### Room Load Summary Report

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

#### 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 = 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.

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-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

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 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/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 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 ft^{3}) = 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

#### Ventilation Air Cooling Load

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/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

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 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