Kylteknik (”KYL”)
Refrigeration course # 424503.0
v. 2014
7. Air conditioning, cooling towers
Ron Zevenhoven Åbo Akademi University Thermal and Flow Engineering Laboratory / Värme- och strömningsteknik tel. 3223 ;
[email protected]
Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
1/36
ÅA 424503 Refrigeration / Kylteknik
7.1 Humid air
Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
2/36
”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 23.11.2014
Picture:http://p.vtourist.com/2377347-Maid_of_the_Mist-Niagara_Falls.jpg
Humid air /1
H2O
Picture: ÇB98
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
3/36
Humid air /2
The amount of water vapour in air can be specified as absolute (or specific) humidity, ω (kg water/kg dry air): ω
masswater vapour massdry air
(kg water/kg dry air)
Mwater ( g / mol ) nwater ( mol ) p ( Pa ) 0.622 water Mair ( g / mol ) nair ( mol ) pair ( Pa )
0.622
p water ( Pa ) ; p total (Pa) - p water ( Pa )
i.e., p water ( Pa )
ω p total (Pa) 0.622 ω
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: hhumid air hdry air hwater,sat vapour
(kJ/kg)
1.005 T(C) (2500 1.87 T(C)) (kJ/kg) 23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
Picture: ÇB98 4/36
Humid air /3
RH
Picture: http://tokerud.typepad.com/photos/mv/goldengatebridge.jpg
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 %) : masswater vapour ( 100% ) masswater vapour ,max
p water ( Pa ) pwater ,sat ( Pa )
ω p total (Pa) ( 0.622 ω ) pwater ,sat ( Pa )
i.e. ω
0.622 RH pwater ,sat ( Pa ) Ptotal ( Pa ) RH pwater ,sat ( Pa ) 23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
5/36
Picture: http://www.bu.edu/ocs/castle/images/musicroom.jpg
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 kg with V = 75 m3 → mair = 85.10 kg Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
6/36
Humid air /4 The dew point, Tdp, is the temperature at which condensation 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 inside temperature will moisture condense 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 23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
Adiabatic saturation
Symbol Φ = relative humidity RH
The adiabatic saturation temperature is the temperature when reaching saturation in an air stream at equilibrium with liquid water in an adiabatic process
ω1 with
7/36
c p ,air (T2 T1 ) ω2 hfg ,2 hg 1 hf 2 0.622 psat ,2 ω2 p2 psat ,2
T2 T1
Pictures: ÇB98 23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
8/36
Adiabatic saturation (derivation)
c (T T1 ) ω2 hfg ,2 ω1 p ,air 2 hg 1 hf 2 with ω2
T2 T1
0.622 psat ,2 p2 psat ,2
23.11.2014
Pictures: ÇB98 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
9/36
ÅA 424503 Refrigeration / Kylteknik
Adiabatic saturation – example
Source: SB01 Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
10
Wet bulb temperature Picture: http://www.idalex.com/images/Dry_Wet_Dew.jpg
The wet bulb temperature is the temperature reached as a result of heat withdrawal for evaporation of water; Twet bulb < Tdry bulb if the relative humidity RH < 100%
23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
11/36
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
Source: T06
Answer: the dewpoint gives the water vapour pressure From tables: pwater vapour = psat @Tdewpoint = 1.073 kPa, and psat @ 18°C = 2.065 kPa RH = 1.073 kPa / 2.065 kPa = 0.52 = 52% ω=0.622· pwater vapour /(ptotal – pwater vapour) = 6.83 g water/kg dry air Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
12/36
ÅA 424503 Refrigeration / Kylteknik
7.2 The psychrometric chart
Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
13/36
Piispankatu 8, 20500 Turku
23.11.2014
14/36
Psychrometric chart /1
Åbo Akademi Univ - Thermal and Flow Engineering
Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
Source: http://www.isa.org/Content/ContentGroups/InTech2/ Departments/Control_Fundamentals/200217/20020857.gif
g water / kg dry air
Psychrometric chart /2a Climate
15/36
Sources: http://www.flickriver.com/photos/mitopencourseware/3038898862/ http://www.dvlnet.com/blog/bid/77642/ Allowable-vs-Recommended-ASHRAE-Guidelines-Design-Your-Data-Center
Psychrometric chart /2b ”Comfort zone”
ASHRAE = American Society of Heating, Refrigerating and Air Conditioning Engineers
Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
16/36
Psychrometric chart /3 Picture: ÇB98
Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
17/36
Psychrometric chart /4 *
Lines in the psychrometric chart 23.11.2014
Saturated air at 15°C Tdewpoint = Twet bulb =Tdry bulb Symbol Φ = RH Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
Pictures: ÇB98 18/36
See also course möf-st 424302
Example: psychrometric chart /1 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 and source: BMH99
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 Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
19/36
See also course möf-st 424302
Example: psychrometric chart /2
Question 2: Cool this air to 30°C, how much water will condense per m³ saturated air ?
Picture and source: BMH99
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 condensed 0.105 – 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. Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
20/36
See also course möf-st 424302
Example: psychrometric chart /3 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?
Picture and source: BMH99
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. Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
21/36
ÅA 424503 Refrigeration / Kylteknik
7.3 Air conditioning
Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
22/36
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
23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
23/36
Air conditioning /2 Systems based on active cooling (and heating) using a vapourcompression refrigeration cycle
Pictures: T06
Air conditioning without and with de-humidifying Åbo Akademi Univ - Thermal and Flow Engineering
A residential air conditioning system Piispankatu 8, 20500 Turku
23.11.2014
24/36
Air conditioning /3
Pictures: T06
Evaporative coolers (”swamp coolers”) Humifidifiers work similarly
A residential de-humidifier
Åbo Akademi Univ - Thermal and Flow Engineering
The main difference with standard air conditioning is the condenser in the air stream Piispankatu 8, 20500 Turku
23.11.2014
25/36
Air conditioning /4 Balance equations for air conditioning systems: – (a) control volume – (b) mass balance – (c) energy balance water mass conservation : H2O,vapour,stream i m A ωstream i ) (A dry air, m A 1 m H2O,liquid,3 m A 4 m H2O,liquid,2 m overall mass conservation : A (1 1 ) m H2O,liquid,3 m A (1 4 ) m H2O,liquid,2 m overall energy conservation : W m A (h A1 1 hH2O,vapour,1 ) Q in in H2O,liquid,3 hH2O,liquid,3 Q m out A (h A 4 4 hH2O,vapour,4 ) m H2O,liquid,2 hH2O,liquid,2 m 23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
Picture: T06 26/36
Example: household dehumidifier /1 Picture: http://www.daviddarling.info/images/dehumidifier.gif
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
Source: T06 23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
27/36
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 air ωout = (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/kg mass 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 Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
28/36
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 % Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
29/36
Air washer
(t = temperature)
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*) 23.11.2014
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) Source: A11
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
30/64
ÅA 424503 Refrigeration / Kylteknik
7.4 Cooling towers
Åbo Akademi Univ - Thermal and Flow Engineering
Piispankatu 8, 20500 Turku
23.11.2014
31/36
Cooling towers /1
Åbo Akademi Univ - Thermal and Flow Engineering
Picture: http://www.superstock.com/search/Energy/coal
Wet cooling towers are used for rejecting waste heat from power plants, large air conditioning or refrigeration systems and industries to cooling water from lakes, rivers etc. 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 objective is to cool water, and make-up water must be added to compensate for the evaporated cooling water
Picture: T06 Piispankatu 8, 20500 Turku
23.11.2014
32/36
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 its humidity 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 ! Åbo Akademi Univ - Thermal and Flow Engineering
Pictures: ÇB98 Piispankatu 8, 20500 Turku
23.11.2014
33/36
Example: cooling tower /1 Cooling water from a power plant is cooled in a wet cooling tower, see Figure → 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) 23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
Φ = RH Picture: ÇB98 34/36
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 flow ṁA = 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 air h2= 100.0 kJ/kg; ω2 = 0.0273 kg H2O/kg dry air; from water saturation tables: h3 = hf @ 35°C = 146.68 kJ/kg h4 = hf @ 22°C = 92.33 kJ/kg Åbo Akademi Univ - Thermal and Flow Engineering
Symbol Φ = RH
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 Piispankatu 8, 20500 Turku
23.11.2014
35/36
Sources #7
Picture:http://marvin.kset.org/~draak/Cooling_Fan.jpg
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) 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 23.11.2014
Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku
36/36
