HVAC Cooling Load Calculation

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HVAC Cooling Load Calculation

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Post Date: 19/07/2018, 03:46
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1. Cooling Load ESTIMATION 2. Principles of Heat Transfer • Heat energy cannot be destroyed. • Heat always flows from a higher temperature substance to a lower temperature substance. • Heat can be transferred from one substance to another. 3. Methods of Heat Transfer radiation hot water conduction convection cool air warm air 4. Conduction, Convection & Radiation 5. Measuring Heat Quantity 60°F 61°F 15°C 16°C 1 Btu 1 kcal [4.19 kJ] 1 lb water 1 kg water 6. What is BTU? • BTU refers to British Thermal Unit. • Unit of Heat Energy in Imperial System or I-P System. • 1 BTU is the amount of Heat energy required to raise the temperature of 1 lb water by 1?F. 7. Sensible Vs Latent Heat 60°F [15.6°C] 212°F [100°C] 212°F [100°C] 212°F [100°C] Sensible Heat Latent Heat 8. Human Comfort • Conditions at which most people are likely to feel comfortable most of the time. • Also called as Thermal Comfort. • Temperature: 78?F (Summer) – 68?F(Winter). • Relative Humidity: 30 %– 40%. 9. Factors Affecting Human Comfort • Dry-bulb temperature • Humidity • Air movement • Fresh air • Clean air • Noise level • Adequate lighting • Proper furniture and work surfaces 10. Dry-bulb Temperature HumidityRatio Indoor Design Conditions 80°F [26.7° C] 70°F [21.2° C] comfort zone A 11. Cooling Load Estimation Procedure 12. Cooling Load Components roof lights equipme nt floor exterior wall glass solar glass conductio n infiltration people partition wall 13. Cooling Load Components Sensible Load latent Load Conduction through roof, walls, windows, and skylights Solar radiation through windows, skylights Conduction through ceiling, interior partition walls, and floorPeople Lights Equipment/ Appliances Infiltration Ventilation System Heat Gains space Load coil Load Cooling Load components 14. Time of Peak Cooling LoadHeatgain roof East-facing window 12 6 12 6 12 noona.m. p.m. midmid 15. Example Office Space (Room 101) Plan view Elevation view (Room 101) Room 101 North Room 102 16. Outdoor Design Conditions St. Louis, Missouri DB WB DB WB DB WB 0.4% 1% 2% 95°F [35°C] 76°F [25°C] 93°F [34°C] 75°F [24°C] 90°F [32°C] 74°F [23°C] 17. Heat Conduction through Surfaces 18. Conduction through a Shaded Wall Q = U ? A ? ?T U – Overall heat transfer coefficient of the surface A – Area of the surface ? T – Dry bulb temperature difference across the surface 19. U-factor wood studs insulation gypsum board concrete block aluminum siding 20. U-factor for Example Wall thermal resistance (R) Routdoor-air film 0.25 [0.04] Rsiding 0.61 [0.11] Rconcreteblock 2.00 [0.35] Rinsulation 13.00 [2.29] Rgypsumboard 0.45 [0.08] Rindoor-air film 0.68 [0.12] Rtotal 16.99 [2.99] ] U = Rtotal 1 U = 0.06 Btu/hr•ft2•°F [ U = 0.33 W/m2•°K ] 21. Conduction through a Shaded Wall Qwall = 0.06 ? 380 ? (95 – 78) = 388 Btu/hr [ Qwall = 0.33 ? 36.3 ? (35 – 25.6) = 113 W ] 22. Sunlit Surfaces sun rays solar angle changes throughout the da 23. Time LagSolarEffect 12 6 12 6 12 noona.m. p.m. midmid A B Time Lag 24. Q = U ? A ? CLTD Conduction through Sunlit Surfaces CLTD : Term used to account for the added heat transfer due to the sun shining on exterior walls, roofs, and windows, and the capacity of the wall and roof to store heat. 25. CLTD Factors for West-Facing Wall hou r 2117 14 11 8 7 6 6 7 CLTD (°F) 35 30 25 6 7 8 9 10 11 121 2 3 4 5 1314 1516 181917 20 21 22 23 24 CLTD (°C) 8 10 121622 303744 4848 45 41 12 9 8 6 4 4 3 3 419 17 14 4 6 7 9 12 172124 2727 25 23 26. Conduction through Sunlit Surfaces Qwall = 0.06 ? 380 ? 22 = 502 Btu/hr Qroof = 0.057 ? 2700 ? 80 = 12312 Btu/hr [ Qwall = 0.33 ? 36.3 ? 12 = 144 W ] [ Qroof = 0.323 ? 250.7 ? 44 = 3563 W ] 27. U-factors for Windows fixed frames, vertical installation single glazing 1/8 in. [3.2 mm] glass double glazing 1/4 in. [6.4 mm] air space 1/2 in. [12.8 mm] air space 1/4 in. [6.4 mm] argon space 1/2 in. [12.8 mm] argon space triple glazing 1/4 in. [6.4 mm] air spaces 1/2 in. [12.8 mm] air spaces 1/4 in. [6.4 mm] argon spaces 1/2 in. [12.8 mm] argon spaces 1.13 [6.42] aluminum without thermal break wood/viny l 0.69 [3.94] 0.64 [3.61] 0.66 [3.75] 0.61 [3.47] 0.49 [2.76] 0.55 [3.10] aluminum with thermal break 0.47 [2.66] 0.51 [2.90] 1.07 [6.07] 0.63 [3.56] 0.57 [3.22] 0.59 [3.37] 0.54 [3.08] 0.42 [2.39] 0.48 [2.73] 0.40 [2.30] 0.45 [2.54] 0.98 [5.55] 0.56 [3.17] 0.50 [2.84] 0.52 [2.98] 0.48 [2.70] 0.35 [2.01] 0.41 [2.33] 0.34 [1.91] 0.38 [2.15] 28. Conduction through Windows Qwindows = U ? A x CLTD Qwindows = 0.63 ? 160 ? 13 = 1310 Btu/hr [ Qwindows = 3.56 ? 14.4 ? 7 = 359 W ] 29. Solar Radiation through Glass 30. Solar Heat Gain through Glass Q = A ? SC ? SCL Where, SC – Shading Coefficient SCL – Solar Cooling Load Factor 31. Solar Cooling Load Factor (SCL) • Direction that the window faces • Time of day • Month • Latitude • Construction of interior partition walls • Type of floor covering • Existence of internal shading devices SCL: A factor used to estimate the rate at which solar heat energy radiates directly into the space, heats up the surfaces and furnishings, and is later released to the space as a sensible heat gain. 32. Shading Coefficient (SC) ? It is an expression used to define how much of the radiant solar energy, that strikes the outer surface of the window, is actually transmitted through the window and into the space. 33. Shading Coefficient (SC) shading coefficient at normal incidence uncoated single glazing 1/4 in. [6.4 mm] clear 1/4 in. [6.4 mm] green reflective single glazing 1/4 in. [6.4 mm] SS on clear 1/4 in. [6.4 mm] SS on green uncoated double glazing 1/4 in. [6.4 mm] clear - clear 1/4 in. [6.4 mm] green - clear reflective double glazing 1/4 in. [6.4 mm] SS on clear - clear 1/4 in. [6.4 mm] SS on green - clear 0.82 aluminum frame other frames operable fixed 0.85 0.69 0.82 0.59 0.61 0.49 0.59 0.26 0.28 0.22 0.25 0.26 0.28 0.22 0.25 0.70 0.74 0.60 0.70 0.48 0.49 0.40 0.47 0.18 0.18 0.15 0.16 0.20 0.18 0.15 0.17 SS = stainless-steel reflective coating operable fixed 34. Solar Radiation through Windows Qwindows = 160 ? 0.74 ? 192 = 22733 Btu/hr [ Qwindows = 14.4 ? 0.74 ? 605 = 6447 W ] 35. Internal Heat Gains People Equipment Appliances Lights 36. Heat Generated by People • Metabolism of the human body normally generates more heat than it needs • 60% heat is transferred by convection and radiation to the surrounding environment. • 40% is released by perspiration and respiration. 37. Heat Generated by People (Chart) Level Of Activity Sensible Heat Gain Latent Heat Gain Moderately active work (Office) 250 BTU/hr (75W) 200 BTU/hr (55W) Standing, light work, walking (Store) 250 BTU/hr (75W) 200 BTU/hr (55W) Light bench work (Factory) 275 BTU/hr (80W) 475BTU/hr (140W) Heavy work (Factory) 580BTU/hr(170W) 870BTU/hr (255W) Exercise (Gymnasium) 710BTU/hr (210W) 1090BTU/hr (315W) 38. CLF Factors for People Hours after people enter space 0.11 0.08 0.06 0.05 0.04 0.03 0.02 0.02 0.010.65 0.74 0.16 6 7 8 9 10 11 121 2 3 4 5 Total hours in space 2 4 6 8 10 0.65 0.65 0.65 0.65 0.85 0.24 0.17 0.13 0.10 0.07 0.06 0.04 0.030.75 0.81 0.85 0.89 0.91 0.29 0.20 0.15 0.12 0.09 0.070.75 0.81 0.85 0.89 0.91 0.93 0.95 0.31 0.22 0.17 0.130.810.75 0.85 0.89 0.91 0.93 0.95 0.96 0.97 0.33 0.240.810.75 Note: CLF – Cooling Load Factor Capacity of a space to absorb and store heat. 39. Heat Gain from People QS = No: of people x Sensible heat gain per person x CLF Qsensible = 18 ? 250 ? 1.0 = 4500 Btu/hr QL = No: of people ? Latent heat gain/ person Qlatent = 18 ? 200 = 3600 Btu/hr [ Qsensible = 18 ? 75 ? 1.0 = 1350 W ] [ Qlatent = 18 ? 55 = 990 W ] 40. Heat Gain from Lighting Q = Btu/hr ? Ballast factor ? CLF [ Q = watts ? Ballast factor ? CLF ] Ballast factor = 1.2 for fluorescent lights Ballast factor = 1.0 for incandescent lights 41. Heat Gain from Lighting Qlights = 5400 ? 3.41 ? 1.2 ? 1.0 = 22097 Btu/hr [ Qlights = 5400 ? 1.2 ? 1.0 = 6480 W ] 42. Heat generated by equipment Equipment Sensible Heat Gain Latent Heat Gain Coffee maker 3580 BTU/hr (1050W) 1540 BTU/hr (450W) Printer 1000 BTU/hr (292W) Typewriter 230 BTU/hr (67W) 43. Infiltration 44. Methods of Estimating Infiltration • Air change method • Crack method • Effective leakage-area method 45. Infiltration Airflow Infiltration airflow 32400 ? 0.3 60 = = 162 CFM Infiltration airflow 927.6 ? 0.3 3600 = = 0.077 m3/s 46. Heat Gain from Infiltration Qsensible = 1.085 ? airflow ? ?T Qlatent = 0.7 ? airflow ? ?W [ Qsensible = 1210 ? airflow ? ?T ] [ Qlatent = 3010 ? airflow ? ?W ] ?W = (Outdoor Humidity Ratio – Indoor Humidity Ratio) Air Flow – Quantity of air infiltrating the place ?T = (Outdoor D.B.T – Indoor D.B.T) Density x Specific Heat = 1.085 (1210) Btu•min/hr•ft3•ºF [J/m3•ºK] Latent Heat Factor = 0.7 (3010) Btu•min•lb/hr•ft3•gr [J•kg/m3•g] 47. Heat Gain from Infiltration QS = 1.085 ? 162 ? (95 – 78) = 2,988 Btu/hr [ QS = 1,210 ? 0.077 ? (35 – 25.6) = 876 W ] QL = 0.7 ? 162 ? (105 – 70) = 3,969 Btu/hr [ QL = 3,010 ? 0.077 ? (15 – 10) = 1,159 W ] 48. sensible Load Btu/hr [W] conduction through roof solar radiation through windows people lights equipment infiltration conduction through windows conduction through exterior wall 12,312 [3,563] 4,500 [1,350]22,097 [6,480]8,184 [2,404]2,988 [876] 74,626 [21,623]Total space Cooling Load 3,600 [990] 1,540 [450]3,969 [1,159] latent Load Btu/hr [W] 9,109 [2,599] space Load components 502 [144]1,310 [359]22,733 [6,447] Summary of Space Cooling Loads 49. Ventilation air handler with fan and Cooling coil supply duct diffuser outdoor-air intake 50. Outdoor Air Requirements Type of Space Outdoor Air/ person Outdoor Air/ ft2 (m2) Auditorium 15 CFM (0.008 m3/s) Class rooms 15 CFM (0.008 m3/s) Locker rooms 0.5 CFM (0.0025 m3/s) Office space 20 CFM (0.01 m3/s) Public restrooms 50 CFM (0.025 m3/s) Smoking lounge 60 CFM (0.03 m3/s) 51. Cooling Load Due to Ventilation QS = 1.085 ? 360 ? (95 – 78) = 6640 Btu/hr QL = 0.7 ? 360 ? (105 – 70) = 8820 Btu/hr [ QS = 1210 ? 0.18 ? (35 – 25.6) = 2047 W ] [ QL = 3010 ? 0.18 ? (15 – 10) = 2709 W ] 52. System Heat Gains air handler fan motor 53. Components of Fan Heat blow-through configuration draw-through configuration 54. Heat Gain in Ductwork 55. sensible Load Btu/hr [W] conduction through roof solar radiation through windows people lights equipment infiltration conduction through windows conduction through exterior wall 4,500 [1,350]22,097 [6,480]8,184 [2,404]2,988 [876] total space Cooling Load 3,600 [990] 1,540 [450]3,969 [1,159] latent Load Btu/hr [W] 9,109 [2,599] 502 [144]1,310 [359]22,733 [6,447] ventilation 6,640 [2,047] 8,820 [2,709]81,266 [23,670]total coil Cooling Load 17,929 [5,308] Summary of Cooling Loads 12,312 [3,563] 74,626 [21,623] 56. Psychometric Analysis © American Standard Inc. 2000 Air Conditioning Clinic TRG-TRC002-EN 57. Space Load versus Coil Load space Load coil Load conduction through roof, walls, windows, and skylightssolar radiation through windows, skylights conduction through ceiling, interior partition walls, and floorpeople lights equipment and appliances infiltration ventilation system heat gains 58. Space Sensible and Latent Loads sensible Load Btu/hr [W] conduction through roof solar radiation through windows people lights equipment infiltration conduction through windows conduction through exterior wall 12,312 [3,563] 4,500 [1,350]22,097 [6,480]8,184 [2,404]2,988 [876] 74,626 [21,623]total space Cooling Load 3,600 [990] 1,540 [450]3,969 [1,159] latent Load Btu/hr [W] 9,109 [2,599] space Load components 502 [144]1,310 [359]22,733 [6,447] 59. Sensible Heat Ratio (SHR) SHR sensible heat gain sensible heat gain + latent heat gain = = 0.89 74,626 74626 + 9109 SHR = = 0.89 21623 21623 + 2599 SHR = 60. Single-Space Analysis space supply fan Cooling coil outdoor air return air supply air exhaust air 61. Determine Supply Airflow sensible heat gainsupply airflow = 1.085 × (room DB – supply DB) sensible heat gainsupply airflow = 1,210 × (room DB – supply DB) 62. Determine Supply Airflow 74,626 2,990 cfm= 1.085 × (78 – 55) supply airflow = 21,623 1.40 m3/s= 1,210 × (25.6 – 12.8) supply airflow = 63. Calculate Entering Coil Conditions ventilation airflow % outdoor air = total supply airflow 360 cfm %OA = 2990 cfm = 0.12 0.18 m3/s %OA = 1.40 m3/s = 0.12 64. Calculate Entering Coil Conditions B A C 95°F × 0.12 = 11.4°F 78°F × 0.88 = 68.6°F mixture = 80.0°F 35°C × 0.12 = 4.2°C 25.6°C × 0.88 = 22.5°C mixture = 26.7°C dry-bulb temperature humidityratio 95°F [35°C] 76°F [24.4°C] 80°F [26.7°C] 78°F [25.6°C] 66.5°F [19.2°C] 65. Determine Supply Air Temperature dry-bulb temperature sensibleheatratio D 59°F [15°C] B A 1.0 0.8 0.6 0.4 C 66. Recalculate Supply Airflow 21,623 1.69 m3/s= 1,210 × (25.6 – 15) supply airflow = 74,626 3,620 cfm= 1.085 × (78 – 59) supply airflow = 67. Room 101 Btu/hr [W] total coil Cooling Load 99,195 [28,978] ventilation 15,460 [4,756] Total Cooling Load on Coil total space sensible Load 9,109 [2,599] 74,626 [21,623] total space latent Load 68. Multiple-Space Analysis supply fan Room 101 Room 102 Cooling coil 69. Room 101 (Faces West) 8 a.m. Btu/hr [W] conduction through roof solar radiation through windows people lights equipment infiltration conduction through windows conduction through exterior wall160 [48] 2,616 [740] 202 [51]3,552 [1,012]4,500 [1,350]22,097 [6,480]8,184 [2,404]2,988 [876] 44,299 [12,961]total space sensible Load 4 p.m. Btu/hr [W] 4,500 [1,350]22,097 [6,480]8,184 [2,404]2,988 [876] space sensible Load components 74,626 [21,623] 502 [144]1,310 [359]22,733 [6,447] 12,312 [3,563] 70. Room 102 (Faces East) 8 a.m. Btu/hr [W] conduction through roof solar radiation through windows people lights equipment infiltration conduction through windows conduction through exterior wall160 [48] 21,667 [6,138]4,500 [1,350]22,097 [6,480] 2,988 [876] 62,414 [18,087]total space sensible Load 4 p.m. Btu/hr [W] 844 [252]1,310 [359]3,078 [874]4,500 [1,350]22,097 [6,480] 2,988 [876] 55,313 [16,158] 202 [51] space sensible Load components 8,184 [2,404] 8,184 [2,404] 2,616 [740] 12,312 [3,563] 71. “Sum-of-Peaks” versus “Block” Room 101 (faces west) Room 102 (faces east) space sensible Load sum-of-peaks = 74626 + 62414 = 137040 Btu/hr [21623 + 18087 = 39710 W] block = 74626 + 55313 = 129939 Btu/hr [21623 + 16158 = 37781 W] 8 a.m. Btu/hr [W] 4 p.m. Btu/hr [W]44299 [12961] 74626 [21623]62414 [18087] 55313 [16158] 72. “Sum-of-Peaks” versus “Block” • Sum-of-peaks supply airflow = 6,648 CFM [3.10 m3/s] • Block supply airflow = 6,303 CFM [2.95 m3/s] 73. Room 101 Btu/hr [W] Total coil Cooling Load Room 102 Btu/hr [W] 99195 [28978] ventilation 15460 [4756] “Block” Cooling Load 79882 [23513] 15460 [4756] total space sensible Load 9109 [2599] 74626 [21623] total space latent Load 55313 [16158] 9109 [2599] Block Cooling Load (4 p.m.) = 99195 + 79882 = 179077 Btu/hr [28978 + 23513 = 52491 W] Loads at 4 p.m.
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