## Heat Load Calculations

Building Specs

- Building location: Jacksonville, FL
- Building area: 2,420 sf
- Occupants allowed: 92
- Infiltration loss: 1 – 2 ACH (air changes per hour) is typical
- Outdoor air required: 15 – 25% of ventilation air
- cedengineering.com

__Breathing zone__

- 3″ – 72″ from the floor
- 24″ from the walls or AC equipment
- Humidity: 30% – 35% (< 20% or > 60% is problematic)
- Inside design temperature: 70 – 72˚F (cooling set point)
- T_o = 92.8˚F (1% cooling DB)
- T_o = 28.6˚F (99.6% heating DB)
- ASHRAE Climatic Design Conditions

Rough Estimation

- 20 cfm/person (common minimum design standard) and a reheat system
- The estimated cooling load is 0.25 – 0.35 tons per 100 sf of total building area

Q = 2,420 sf x 0.35 tons / 100 sf = 8.47 tons

Cooling Load

- Q_dot = U x A x (T_i – T_o) or more accurate Q_dot = U x A x (CLTD)
- Cooling Load Temperature Difference gives ~15% error
- T_i = 70˚F
- T_o = 92.8˚F
- ∆T = 22.8

Roof

- A = 2,420 sf with 3′ attic air space
- R = 1.79 (1/2” acoustical ceiling tile)
- R = 30 (9-1/4” R-19 insulation)
- U = 0.03
- CLTD = 28 (L- light construction &; ∆T=95˚C)

Q_roof = 0.03 x 2,420 x 28 = 2,130 btuh

Doors

- Glass door in the front (west wall)
- Qty (2) 1-¾” insulated metal doors in the back (east wall)
- A = 80” x 36” = 20 sf
- U = 0.40 (Btu/hr-sf-°F)
- CLTD = 16 (light construction &; 95˚C)

Q_doors = U x A x #_doors x CLTD = 0.40 x 20 x 2 x 16 = 1,220 btuh

Concrete Slab

- 4” thick slab = 0.333’
- ∆T = 5˚F
- Slab edge of N & S walls have zero heat transfer
- U_slab_face = 0.05
- U_ slab_edge = 0.81
- A_slab_face = 2,420 sf
- A_ slab_edge = 0.333’ x (41’ x 2) = 27.33 sf

Q_slab = (0.05 x 2,420 + 0.81 x 27.33) x 5 = 715 btuh

Exterior Walls

Zero heat transfer from N & S facing walls

__East wall__

- Height is 128” & length is 41’
- Made of 12” concrete block, ¾” plywood, ½” sheetrock, 1″ stucco
- R = 0.33 (7.5 mph wind outside)
- R = 0.68 – 0.69 (inside vertical air film)
- R = 4.76 (1” stucco)
- R = 2.04 – 2.56 (12” LW (light weight) block)
- R = 1.08 (3/4” plywood)
- R = 2.22 (1/2” sheetrock)
- CLTD = 16 (light construction & 95˚C)
- Area of East wall = 128’ x 41” = 200 sf
- A_wall = (12’ x 200’) – A_windows – A_E_door = 2,400 – 80 – 60 = 2,260 sf
- U_total = 1/R_tot = 0.09
- Q_E_wall = 0.09 x 2,260 x 16 = 3,255 btuh

__West wall__

- U_total = 1/R_tot = 0.09
- Area of West wall = 24” (above window) x 41’ + 16.5” x 104” x 3 (between windows) = 82 + 36 = 118 sf
- Q_W_wall = 0.09 x 118 x 16 = 170 btuh

Q_wall_tot = 3,255 + 170 = 3,425 btuh

Windows

- Solar radiation through glass
- 3 – 5 hours of peak sun per day
- 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}(use latitude 32˚N & June) - SHGF_sh = Shaded Solar Heat Gain Factor (East/West & May)
- A_s = Unshaded Area of Window Glass
- A_sh = Shaded Area of Window Glass
- SC = Shading Coefficient
- SL = Shade Line = Shade Line factor x shadow width beneath edge of the overhang
- SCL = Solar Cooling Load (SCL = SHGF x CLF, where CLF takes into account time lag)
- GLF = Glass Load Factor (GLF = SCL x SC)

__Aluminum frame single-pane glass door & windows__

- U = 1.27 Btu/hr-sf-°F
- Area of door = 72” x 104” = 52 sf
- Area of window = 66” x 104” = 48 sf
- 6 windows
- Area_tot = 52 + (48 x 6) = 340 sf
- CLTD = 16 (light construction & 95˚F)

Q_windows = 1.27 x 16 = __6,910 btuh (less accurate)__

- Width of the overhang = 101”
- SLF = 0.8 | SL = 0.8 x 101” = 81”
- A_s = (1 – (81/104) x 340) = 75 sf
- A_sh = 340 – 75 = 265 sf
- SHGF = 1,169 Btu/sf-day / 8 hours = 146
- SHGF_sh = 142 W/m
^{2}x 0.0929 m^{2}/sf x 3.41 Btu/h-W = 45 - SC = 0.50 (blinds or translucent roller shade for single pane)
- 0.25 for white shades, 1.0 for no shades

Q_windows = (75 x 146 + 265 x 45) x 0.50 = 11,440 btuh (more accurate)

Lighting

- Q = 3.412 x W x BF x CLF
- 3.412: converts watts to btuh
- W: lighting capacity
- BF: Ballast Factor (heat loss in the ballasts of fluorescent lights)
- CLF: Cooling Load Factor (heat storage in the lighting fixtures)
- Fluorescent lights
- Qty (25) 4’X2’ fixtures with qty (4) 48” T12 lamps, 32 W per bulb, BF = 0.92
- Qty (6) 2’X2’ fixtures with qty (2) 24” T12 u-shaped lamps, 32 W per bulb, BF = 0.94
- CLF (1.0 is often used)

Q_lighting = 3.412 x [(25 x 4 x 32 W x 0.92 x 1.0) + (6 x 2 x 32W x 0.94 x 1.0)] = 11,270 btuh

Occupants

- Qs = qs x n x CLF | Ql = ql x n
- Qs & Ql = Sensible and Latent heat gains, btuh
- qs & ql = Sensible and Latent heat gains per person, btuh-person
- n = Number of People
- CLF = Cooling Load Factor (capacity of a space to absorb and store heat is 0.91 – 1.0)
- Activity type: office work
- Activity level: moderate
- Sensible heat gain is 250 btuh-person
- Latent heat gain is 200 btuh-person

Q_occupants: 450 btuh-person x 92 people = 41,400 btuh

Equipment

- Qty (59) computers (21” monitors) 30 W at idle & 130 W continuous
- Fridge (15 ft
^{3}) = 150 W - Laser Printer is 70 W
- Coffee maker is 2,590 btuh
- Qty (2) 50” tv (Westinghouse) = 151.3 kWh/yr
- Any other office equipment = 25% nameplate
- Q = 3.41 x (59 computers x 130 W) = 26,150 btuh (21” monitors)
- Q = 3.41 x (1 computer x 110 W) = 375 btuh (15” monitor)
- Q = 2 x 151.3 kWh/yr x 0.3895 [(btuh) / (kWh/yr)] = 120 btuh (tv)
- Q = 3.41 x 150 W = 510 btuh (fridge)
- Q = 3.41 x 70 W = 240 btuh (laser printer)
- Q = 2,590 btuh (coffee maker)
- Q = 2,185 btuh (8 head soda machine)

Q_equipment_undiversified = 32,170 btuh

Q_equipment_diversified = 32,170 x 0.60 = __19,300__

Q_l+s_r = 9,800 + 2,130 + 1,220 + 715 + 3,425 + 11,440 + 11,270 + 41,400 + 19,300 = 100,700 btuh

Ventilation

- Rp: 5 cfm/person
- Pz (no. of people): 92
- Ra: 0.06 cfm/ft
^{2} - Az (floor area): 2,420 ft
^{2} - Ez (distribution effectiveness): 1.00
- V_bz_dot = Rp x Pz x Ra x Az (ventilation rate, breathing zone outdoor air)
- Voz_dot = V_bz_dot / Ez (zone outdoor airflow)
- Voz = (5 x 92 + 0.06 x 2,420) / 1.0 = 605 cfm

Ventilation Air Cooling Load

- engineeringtoolbox.com/cooling-heating-equations-d_747.html
- Load on the coil due to leakage in the return air duct and the 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 = 92.8˚F (dry-bulb) | T_i = 70˚F (dry-bulb) | ∆T = (T_o – T_i) = 22.8
- h_s = 1.08 x 605 ft
^{3}/min x 22.8 = 14,900 btuh

Latent heat due to 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)
- dw_lb = 0.0206
- h_l = 4,840 x 605 ft
^{3}/min x 0.0206 = 60,320 btuh - h_t = h_s + h_l = 14,900 + 60,320 = 75,220 btuh

Q_t = Q_l+s_r + h_t = 105,740 + 76,395 = 182,135 btuh / 12,000 = 15.18 tons

Alternative 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 = 92.8˚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 btuh

Q_t = Q_l+s_r + h_t = 100,700 + 63,980 = 156,850 btuh / 12,000 = 13.07 tons