ASHRAE Standard 62.1 Update
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Dennis Stanke of Trane presents ASHRAE Standard 62.1 Update. Dennis is the Chair of the Standard 62.1 committee. Presented at the 2008 ASHRAE Region VI CRC in Chicago, Illinois. May 15, 16, 17, 2008
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ASHRAE Standard 62.1 Update
- 1. ASHRAE Std 62.1 Update Where Are We Now? May 2008 Dennis A. Stanke Trane • La Crosse, WI
- 4. ASHRAE Standard 62.1 What’s Its History? 62-2001 a little more change 62-1999 a little change 62-1973 first issued 1990 2000 2010 1970 1980 62.1-2004 new VRP, many lower rates 2006 Supplement ETS separation 62.1-2007 corrections, clarifications 62-1989 higher rates 62-1981 lower rates
- 26. non-attainment areas PM 10 (size ≤ 10 microns) US EPA AQS Database January 17, 2007 6.2.1 OA Treatment: Need MERV 6 filter
- 27. non-attainment areas ( future ) PM 2.5 (size ≤ 2.5 microns) US EPA AQS Database January 17, 2007 Future: Addendum 62.1c would require MERV 11 in many locations
- 28. non-attainment areas Ozone (8-hour) US EPA AQS Database January 17, 2007 6.2.1 OA Treatment: Need 40% air cleaner where 1-hour peak exceeds 160 ppb Fresno, Riverside, Long Beach
- 29. non-attainment areas ( future ) Ozone (8-hour) US EPA AQS Database January 17, 2007 Future: Addendum 62.1c require 40% air cleaners in many more locations
- 32. ventilation rate procedure Effective Minimum Rates Office (5p) 100 20.0 85 17.0 Classroom (ages 5-8) (25p) 375 15.0 370 15.0 Lecture classroom (65p) 975 15.0 550 8.5 Retail sales (20p) 300 15.0 270 14.0 Auditorium (150p) 2250 15.0 810 5.4 Occupancy category (default density/1000 ft²) Std 62-2001 Vbz cfm Effective cfm/p Std 62.1-2007 70% of OA rates drop; 30% stay same or rise Comparison of breathing-zone OA flow Vbz cfm Effective cfm/p
- 33. ventilation rate procedure Effective Minimum Rates Daycare sickroom (25p) 10 0.18 430 17.2 Univ/Col laboratory (25p) 10 0.18 430 17.2 Central laundry rm (10p) 5 0.12 170 17.0 Res dwelling unit (5p) 5 0.06 85 17.0 Res corridor (--) 0 0.06 60 NA Occupancy category (default density/1000 ft²) Std 62.1-2007 Vbz cfm Effective cfm/p Check Standard for complete list of rates New rates and breathing-zone OA flow Rp cfm/p Ra cfm/ft 2 2007: Added some occupancy categories/rates, for example …
- 38. VRP 6-zone school example Zone-Level Calculations * Average (81% of 40 peak) Step 1: Vbz = Rp*Pz + Ra*Az Step 2: Look up Ez Step 3: Voz = Vbz/Ez 680 1.0 2,000 0.18 32* 10 North art classrm 85 1.0 1,000 0.06 5 5 Interior offices 2,190 1.0 4,000 0.06 260 7.5 East lecture class 2,190 1.0 4,000 0.06 260 7.5 North lecture class 1,880 1.0 4,000 0.12 140 10 West classrms (9+) 1,880 1.0 4,000 0.12 140 10 South classrms (9+) cfm -- ft 2 cfm/ft 2 people cfm/per Voz Ez Az Ra Pz Rp Example School
- 39. VRP – system calculations Six Single-Zone Systems - Clg * Average (81% of 40 peak) Step 1: Vbz = Rp*Pz + Ra*Az Step 2: Look up Ez Step 3: Voz = Vbz/Ez Step 4: Vot = Voz each zone 8,900 Vot = Total zone OA flow 680 1.0 2,000 0.18 32* 10 North art classrm 85 1.0 1,000 0.06 5 5 Interior offices 2,190 1.0 4,000 0.06 260 7.5 East lecture class 2,190 1.0 4,000 0.06 260 7.5 North lecture class 1,880 1.0 4,000 0.12 140 10 West classrms (9+) 1,880 1.0 4,000 0.12 140 10 South classrms (9+) cfm -- ft 2 cfm/ft 2 people cfm/per Voz Ez Az Ra Pz Rp Example School
- 40. VRP – system calculations 100% OA Systems – CV *Average (81% of 40) Step 4: Vot = Voz Step 1: Vbz = Rp*Pz + Ra*Az Step 2: Look up Ez Step 3: Voz = Vbz/Ez 8,900 Vot = OA intake flow (Vot) 680 1.0 2,000 0.18 32* 10 North art classrm 85 1.0 1,000 0.06 5 5 Interior offices 2,190 1.0 4,000 0.06 260 7.5 East lecture class 2,190 1.0 4,000 0.06 260 7.5 North lecture class 1,880 1.0 4,000 0.12 140 10 West classrms (9+) 1,880 1.0 4,000 0.12 140 10 South classrms (9+) cfm -- ft 2 cfm/ft 2 people cfm/per Voz Ez Az Ra Pz Rp Example School
- 41. VRP – system calculations 100% OA Systems – CV *Average (81% of 40) Note: Must assume peak (or average) population in every zone because ventilation airflow per zone is constant. 8,900 Vot = OA intake flow (Vot) 680 1.0 2,000 0.18 32* 10 North art classrm 85 1.0 1,000 0.06 5 5 Interior offices 2,190 1.0 4,000 0.06 260 7.5 East lecture class 2,190 1.0 4,000 0.06 260 7.5 North lecture class 1,880 1.0 4,000 0.12 140 10 West classrms (9+) 1,880 1.0 4,000 0.12 140 10 South classrms (9+) cfm -- ft 2 cfm/ft 2 people cfm/per Voz Ez Az Ra Pz Rp Example School
- 42. multiple-zone system calculations Single-Path VAV: Calc Ev Step 4: Find outdoor air fraction for each zone: Zd = Voz/Vdzm = 1880/4000 = 0.47 * System population Ps = 550; Load diversity factor = 0.70 0.52 1,300 680 1,700 32* North art classrm 0.28 300 85 500 5 Interior offices 0.55 4,000 2,190 7,900 260 East lecture class 0.55 4,000 2,190 5,500 260 North lecture class 0.47 4,000 1,880 6,700 140 West classrms (9+) 0.47 4,000 1,880 6,500 140 South classrms (9+) (5,6,7) -- cfm cfm cfm people Evz Zd Vdzm Voz Vdz Pz Example School
- 43. multiple-zone system calculations Single-Path VAV: Calc Ev Step 5a: Find occupant diversity: D = Ps/ Pz = 550/837 = 0.66 Step 5b: Find uncorrected outdoor air intake: Vou = D* (Rp*Pz) + (Ra*Az) = 0.66*7,000 + 1,900 = 6,500 cfm * System population Ps = 550; Load diversity factor = 0.70 6,500 Vou = Step 5 -- -- -- Uncorrected OA flow -- 0.52 1,300 680 1,700 32* North art classrm -- 0.28 300 85 500 5 Interior offices -- 0.55 4,000 2,190 7,900 260 East lecture class -- 0.55 4,000 2,190 5,500 260 North lecture class -- 0.47 4,000 1,880 6,700 140 West classrms (9+) -- 0.47 4,000 1,880 6,500 140 South classrms (9+) (5,6,7) -- cfm cfm cfm people Evz Zd Vdzm Voz Vdz Pz Example School
- 44. multiple-zone system calculations Single-Path VAV: Calc Ev * Average (81% of 40 peak) Step 6a: Find system primary airflow: Vps = LDF* Vpz = 0.70*28,800 = 20,200 Step 6b: Find average outdoor air fraction: Xs = Vou/Vps = 6,500/20,200 = 0.32 Step 6c: Find ventilation efficiency for each zone: Evz1 = 1+Xs–Zd = 1+0.32–0.47 = 0.85 Step 6d: Find system ventilation efficiency Ev = min(Evz) = 0.77 0.77 Ev = -- -- -- Sys vent eff 0.32 Xs = Step 6 Uncorrected OA frac 6,500 Vou = Step 5 -- -- -- Uncorrected OA flow 0.80 0.52 1,300 680 1,700 32* North art classrm 1.04 0.28 300 85 500 5 Interior offices 0.77 0.55 4,000 2,190 7,900 260 East lecture class 0.77 0.55 4,000 2,190 5,500 260 North lecture class 0.85 0.47 4,000 1,880 6,700 140 West classrms (9+) 0.85 0.47 4,000 1,880 6,500 140 South classrms (9+) (5,6,7) -- cfm cfm cfm people Evz Zd Vdzm Voz Vdz Pz Example School
- 45. multiple-zone system calculations Single-Path VAV: Calc Ev Step 7: Find outdoor air intake flow: Vot = Vou/Ev = 6,500/0.77 = 8,400 * System population Ps = 550; Load diversity factor = 0.70 0.32 Xs = Step 6 Uncorrected OA frac 0.77 Ev = -- -- -- Sys vent eff 8,400 Vot = Step 7 -- -- -- Outdoor air intake 6,500 Vou = Step 5 -- -- -- Uncorrected OA flow 0.80 0.52 1,300 680 1,700 32* North art classrm 1.04 0.28 300 85 500 5 Interior offices 0.77 0.55 4,000 2,190 7,900 260 East lecture class 0.77 0.55 4,000 2,190 5,500 260 North lecture class 0.85 0.47 4,000 1,880 6,700 140 West classrms (9+) 0.85 0.47 4,000 1,880 6,500 140 South classrms (9+) (5,6,7) -- cfm cfm cfm people Evz Zd Vdzm Voz Vdz Pz Example School
- 46. VRP 6-zone school example OA Intake Flow Summary No population diversity credit Penalty for “too warm” htg air No population diversity credit No population diversity credit Conservatively low default Ev value Equations for more accurate Ev Two ventilation paths, highest Ev 10,800 VAV Default Ev 7,800 Series FP VAV 8,900 100% OA – VAV 8,400 VAV Calculated Ev 8,900 100% OA – CV 11,100 Single-Zone Htg 8,900 Single-Zone Clg OA Intake Vot Ventilation System
- 47. VRP 6-zone school example OA Intake Flow Summary 7,800 8,400 10,800 8,900 8,900 11,100 8,900 OA Intake (2007 Vot) 10,900 10,900 10,900 12,600 12,600 15,800 12,600 OA Intake (2001 Vot) -1 MZS-VAV Default Ev -28 MZS-VAV Series FP -29 100% OA – VAV -23 MZS-VAV Calc Ev -29 100% OA – CV -30 Single-Zone Htg -29 Single-Zone Clg % Chg Ventilation System
- 51. zone-level CO 2 -based DCV (Users Manual) Modulate Vbz CO 2 10 20 200 400 600 800 1000 zone population, Pz breathing zone OA, Vbz 30 40 50 60 differential CO 2 , ppm 200 400 600 800 1000 1200 0 0 0 1200 First, find max and min values for CO 2 Vbz-min = 60 cfm C-min = 0 ppm Vbz-des = 548 cfm C-max = k*m*Pz/Vbz = 10,500*65/548 = 1240 ppm Vbz = 7.5 × Pz + 0.060 × Az
- 52. zone-level CO2-based DCV (Users Manual) Modulate Vbz CO 2 C (CO 2 , ppm) Vbz (cfm) 60 548 0 1240 Vbz = 0.393 × C + 60 The Controller Second, define the proportional Controller
- 53. zone-level CO 2 -based DCV (Users Manual) Modulate Vbz CO 2 10 20 200 400 600 800 1000 zone population, Pz breathing zone OA, Vbz 30 40 50 60 differential CO 2 , ppm 200 400 600 800 1000 1200 0 0 0 1200 Controller adjusts Vbz in direct proportion to sensed CO 2 Optional CO 2 DCV can save operating energy For single zone systems, Vbz => min Vbz req’d To analyze Controller operation: … 1. Assume initial C-int, find: Vbz = 0.393* C-int + 60 2. Given Pz and Vbz, find: C = Pz*k*m/Vbz 3. Repeat until C-int = C
- 56. VAV ventilation reset control (no DCV) Single-Duct VAV System 100% system load (20,200 cfm) VRC only: low D, constant Vou, reduces Vot 8, 810 © 2005 American Standard Inc. Vot w/vent reset disc airflow Vdz 4,000 4,100 4,200 4,300 300 1,300 vent rate Voz 1,880 1,880 2,190 2,190 85 680 vent fraction Zdz 0.470 0.459 0.521 0. 509 0.283 0.585 Vou = 6,500 Xs = Vou/Vps = 6,500/18,200 = 0.357 Ev = 1 + 0.357 – 0.585 = 0.772 Vot = Vou/Ev = 6,500/0.772 = 8,410 90% system load (18,200 cfm) 8,810 Vot req’d @ design 8,810 8,410 disc airflow Vdz 5,000 5,400 4,000 4,000 500 1,300 vent rate Voz 1,880 1,880 2,190 2,190 85 760 vent fraction Zdz 0.376 0.351 0.548 0.548 0.170 0.585 Vou = D* Rp*Pz + Ra*Az = 0.65 *7,130 + 1860 = 6,500 Xs = Vou/Vps = 6,500/20,200 = 0.322 Ev = 1 + 0.322 – 0.585 = 0.738 Vot = Vou/Ev = 6,500/0.738 = 8,810
- 58. VAV ventilation reset control (with DCV zones) Single-Duct VAV System disc airflow Vdz 5,000 5,400 4,000 4,000 500 1,300 vent rate Voz 1,880 1,880 2,190 2,190 85 760 vent fraction Zdz 0.376 0.351 0.548 0.548 0.170 0.585 Vou = D* Rp*Pz + Ra*Az = 0.65*7,130 + 1860 = 6,500 Xs = Vou/Vps = 6,500/20,200 = 0.332 Ev = 1 + 0.332 – 0.585 = 0.738 Vot = Vou/Ev = 6,500/0.738 = 8,810 100% VRC w/zone-level DCV reduces Vot even more © 2005 American Standard Inc. 8,810 Vot w/vent & DCV 8,810 8,410 5 7,040 Vot w/vent reset Pz disc airflow Vdz 4,000 4,100 4,200 4,300 300 1,300 vent rate Voz 1,880 1,880 2,190 2,190 85 680 vent fraction Zdz 0.470 0.459 0.146 0. 509 0.283 0.431 360 0.277 915 0.218 Sense CO 2 , find new Voz Sense motion Vou = D*( NON Rp*Pz-des) + NON (Ra*Az) + DCV-NON-CO2 (Rp*Pz-est + Ra*Az) + CO2 (Vbz-est) = 0.65*(4,780) + 1260 + 0 + 360 + 915 = 5,640 Xs = Vou/Vps = 5,640/18,200 = 0.310 Ev = 1 + 0.310 – 0.509 = 0.801 Vot = Vou/Ev = 5,640/0.801 = 7,040 CO 2 OCC 260 140 140 260 40 90% 260 140 140 ?? 5 0 Pz’
- 80. Questions? ASHRAE 62.1 update: Where are we now?
Notas do Editor
- Step 1: Vot = Voz = Vbz = 7.5*65 + 0.06*1000 = 547.5 at design at population Step 2: Vat = 0 + 0.06*1000 = 60 at zero population Step 3: Cr-max = Co + 0.0105*65/550 = 0.000400 + 0.001240 = 0.001640 at design population and design OA Step 4: Max signal when Cr = Cr-max or C-max = 1240 ppm, and min signal when Cr = Cr-min or C-min = 0 Step 5: Adjust damper so Vot = Vot-des at max signal, Step 6: Adjust damper so Vot = Vot-min at min signal
- The space requires Vot = 60 cfm @ Pz = 0, Vot = 547.5 cfm @ Pz = 65. If Co = 400 ppm, then indoor concentration ranges from 400 to 1640, and C ranges from 0 to 1240. With the Users Manual approach, the controller simply adjusts the outdoor-air intake proportionally between 60 and 550 cfm, as Ci varies between 400 and 1640 (or C varies between 0 and 1240). This equation defines outdoor air intake flow as a function of indoor CO2 concentration. m = (547.5 - 60)/(1640 - 400) = 487.5/1240 = 0.393 b = Vot - m*Crz = 547.5 - 0.393*1640 = - 97 Vot = 0.393*Crz – 97 (assuming a constant Co = 400 ppm) or m = (547.5 - 60)/(1240 - 0) = 487.5/1240 = 0.393 b = Vot – m*(Crz – Co) = 547.5 - 0.393*1240 = + 60 Vot = 0.393*(Crz – Co) + 60 (assuming Co = unknown)
- If you do these steps, the result of maintaining C (see Article #5) will be a Vbz level somewhat higher than the required minimum. “Using a spreadsheet at each given zone population, we assumed an initial value for zone CO2 level ( Cr ), solved Equation 8 for intake airflow, then solved Equation 2 for the differential CO2 expected for the given zone population and calculated intake airflow. This process was repeated until the expected CO2 level matched the initially assumed level.” Vbz = 0.393 · ( Cr – 400) + 60 (8) ( Cr – Co ) = Pz*k*m/Vbz = Pz*8400*1.25/Vbz = Pz*10,500/Vbz (2) So, if Co = 400 and Pz = 30, what will we sense for C and what will we find for Vbz? Solve equations repeatedly, first assuming initial C-int value, then using Vbz from Eq 8 to find C. Repeat until C-int = C. If C-int = 1200, Controller would find Vbz = 0.393*1200+60 = 529, so C = 30*10500/529 = 495, so 1200 was too hi If C-int = 1000 ppm, Controller would find Vbz = 0.393*1000+60 = 451, so C = 30*10500/451 = 698, so 1000 was too hi If C-int = 800 ppm, Controller would find Vbz = 0.393*800+60 = 373, so C = 30*10500/373 = 845, so 800 was too lo If C = 850 ppm, Controller would find Vbz = 0.393*850+60 = 392, so C = 30*10500/392 = 803, so 850 was too hi If C = 824 ppm, Controller would find Vbz = 0.393*824+60 = 382, so C = 30*10500/382 = 824, so 824 was “right”
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