__Building Specifications__

Location: Jacksonville, FL

Type of facility: electronic casino

Area: 2,420 sf

Occupants: 59__Design Conditions__

Infiltration loss: typical range of 1 – 2 ACH (air changes per hour)

Outdoor air requirement: 15 – 25% of ventilation air

Breathing zone: 3″ – 72″ height from the floor and 24″ distance from the walls or AC equipment

Relative Humidity: 30% heating (30% – 40%), 50% cooling (40% – 60%); in general 30% – 35% (< 20% or > 60% is problematic)

Inside heating design temperature: 70˚F per ACCA Manual J and 72˚F per NEC (68˚F – 72˚F)

Inside cooling design temperature: 75˚F (72˚F – 76˚F)

Outdoor dry-bulb temperature (T_o): 92.8˚F (1% cooling), 28.6˚F (99.6% heating)

∆T = T_o – T_i = 92.8˚F – 75˚F = 17.8˚F

ASHRAE Climatic Design Conditions__Cooling Load__

To calculate the cooling load, the formula commonly used is Q_dot = U x A x ∆T.

However, a more accurate approach is to use the Cooling Load Temperature Difference (CLTD) instead of ∆T.

It’s important to note that it may still result in approximately a 15% error margin.__Rough estimation__

Cooling load of 20 btuh per sf of building area

Q = 2,420 sf x 20 btuh x (1 ton / 12,000 btuh) = 4.0 tons

Minimum ventilation design standard: 20 cfm/person with a reheat system__Roof__

Area (A) = 2,420 sq ft with 3′ attic air space

R-value (R) = 1.79 (1/2” acoustical ceiling tile), 30 (9-1/4” R-19 insulation)

Overall U-value (U) = 1/R_tot = 0.03

CLTD = 28 (L- light construction and 95˚F)

Q_roof = 0.03 x 2,420 x 28 = __2,130 btuh____Doors__

Qty (2) 1-¾” insulated metal doors in the east wall

A = 80” x 36” = 20 sf

U = 0.40 (Btu/hr-sf-°F)

CLTD = 16 (light construction and 95˚F)

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 and S walls have zero heat transfer

A_slab_face = 2,420 sf

A_ slab_edge = 0.333’ x (41’ x 2) = 27.33 sf

U_slab_face = 0.05

U_ slab_edge = 0.81

Q_slab = (0.05 x 2,420 + 0.81 x 27.33) x 5 = __715 btuh____Exterior Walls__

Zero heat transfer from N and S facing walls

East wall

Height = 128”, Length = 41’

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

Construction materials: 12” concrete block, ¾” plywood, ½” sheetrock, 1″ stucco

R-values:

Outside wind: 0.33 (for 7.5 mph wind)

Inside vertical air film: 0.68 – 0.69

Stucco: 4.76 (1”)

Light-weight (LW) block: 2.04 – 2.56 (12″)

Plywood: 1.08 (3/4”)

Sheetrock: 2.22 (1/2”)

U_tot = 0.09

CLTD = 16 (light construction and 95˚F)

Q_E_wall = 0.09 x 2,260 x 16 = 3,255 btuh

West wall

Area of west wall: 24” (wall above window) x 41’ + 16.5” x 104” x 3 (wall 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 passing through glass affects the heat gain of a space

Peak sun per day: 3 – 5 hours

Q_fes = (A_s x SHGF + A_sh x SHGF_sh) x SC

Q_fs = Q_fes x CLF (space cooling load)

A_s (Unshaded Area of Window Glass) = 1 – (SL/window width) x A_tot

A_sh (Shaded Area of Window Glass) = A_tot – A_s

SHGF (Solar Heat Gain Factor), Btu/hr-ft^{2}, using a latitude of 32˚N and the month of June

SHGF_sh (Shaded Solar Heat Gain Factor) accounts for shading due to orientation (East/West) and the month of May

SC = Shading Coefficient

SLF = Shade Line factor

SL (Shade Line) = SLF x shadow width beneath the edge of the overhang

SCL (Solar Cooling Load) = SHGF x CLF, where CLF takes into account the time lag

GLF (Glass Load Factor) = SCL x SC

Number of windows: 6

Area of (aluminum frame single-pane glass) door = 72” x 104” = 52 sf

Area of window = 66” x 104” = 48 sf

Area_tot = 52′ + (48′ x 6) = 340 sf

U = 1.27 Btu/hr-sf-°F

CLTD = 16 (light construction and 95˚F)

Q_windows = 1.27 x 16 x 340 = 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 shades for single-pane windows), 0.25 (white shades), 1.0 (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 = a conversion factor from watts to btuh

W = watts, which represents the lighting capacity

BF = Ballast Factor, accounting for heat loss in the ballasts of fluorescent lights

CLF = Cooling Load Factor (heat storage in the lighting fixtures)

(25) fixtures measuring 4’X2’ each, with (4) T12 lamps (48” length) per fixture, consuming 32 W per bulb, with a BF = 0.92

(6) fixtures measuring 2’X2’ each, with (2) u-shaped T12 lamps (24″ length) per fixture, consuming 32 W per bulb, with a BF = 0.94

CLF of 1.0 (typical value)

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__

Activity type: Office work

Activity level: Moderate

Sensible heat gain per person (qs): 250 btuh

Latent heat gain per person (ql): 200 btuh

Number of people (n): 59

CLF (capacity of space to absorb and store heat): 0.91 – 1.0

Q_s = qs x n x CLF

Q_l = ql x n

Q_occupants_undiversified = (qs + ql) x n = 450 btuh-person x 59 people = 26,550 btuh

Q_occupants_diversified = 26,550 x 0.9 = __23,895 btuh____Equipment__

(59) computers with 21” monitors consuming 130 W continuously

(1) computer with 15″ monitor consuming 110 W continuously

(2) 50” televisions (Westinghouse) consuming 151.3 kWh/yr

(1) refrigerator (15 ft^{3}) consuming 510 btuh

(1) laser printer consuming 240 btuh

(1) coffee maker consuming 2,590 btuh

(1) 8 head soda machine consuming 2,185 btuh

Other office equipment accounts for 25% of the nameplate power

Q = 3.41 x (59 computers x 130 W) = 26,150 btuh

Q = 3.41 x (1 computer x 110 W) = 375 btuh

Q = 2 x 151.3 kWh/yr x 0.3895 [(btuh) / (kWh/yr)] = 120 btuh

Q_equipment_undiversified = 32,170 btuh

Q_equipment_diversified = 32,170 x 0.7 = __22,520 btuh__

Q_l+s_r = 9,800 + 2,130 + 1,220 + 715 + 3,425 + 11,440 + 11,270 + 23,895 + 22,520 = __86,415 btuh____Ventilation Air__

Ventilation rate per person (Rp): 5 cfm/person

Number of people (Pz): 59

Ventilation rate per sf (Ra): 0.06 cfm/ft^{2}

Floor area (Az): 2,420 ft^{2}

Distribution effectiveness (Ez): 1.00

Ventilation rate in the breathing zone outdoor air: V_bz_dot = Rp x Pz x Ra x Az

Zone outdoor airflow: Voz_dot = V_bz_dot / Ez

Voz = (5 x 59 + 0.06 x 2,420) / 1.0 = 440 cfm__Sensible Cooling of Ventilation Air__

Sensible heat (h_s) = 1.08 x V_oz x ∆T = 1.08 x 440 ft^{3}/min x 17.8 = __8,460 btuh__

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__Latent Cooling of Ventilation Air__

Latent heat (h_l) = 4,840 x V_oz x dw_lb

Humidity ratio difference (dw_lb) = 0.0206 lb water/dry air

Mean coincident wet-bulb temperature (T_wb) = 25.4˚C = 77.7˚F

Latent heat (h_l) = 4,840 x 440 ft^{3}/min x 0.0206 = __43,870 btuh__

Total heat (sensible + latent): h_t = h_s + h_l = 8,460 + 43,870 = __52,330 btuh____Total Cooling Load__

(Q_t) = Q_l+s_r + h_t = 86,415 + 52,330 = 138,745 btuh / 12,000 = **11.6 tons**__Alternative Method__

Total heat (h_t) = 4.5 x V_oz x dh

Enthalpy difference (dh) = h_o – h_i

h_o = 45.5 (using psychrometric chart at T_db = 92.8˚F and dw_lb = 0.0206)

h_i = 28 (using psychrometric chart at T_db = 75˚F and RH = 50%)

h_t = 4.5 x 440 x (45.5 – 28) = 34,650 btuh

Q_t = Q_l+s_r + h_t = 86,415 + 34,650 = 121,065 btuh / 12,000 = 10.1 tons