

100xW

W

s

airdry kg/kg 0.02133..101

C)26 T (at3.36.

P-P

P0.622 W

airdry kg/kg 0.00762.1.101

C)10 T (at1.227.

P-P

P0.622 W

o

satm

ss

o

vatm

v

3633256220

2273256220

Then

%7.3510002133.0

00762.0 x

Dry Bulb Temperature (DB)

Is the normally measured temperature.


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Wet Bulb Temperature (WB)

Is the temperature that a temperature measuring device, kept enclosed in a wetted environment (Wick), comes to exposed to an air stream, provided the air velocity over the device is significant (over 3m/s is recommended).

Enthalpy Deviation

Is the difference between the actual or true specific enthalpy of the air at any given condition and the specific enthalpy of saturated air at the same WB temperature.

1.7- The Heat Content of Enthalpy of Air

Air has both sensible heat and latent heat. The total heat of the air at any condition is the sum of sensible heat and latent heat.

Sensible heat of the air is a function of the DB temperature.

Latent heat of the air is a function of the DP temperature.

Total heat of the air is a function of the WB temperature

The enthalpy of a mixture of perfect gases is equal to the sum of the enthalpies of each constituent and is usually referenced to a unit mass of one constituent. For the air-water vapour mixture dry air is used as the reference because the amount of water vapour may vary during some processes

h = ha +W x hw,v

ha =Cpa x t, hw,v = hg +Cpv x t , and hg =2501.3 at zero oC

h = t + W (2501.3 +1.88t) (kJ/kga)

Sensible Heat ( sQ )

For any given DB temperature, the sensible heat of air is taken as the enthalpy of dry air at that temperature as calculated from 0 oC and can be computed from

aaapapas hmTCmTCmQ

,,

Example 1.6

The temperature of 10 kg of air is increased from 30 oC to 50 oC by addition of heat. Determine the quantity of sensible heat supplied.

Solution


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aapas TCmQ , = 10 * 1.005 * 20 = 200 kJ.

Latent Heat ( Q )

Since all the components of dry air are non-condensable at normal temperatures and pressures, for all partial pressure purposes the only latent heat in the air is the latent heat of the water vapour in the air. Therefore, the amount of the latent heat in any given quality of air depends on the mass of water vapour in the air and on the latent heat of vaporization of water corresponding to the saturation temperature of the water vapour.

lQ = am [W2 hw,v2- W1 hw,v1]

= am [ hl,2- hl,1]

where hw,v is the specific enthalpy of the water vapour in the air, usually taken as the enthalpy of saturated water vapour at the saturation temperature equal to the air DP temperature

= hg + Cpv (DB-DP), Cpv = 1.88 kJ/kg

Convenient empirical equation for determining the enthalpy of low pressure, low temperature steam (such as that found in the air) is,

hw,v = hg + Cpv DB, hg = 2501.3 kJ/kg

Example 1.7

Fourty five kg/min of air having DB and DP temperature of 35 oC and 20 oC respectively are passed across a cooling coil and cooled to a final DB temperature of 12 oC. Assuming that the air leaving the cooling coil is saturated (DB, WB, and DP temperatures are the same). Calculate

a- the Mass of water vapour condensed from the air in kg/s.

b- The latent heat transferred in kJ/s. (kW).

Solution


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1.8- Total Heat of the Air

ht = hs+ hl

Ht = am ht

tQ = am [ ht,2- ht,1]

1.9- Standard Air

Is defined as air having a density of 1.2kg/m3 or a specific volume of (1/1.2) = 0.833 m3/kg.

sQ = am Cp,a T

= 1.2 aV T


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1.10- Classic Moist Air Processes

The most powerful tools of the air conditioning design engineer are, the first law of thermodynamics or energy balance and the conservation of mass or mass balance.

Heating and Cooling of Moist Air

Figure 1.1 shows a schematic of heating or cooling device and Figure 1.2 shows the process on the Psychrometric chart.

Energy Balance between 1 and 2

Figure 1.1 Schematic of heating or cooling device

Figure 1.2 Sensible heating and cooling process

am h1 = tQ + am h2

tQ = am (h1 h2)

h1 = ha,1 + W1 hw,1

h2

h1

h1 h2


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h2 = ha,2 + W2 hw,2

tQ = am Cp,a (T1 T2)

Where Cp = Cp +W Cp,v

Cp,a = 1.0035 kJ/kg.K

Cp,v = 1.88 kJ/kg.K

Example 1.8

2 kg/s of air having initial dry bulb temperature and wet bulb temperature of 21 oC and 16 oC are passed across a heating coil and dry bulb temperature is raised to 33 oC. Plot the process on a Psychrometric chart and determine:

a-The final WB temperature of the air. b-The sensible heat transferred in kW. c-The total heat transferred in kW.

Solution


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Cooling and Dehumidifying of Moist Air

When moist air is cooled to a temperature below its dew point temperature some of the water vapour will condense and leave the stream. Figure 1.3 shows a schematic of cooling and dehumidifying device and Figure 1.4 shows the process on the psychrometric chart.

Figure 1.3 Schematic of a cooling and dehumidifying device

Figure 1.4 Cooling and dehumidifying process


am h1 = tQ + am h2 + wm hw

tQ = am (h1 h2) - wm hw (1.1)

Mass Balance on the water vapour between point 1 and 2

am W1 = mw + am W2

wm = am (W1 W2) (1.2)

ha

h2

h1

h1

hw

h2


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Substituting equation 1.2 into equation 1.1, we have

tQ = am (h1 h2) - am (W1 W2) hw (1.3)

Where hw is the water enthalpy at the condensate water temperature

Equation (3) represents the total amount of heat transfer from the moist air. The last term on the right hand side of equation (3) is usually small compared to the others and is often neglected.

The cooling and dehumidifying process involves both sensible and latent heat transfer where sensible heat is associated with the decrease in dry bulb temperature and latent heat is associated with the decrease in humidity ratio.

sQ = am Cp,a (T1-T2) , lQ

= am (W1-W2) hw,v , hw,v = hg + Cp,v db

(1.4)

By referring to psychrometric chart, we may also express the latent heat transfer as

lQ = am ( h1-ha) (1.5)

And the sensible heat is given by

sQ = am ( ha-h2) (1.6)

The energy of the condensate has been neglected. Obviously

tQ = sQ

+ lQ (1.7)

The sensible heat factor SHF is defined as sQ / tQ

. This parameter is shown on the

semicircular.


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Example 1.9

Conditioned air is supplied to a space at 15 oC DB temperature and 14 oC WB temperature at the rate of 0.5 m3/s. The sensible heat factor for the space is 0.7 and the space is to be maintained at 24 oC DB temperature. Determine the sensible and latent cooling loads for the space.

Solution


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Example 1.10

4 kg/s of air, having initial DB and WB temperatures of 30 oC and 21 oC respectively, are passed across a cooling coil whose mean effective surface temperature is 10 oC. Assuming that all parts of the air contact the cooling surface so that the air leaves the coil saturated at the temperature of the coil surface. Plot the process on the psychrometric chart and determine

a- Total heat transferred in kW. b- The sensible heat transferred in kW. c- The latent heat transferred in kW.

d- wm .

e- Sensible Heat Factor

Solution


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Heating and Humidifying of Moist Air

Air conditioning which human comfort is the prime consideration may be described as the control of temperature and humidity. The process of heating and humidifying of moist air is generally required during cold months of the year. Figure 1.5 shows a schematic of heating and humidifying device and Figure 1.6 shows the process on the psychrometric chart.

Figure 1.5 Schematic of a heating and humidifying device

Figure 1.6 Heating and humidifying process

Energy Balance and Mass Balance

The most powerful tools of the air conditioning design engineer are the first law of thermodynamics or energy balance and the conservation of mass or mass balance.


am h1 + tQ + wm hw = wm h2

tQ = am (h2 h1) - wm hw (1.8)

hw=(hg)at Temperature of steam

h2

h1

h1 h2


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Mass Balance on the water vapour between point 1 and 2

am W1 + mw = wm W2

wm = am (W2 W1) (1.9)

Substituting equation 1.9 into equation 1.8, we have

tQ = am (h2 h1) - am (W2 W1) hw (1.10)

where hw is the water vapour enthalpy at inlet temperature which is normally 100 oC

(2676 kJ/kg).

Dividing by am (W2 W1)

w

a

t hWW

hh

)W(Wm

Q

12

12

12

(1.11)

w

w

t hm

Q

WW

hh

12

12 (1.12)

Humidifying moist air

Moisture is frequently added to moist air without the addition of heat. Equation (1.12) then becomes

whWW

hh

12

12

hw=hg

hwhg

Tw=Twb

W

Tdb

h

hw=hg

hwhg

W

h

Protractor

SHF=1

24 oC

17 oC

SHF=0

SHF


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Example 1.11

Moist air at 16 oC DB temp. and 20% relative humidity flows through a heater and humidifier at the rate of 0.75 m3/s. Heat is transferred to the air at the rate of 30 kW. Saturated water vapour at 100 oC is injected at the rate of 0.00625 kg/s. Determine the final state of the moist air if the process occurs at 100 kPa.

Solution


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1.11-Adiabatic mixing of two streams of moist air

The mixing of air streams is quite common in air conditioning systems. The mixing process usually occurs under adiabatic conditions and steady state steady flow. Figure 1.7 illustrates the mixing of two air streams and Figure 1.8 shows the process on the Psychrometric chart.

Figure 1.7 Schematic adiabatic mixing of two air streams

Figure 1.8 Adiabatic mixing process

Mass balance on the dry air is

213 aaa mmm (1.13)

Mass balance on the water vapour

2a1a3a WmWmWm 213 (1.14)

3

21

a

2a1a

3m

WmWmW

(1.15)

Since the mixing of the airstreams occurs adiabatically (without gain or loss of heat) and without gain or loss of moisture, it follows that the enthalpy of the mixture is the sum of the enthalpies of the components, that is

h2

h1

h3

h1

h2

h3


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Hs,3 = Hs,1+ Hs,2 (1.16)

where s denotes to sensible heat

Hl,3 = Hl,1+ Hl,2 (1.17)

where l denotes to latent heat

Then the energy balance is,

Ht,3 = Ht,1+ Ht,2 (1.18)

221133 hhh mmm (1.19)

In accordance with equation (4)

2pa21pa13pa3 TCmTCmTC m (1.20)

3

21

a

2a1a

3m

TmTmT

(1.21)

By combining equations (1.19), (1.15) and (1.13) and eliminating am , we obtain the

following result.

2

1

a

a

13

32

13

32

m

m

WW

WW

hh

hh

(1.22)

Example 1.12

0.575 kg/s of air at 38 oC DB temperature and 24 oC WB temperature are mixed with 1.15 kg/s of air at 15 oC DB temperature and 10 oC WB temperature. The process is adiabatic, at a steady state steady flow rate and standard sea level pressure. Find the condition of the mixed stream.

Solution


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1.12- Coil Bypass Factor

An alternative approach to the analysis of the cooling coil in example 1.13 uses the so-called coil bypass factor (BPF). Note that when line 1-2 of example 1.13 figure is extended, it intersects the saturation curve at point d. This point represents the apparatus dew point temperature of the cooling coil. The coil cannot cool all of the air passing through it to the coil surface temperature. This fact makes the coil perform in a manner similar to what would happen if a portion of the air was brought to the coil temperature and the remainder bypassed the coil unchanged. A humidifying coil thus produces unsaturated air at a higher temperature than the coil temperature.

Again referring to example 1.13 figure, notice that in terms of the length of the line d-2, the length d-1 is proportional to the air bypassed, and the length 1-2 is proportional to the air not bypassed. Because of the chart construction, it is approximately true that

d

d

TT

TTBPF

1

2 (1.23)

and

dTT

TTBPF

1

211 (1.24)

Where BPF is the fraction of air bypassed, or coil bypass factor, expressed as a decimal, and where the temperatures are dry bulb values. Now

sQ = 1m cpa (T1-T2) (1.25)

or

lQ = 1m cpa ( Tt1-T2) (1-BPF) (1.26)

Example 1.15

Find the bypass factor for the coil of example 1.13 and compute the sensible and latent heat transfer rates.

Solution


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1.13- Contact Factor

dtt

ttCF

1

21 (1.27)

1.14- Revision Questions

1- A mixture of dry air and water vapor is at a temperature of 20 oC under a total pressure of 101.325 kPa. The dew point temperature is 15 oC. Calculate

Partial pressure of water vapor.

Relative humidity

Humidity ratio or specific humidity or absolute humidity or moisture content.

Specific enthalpy of water vapor

Enthalpy of air per kg of dry air

Specific volume of air per kg of dry air.

2- Calculate (i)- RH, (ii)- W, (iii)- dew point temperature, (iv)- density and (v)- enthalpy of

atmospheric air when the DBT is 35 oC and WBT is 23 oC.

3- A certain quantity of air has a temperature of 25 oC and that the water vapor in the air exerts a partial pressure of 1.7051kPa. Determine the DP temperature and RH

4- Determine the humidity ratio, enthalpy, and specific volume for saturated air at 101.325 kPa using perfect gas relations and Thermodynamic Tables for temperatures of,

a- 25 oC b- -5 oC

5- The temperature of a certain room is 20oC and the relative humidity is 50%. The barometric pressure is 100kPa. Find,

a- the partial pressures of the air and water vapor, b- the vapor density, and c- the humidity ratio of the mixture

6- 40 m3/min of a mixture of recirculated room air and outdoor air inter a cooling coil at 31 oC DB and 18.5 oC WB temperatures. The effective surface temperature of the coil is 4.4 oC. The surface area of the coil is such as would give 12.5 kW of refrigeration with the given entering air state. Determine the dry and wet bulb temperatures of the air leaving the coil and the coil bypass factor.

7- Moist air enters a chamber at 10 oC DBT and 5 oC WBT at a rate of 1.5 m3/s. The barometric pressure is 101.325 kPa. While passing through the chamber, the air absorbs sensible heat at a rate of 40 kW and picks up 40 kg/h of saturated steam at 100 oC. Determine the dry and wet temperatures of the leaving air.


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