Ventilation. 6 Heat Exchangers. Air-conditioning processes. Vladimír Zmrhal (room no. 814)

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1 Ventilation 6 Heat Exchangers Vladimír Zmrhal (room no. 814) Dpt. Of Environmental Engineering 1 Air-conditioning processes comfort 2 1

2 Dimensioning of air-conditioning Outdoor air conditions (Czech Republic) WINTER t e = t e,loss 3 C, ϕ e = 100 % (t e,loss = -12, -15, -18 C) SUMMER t e = 32 C, h e = 58 kj/kg Indoor air conditions (for air-conditioned spaces) Indoor air temperature t i = 18 to 26 C Relative humidity ϕ i = 30 to 70 % Example 1: Find the summer and winter extreme in the diagram 3 Air-conditioning processes Typical climatic data (Czech Republic) TRY test reference year t e ambient air temperature [ C] x e humidity ratio [g/kg] 4 2

3 Air heating Heating coil water coil heat exchangers electric heaters direct heaters 5 Air heating Heat recovery heat recuperation heat regeneration Example 2: Φ = 65 % t e = -15 C t i = 20 C t sup =? Recovered heat coefficient t Φ = t t t e2 e1 o1 e1 6 3

4 Air heating Heat flow rate heating coil capacity h x = 0 ( ) Q = V ρc t t δ = 2 1 Example 2: V = m 3 /h t 1 = -15 C t 2 = 20 C Q h =? 7 Air heating 8 4

5 Air cooling Sensible cooling - without condensation ( dry cooling ) the temperature on a cooling surface is above or equal to the dew point temperature t c > t dp Latent cooling - dehumidifying moist air the temperature on a cold surface is lower than the dew point temperature t c < t dp Cold surface temperature water exchangers e.g. 6/12 C t c = (t w1 + t w2 )/2 = 9 C evaporators t c 4-5 C 9 Air cooling Cooling coil capacity total heat flow rate through the cooling coil c sen lat c ( ) 1 2 ( ) ρ ( ) Q = Q + Q = V ρc t t + V l x x Q = V ρ h h h = c t + l x 10 5

6 Air cooling Cooling without condensation t c > t dp cooling coil capacity sen lat c = 0 ( ) Q = V ρc t t Q 1 2 ρ ( 1 2 ) ( ) Q = V h h = = V ρc t t Air cooling Dehumidifying moist air t c < t dp cooling coil capacity ( ) Q = V ρ h h c

Air cooling Cooling without condensation Example 3a: V = 10 000 m 3 /h t 1 = 32 C h 1 = 58 kj/kg t 2 = 20 C t c = 16 C Q c =?

7 Air cooling Cooling without condensation Example 3a: V = m 3 /h t 1 = 32 C h 1 = 58 kj/kg t 2 = 20 C t c = 16 C Q c =? Cooling with dehumidifying Example 3b: V = m 3 /h t 1 = 32 C h 1 = 58 kj/kg t 2 = 20 C t w1 /t w2 = 6/12 C Q c =? 13 Air cooling Direct air cooling cooling coil = evaporator t c = 4-5 C 14 7

8 Air cooling 15 Control of water exchangers Supply temp. control mixing of the flow V = const. t w = var. heating/cooling coils Heat mass flow control dividing of the flow V = var. t w = const. cooling coils 16 8

9 Humidifying (adding steam) Moisture balance M x + M = M x 1 v 2 Heat balance M h + M h = M h 1 v v 2 17 Humidifying (adding steam) Enthalpy of water vapor h = l + c t = t 2540 v v [kj/kg] Direction of the process δ h h h x x x 2 1 = = = 2 1 h v δ = h v = 2.5 [kj/g] 18 9

Humidifying (adding steam) Enthalpy of water vapor reality l = 2500

2 0 2680 [kj/kg] l = 2257 kj/kg for 100 C hv = l + cvt + cwt = 2257 +

68 [kj/kg] [kj/g] In reality the air is also heated!

10 Humidifying (adding steam) Enthalpy of water vapor reality l = 2500 kj/kg for 0 C hv = l + cvt + cwt = [kj/kg] l = 2257 kj/kg for 100 C hv = l + cvt + cwt = δ = h v = 2.68 [kj/kg] [kj/g] In reality the air is also heated! 19 Humidifying (adding steam) Distribution of steam (sprays) Example 4: V = m 3 /h x 1 = 1 g/kg x 2 = 5 g/kg M v =? Q hum =? 20 10

11 Humidifying (adding water - injection) Adding water (100% evaporation) δ h h h x x x 2 1 = = = 2 1 Enthalpy of water h w h = c t = 4.187t = 0 to 420 w w [kj/kg] Direction of the process δ = h w 0 [kj/g] 21 Humidifying (adding water) Adiabatic cooling (air washer) t = t w1 w 2 Q = 0 δ = c t = 4.187t 0 kj/g w w w Polytropic change t w1 t w2 Q > 0 δ > 0 Q < 0 δ <

12 Humidifying (adding water) Legionella Legionella acquired its name after a July, 1976 outbreak of a then-unknown "mystery disease" sickened 221 persons, causing 34 deaths. The outbreak was first noticed among people attending a convention of the American Legion - an association of U.S. military veterans. The convention in question occurred in Philadelphia during the U.S. Bicentennial year. This epidemic among U.S. war veterans, occurring in the same city as and within days of the 200th anniversary of the signing of the Declaration of Independence, was widely publicized and caused great concern in the United States. On January 18, 1977 the causative agent was identified as a previously unknown bacterium (from air-conditioning) subsequently named Legionella. See Legionnaires' Disease for full details Humidifying (adding water) Efficiency of air washer x η = x 2 1 wb x x 1 Example 5: η = 60 % t 1 = 27 C ϕ 1 = 40 % x 2 =? ϕ 2 =? 24 12

Dehumidifying Cooling of air - vapor condensation t c < t dp Adsorption dehumidifying rotary heat recovery exchangers rotor with hygroscopic material as silica gel or activated alumina the sorbent

13 Dehumidifying Cooling of air - vapor condensation t c < t dp Adsorption dehumidifying rotary heat recovery exchangers rotor with hygroscopic material as silica gel or activated alumina the sorbent material contains a vast number of microscopic pores where water is absorbed moisture is condensed and held on the surface of the material the adsorbent material can be reactivated by heat - regeneration 25 Dehumidifying R1 R2 E2 E

14 Example of winter process Single duct air system heat recovery heater humidifying (steam) conditioning in the room 27 Example of summer process Fan coil units outdoor air cooling recirculation of air mixing in the room conditioning in the room 28 14

15 Thank you for your attention 29 15

Subject: Principles of Refrigeration and Air Conditioning Lecturer: Assistant Professor Dr. Waheed Shaty Mohammed

Subject: Principles of Refrigeration and Air Conditioning Lecturer: Assistant Professor Dr. Waheed Shaty Mohammed Refrences: 1-A. R. Trott and T. Welch " Refrigeration and Air conditioning ",Third Edition