MECH 3340 Semester Project Phase 4 In paragraph form provide an introduction, methods, conclusions, and recommendations section for the semester project. Include figures, tables and graphs. I. Introduction: Discuss the overall objectives of each Phase of the semester project. Provide and explain a floor plan of the building. Also show and explain the wall, roof, and foundation construction. II. Methods: (combine your work from Phase 1, 2, 3) a. Phase 1: i. Heating Load, Summary of procedures and results ii. Heating load for each room. iii. Heating load for your zone. The heating load for your zone can be found by summing the heating load of the rooms in your zone. See Phase 3 for zone information. b. Phase 2: i. Cooling load, Summary of procedures and results ii. Use your room as an example. iii. Provide the overall cooling load requirement for your zone based on the data provided in Phase 3. c. Phase 3: i. RTU selection based on cooling requirements for your zone from the Trane Catalogs. ii. Duct work sizing for your room, Summary of procedures and results III. Conclusions: Provide a summary of heating and cooling load for your zone. Provide specifications for the roof top unit (RTU), heater, and flow rate for your zone. The heater can be specified based on the Trane Catalog. IV. Recommendations: Discuss addition design problems that would need to be addressed. Possible HVAC system improvement. Possible energy savings options. SEMESTER PROJECT PHASE -1 Load Calculation WINTER HEATING LOAD FOR THE COMMERCIAL BUILDING CALCULATING THE WINTER HEAT LOAD FOR THIS COMMERCIAL BUILDING HERE THE TOTAL BUILDING IS CONSISTING OF 20 ROOMS WITH THE FOLLOWING DIMENSIONS WHERE THE HEAT LOAD IS REQUIRED: SL NO. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 NAME OF THE ROOM OFFICE ROOM OFFICE ROOM OPEN OFFICE SPACE OFFICE ROOM OFFICE ROOM OFFICE ROOM OPEN OFFICE SPACE OFFICE ROOM IT ROOM WOMEN ROOM MEN ROOM CLOSET CONFERENCE ROOM RECORD STORAGE ROOM RECORD STORAGE ROOM AREA DIMENSIONS 17.33 *12.75 16.833*12.75 29.4167*12.75 TOTAL AREA 220.9575 214.62075 375.06292 REMARKS 200.3332 15*10 9.67*14.583 87.58*37.437 200.332 150 141.01761 3278.7326 (15*10)+(4*12.5833) 12.5*11.5 20.83*14.5 7.33*14 7.67*14.16 143.75 302.035 102.62 108.6072 NOT REQUIRED 15.58*23.5 366.13 20.21*9.417 190.317 12.21*8.21 100.244 (32.958*12.542)+(54.622*16.79) 16 17 18 19 20 RECORD STORAGE ROOM OFFICE ROOM OFFICE ROOM COPY ROOM 7.21*4.75 34.2475 14*14 14*23 16.79*13.625 196 322 228.764 OFFICE ROOM 12.41*12.5 155.125 BUILDING VIEW WITH ROOM NUMBERS For office room 1 The heat transfer for room 1 is obtained from various heat losses such as : Heat lost from walls, Roofs, Slabs, Windows all the equations pertaining these heat loads are taken from figure 8.2 heat loss sources as shown below FIGURE REPRESENTING THE VARIOUS HEAT LOSS SOURCES So the climatic information is obtained first assuming the city to be BALTIMORE, MD the climatic conditions, latitude, Elevation, Heating Temperature, Humidification & Dry bulb Temperature range are taken from the below table: BALTIMORE , MD LATITUDE – 39 N ELEVATION – 154 FT HEATING TEMPERATURE – 11 F EVAPORATION – 78 F, 88 F DRY BULB RANGE -19 F HEAT LOSS THROUGH WALLS: Wall detail is assumed to be 5 inches concrete slab along with ¾ inches Gravel The heat lost through the wall is given as Q = U x A x(t0-ti) = 1.61 x 220.9575 x (11- 70) = -20988.753 Btu / Hr Value of U is obtained from Masonory split face block with Perlite cores where R =5.50 so U = 0.182 For concrete slab R= 0.14/inch = (5x 0.14) = 0.7 ,so U= 1/R = 1/0.7 =1.428 U total= 0.182 + 1.428 = 1.61 Heat lost through roof is given as: Q=Ux Ax(to-ti) Q= 1.61 x 220.9575 x (11-70) = -20988.753 Btu/hr Heat lost through Slabs is: Q= F x P bldg. x (t0-ti) F (Slab perimeter heat transfer coefficient) is obtained from table 7.5 Considering horizontal + vertical fully insulated condition F = 0.46 (since 3” rigid insulation is given in data) Q= 0.46 x (2(17.33 +12.75)) x (11-70) = -1632.7424 Btu/Hr Heat transfer through windows is: Q=U x A x (to – ti) U= 0.83 (Considering double panel with Aluminium frame) A= 30 sq. ft Q= 0.83 x 30 x (11-70) Q= -1469.1 Btu/hr So for the office room 1 we have the total heat load to be : Qtotal = 20988.753 +20988.753+ 1632.7424+1469.1 =45079.3484 BTU/hr The Total Heat Load through Room 1 is 45079.3484 BTU/Hr . The Following Assumptions are made for OFFICE ROOM 1 1. Assuming no heat transfer through interior walls. 2. Wall detail is assumed to be 5 inches concrete slab along with ¾ inches Gravel for which the value of U= 1.61 3. Value of U is obtained from Masonory split face block with Perlite cores where R =5.50 so U = 0.182 ,For concrete slab R= 0.14/inch = (5x 0.14) = 0.7 ,so U= 1/R = 1/0.7 =1.428 ,U total= 0.182 + 1.428 = 1.61 4. Value of F (Slab perimeter heat transfer coefficient) is obtained from table 7.5=0.46 For Window the value is assumed as U= 0.83 (Considering double panel with Aluminium frame) 5. The indoor temperature is maintained at 70 0 F 6. No heat loss through crawlspace & Basement , Doors FOR OFFICE ROOM 2 The heat transfer for room 2 is obtained from various heat losses such as : Heat lost from walls, Roofs, Slabs, Windows: HEAT LOSS THROUGH WALLS: Wall detail is assumed to be 5 inches concrete slab along with ¾ inches Gravel The heat lost through the wall is given as Q = U x A x(t0-ti) = 1.61 x 214.62075x (11- 70) = -20386.825 Btu / Hr Value of U is obtained from Masonory split face block with Perlite cores where R =5.50 so U = 0.182 For concrete slab R= 0.14/inch = (5x 0.14) = 0.7 ,so U= 1/R = 1/0.7 =1.428 U total= 0.182 + 1.428 = 1.61 Heat lost through roof is given as: Q=Ux Ax(to-ti) Q= 1.61 x 214.620 x (11-70) = -20386
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