Calculations

Building area is 2,420 sf
Number of Occupants is 92 people (per fire code)
Infiltration loss is typically 1 – 2 ACH (air changes per hour)
Outdoor Air required is 15 – 25% of ventilation air

Breathing Zone:
3 – 72 inches from the floor
24 inches from walls or AC equipment

Inside design temperature of 70 – 72˚F (cooling set point).
Humidity is 30 – 35% (< 20% or > 60% is problematic).

Climatological data for Jacksonville, Florida:
http://cms.ashrae.biz/weatherdata/STATIONS/722060_s.pdf
T_db is 34.8˚C or 94.6˚F (0.4% extreme)
Latitude is 30.50 N

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

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 = 94.6˚F
∆T = 24.6
U = 1/R

Roof

3 feet of air space in attic
R = 1.79 (1/2” acoustical ceiling tile)
R = 30 (9-1/4” thick R-19 insulation)

U_total = 1/R_total = 0.03
A = 2,420 sf
CLTD = 28 (L- light construction & ∆T=35˚C, since the worst case for Jacksonville, FL is 36.6˚C).
Q_roof = 0.03 x 2,420 x 28 = 2,130 Btu/hr

Doors

Front west facing glass door is treated as a window
Number of 1-¾ inch insulated metal doors (in east wall) is 2
A = 80” x 36” = 20 sf
U = 0.40 (Btu/hr-sf-°F)
CLTD = 16 (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.
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 Btu/hr

Exterior Wall

Zero heat transfer from N & S facing walls since they abut against other business spaces.
East wall height is 128” high and 41’ long; made of 12” CMU (concrete masonry unit), ¾” wood, and ½” sheetrock.
West wall is a window except parts comprised of 12” CMU, ¾” wood, ½” sheetrock, and 1” stucco.

East wall

R = 0.33 (7.5 mph wind outside)
R = 2.04 – 2.56 (12” CMU- LW block not HW block)
R = 0.68 – 0.69 (Inside Air)
R = 1.08 (3/4” plywood)
R = 2.22 (1/2” sheetrock)
CLTD = 16 (light construction & 35˚C)
Area of East wall = 41’ x 128” = 200 sf

U_total = 1/R_tot = 0.15
A_wall = (12’ x 200’) – A_windows – A_E_door = 2,400 – 80 – 60 = 2,260 sf

West wall

The east wall adds stucco: R = 4.76 (1” stucco)
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_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_wall_tot = 5,425 + 170 = 5,595 Btu/hr

Windows

3 – 5 hours of peak sun per day.
Conduction heat gain through fenestration areas: Q = A x U x CLTD.

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-ft2 (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
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 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
Number of windows is 6.
Area_tot = 52 + 48 x 6 = 340 sf
CLTD = 16 (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
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/m2 x 0.0929 m2/sf x 3.41 Btu/h-W = 45
SC = 0.50 (blinds or translucent roller shade for single pane).

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
W – lighting capacity, Watts
BF – Ballast Factor (heat loss in the ballasts of fluorescent lights)
CLF – Cooling Load Factor (heat storage in the lighting fixtures).
Fluorescent lights:
(25) 4’X2’ fixtures with (4) 48” lamps of T12 diameter, 32 W per bulb, BF = 0.92
(6) 2’X2’ fixtures with (2) 24” u-shaped lamps of T12 diameter, 32 W per bulb, BF = 0.94
CLF (1.0 is often used)

Q_lighting = 3.41 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 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 is 0.91 – 1.0).
Activity – office work
Level – moderate

Sensible heat gain is 250 Btu/hr-person
Latent heat gain is 200 Btu/hr-person
Q_occupants: 450 Btu/hr-person x 92 people = 41,400 Btu/hr

Equipment

59 computers (21” monitors)
Running at idle is 30 W and continuously is 130 W.
Refrigerator (15 ft3) = 300 W (there is a small fridge only, so halving this value).
Laser Printer is 70 W
Coffee maker is 2,590 Btu/hr
(2) 50” televisions (Westinghouse) = 151.3 kWh/yr
Any other office equipment = 25% nameplate

Q = 3.41 x (59 computers x 130 W) = 26,150 Btu/hr (21” computers)
Q = 3.41 x (1 computer x 110 W) = 375 Btu/hr (15” computer)
Q = 2 x 151.3 kWh/yr x 0.3895 [(Btu/hr) / (kWh/yr)] = 120 Btu/hr (television)
Q = 3.41 x 150 W = 510 Btu/hr (refrigerator)
Q = 3.41 x 70 W = 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): 5
Pz (Number of People): 92
Ra (cfm/ft2): 0.06
Az (Floor Area, ft2): 2,420
Ez: 1.00

Single Zone and Dedicated OA (outdoor air) Systems (DOAS)
V_bz_dot = Rp x Pz x Ra x Az (ventilation rate, breathing zone outdoor air)
Ez = distribution effectiveness
Voz_dot = V_bz_dot / Ez (zone outdoor airflow)
Voz = (5 x 92 + 0.06 x 2,420) / 1.0 = 605 cfm

http://www.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/ft3
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 ft3/min x 24.6 = 16,075 Btu/hr

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 ft3/min x 0.0206 = 60,320 Btu/hr
h_t = h_s + h_l = 16,075 + 60,320 = 76,395 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

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 = 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 + 63,980 = 169,720 Btu/hr
Q_t = 169,720 / 12,000 = 14.14 tons