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  • 1. AIR-CONDITIONING-SYSTEM HEAT-LOAD DETERMINATIONNUMERICAL COMPUTATIONReference: Handbook of Mechanical Engineering Calculations by Tyler G. HicksDetermine the required capacity of an air-conditioning system to serve the industrial building shown. Outside walls: 8-in brick with interior finish of 3/8-in gypsum lath plastered and furred 6-in plain poured-concrete partition separates the machinery room from the condition space. Roof: 6-in concrete covered with ½-in thick insulating board. Floor: 2-in concrete Windows: Double-hung, metal frame locked units with light-colored shades three-quarters drawn. Internal heat loads: 100 people doing slightly assembly work 25, 1-hp motors running continuously at full load 20,000 W of light kept on at all times Location: Port Arthur, Texas, at about 30o north latitude. Desired indoor design conditions: 80oF DB, 67oF WB and 51% relative humidity. Air-conditioning equipment will be located in the machinery room.If the desired indoor conditions were not given, the recommended conditions given in Table 5 – Typical Inside Design Conditions – Industrial(Dr. Carrier Handbook) would be used. 1. Determine the design outdoor and indoor conditionsUsing Table 1 – Outdoor Design Conditions – Summer and Winter by Dr. Carrier HandbookPort Arthur, Texas: DB: 95oF WB: 79oF Average wind velocity: 10.7 mphWhere the exact location of a plant is not given in the handbook, consult the nearest location of the weather bureau for information on theusual summer outdoors high and low DB and WB temperatures, relative humidity, and velocity.
  • 2. 2. Compute the sunlight heat gainThe sunlight heat gain results from the solar radiation through the glass in the building’s windows and materials of construction in certain ofthe building walls. If the glass or wall of a building is shaded by an adjacent solid structure, the sunlight heat gain for that glass and wall isusually neglected. The same is true for the glass and wall of the building facing the north.Table 15 – Solar Heat Gain thru Ordinary Glass by Dr. Carrier Handbook Btu Maximum radiation is 165 , occurs through east (sunrise) and west (sunset) walls at 8:00 AM for the east wall and the same hr − ft 2 for the west wall at 4:00 PM. sunlight heat gain = (glass area)(solar heat gain TABLE 15)(factor for shades, if any are used)Where shades are used in the building, choose a suitable shade factor from the appropriate Guide table, (Table 16 – Over-all Factors for SolarHeat Gain thru Glass by Dr. Carrier Handbook) and insert it in the equation above.For metal frame locked units with light-colored shades three-quarters drawn, we use 0.60.West-glass sunlight heat gain  Btu  west sunlight heat gain = (22 windows)(5 ft x 8 ft)165  (0.60)   hr − ft 2  Btu west sunlight heat gain = 87120 hrEast-glass sunlight heat gain  Btu  east sunlight heat gain = (20 windows)(5 ft x 8 ft)11  (0.60) 2   hr − ft  Btu east sunlight heat gain = 5280 hrSouth-glass sunlight heat gain  Btu  south sunlight heat gain = (10 windows)(5 ft x 8 ft)13  (0.60)  hr − ft2   Btu south sunlight heat gain = 3120 hrNorth-glass sunlight heat gain  Btu  north sunlight heat gain = (15 windows)(5 ft x 8 ft)11  (0.60)  hr − ft2   Btu north sunlight heat gain = 3960 hrThe weight per sq. ft of given material is shown in TABLE 21 to Table 33 - Transmission Coefficient UFor the walls: 8-in brick with interior finish of 3/8-in gypsum lath plastered and furred: lb Btu Weight: 80 2 U = 0.29 ft hr − ft 2 −o FFor the roof: 6-in concrete covered with ½-in thick insulating board: lb Btu Weight: 70 2 U = 0.27 ft hr − ft 2 − o FThe same three walls, and the roof, are also subject to sunlight heat gains. Table 19 and Table 20 – Equivalent Temperature Difference (DegF) shows the temperature difference based on the weight of wall and roof resulting from sunlight heat gains.For west wall: Temp. Difference: 12 oFFor east wall: Temp. Difference: 18 oFFor south wall: Temp. Difference: 16 oFFor north wall: Temp. Difference: 4 oFFor the roof: Temp. Difference: 32 oF sunlight heat gain = (wall area)(wall transmission coefficient)(temperature difference)
  • 3. Hence, sunlight heat gains:For west wall:  Btu  west sunlight heat gain = (1220 ft 2 ) 0.29  o (12 F)   hr − ft 2 − o F  Btu west sunlight heat gain = 4245.6 hrFor east wall:  Btu  east sunlight heat gain = (1300 ft 2 ) 0.29  (18 o F)   hr − ft 2 − o F  Btu east sunlight heat gain = 6786 hrFor south wall:  Btu  south sunlight heat gain = (1100 ft 2 ) 0.29  (16 o F)   hr − ft 2 − o F  Btu south sunlight heat gain = 5104 hrFor north wall:  Btu  o north sunlight heat gain = (900 ft 2 ) 0.29  (4 F)  hr − ft 2 −o F   Btu north sunlight heat gain = 1044 hrFor the roof:  Btu  roof sunlight heat gain = (21875 ft 2 ) 0.27  o (32 F)   hr − ft 2 − o F  Btu roof sunlight heat gain = 189000 hr BtuThe sum of the sunlight heat gains gives the total sunlight gain, or 305659.6 hrUse design factor of 1.5: Btu Sunlight heat gain = 458489.4 hr Sunlight heat gain = 134.3152 kW 3. Compute the glass transmission heat gainAll the glass in building windows is subject to transmission of heat from outside to the inside as a result of the temperature differencebetween the outdoor and indoor dry-bulb temperatures. This transmission gain is commonly called the all-glass gain.Glass transmission heat gainglass transmission heat gain = (total glass area)(coefficient of glass heat transmission)(outdoor DB temperature − indoor DB temperature)Where coefficient of glass heat transmission using Table 33 – Transmission Coefficient U – Windows, skylights, doors and glass block walls by BtuDr. Carrier Handbook. For single glass without storm windows: 1.13 hr − ft 2 − o F  Btu  glass transmission heat gain = (2680ft 2 )1.13  (95 − 80)o F   hr − ft2 − o F  Btu glass transmission heat gain = 45426 hrWall transmission heat gainThe transmission heat gain of the south, east and west walls can be neglected because the heat gain is greater. Hence, only the north-walltransmission heat gain needs to be computed.wall transmission heat gain = (wall area)(coefficient of heat transmission)(outdoor DB temperature − indoor DB temperature)For the walls: 8-in brick with interior finish of 3/8-in gypsum lath plastered and furred: lb Btu Weight: 80 2 U = 0.29 ft hr − ft 2 −o F
  • 4. For west wall:  Btu  west transmission heat gain = (1220 ft 2 ) 0.29  (95 − 80)o F  hr − ft2 −o F   Btu west transmission heat gain = 5307 hrFor east wall:  Btu  east sunlight heat gain = (1300 ft2 ) 0.29  (95 − 80)o F  hr − ft2 −o F   Btu east transmission heat gain = 5655 hrFor south wall:  Btu  south sunlight heat gain = (1100 ft2 ) 0.29  (95 − 80)o F  hr − ft2 −o F   Btu south transmission heat gain = 4785 hrFor north wall  Btu  north transmission heat gain = (900 ft 2 ) 0.29  (95 − 80)o F   hr − ft 2 − o F  Btu north transmission heat gain = 3915 hrThe heat gain from the ground can be neglected because the ground is usually at a temperature than the floor. Thus the total transmissionheat gain is the sum of the individual gains. BtuThe sum of the transmission heat gains gives the total transmission gain, or 65088 hrUse design factor of 1.5: Btu Transmission heat gain = 97632 hr Transmission heat gain = 28.6014 kW
  • 5. 4. Compute the infiltration heat gainThe infiltration heat gain is the heat through cracks caused by the wind acting on the building.Since the wind cannot act on all sides of the building at once, one-half the total crack length is generally used (but never less than one half)in computing the infiltration heat gain.Note that crack length varies with different types of windows.inf iltration heat gain = (window crack length)(window inf iltration)(outdoor DB temperature − indoor DB temperature)(1.08)The factor 1.08 converts the computed infiltration to Btu/h.For general type of windows the crank length is given by: Crack length = (3 x width) + (2 x height) Crank length = (67 windows)(3 * 5ft) + (2 * 8ft) = 2077 ftBy using one-half of the total crank length: 2077 ft Crank length = = 1038.5 ft 2For the window infiltration, use Table 41 – 44 Infiltration Thru Windows and Doors by Dr. Carrier Handbook ft 3 / min Window infiltration for 10 mph wind velocity to a double hung window, metal sash: 0.78 crank lengthHence, the heat gain due to infiltration through the window cracks:  ft 3 / min  inf iltration heat gain = (1038.5 ft) 0.78 (95 − 80) o F(1.08)  crank length    Btu inf iltration heat gain = 13122 .486 hr inf iltration heat gain = 3.8504 kW 5. Compute the outside-air bypass heat loadSome outside air may be needed in the conditioned space to ventilate fumes, odors and other undesirables in the conditioned space. Thisventilation air imposes a cooling or dehumidifying load on the air conditioner because the heat or moisture, or both must be removed fromthe ventilation air. Most air conditioners are arranged to permit some outside air to bypass the cooling coils. The bypassed outdoor airbecomes a load within the conditioned space similar to infiltration air.The recommended ventilation air quantity cfm per person is from Table 45 – Ventilation Standard by Dr. Carrier Handbook cfm For factories the recommended cfm per person is 10 person  cfm  ft 3 Since there are 100 people in this factory, the required ventilation quantity is 10  (100 person) = 1000   person  minTable 61 and 62 shows the Typical Bypass Factors for finned coils and various applications, respectively. For factories the recommended bypass factor ranges from 0.10 to 0.20, we assume 0.10 for this installationbypass heat load = (cfm of ventilation air)(outdoor DB temperature − indoor DB temperature)(air conditioner by pass factor)(1.08)Hence;  ft 3  bypass heat load = 1000  (95 − 80)o F(0.10)(1.08)  min  Btu bypass heat load = 1620 hr bypass heat load = 0.4783 kW
  • 6. 6. Compute the heat load from internal heat sourcesWithin an air-conditioned space, heat is given off by people, lights, appliances, machines, pipes, etc.People sensible heat load = (number of people) (sensible heat release per person) For this building with 80oF indoor dry-bulb temperature and 100 occupants doing light work, the heat load produced is from Btu Table 48 – Heat Gain from People is 220 . hr  Btu  Btu sensible heat load = (100people) 220   = 22000  hr  hrMotors (motor − hp)(2546) motor heat load = motor efficiency For 25, 1-hp motors running continuously at full load, from Table 53 – Heat Gain from Electric Motors, efficiency is 79%. (25)(1 − hp)(2546) Btu motor heat load = = 80569.62 0.79 hrLight bulbs 20,000 W of light kept on at all times Using Table 49 – Heat Gain from Lights heat gain from lights = Total watts * 1.25 * 3.4 Btu heat gain from lights = 20,000W * 1.25 * 3.4 = 85000 hrThus, the total internal heat load is: Btu Internal heat load = 187569.62 hr Internal heat load = 55.38 kW 7. Compute the room sensible heatFind the sum of the sensible heat gains computed is steps 2 (sunlight heat gain), 3 (transmission heat gain), 4 (infiltration heat gain), 5(outside air heat gain) and 6 (internal heat sources). This sum is the room sensible-heat subtotal. Btu sensible-heat subtotal = (458489.4 + 97632 + 13122.486 + 1620 + 187569.62) hr Btu sensible-heat subtotal = 758433.506 hrA further sensible heat gain may result from supply-duct heat gain, supply-duct leakage loss, and air-conditioning-fan horsepower. A designfactor is usually added in the form of percentage. Using an assumed design factor of 5 percent to cover various losses that may beencountered in the system. Btu sensible-heat subtotal = 796355.1813 hr sensible-heat subtotal = 235.1144 kW
  • 7. 8. Compute the room latent heatThe room latent load results from the moisture entering the air-conditioned space with the infiltration and bypass air, moisture given off byroom occupants, and any other moisture source such as open steam kettles, sterilizers, etc.Infiltration Latent Heat inf iltration latent heat load = (cfm inf iltration)(moisture content of outside air − moisture content of inside air; gr / lb)(0.68)Using psychrometric chart or Table 41 – 44 Infiltration Thru Windows and Doors by Dr. Carrier Handbook ft 3 / min Window infiltration for 10 mph wind velocity to a double hung window, metal sash: 0.78 and knowing crank length  ft 3 / min  ( ) 3 Crank length = 1038.5 ft , hence, cfm infiltration is 1038.5 ft  0.78  = 810.03 ft .  crank length  min  Outdoor Design Conditions: DB: 95oF and WB: 79oF Using Table 1 – Outdoor Design Conditions – Summer and Winter by Dr. Carrier Handbook gr Moisture content: 124 lb NOTE: Using Psychrometric chart, we can say that 124 gr/lb = 0.01772 lb/lbIndoor design conditions: 80oF DB, 67oF WB and 51% RH  lb  124 gr / lb  Using psychrometric, chart moisture content:  0.01117     = 78.1648 gr  lb  0.01772 lb / lb   lbHence;Ventilation Latent Heatventilation latent heat load = (cfm inf iltration)(moisture content of outside air − moisture content of inside air; gr / lb)(bypass factor)(0.68)  cfm  ft 3 Since there are 100 people in this factory, the required ventilation quantity is 10  (100 person) = 1000   person  minHence;  ft 3  ventilation latent heat load = 1000 (124 − 78.1648) gr ( 0.10)(0.68) = 3116.7936 Btu  min  lb hr  People people latent heat = (number of occupants)(latent heat gain) For this building with 80oF indoor dry-bulb temperature and 100 occupants doing light work, the latent heat load produced is Btu from Table 48 – Heat Gain from People is 530 . hrHence;  Btu  Btu people latent heat = (100) 530   = 53000  hr  hr BtuFind the latent heat subtotal by taking the sum of the above heat gains: sensible-heat subtotal = 81363.7568 hrUsing an allowance of 5 percent for supply-duct leakage loss and safety margins: Btu latent-heat subtotal = 85431.9446 hr latent-heat subtotal = 25.035 kW
  • 8. 9. Compute the outside heatAir brought in for space ventilation imposes a sensible and latent heat load on the air-conditioning apparatus.Sensible Outside space ventilation: outside ventilation = (cfm outside ventilation air)(design DB − inside DB)(1 − bypass factor)(1.08)  ft 3  outside ventilation = 1000 (95 − 80)o F(1 − 0.10)(1.08) = 14580 Btu  min  hr  Latent Outside space ventilation:outside latent heat load = (cfm out ventilation)(moisture content of outside air − moisture content of inside air; gr / lb)(1 − bypass factor)(0.68)  ft 3 outside latent heat load = 1000 (124 − 78.1648) gr (1 − 0.10)(0.68) = 28051.1424 Btu  min  lb hr   10. Compute the grand-total heat and refrigeration tonnageRoom total heat load = room sensible heat + room latent heat Btu Btu Room total heat load = [796355.1813 + 14580] + [85431.9446 + 28051.1424] hr hr Btu Room total heat load = 924418.2683 hr Room total heat load = 271.1252 kWRefrigeration Load, tons room total heat load, Btu / hr Refrigeration load = 12000, Btu / hr per ton 924418.2683Btu / hr Refrigeration load = 12000, Btu / hr per ton Refrigeration load = 77 tonsQuantity of cooling water required The quantity of cooling water required for the refrigeration-system condenser is Q.  tons of refrigeration  Q = 30    condenser water temperature rise, o F  ; gpm   Assuming at 75oF entering water temperature and 95oF leaving water temperature.  77 tons  Q = 30    95 o F − 75 o F    Q = 115.5 gpm 11. Compute the sensible heat factor and apparatus dew pointSensible heat factor (SHF) room sensible heatsensible heat factor = room total heat Btu 796355.1813 sensible heat factor = hr [796355.1813 + 14580] Btu hr sensible heat factor = 0.9820Apparatus dew point temperature:Using psychrometric chart and known room conditions, we find apparatus dew point temperature. Apparatus dew point temperature = 60oF or 15.76oC
  • 9. 12. Compute the quantity of dehumidified air requireddehumified air temperature rise = (1 − bypass factor)(indoor temperature − apparatus dew po int) dehumified air temperature rise = (1 − 0.10)(80 o F − 60 o F) dehumified air temperature rise = 18oF room sensible heatdehumified air quantity = ( 1.08 dehumified air temperature rise ) ; cfm Btu 796355.1813 dehumified air quantity = hr 1.08 (18 o F) dehumified air quantity = 40964.7727 cfm