Willkommen. Welcome. Bienvenue. Ventilation energy efficiency of fans and drives Energy recovery and energy efficiency in ventilation technology

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and energy efficiency in ventilation technology

1 Willkommen Bienvenue Welcome Ventilation energy efficiency of fans and drives Energy recovery and energy efficiency in ventilation technology Prof. Dr.-Ing. Christoph Kaup

2 Components Ventilation systems tasks: Thermodynamic air treatment Heating Cooling Humidification Dehumidification 2 K

3 Components Ventilation systems tasks: Thermodynamic air treatment Example of a ventilation system (supply air) WP F HRS C H S V S ODA SUP 3 K

4 Drives Ventilation systems tasks: Transport of air fan motor drive control K 4

5 Energy efficiency in ventilation Energy demand in ventilation systems Air treatment heating/cooling/de-/humidification Air transport against p internal/external K 5

6 Drives Power consumption. P m = V Δp 1 / η s P m. V Δp absorbed power consumption [KW] air flow [m³/s] pressure losses of the system [Pa] η s system efficiency [./.] η S = η F η M η D η C fan motor drive control K 6

DIN EN 13779 Ventilation system non-residential General requirements and measured values specifische fan power (SFP) P SFP specific fan power [W/(m³/s)] P

7 DIN EN Ventilation system non-residential General requirements and measured values specifische fan power (SFP) P SFP specific fan power [W/(m³/s)] P Input P SFP = = q V p fan η total P Input q V electric power consumption [W] air flow rate [m³/s] p fan total pressure [Pa] η total system efficiency [-] K 7

8 Specific Fan Power EN 13779: 2007 categoriy PSFP W/(m³/s) η total 0,55 ΔpFan [Pa] η total 0,65 application default value SFP 1 < EXH without HR SFP 2 SFP 2 SFP 3 < 750 < EXH with HR SUP without HR SUP full AC SFP 3 SFP 3 SFP 4 SFP 4 < SFP 5 SFP 6 < < for specific components (e. g. HEPA-Filter, HRS H1 or H2) is a use of additional SFP possible. SFP 7 > K 8

9 Additional fan power EN 13779: 2007 component add. P SFP [W/(m³/s)] add. filterstage HEPA filter gasfilter HRS class H2-H1 high capacity cooler K 9

10 Air velocity classes EN 13053: 2012 class V1 V2 V3 V4 V5 V6 V7 V8 V9 velocity in m/s 1,6 m/s 1,8 m/s 2,0 m/s 2,2 m/s 2,5 m/s 2,8 m/s 3,2 m/s 3,6 m/s > 3,6 m/s K 10

1200 1000 800 600 400 0,1 1000 10000 100000 airflow q [m³/h] P input max p

11 Electric power consumption fans 100,0 max. absorped motorpow er PMmax [kw] 10,0 1,0 p stat [Pa] , airflow q [m³/h] P input max p stat q V (kw) (Pa) (m³/s) class Base equation P Input max = 0,925 0,95 p stat. ( x q + 0,08 v 450 ) ( ) K 11

12 Power consumption classes EN 13053: 2012 class P1 P2 P3 P4 P5 powerconsumption related to Pm ref Pm ref 0.85 Pm ref 0.90 Pm ref 0.95 Pm ref 1.00 Pm ref 1.06 P6 Pm ref 1.12 P7 > Pm ref 1.12 K 12

13 AHU components Energy efficiency reduction of components (e. g. drop eliminator) hybrid components (humidifier) bypasses in components K 13

14 AHU components Energy efficiency Internal pressure losses components (example hybrid humidifier) separate humidifier integrated contact humidifier (hybridsystem) K 14

15 AHU components Energy efficiency Internal pressure losses face velocity (example HRS CC-System) Pressure losses temperature efficiency number of rows number of rows K 15

16 AHU components Energy efficiency Internal pressure losses arrangment of components F7 F7 F7 K 16

17 AHU components Filter stages Single stage first stage F7 (80 % at 1 µm) better protection of the AHU reduction of pressure losses reduction of AHU lenght Two filter stages first stage F7 (80 % at 1 µm) second stage F7 = efficiency (F5 / F9) reduction of pressure losses K 17

18 AHU components Efficiency in % F9 F7 F5 Size in µm F5 + F9 4,5 m² 9,0 m² F5 F9 55 Pa 200 Pa m³/h 140 Pa 250 Pa m³/h 450 Pa F7 + F7 9,0 m² 9,0 m² F7 F7 95 Pa 95 Pa 150 Pa 110 Pa m³/h m³/h 260 Pa K 18

19 AHU components Cooler Suction side use with dehumdification fan waste heat used to rewarm finspace e. g. 2,5 mm Pressure side use with a dry cooling fan waste heat before the coller bigger average log. temperature difference C V V C K 19

20 Air transport Power mechanical electrical K 20FtKηpVP tfpvp MFINηηcos3IUP DMFIKηηηcos3IUP FDMFIVηηηηcos3IUP AHU drivesdftnηηpvp

21 Air transport Power mechanical electrical FUFIMDFtmηηηηpVP 3 cos I U P m FIMηcos3IUP MFINηηcos3IUP DFtNηηpVP MDFtMηηηpVP Motor: K 21 AHU drives

22 Air transport Power Systemefficiency η Syst Pm PV FU FPηηηV IηM F D Pm ηystk 22S

23 Air transport Example: 5,34 KW 3,33 KW FU 100 % 62 % K 23

24 Drives Air transport Fan Types Axial with or without a casing Radial with a casing Radial without a casing K 24

25 Drives Air transport Fan Overview Radialfans Direct driven Belt driven with or without a casing Forward curved backward curved Forward curved backward curved Motor inside Motor external K 25

26 Air transport Fan Selection characteristics K 26

27 Fan concepts K 27

28 Energy efficiency Internal pressure losses inlet losses (example fan with a spiral housing) p EV = 1,5-4,5 p dyn K 28

29 Energy efficiency Internal pressure losses inlet losses (example plugged fan without a housing) p EV = 0,5-1,5 p dyn K 29

30 Energy efficiency Internal pressure losses potential of plugged fans compared to spiral fans Energy savings (%) Stat. fan pressure (Pa) K 30

31 Energy efficiency Internal pressure losses guide vane at a axial inlet situation efficiency turbulence DE K 31

32 Air transport Internal pressure losses practice: m³/h transported in a AHU against a stat. pressure of p stat =1.171 Pa given: η Vt = 80,5 %; belt drive impeller diameter D = 400 mm outlet area mm η M = 88,3 % η B = 94,0 % p EV = 3,0 p dyn asked: motor power P N K 32

33 Air transport Internal pressure losses solution: A R = 0,501 0,501 = 0,251 m² w fan outlet = / / A R = 11,07 m/s p dyn = ρ / 2 w² = 0,6 11,07 ² = 74 Pa p EV = 3 74 = 222 Pa p t = p stat + p EV = = Pa P N = V p t / (η Vt η K ) = / / (0,805 0,94) = 5,113 KW K 33

34 Air transport Power given: m³/h in a AHU against a pressure drop of p t =1.200 Pa. η Ffa = 71,0 %; direct driven Impeller diameter D = 630 mm Impeller width b = 201 mm η M = 88,3 % η FI = 97,0 % calculate: Motorpower P N Systemeff. absorpt motor power η Syst. P m K 34

35 Air transport Power solution: A R = D b = 0,3978 m² w Impeller = / / A R = 6,98 m/s p dyn = ρ / 2 w² = 0,6 6,98² = 29 Pa p stat = = Pa P N = V p stat / η fa = 4,581 KW ; choosen motor 5,5 KW η t = V p t / P N = 72,8 % η Syst. = η M η t η FI = 0,883 0,728 0,97 = 62,4 % P m = p V / η Syst. = / / 0,624 = 5,342 KW K 35

36 Air transport Internal pressure losses solution: A R = D b = 0,3978 m² W fan outlet = / / A R = 6,98 m/s p dyn = ρ / 2 w² = 0,6 6,98² = 29 Pa p stat = = Pa P N = V p stat / η fa = 4,581 KW choosen motor 5,5 KW K 36

37 Air transport Motor Types AC Rotary current with 3 phases alternating current EC electric commutaded continuous current PM permanent magnet Motor K 37

38 Air transport Motor Efficiencies (example n = /min) Power IE2 IE3 1,1 KW 77,0 % 84,0 % 2,2 KW 82,0 % 86,5 % 4,0 KW 85,0 % 88,5 % 7,5 KW 87,0 % 90,3 % 55,0 KW 93,5 % 95,1 % K 38

39 Motors Motorpower in W K 39

40 Air transport Drive Coupling fan / motor overview Drive Direct driven Belt driven Internal motor External motor Internal motor Belt- Flat belt- impeller direct impeller with drive drive mounted Shaft mounted A clutch K 40

41 Air transport Drive Coupling fan / motor indirect via belt (standard or flat belt) direct K 41

42 Air transport Drive Coupling fan / motor Waste power: standard belt 3 12 % flat belt 2 4 % K 42

43 Air transport Control Air flow speed control e. g. via frequency inverter pressure control e. g. via dampers power losses: 3 5 % K 43

44 Energy efficiency Proportional law Air flow Pressure Power consumption V 2 n 2 = V 1 n 1 V 2 p 2 = V 1 p 1 ( ) 2 3 V 1 P 1 = V 2 P 2 ( ) K 44

45 Energy efficiency Air flow rate-measuring device effective pressure to air flow V αε d² π 4 2 Δp ρ Eff. pressure in % to nominal pressure Flow rate in % to nominal flow rate K 45

46 Energy efficiency Proportional law given: V 1 = m³/h transported with a air handling unit against p t1 = Pa with a speed of n 1 = min -1. The absorbed motor power is Pm 1 = 5,7 KW. calculate: Parameter p t2, n 2, Pm 2 at a reduced air flow of 80% ( V 2 = m³ / h ). K 46

47 Energy efficieny Proportional law solution: n 2 = ( V 2 / V 1 ) n 1 = (8.000 / ) = min -1 ( -20,0% ) p t2 = ( V 2 / V 1 ) 2 p t1 = (8.000 / ) = 768 Pa ( -36,0% ) P 2 = ( V 2 / V 1 ) 3 P 1 = (8.000 / ) 3 5,7 = 2,92 KW ( -48,8% ) K 47

48 Thank you Ventilation for your Attention energy efficiency of fans and drives Energy recovery and energy efficiency in ventilation technology Prof. Dr.-Ing. Christoph Kaup

ALUMINIUM CENTRIFUGAL FANS. 4 x M6 x 16

ALUMINIUM CENTRIFUGAL FANS. 4 x M6 x 16 DESCRIPTION Dutair BCT-series aluminium centrifugal fans with open die-cast aluminium impellers and forward curved blades are suitable for material transport and high pressure applications. The asynchronous