Chap 13 Ideal Gass Mix & Psych Rome Try

8/2/2019 Chap 13 Ideal Gass Mix & Psych Rome Try

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Slide 1

Department of Mechanical and Aeronautical EngineeringDepartement Meganiese en Lugvaartkundige Ingenieurswese

MTX311Chapter 13:

Ideal Gas Mixtures and Psychrometry

J. Dirker

Sections: 13.0-13.5 (Borgnakke & Sonntag 7 th Edition)

External References also used: Moran and Shapiro 6 th Edition (chapter 12) Cengel and Boles 7 th Edition (Chapter 14)

These slides will be made available,however students are encouraged to take down notes in class.

Ideale Gasmengsels en Psigrometrie



Slide 2


• WHY?

• Where does the water come from?

• Can we describe this phenomenon?

• Why is it important in Engineering?

– Air-conditioning (control temperature,

humidity etc… to make humanscomfortable)

– Drying of objects

– Condensate might be an unintended by-

product at heat exchangers etc…• We will be considering a basic introduction into

Gas Mixtures and Psychrometry which will

help to address such issues.

13.0. Introduction

Diesel-kringloop

What happens when an ice-cold glass of water is left for a while in a

warmer room?

• Water-vapour will

condense



Slide 3


• Atmospheric air is not dry and contains water vapour:

– Dry Air:

• 78.1% per volume Nitrogen

• 20.9% per volume Oxygen• 0.9% per volume Argon

• Other gasses: Neon, Helium, Methane, Sulphur Dioxide,

Hydrogen (all in small quantities)

– Moist Air:

• Water content varies from 0% (dry air) to a maximum

depending on temperature and pressure.

• Partial Pressure = is the pressure which a gas (in a mixture of

gases) would have if it alone occupied the volume.

13.1 General Considerations and Mixtures of Ideal Gasses

Algemene oorwegings en Ideaal-gas mengsels

....21 ++= gas gastot P P P

droë lug

vogtige lug

Parsiële druk



Slide 4


• Consider a simplification of the problem involving

a mixture of ideal gasses which are in contact with

the liquid or solid phase of one of the mixture

components.

• Assumption needed:

– The solid or liquid contains no dissolved gasses.

– Gaseous phase is treated as a mixture of ideal

gasses.

– The liquid-solid-vapour equilibrium of one

component is not influenced by the presence

any other component in the mixture.

13.2 Simplified Model of a Mixture Involving Gases & a Vapor

Vereenvoudigde Mengselmodel vir gasse en dampe

Fase-ewewig van ‘n komponent onafhanklik van teenwoordigheid van

‘n ander komponent

Geen opgeloste gasse

Gasfase is ideaal

Aannames:



Slide 5


• Dew point: The temperature at which condensation begins when air is

cooled at constant pressure.

• 1→ 2 : Const. Pressure

• 1→ 3 : Const. Volume (Condensation starts at lower temperature) – At the condensation point this is called a saturation mixture

(saturated air)



Doupunt

(Versadigde lug)



Slide 6


• Relative Humidity : Φ is defined as the as the mole fraction of water

vapour in moist air to the mole fraction of water vapour in saturated air

for a given temperature and total pressure. (sum of partial pressures)

– From perfect gas relation : [%]

– In Fig 13.3:

Φ = 100% on Saturation vapour line



Relatiewe humiditeit

(Versadigde lug)

retemperatusameatwaterpureof pressuresaturation

ourwater vapof pressurepartialcurrent=φ

%1004

1 ≤= P

P φ



Slide 7


• Humidity Ratio: ω, is the mass of water

distributed within 1 kg of dry air.

– [kg/kg]

– When ideal gas behaviour is assumed

for both air and water vapour:

– For air / water:



Humiditeits-verhouding

1kg air

ω kgvapour

airdryof kg

ourwater vapof kg=ω

( ) avtot

vv

aa

vv

R P P

R P

T RV P

T RV P

/

/

/

/

−==ω

vtot

v

P P

P

−= 622.0ω



Slide 8


• Degree of Saturation: Compares the current

water content of air with the maximum watercontent for a given temperature.



Versadigings-graad

1kg air ω kg

vapour

vtot

v

P P

P

−= 622.0ω

g tot

g

P P

P

−= 622.0maxω

The warmer the air, the more moisture it can carry.

By dropping

temperature, some

of the vapour might

condense once the

max. ω is reached.



Slide 9



Consider 2 m3 of an air-water-vapour

mixture at 80 kPa ,50ºC and 60%relative humidity. It is cooled to

20ºC in a constant pressure

process. Determine the following:

• Initial Humidity Ratio.

• Dew Point .

• Final Humidity Ratio.

• Mass of water condensed.

Beskou ‘n 2 m3 lug-waterdamp

mengsel teen 80 kPa ,50ºC en60% relatiewe humiditeit. Dit word

verkoel tot 20ºC in ‘n konstante

druk proses. Bepaal die

volgende:

• Aanvangs-Humiditeitsverhouding.

• Doupunt.

• Eind-Humiditeitsverhouding.

• Massa water gekondenseer.

Humidity Ratio, Relative Humidity [Calculation] Example

EX 13.3 & EX 13.4 Equivalent (Values are changed) .




Slide 10


13.3 First Law Applied to Gas Vapour Mixtures

An Air-conditioning unit is shown with

relevant temperature, pressure

and relative humidity data. Water

condensate forms.

Determine the heat transfer per

kilogram dry air. Ignore kineticenergy changes.

‘n Lugreëlling eenheid word getoon

met betreklike temperatuur, druken relatiewe humiditeit data.

Water kondensaat vorm.

Bepaal die hitte-oordrag per

kilogram droë lug. Ignoreer kinetiese energie verandering.

First Law Gas-Mixture [Calculation] Example

EX 13.5 EQUIVALENT - Values are changed

Eerste wet toegepas tot gas-damp mengsels



Slide 11


13.3 First Law Applied to Gas Vapour Mixtures

Eerste wet toegepas tot gas-damp mengsels

EX 13.6 SELF STUDY



Slide 12


13.4 Adiabatic Saturation Process.Adiabatiese Saturasie Proses

• How do we measure partial pressure of water vapour to

determine ω [kg/kg] and Φ [%]?

• There is another route to follow (later), but first we need

to define the Adiabatic saturation process:

• Consider an air-vapour mixture in

contact with liquid water.

• Duct insulated.

• If humidity is less than 100% at inlet,

water will evaporate into the airstream.

• Due to evaporation, air and water temp

will decrease.

• If exit air is saturated (ie 100%

humidity) the exit temp = adiabatic

saturated temperature.

• For steady state, water is to be

added.



Slide 13


13.4 Adiabatic Saturation Process.Adiabatiese Saturasie Proses

Adiabatic saturation process….

• Energy Balance:

• Mass Balance:

( ) 222212111 val va hhhhh ω ω ω ω +=−++

222221111 vvaal wvvaa hmhmhmhmhm &&&&& +=++ E

aaa

mmm &&& ==21

avwv mmm 21&&& =+

m

Air

H2O a

v

mm&

&

=ω

( )

21

2212

1

l v

fg pa

hh

hT T C

−

+−=

ω ω

E

2

2

2 622.0 g tot

g

P P

P

−=ω

m

Thusω

1 can be determined from Temp. and Pressure measurements:

•Exit temperature = Thermodynamic Wet-Bulb Temperature



Slide 14


13.4 Adiabatic Saturation Process.

DO EXAMPLE 13.7 SELF-STUDY

Adiabatiese Saturasie Proses



Slide 15


13.5 Engineering Applications

• The adiabatic saturation process is

not practical to determine the

absolute and relative humidity of air.

• A more practical approach is to use a

thermometer whose bulb is coveredwith a cotton wick saturated with

water and to blow air over the wick.

Ingenieurswese-toepassings

• The temperature measured is the wet-bulb temperature T wb and it is

commonly used in air conditioning (A-C) applications.

• For air–water vapor mixtures at atmospheric pressure, T wb isapproximately equal to the adiabatic saturation temperature.

(Thermodynamic Wet-Bulb Temperature).

Natbol temperatuur



Slide 16


13.5 Engineering Applications

Psychrometric Chart

• Present moist air properties

in a convenient form.

• It is used extensively in A-Capplications.

• The psychrometric chartserves as a valuable aid in

visualizing the A-C

processes such as heating,

cooling, and humidification.


Psigrometrie-kaart

Human comfort is approximately between

20ºC and 27ºC and between 40% and 60%

relative humidity. (Shown in blue-shade)



Slide 17


13.5 Engineering ApplicationsIngenieurswese-toepassings

Psigrometrie-kaart Psychrometric Chart

• Consider hand-out (FIG E.4).

• Note: The enthalpy is of the mixture of air

and water and is per kg of dry air.

• WHY per kg dry air? – In A-C the amount

of air flow is constant, but the mass of

water may change.

HAND-OUT ONLYVALID FOR 100 kPa!

Cengel and Boles 7th Edition (Chapter 14)



Slide 18


13.5 Engineering ApplicationsIngenieurswese-toepassings

Quick Test:

For an air-water-vapour mixture at

100kPa, 20ºC DB and 50%

relative humidity find for themixture using Fig E.4:

•Wetbulb temp.

•Enthalpy.

•Humidity ratio.

Vinnige Toets:

Vir ‘n lug-waterdamp mengsel teen

100kPa, 20ºC DB en 50% relatiewe

humiditeit, vind vir die mengsel dmv.

Fig E.4:

•Natbol temperatuur.

•Entalpy.

•Humiditeitsverhouding.

Approximate Answers:13.5 ºC

0.0076 kg water /kg air

58 kJ/(kg dry air)



Slide 19


Engineering ApplicationsIngenieurswese-toepassings

Air Conditioning Processes

•Maintaining a space at the desired temperature and humidity requires

some processes called air-conditioning processes:

•simple heating (raising the temperature),•simple cooling (lowering the temperature),

•humidifying (adding moisture),

•dehumidifying (removing moisture).

•In winter: Air is normally heated and humidified.•In Summer: Air is normally cooled and de-humidified.

Not in

Textbook




Slide 20



1. Simple Heating and Cooling (ω = constant)

• Heating systems such as a stove, a heat pump,

or an electric resistance heater.

• Air circulating over a tube containing hot gas or over electric resistance wires to heat it.

• Similar for cooling – air circulated over a coil

carrying evaporating refrigerant or chilled water.

• No moisture added or removed ie: ω = const.

• During simple heating, relative humidity (Φ)

decreases (moving away from saturation line on

the Psycro Chart).

• During simple cooling, relative humidity (Φ)

increases (moving closer to saturation line on

the Psycro Chart).

Not in

Textbook




Slide 21


1. Simple Heating and Cooling… (ω = constant)

• Energy and Mass Balance:

• Thus, using the enthalpy of the air-water

mixture:


Not in

Textbook

( ) ( )[ ]111222

11112222

vavaaCV

vvaavvaaCV

eeiiCV

hhhhmQ

hmhmhmhmQ

hmhmQ

ω ω +−+=

+−+=

−= ∑∑&&&&&

&&

&

( )12hhmQ

aCV

−= &&

E

m

E

E m Air




Slide 22


Engineering Applications

Moist air at 100 kPa enters a duct at

10ºC, 80% relative humidity at2.5 m3/s. The mixture is heated

until 30ºC. No moisture is added

or removed. Determine using the

Psychrometric Chart:• Heat added to the air.

• Relative humidity at the exit.

Vogtige lug teen 100 kPa gaan ‘n

kanaal in teen 10ºC, 80% relatiewe

humiditeit en 2.5 m3/s. Die mengsel

word verhit tot 30ºC. Geen vog is

toegevoeg of verwyder nie. Bepaal

met die Psigrometrie Kaart:• Hitte bygevoeg tot lug.

• Relatiewe humiditeit by die uitgang.

Simple Heating [Chart] Example.

– Adapted from Moran and Shapiro, 6th Edition

Ingenieurswese-toepassings Not in

Textbook



Slide 23



2. Heating with Humidification

• Simple heating may result in problems

with the low relative humidity of heated

air.• Humidification may be needed.

• Pass the heated air through a

humidifying section after simple heating.

• Steam or water droplets may be injected

into the air-stream.

Not in

Textbook

Steam

Water

droplets


I i i



Slide 24



3. Cooling with Dehumidification

• If relative humidity is too high, some

moisture has to be removed

• Cool the air/water mixture to below thedewpoint.

• The air/water may be heated again to

the appropriate dry bulb temperature.

Not in

Textbook


S 25 I i t i



Slide 25



4. Evaporative Cooling

• In dry-climates, water may be

evaporated into the dry-air to reduce its

dry-bulb temperature.

• During water vaporization, latent heat isabsorbed from both the water and the

air.

Not in

Textbook


Slid 26 I i t i



Slide 26



5. Adiabatic Mixing of Airstreams

• Frequently in A-C applications, two

streams of air/water mixtures are to be

mixed.

Not in

Textbook


Slide 27 Ingenieurswese toepassings



Slide 27



6. Wet Cooling Towers

• Power plants, large air-conditioning

systems, and some industries generate

large quantities of waste heat that is often

rejected to cooling water from nearby lakesor rivers.

• In some cases, however, the cooling watersupply is limited or thermal pollution is a

serious concern.

Not in

Textbook

• Thus, waste heat must be rejected to the atmosphere

• Recirculating cooling water serve as a transport medium for heat

transfer between the source and the sink (the atmosphere).• Can be done via a wet cooling tower.

• A wet cooling tower is essentially a semi-enclosed evaporativecooler.





Slide 28



6. Wet Cooling Towers…

Natural-draft cooling tower:

• large chimney

• air in the tower has a high warm water-vapor content, - lighter than the outside air.

• the light air in the tower rises, and the

heavier outside air fills the vacant space

• airflow from the bottom of the tower to thetop is created.

Spray pond :

• warm water is sprayed into the air and is

cooled by the air as it falls into the pond.Cooling pond :

• Dumping the waste heat into a still large

artificial lake open to the atmosphere.

Not in

Textbook

www.nucleartourist.com

www.wikipedia.com




Slide 29



Water exits the condenser of a power plant at 38ºC at a flow rate of

4.5 x 107 kg/h. The water is cooled by a cooling tower via evaporation.Make-up water at 20ºC is added in the cooling tower such that a

combined stream of water at 30ºC is returned to the condenser.

Atmospheric air at 100kPa enters the cooling tower at 25ºC DB and

30% relative humidity. Moist air exits the tower at 30ºC DB and 90%relative humidity.

Determine:

• Mass flow rate of air.

• Mass flow rate of make-up water.

Ignore heat transfer to the surroundings, pump power, kinetic energy

changes and potential energy changes.

Cooling Tower [Chart] Example.

– Adapted from Moran and Shapiro, 6th Edition

Not in

Textbook


Afrikaans of volgende skyfie

Slide 30 Ingenieurswese-toepassings



Slide 30



Water verlaat die kondensator van ‘n kragstasie teen ‘n vloeitempo van

4.5 x 107 kg/h en 38ºC. Die water word deur ‘n koeltoring verkoel viaverdamping. Water teen 20ºC word toegevoeg in die koeltoring sodat

‘n gekombineerde waterstroom teen 30ºC en dieselfde tempo

terugvloei na die kondensator. Atmosferiese lug teen 100kPa gaan die

koeltoring in teen 25ºC DB en 30% relatiewe humiditeit. Vogtige lugverlaat die toring teen 30ºC DB en 90% relatiewe humiditeit.

Bepaal:

• Massavloei tempo van die lug.

• Massavloei tempo van die water toegevoeg.

Ignoreer hitte-oordrag aan die omgewing, pomp-drywing, kinetiese energie

verandering en potensiele energie verandering.

Koeltoring [Kaart] Voorbeeld.

– Aangepas vanuit Moran en Shapiro, 6de Uitgawe

Not in

Textbook

Ingenieurswese toepassings

English on previous slide

Chap 13 Ideal Gass Mix & Psych Rome Try

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