WC Engineering Corp Touchton Rd., Bldg. 100, Suite 150 Jacksonville, FL Attn: Michael Jones May 25, 2017

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WC Engineering Corp. 4600 Touchton Rd., Bldg. 100, Suite 150 Jacksonville, FL 32246 Attn: Michael Jones May 25, 2017 Subject: HVAC Calculations per elegantcafedwg.pdf Quote #: QU-51217

Building area: 2,420 sq. ft. No. of Occupants = 92 people (per the fire code sign on the entry wall) Typical Infiltration loss: 1-2 air changes/hr (assume 2) Outdoor Air: 15-25% of ventilation air The breathing zone is defined as between 3 and 72 from the floor and 24 inches from walls or AC equipment. Inside design temp.: 70-72 F (assume 70 F cooling set point) Humidity: 30-35% (<20% and >60% is bad) Climatological data for Jacksonville, FL: http://cms.ashrae.biz/weatherdata/stations/722060_s.pdf From the link: Annual Cooling Design Conditions give T_db is 34.8 C or 94.6 F (0.4% extreme) & Latitude is 30.50 N Rough Estimation: Using 20 cfm/person (20 cfm is a common minimum design standard - new, outside air) and a reheat system the estimated cooling load is 0.25 0.35 tons per 100 sq. ft. of total building area. Q = 2,420 sq. ft. x 0.35 tons / 100 sq. ft. = 8.47 tons Cooling Load: Q_dot = U x A x (T_i T_o) Q_dot = U x A x (CLTD) T_i = 70 F T_o = 94.6 F T = 24.6 U = 1/R (cooling load temperature difference, more accurate, ~15% error) Roof: 3 of air space in attic R = 1.79 (1/2 acoustical ceiling tile) (see Excel) R = 30 (9-1/4 thick R-19 insulation) (see Excel) U_total = 1/R_tot = 0.03 A = 2,420 ft 2 CLTD = 28 (see Figure 2.0 Roofs & Ceilings; L- light construction & T 35 C since worst case for Jacksonville, FL is 36.6 C) Q_roof = 0.03 x 2,420 x 28 = 2,130 Btu/hr Doors: Treat the front W facing glass door as a window No. of 1-¾ inch insulated metal doors (in E wall) = 2 A = 80 x 36 = 20 sq. ft. U = 0.40 (Btu/hr-ft 2 - F) (see R-U sheet in Excel) CLTD = 16 (see Figure 2.0 All Walls and Doors; East and West facing wall, light construction & 35 C) Q_doors = U x A x #_doors x CLTD = 0.40 x 20 x 2 x 16 = 1,220 Btu/hr Concrete Slab:

4 thick slab = 0.333 T = 5 F Slab edge of N & S walls have zero heat transfer since they abut against neighboring business spaces which are at 70 F. U_slab_face = 0.05 (see Figure 8.1) U_ slab_edge = 0.81 (see Figure 8.0) A_slab_face = 2,420 sq. ft. A_ slab_edge = 0.333 x (41 x 2) = 27.33 sq. ft. Q_slab = (0.05 x 2,420 + 0.81 x 27.33) x 5 = 715 Btu/hr Exterior Wall: Zero heat transfer from N & S facing walls since they abut against other business spaces at 70 F Wall height E wall is 128 high and 41 long; comprised of 12 CMU, ¾ wood & ½ sheetrock W wall is considered a window with the exception of parts which are comprised of 12 CMU, ¾ wood & ½ sheetrock (also referred to as drywall & gypsum board) & 1 stucco East wall: R = 0.33 (outside air 7.5 mph wind) Excel: U = 0.25 R = 2.4 (12 CMU) R = 2.04 2.56 (12 CMU concrete masonry unit, or concrete block) (LW block not HW block) R = 0.69 (Inside air) Excel: R = 0.68 R = 1.08 (3/4 plywood) R = 2.22 (1/2 sheetrock) CLTD = 16 (see Figure 2.0 All Walls and Doors; East and West facing wall, light construction & 35 C) Area of E wall = 41 x 128 = 200 U_total = 1/R_tot = 0.15 A_wall = (12 x 200 ) A_windows A_E_door = 2,400 80 60 = 2,260 sq. ft. West wall: The same as the east wall except stucco is added R = 4.76 (1 stucco) U_total = 1/R_tot = 0.09 Area of W wall = 24 (above glass window) x 41 + 16.5 x 104 x 3 (parts between windows) = 82 + 36 = 118 sq. ft. Q_E_wall = 0.15 x 2,260 x 16 = 5,425 Btu/hr Q_W_wall = 0.09 x 118 x 16 = 170 Btu/hr Q_tot = 5,425 + 170 = 5,595 Btu/hr Windows:

Most areas average 3-5 peak sun hours Conduction heat gain through fenestration areas: Q = A x U x CLTD Solar Radiation through glass: Q_fes = (A_s x SHGF + A_sh x SHGF_sh) x SC Q_fs = Q_fes x CLF (space cooling load) SHGF = Maximum solar heat gain factor, Btu/hr-ft 2 (The SHGF is based on orientation and time of year) https://engineering.dartmouth.edu/~d30345d/courses/engs44/shgf-daily-totals.pdf (use latitude 32 N & June) SHGF_sh = shaded solar heat gain factor http://personal.cityu.edu.hk/~bsapplec/cooling.htm (use table 5, East/West & May) A_s = Unshaded area of window glass, ft 2 A_sh = shaded area of window glass, ft 2 SC = Shading coefficient SL = Shade line = Shade line factor x width of the overhang (the distance a shadow falls beneath the edge of an overhang) SCL = Solar Cooling load (SCL = SHGF x CLF, where CLF is cooling load factor; CLF takes into account time lag) GLF = Glass load factor (GLF = SCL x SC, see figure 3.1) Aluminum frame single glass door & windows U = 1.27 Btu/hr-ft 2 - F (see Excel, for front door and windows) Area of door = 72 x 104 = 52 sq. ft. Area of window = 66 x 104 = 48 sq. ft. No. of windows = 6 Area_tot = 52 + 48 x 6 = 340 sq. ft. CLTD = 16 (see Figure 2.0 All Walls and Doors; East and West facing wall, light construction & 35 C) Q_windows = 1.27 x 16 x 340 = 6,910 Btu/hr (least accurate method) Width of overhang = 101 SLF = 0.8 (see Figure 3.0) SL = 0.8 x 101 = 81 (see Figure 3.0) A_s = (1- (81/104) x 340) = 75 sq. ft. A_sh = 340 75 = 265 sq. ft. SHGF = 1,169 Btu/ft 2 -day / 8hrs = 146 SHGF_sh = 142 W/m 2 x 0.0929 m 2 /ft 2 x 3.41 Btu/h-W = 45 SC = 0.50 (blinds or translucent roller shade for single pane) (see Figure 3.0) 0.25 if using white shades, 1.0 if no shades http://personal.cityu.edu.hk/~bsapplec/cooling.htm Q_windows = (75 x 146 + 265 x 45) x 0.50 = 11,440 Btu/hr Lighting: Q = 3.41 x W x BF x CLF

3.41 = conversion coefficient between Watts and Btu/hr Q = Cooling load from lighting, Btu/hr W = Lighting capacity, Watts BF = Ballast factor CLF = Cooling load factor for lighting BF accounts for heat losses in the ballasts of fluorescent lights and CLF accounts for heat storage in the lighting fixtures. Fluorescent lights Quantity of (25) 4 X2 fixtures with (4) 48 lamps of T12 diameter W = 32W per bulb, BF = 0.92 (see HeatGain sheet in Excel) Quantity of (6) 2 X2 fixtures with (2) 24 u-shaped lamps of T12 diameter W = 32W per bulb, BF = 0.94 CLF (1.0 is often used) Q_lighting = 3.41 x [(25 x 4 x 32W x 0.92 x 1.0) + (6 x 2 x 32W x 0.94 x 1.0)] = 11,270 Btu/hr Occupants: Qs = qs x n x CLF, Ql = ql x n Qs & Ql = Sensible and latent heat gains, Btu/hr qs & ql = Sensible and latent heat gains per person, Btu/hr-person n = Number of people CLF = Cooling load factor for people (capacity of a space to absorb and store heat, 0.91 could be used, however 1.0 will be used) Activity: Office work Level: moderate Sensible heat gain: 250 Btu/hr-person Latent heat gain: 200 Btu/hr-person Q_occupants: 450 Btu/hr-person x 92 people = 41,400 Btu/hr Equipment: Quantity of 59 computers (21 monitors) at idle is 30W and continuous running is 130W Refrigerator 15ft 3 = 300W (small fridge so halving value) Laser Printer Desktop at idle = 70W Coffee maker = 2,590 Btu/hr No. of (2) Westinghouse 50 televisions = 151.3 kwh/yr Any other office equipment = 25% Nameplate (Watts volts amps) All 59 computers are running Q = 3.41 x (59 computers x 130W) = 26,150 Btu/hr Q = 3.41 x (1 computer x 110W) = 375 Btu/hr (21 computers) (15 computer)

Q = 2 x 151.3 kwh/yr x 0.3895 [(Btu/hr) / (kwh/yr)] = 120 Btu/hr (television) Q = 3.41 x 150W = 510 Btu/hr (refrigerator) Q = 3.41 x 70W = 240 Btu/hr (laser printer) Q = 2,590 Btu/hr (coffee maker) Q = 2,185 Btu/hr (8 head soda fountain machine) Q_equipment = 32,170 Btu/hr Q_l+s_r = 9,800 + 2,130 + 1,220 + 715 + 5,595 + 11,440 + 11,270 + 41,400 + 32,170 = 105,740 Btu/hr Ventilation: Rp (cfm/person) Pz (No. of People) Ra (cfm/ft 2 ) Az (Floor Area, ft 2 ) Ez 5 92 0.06 2,420 1.00 Single Zone and Dedicated OA (outdoor air) Systems (DOAS) V_bz_dot = Rp x Pz x Ra x Az Ez = distribution effectiveness (see Figure 1.1) Voz_dot = V_bz_dot / Ez Voz = (5 x 92 + 0.06 x 2,420) / 1.0 = 605 cfm Ventilation Air Cooling Load: http://www.engineeringtoolbox.com/cooling-heating-equations-d_747.html Load on the coil due to leakage in return air duct and return air fan is negligible. Sensible heat in a cooling process of air: h_s = 1.08 x V_oz x T ρ = 0.075 lbm/ft 3 T_o = 94.6 F (dry bulb) T_i = 70 F (dry bulb) T = (T_o T_i) = 24.6 h_s = 1.08 x 605 ft 3 /min x 24.6 = 16,075 Btu/hr (ventilation rate, breathing zone outdoor air) (zone outdoor airflow) Latent heat due to the moisture in the air, in a humidification process of air: h_l = 4,840 x V_oz x dw_lb dw_lb = humidity ratio difference (lb water/dry air) T_wb = 25.4 C = 77.7 F (mean coincident wet bulb; wet bulb at the cooling 0.4% dry bulb temp. time of yr) dw_lb = 0.0206 (from design conditions for Jacksonville, FL) h_l = 4,840 x 605 ft 3 /min x 0.0206 = 60,320 Btu/hr h_t = h_s + h_l = 16,075 + 60,320 = 76,395 Btu/hr

Another method: h_t = 4.5 x V_oz x dh dh = h_o h_i (enthalpy difference) h_o = 45.5 (using psychrometric chart at T_db = 94.6 F & dw_lb = 0.0206) h_i = 22 (using psychrometric chart at T_db = 70 F & RH = 30%) h_t = 4.5 x 605 x (45.5 22) = 63,980 Btu/hr Q_t = Q_l+s_r + h_t = 105,740 + 76,395 = 182,135 Btu/hr Q_t = 182,135 / 12,000 = 15.18 tons Using h_t from the other method yields 14.14 tons Q_t = Q_l+s_r + h_t = 105,740 + 63,980 = 169,720 Btu/hr Q_t = 169,720 / 12,000 = 14.14 tons