HEATING & COOLING LOAD CALCULATIONS

Heating and Cooling Load CalculationsRoom SpecificationsRoom name: Sir Rizwan Saeed Chaudhrys Office

NOTE: The directions are taken for reference. They dont represent the actual direction. The diagram is not up to scaleThis room has 2 windows, 1 door (which also has a window pane above it 38 x 16) and a heater place (28 x 29). The height of the walls is 149.The room is located on ground floor of a two floor building. It has 4 tubelights.

Composition of Walls and RoofThe walls are composed of 9 common bricks and 1 plaster 0.5 on both sides. While roof is composed of 9 concrete and 1 plaster 0.5 on both sides.

AreasArea of the North wall (without window and door) = 106.118 Area of South wall (without window) = 126.576 Area of East wall = 175.91 Area of West wall (less thickness) = 134.52 Area of West wall (more thickness without heater place) = 35.75 Area of Door without Window pane = 21.51 Area of Window pane above door = 4.23 Area of Window 1 = 32.77 Area of Window 2 = 27.50 Area of Heater Place= 5.64 Area of Floor = 179.03 Area of Roof = 179.03

Heating Load CalculationsOutdoor ConditionsD.B. Temperature: DB = = D.B. Temperature of the areas around room which is not exposed to atmosphere= = Daily Temperature Range: = Relative Humidity: RH = 60% From psychometric chart we have; Specific Humidity, W1 = 16gr/lb Specific Humidity at W1 = 32gr/lb Room ConditionsD.B. Temperature: DB = = Daily Temperature Range: = Relative Humidity: RH = 50% From psychometric chart we have; Specific Humidity, W2 = 55gr/lb = .008 lbs. water/lbs. dry air

Heat Transfer Co-efficient through various elementsWalls: Thermal resistance of 9 Common Brick == 1.8 hr.F. /BTU Thermal resistance of 1 Plaster = = 0.2 hr.F. /BTU Thermal resistance of Air Film (Inside) = = 0.68 hr.F. /BTU Thermal resistance of Air Film (Outside) [For wall exposed to ambient) = = 0.17 hr.F. /BTU Thermal resistance of Air Film (Outside) [For wall not exposed to ambient) = = 0.68 hr.F. /BTU Total R (For wall exposed to ambient) = 2.85 hr.F./BTU U (for wall exposed to ambient) = 1/R= 0.351 (BTU/hr..oF) Total R (For wall not exposed to ambient) = 3.36 hr.F./BTU U (for wall not exposed to ambient) = 1/R= 0.298 (BTU/hr.. F)Roof: Thermal resistance of 9 concrete = = 1.8 hr.F./BTU Thermal resistance of 1 Plaster = = 0.2 hr.F. /BTU Thermal resistance of Air Film (Inside) = = 0.61 hr.F./BTU Thermal resistance of Air Film (Outside) = = 0.61 hr.F./BTU Total R = 3.22 hr.F./BTU U = 1/R = 0.311 (BTU/hr.. F) Glass Window: For Single Glass Aluminum frame (no air space), U = 1.10 (BTU/hr.. F) For Single Glass wooden frame for window pane above door, U = 0.98 (BTU/hr.. F)

Door: For solid wood door 1.5 thickness U = 0.49 (BTU/hr.. F) Floor: U = 0.04 (BTU/hr.. F) Same as for basement floor. Extended West Wall (minus Heater Place): Thermal resistance of 19 Common Brick == 3.8 hr.F /BTU Thermal resistance of 1 Plaster = = 0.2 hr.F. /BTU Thermal resistance of Air Film (Inside) = = 0.68 hr.F. /BTU Thermal resistance of Air Film (Outside) = = 0.68 hr.F. /BTU Total R = 5.36 hr.F./BTU U = 1/R= 0.187 (BTU/hr.. F)Heater Place: Thermal resistance of 9 Common Brick == 1.8 hr.F. /BTU Thermal resistance of 1 Plaster = = 0.2 hr.F. /BTU Thermal resistance of Air Film (Inside) = = 0.68 hr.F. /BTU Thermal resistance of Air Film (Outside) = = 0.68 hr.F. /BTU Total R (For wall not exposed to ambient) = 3.36 hr.F./BTU U (for wall not exposed to ambient) = 1/R= 0.298 (BTU/hr.. F)

Heat Loss Calculation Heat transfer equation used is: Q = U*A*TDHeat Loss through Walls by Conduction Heat Loss through EAST Wall; = U*A*TD= 0.298*175.91*(69.8-50.9)= 990.77 BTU/hr Heat Loss through WEST Wall; Q = Q through west wall (less thickness) + Q through west wall (more thickness) + Q through Heater Place = U*A*TD = 0.298*134.52*(69.8-50.9) + 0.187*35.75*(69.8-50.9) +0.298*5.64*(69.8-50.9)= 915.77 BTU/hr Heat Loss through SOUTH Wall (without window); = U*A*TD= 0.351*126.576*(69.8-32)= 1679.39 BTU/hr Heat Loss through NORTH Wall (without door and window); = U*A*TD= 0.298*106.118*(69.8-50.9)= 597.68 BTU/hr Heat Loss through Glass by Conduction The heat lost through the glass can be found by the following: Q= Q through window1 + Q through window2 + Q through window paneQ = U*A*TD= 1.1*(27.50)*(69.8-50.9) + 0.98*4.23(69.8-50.9) + 1.1*32.77*(69.8-32)= 2012.65 BTU/hr Heat Loss through Floor by Conduction When a floor is on the ground, the heat loss is greatest near the outside edges (perimeter) of the building & is proportional to length of these edges, rather than the area of the floor. This is the only case where heat transfer is not calculated using equation Q = U*A*TD. The following equation is used: Q = E*L*TDWhere; E= edge heat loss coefficient, Btu/hr-ft-F. Values of E for various wall constructions can be taken from standard tables. The edge loss method is recommended for buildings with small floor slab areas. The edge loss method is recommended only for perimeter rooms. For the interior areas (Like in this case where only one side of the room is exposed to atmosphere) equation Q = U*A*TD should be used, with the U & TD values for basement. The heat lost through the floor can be found by the following: Q = U*A*TD= 0.04*179.03*(69.8-50.9)= 135.35 BTU/hr Heat Loss through Roof by Conduction The heat lost through the roof can be found by the following Q = U*A*TD= 0.311*179.03*(69.8-50.9)= 1052.33 BTU/hr Heat Loss through Door by Conduction The heat lost through the roof can be found by the following Q = U*A*TD= 0.49*21.51*(69.8-50.9) = 199.21 BTU/hr Infiltration Load: = 1.1*CFM*TC = 0.68*CFM*(-)

DoorWindows (1,2 and pane)

CFM/ of door = 0.5CFM/ Door Area = 21.51Total CFM = 0.5*21.51= 10.76 = 1.1*10.76*(69.8-50.9) = 223.71 BTU/hr = 0.68*10.76*(55-32) = 168.29 BTU/hrCFM = 0.37 CFM/ft Length of crack (win 2 and pane) =34.67 ft CFM = 34.67*0.37 = 12.83 = 1.1*12.83*(69.8-50.9) = 266.74 BTU/hr = 0.68*12.83*(55-32) = 200.67 BTU/hrLength of crack (win 1) =24.92 ft CFM = 24.92*0.37 = 9.23 = 1.1*9.23*(69.8-32) = 383.79 BTU/hr = 0.68*9.23*(55-16) = 244.78 BTU/hr

Total Heating Load Total Net Load =990.77+915.77+1679.39+597.68+2012.65+135.35+1052.33+199.21+223.71+266.74+383.79+168.29+200.67+244.78= 9071.13 Btu/hr.= 0.76 tons15 % losses =1360.67 Btu/hrSo, Total Load = 10431.80 BTU/hr= 0.87 Ton

Cooling Load CalculationsOutside ConditionsD.B. Temperature: DB = = D.B. Temperature of the areas around room which is not exposed to atmosphere= = Daily Temperature Range: = Latitude = Approximately Date: 21st June RH%=70% W1 at 113F=295.74 gr/lb W2 at 108F=255.12 gr/lb Room ConditionsD.B. Temperature: DB = = Daily Temperature Range: = RH%=50%W=57 gr/lb External Loads: Heat Transfer through Walls: South facing wall: Heat Transfer through wall only: As QCond, Wall =U*A*CLTDCCLTDC=CLTD+LM+ (78-TR) + (TA-85)Latitude=32o (approximately)Maximum CLTD for south facing wall is at 2300 hours at latitude, 32o=22LM factor for month of June and south facing wall= -4TR=71.6 OFTA=TO + (DR/2) where TO =113OF & DR=32.4OFTA=96.8OF HenceCLTDC = 22-4+ (78-71.6) + (96.8-85) =36.2OF R= (.20*9) + (.20*1) + 0.68+0.25 =2.93hr.F./BTU U=0.342 (BTU/hr.. F) A=126.576 QCond, Wall =0.342*126.576*36.2=1567.04 BTU/hrConduction through Window:QCond, Window=U*A*CLTDC U=1.01 A=32.77 CLTDC = CLTD+ (78-TR) + (TA-85) CLTD at 2300 hours for window =3OF CLTDC=1.01+ (78-71.6) + (96.8-85) =21.2 OFQCond, Window =1.01*32.77*21.2=701.68 BTU/hrTotal Conduction heat Transfer through Wall and Window, at 2300 hours =2268.720 BTU/hrNow Consider Maximum CLTD for window which is 15OF at 1600 hours CLTDC = 15+ (78-71.6) + (96.8-85) =32.2OF QCond, Window =U*A*CLTDC QCond, Window =1.01*32.77*32.2=1065.75 BTU/hr Now Consider CLTD at 1600 hours for wall which is 15OFthen CLTDC = 15-4+ (78-71.6) + (96.8-85) =29.2OFQCond, Wall =0.342*136.576*29.2=1261.021 BTU/hr Total heat Transfer though wall and window without radiation heat transfer, at 1600 hours=2326.771 BTU/hrHeat Transfer through Wall and Window is higher at 1600 hrs so,QCond =2326.771 BTU/hrRadiation heat transfer through Window:Radiation heat transfer at 1600QRadiation=SHGF*A*SC*CLFSHGF=60, SC=0.74, CLF=0.35, A=32.77QRadiation =60*32.77*0.74*0.35=509.245 BTU/hrNow Consider Maximum Cooling Load Factor (CLF), which is 0.83 at 1200 hours QRadiation =SHGF*A*SC*CLF SHGF=60, SC=0.74, CLF=0.83, A=32.77 QRadiation =60*32.77*0.74*0.85=1207.65 BTU/hrNow well calculate CLTD for window and wall at 1200 hours CLTD for wall at 1200 is 11So, CLTDC = 11-4+ (78-71.6) + (96.8-85) =25.2QCond, Wall =0.342*126.576*25.2=1090.89CLTD for glass at 1200 is 9CLTDC = 9+ (78-71.6) + (96.8-85) =27.2 OFQCond, Window =1.01*32.77*27.2 = 900.26 BTU/hrQCond+Radation at 1600 hours = 2836.015 is less than QCond+Radation at 1200 hours = 3198.8 BTU/hr which is more than the total heat at 1600 hoursSo,Total Heat Transfer through South facing wall is = 3198.8 BTU/hr

East Facing Wall:Heat Transfer through east facing wall is: Cond=A*TD*U A=175.91 TD= (108-71.6) =36.40OF R= (9*0.20) + (1*.20) +0.68+0.68=3.360 hr.F./BTU U=0.298 BTU/hr.F. Cond=175.91*36.40*0.298=1908.131 BTU/hrWest Facing Wall:Heat Transfer through west facing wall excluding Extended Area: Qcond, 1=A*TD*U A=134.52+5.68=140.20 TD= (108-71.6) =36.40OF R= (9*0.20) + (1*.20) +0.68+0.68=3.360 hr.F./BTU U=0.298 BTU/hr.F. Qcond, 1=140.20*36.40*0.298=1520.77 BTU/hrHeat Transfer through Extruded Area: Qcond, 2=A*TD*U A=35.75 TD= (108-71.6) =36.40OF R= (18*0.20) + (1*.20) +0.68+0.68=3.36 hr.F./BTU U=0.194 BTU/hr.F. Qcond, 2=35.75*36.40*0.194=252.45 BTU/hrTotal heat Transfer through West Facing wall = Qcond=1520.77+252.45 =1773.22 BTU/hrNorth Facing Wall:North Facing Wall has one window of Aluminum frame and other of wooden frame along with the Entrance door.Heat Transfer through wall is:Qcond, 1=A*TD*U TD= (108-71.6) =36.40OF R= (9*0.20) + (1*.20) +0.68+0.68=3.360 hr.F./BTU U=0.298 BTU/hr.F. A=106.118Qcond, 1=106.118*0.298*36.40=1151.08 BTU/hrHeat Transfer through Window with Aluminum frame is:Qcond, 2=A*TD*U TD= (108-71.6) =36.40OF A=27.50 U=1.01 BTU/hr.F.Qcond, 2=27.50*1.01*36.40=1011.01 BTU/hrHeat Transfer through Window pane with wooden frame is:Qcond, 3=A*TD*U TD= (108-71.6) =36.40OF A=4.23 U=0.901.01 BTU/hr.F.Qcond, 3=4.23*.92*36.40=141.65 BTU/hrHeat Transfer through door is:Qcond, 4=A*TD*U TD= (108-71.6) =36.40OF A=21.51 U=0.47 BTU/hr.F.Qcond, 4=21.51*0.47*36.40=367.993 BTU/hrTotal heat Transfer through North Facing wall is:Qcond=Qcond, 1+Qcond, 2+Qcond, 3+Qcond, 4=2671.73 BTU/hr

Heat Transfer through Roof:Roof is made of 9 concrete and 1 plaster on both sides Qcond=A*TD*U R= (9*0.200) + (1*0.20) +0.61+0.61=3.22 hr.F./BTU U=0.311 BTU/hr.F. Qcond, 2=A*TD*U TD= (108-71.6) =36.40OF A=179.03 Qcond=179.03*0.310*36.40=2020.18 BTU/hrHeat Transfer through Floor:Heat transfer through floor is: Qcond=A*TD*U U = 0.04 (BTU/hr.. F) Same as for basement floor. Qcond=A*TD*U TD= (108-71.6) =36.40OF A=179.03 Qcond=179.03*0.04*36.40=260.67 BTU/hr

Infiltration Load: = 1.1*CFM*TC = 0.68*CFM*(-)

DoorWindows (1,2 and pane)

CFM/ of door = 0.5CFM/ Door Area = 21.51Total CFM = 0.5*21.51= 10.76 = 1.1*10.76*(108-71.6) = 430.84 BTU/hr = 0.68*10.76*(255-57) = 1448.72 BTU/hrCFM = 0.37 CFM/ft Length of crack (win 2 and pane) =34.67 ft CFM = 34.67*0.37 = 12.83 = 1.1*12.83*(108-71.6) = 513.71 BTU/hr = 0.68*12.83*(255-57) = 1727.44 BTU/hrLength of crack (win 1) =24.92 ft CFM = 24.92*0.37 = 9.23 = 1.1*9.23*(113-71.6) = 420.34 BTU/hr = 0.68*9.23*(296-57) = 1500.06 BTU/hr

Lighting:There are four fluorescent tube lights of 30 watts each in the room Q = 3.4*W*BF*CLF BF = 1.25 CLF =1Q = 3.4*120*1.25*1=510 BTU/hrPeopleThe room in our calculations is an office room that accommodates 2 persons on average= *n*CLF = 250 BTU/hr CLF=0.84 approximately=250*2*0.84=420 BTU/hr= *n =200 BTU/hr = 200*2=400 BTU/hrTotal Cooling LoadTotal Net Load =3198.8+1908.131+1773.22+2671.73+2020.18+260.67+510+400+420+430.84+513.71+420.34+4676.22=19203.84 BTU/hr= 1.6tons10 % losses =1920.38 Btu/hrSo, Total Load = 21124.23 BTU/hr= 1.76 Ton

ReferencesAir Conditioning Systems and Principles, Edward G. Pita 2nd Edition.