Refrigeration (Kylteknik) - Åbo Akademi | Startsidausers.abo.fi/rzevenho/REF17-OH7.pdf · 0.105 – 0.027 = 0.078 kg water per kg dr y air. With a density of 0.88 kg /m³ saturated

7. Air conditioning, cooling towers

Ron ZevenhovenÅbo Akademi University

Thermal and Flow Engineering Laboratory / Värme- och strömningstekniktel. 3223 ; [email protected]

Refrigeration (Kylteknik) course # 424519.0 v. 2017

ÅA 424519 Refrigeration / Kylteknik

17.2.2017Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 2/40

7.1 Humid air

17.2.2017 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku

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Humid air /1

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”Air” can be considered as a mixture of dry air + moisture

For the range -10 ~ +50°C, dry air can be treated as an ideal gas with cp,air ≈ 1.005 kJ/(kg· K)

The saturation pressure of water at 50°C is 12.3 kPa, behaving also as an ideal gas; lines for h = constant in T, s diagram ≈ isotherms

For air = dry air + moisture: pressure p = pdry air + pmoisture

h(T)water vapour in air ≈ h(T)saturated vapour

h(0°C) water vapour ≈ 2501.3 kJ/kg cp water vapour ≈ 1.82 kJ/(kg· K) -10 ... +50°C cp water vapour ≈ 1.87 kJ/(kg· K) at ~ 25°C


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Humid air /2 absolute humidity, enthalpy The amount of water vapour in air can be specified as

absolute (or specific) humidity, ω (kg water/kg dry air):

For saturated air, pwater = pwater,sat (T) with saturation pressure pwater,sat (T) from tables

The enthalpy of humid air is the sum of the enthalpies of the dry air and water vapour:

Picture: ÇB98(kJ/kg)

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Humid air /3 relative humidity

The amount of water vapour in air relative to the maximum amount of vapour the air can contain at a certain temperature is known as relative humidity, RH (-, or %) :

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Example: water in air For a 5x5x3 m3 room @ 25°C,

100 kPa, RH = 75% (= 0.75) calculate:

a) partial pressure of dry air

b) absolute humidity, ω

c) mass of dry air and water vapour

pwater,sat @ 25°C = 3.17 kPa → pwater = 0.75· 3.17 kPa= 2.38 kPa → pdry air = 100 - 2.38 = 97.62 kPa

ω = 0.622· pwater/pdry air = 0.0152 kg water/kg dry air

mwater= Mwater· nwater = Mwater· pwater· V/(R·T) = 1.30 kgwith V = 75 m3 → mair = 85.10 kg

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Humid air /4 dew point

The dew point, Tdp, is the temperature at whichcondensation begins when air is cooled at constant pressure

For example, air in a house at 20°C, 1 atm, RH = 75 %, surrounded by cold air; at what window glass insidetemperature will moisturecondense on the glass?

Answer: pwater= 0.75· psat = 0.75·2.34 kPa = 1.75 kPa; Tsat @ 1.75 kPa = 15.3 °C

Pictures: ÇB98


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Adiabatic saturation The adiabatic saturation

temperature is the temperature when reaching(or having) saturation in an air stream at equilibrium with liquid water in an adiabaticprocess

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Adiabatic saturation (derivation)

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Adiabatic saturation – example

17.2.2017Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 10

Source:SB01


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Wet bulb temperature The wet bulb

temperature is the temperature reached as a result of heat withdrawalfor evaporation of water;

Twet bulb < Tdry bulb

if the relative humidityRH < 100%

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Example relative humidity A weather station measurement

gives an ambient temperature of 18°C and a dewpoint of 8°C. The barometer reads 98.75 kPa.

Calculate: the water vapour pressure, relative humidity, and absolute humidity

Answer: the dewpoint gives the water vapour pressureFrom tables: pwater vapour = psat @Tdewpoint = 1.073 kPa, andpsat @ 18°C = 2.065 kPaRH = 1.073 kPa / 2.065 kPa = 0.52 = 52%ω=0.622· pwater vapour /(ptotal – pwater vapour) = 6.83 g water/kg dry air

Source: T06



7.2 The psychrometric chart


Psychrometric chart /1


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Psychrometric chart /3Picture: ÇB98


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Psychrometric chart /4

Lines in the psychrometric chart

Pictures: ÇB98

Saturated air at 15°CTdewpoint = Twet bulb =Tdry bulb

Symbol Φ = RH

*


Example: psychrometric chart /1

Answer: absolute saturation line crosses saturation line at 55°C. The right axis gives absolute humidity 0.105 kg water / kg dry air. Relative humidity is 38-39 %, and cooling the gas will make it saturated at around 53.5 °C

Question 1: give the absolute and relative humidity of air at 70°C that has a wet-bulb temperature of 55°C, and what is its dewpoint ?

Picture andsource: BMH99

See also course möf-st 424302



Answer: the saturation line gives for 30°C 0.027 kg water / kg dry air (right axis). Starting with 0.105 kg water per kg dry air gives condensed0.105 – 0.027 = 0.078 kg water per kg dry air. With a density of 0.88 kg /m³ saturated air at 30°C this gives 0.078/0.88 kg water /m3 saturated air.

Question 2: Cool this air to 30°C, how much water will condense per m³ saturated air ?





Answer: The air is saturated at 30°C, the dewpoint is then of course 30°C. At 70°C, the relative humidity is 10% as the diagram shows.

Question 3: The saturated air of 30°C is heated to 70°C. What are the relative humidity an dewpoint of the gas at 70°C?




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7.3 Air conditioning


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Air conditioning /1

Air conditioning processes can be classified as

– Simple heating, raising (dry bulb) temperature

– Simple cooling, lowering (dry bulb) temperature

– Humidifying, adding moisture– De-humidifying, removing moisture& combinations of these

Process calculations are based on– The mass balance for air– The mass balance for water– The energy balance, including heat

exchange and (fan) work input

Air conditioning processes in the psychrometric chart

Picture: ÇB98

Air conditioning /2 evaporative cooling


Pictures: DHW08


Air conditioning /3

Air conditioning without and with de-humidifying

A residential air conditioning system

Pictures: T06

Systems based on active cooling (and heating) using a vapour-compressionrefrigeration cycle


Air conditioning /4

Evaporative coolers (”swamp coolers”)

Humifidifiers work similarly

A residential de-humidifier The main difference with standard air conditioning is the condenser in the air stream

Pictures: T06


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Air conditioning /5 Balance equations for air

conditioning systems: – (a) control volume– (b) mass balance– (c) energy balance

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Example: household dehumidifier /1 A household dehumidifier operates at 0.15 kg/s humid air inlet, inlet

pressure, temperature and RH are 98 kPa, 295 K and 85%. Outlet pressure is 100 kPa, condensate flow and temperature are 0.33 g/s and 281K. Electric power input is 689 W, heat losses to the surroundings are 15 W.

Calculate for the air outlet 1) the air mass flow rate, 2) absolute humidity, 3) temperature and 4) relative humidity

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Example: household dehumidifier /2

Answer: 1- the air mass flow rate:

The overall mass (i.e. mass streams) balance gives: mass air in = mass air out + condensate flow → mass air out = 150 - 0.33 g/s = 149.67 g/s

2 - absolute humidity: The mass balance for water also gives that mass dry air in · ωin = mass dry air out · ωout + condensate flow

note: mass dry air in = mass dry air out = mass dry airωout = (mass dry air· ωin – condensate flow ) / mass dry air ωin = 0.622· pwater / (ptotal - pwater) and pwater = 0.85· pwater,sat @ Tinlet

→ pwater = 0.85·2.621 kPa = 2.228 kPa and ωin = 0.0145 kg/kgmass dry air = mass air in / (1 + ωin) = 0.15/1.0145 = 0.148 kg/s→ ωout = (0.148· 0.0145 - 0.00033)/0.148 = 0.0123 kg/kg


Example: household dehumidifier /3

3 – temperature: Temperature follows from the enthalpy balance:

mass dry air · [ cp air· (Tout-Tin) + hH2O(Tout)· ωout – hH2O(Tin)· ωin ]

+ mass condensate· hH2O condensate(Tcondensate) + Qloss = W

hH2O(Tin) @ Tin = 295K = 2541 kJ/kg, cp air = 1.005 kJ/kg

hH2O condensate (Tcondensate) @ Tcondensate = 281K = 33.1 kJ/kg

→ 0.149·Tout – 43.878 + 0.0018· hH2O(Tout) – 5.453 + 0.011 +

0.015 = 0.689 (kW), note that hH2O(T) = hH2Osat vap(T)

→ 0.149·Tout + 0.0018· hH2Osat vap (Tout) = 49.994, or :

Tout = 335.5 – 0.012 · hH2Osat vap (Tout) (K)

iteration: Tout = 298 K → hH2Osat vap (Tout) = 2547 kJ/kg → Tout = 304.9 K →

hH2Osat vap (Tout) = 2560 kJ/kg → Tout = 304.8 K. result: Tout = 305 K = 32°C

4 relative humidity: pH2O,sat @32°C = 4.693 kPa ; pH2O outlet = ωout· pout / (0.622+ ωout) = 1.931 kPa → RH = 1.931/4.693 = 0.41

= 41 %

Air washer

t1* = wet bulb temperature of inlet air 1 2A: heating & humidification (tw > t1) 1 2B: pure humidification (tw= t1) 1 2C: cooling & humidification (tw< t1) 1 2D: adiabatic saturation (tw= t1*)


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td = dew point temperature of inlet air 1 2E: cooling & humidification

(td < tw < t1*) 1 2F: cooling (td = tw) 1 2G: cooling & dehumidification

(td > tw)

(t = temperature)

Source: A11

Example – using psychrometric chart

It is desired to provide air for space heating at 22°C and 60% relative humidity. Air is available at 10°C and 10% relative humidity. To achieve the desired air conditioning it is proposed to 1) heat the air with heating coils followed by 2) a water spray at 15°C(evaporative cooling) to obtain the desired exit relative humidity.

Determine (using the psychrometric chart given below) for the initial situation (1), the situation after heating (2) and the final situation (3) the values for temperature, relative humidity, enthalpy and absolute humidity and complete the table below.

Determine the heating requirements, in kJ/kg dry air (for the first stage), and the mass of liquid water, in g/kg dry air, required in the evaporative cooler (the second stage)

For the water spray, water/steam tables give (for saturated liquid) enthalpy h = hf = 63 kJ/kg at 15°C


424503 Xm 206 230 238 266

Example – using psychrometric chart

ω3 - ω2 = 9.25 kg/kg d.a.

h3 - h2 = (ω3 - ω2)·hf ≈ 0.6 kJ/kg d.a.

Energy in: ≈ (48 – 12) = 36 kJ/kg d.a.

(more accurate: 35.4 kJ/kg d.a.)

Water in: 9.25 kg / kg d.a.


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7.4 Cooling towers


Cooling towers /1 Wet cooling towers are used for

rejecting waste heat from power plants, large air conditioning or refrigeration systems and industriesto cooling water from lakes, riversetc. if seawater is not within reach

Power plant cooling water from condensers typically enters at ~ 40°C

Rather large: 15 ~ 175 m Similar to evaporative cooling for air

conditioning, but here the objectiveis to cool water, and make-up water must be added to compensate for the evaporated cooling water

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Cooling towers /2 Air flow through the cooling tower, in

counter-flow to the liquid spray can be by natural draft (similar to a chimney) or (using a fan) induced draft

The water is cooled; the air heats up while itshumidity increases

Drift eliminators minimise carry-over of water from the top of the tower

An (earlier) alternative is to use a spray pond to cool water; disadvantages are a large area size, large drift losses and contamination with dirt

Performance depends on the weather! Height calculations can be made using

combined heat/energy and mass transfer equations – see Ö96 p. 36-39; but since the flow field is also important: use CFD !

Pictures: ÇB98


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Example: cooling tower /1 Cooling water from a power plant is

cooled in a wet cooling tower, seeFigure →

Neglecting fan power, calculate– The volume flow rate of air into

the tower– The mass flow rate of

make-up water

Answer:dry air mass balance:

ṁA1 = ṁA2 = ṁA

water mass balance: ṁ3 + ṁA· ω1 = ṁ4 + ṁA· ω2

ṁmake-up = ṁ3 – ṁ4 = ṁA· (ω2-ω1) Picture: ÇB98

Φ = RH


Example: cooling tower /2

Energy balance:ṁA· h1 + ṁ3· h3 = ṁA· h2 + ṁ4· h4

ṁ3· h3 = ṁA· (h2 - h1) +(ṁ3 - ṁmake-up)· h4

This gives for the (dry) air flowṁA = ṁ3· (h3 - h4)/((h2 - h1) - (ω2 - ω1)· h4)

data from the Psychrometric chart, for example:h1= 42.2 kJ/kg; ω1 = 0.0087 kg H2O/kg dry air; v1 = 0.842 m3/kg dry airh2= 100.0 kJ/kg; ω2 = 0.0273 kg H2O/kg dry air; from water saturation tables:h3 = hf @ 35°C = 146.68 kJ/kgh4 = hf @ 22°C = 92.33 kJ/kg

This gives ṁA = 96.9 kg/s Volume flow V1 = ṁA· v1 =

81.6 m3/s ṁmake-up= ṁA· (ω2-ω1)

= 1.8 kg/s

Symbol Φ = RH


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Sources #7 A11: R. C. Arora ”Refrigeration and air conditioning”, 2nd. Ed. PHI

Learning Private Limited , New Delhi (2011) Chapters 15 - 18 BMH99: Beek, W.J., Muttzall, K.M.K., van Heuven, J.W. ”Transport phenomena”

Wiley, 2nd edition (1999) CB98: Y.A. Çengel, M.A. Boles “Thermodynamics. An engineering approach”,

McGraw-Hill (1998) HTW08: G.F. Hundy, A.R. Trott, T.C. Welsh “Refrigeration and air

conditioning 4th ed. Butterworth-Heinemann (2008) Chapters 20-26 R04: He-Sheng Ren “Construction of a generalized psychrometric chart for

different pressures” Int. J. Mech. Eng. Educ. 32(3) (2004) 212-222 SB01: R.E. Sonntag, C. Borgnakke “Introduction to engineering

thermodynamics” Wiley, (2001) Chapter 10 T06: S.R. Turns ”Thermal – Fluid Sciences”,

Cambridge Univ. Press (2006) TW00: A.R. Trott, T.C. Welsh ”Refrigeration

and Air-Conditioning” 3rd Ed. Butterworths- Heineman(2000)

Ö96: G. Öhman ”Kylteknik”, Åbo Akademi University (1996) See also: Martinez, I. ”Lectures on Thermodynamics” – lecture 18 (English or Spanish) and lecture 8

http://webserver.dmt.upm.es/~isidoro/bk3/index.html updated and based on “Termodinámica básica y aplicada", Ed. Dossat, Madrid (1992) ISBN 84-237-0810-1

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Documents

Refrigeration (Kylteknik) - Åbo Akademi | Startsidausers.abo.fi/rzevenho/REF17-OH7.pdf · 0.105 – 0.027 = 0.078 kg water per kg dr y air. With a density of 0.88 kg /m³ saturated