CLTD/SCL/CLF Method (manual method)
ASHRAE Fundamentals 1997 in Chapter 28.39-28.56.
• External cooling load
Roofs, walls, and conduction through glass:
Q = U A (CLTD)
Where U = design heat transfer coefficient (Table A24-4, A29-5)
A = area of roof, wall, or glass
CLTD = cooling load temperature difference (Table A28-32, A28-34)
Solar load through glass:
Q = A (SC) (SCL)
Where SC = shading coefficient
SCL = solar cooling load factor (Table A28-36)
Partitions, ceilings, and floors:
Q = UA(Tb- Ti)
Where U = design heat transfer coefficient for partition, ceiling, or floor (Table A24-4) A = area of partition, ceiling, or floor Tb = adjacent space temperature Ti = inside air temperature
• Internal cooling load People:
Qsensibte = N (sensible heat gain) CLF Qiatent = N (latent heat gain)
Where N = number of people in space (Table A28-3)
CLF = cooling load factor (Table A28-37)
Lights:
Qet = W Ful Fsa (CLF) where W = watts input from electrical plans or lighting fixture
F„i = lighting use factor Fsa = special allowance factor
CLF = cooling load factor (a schedule factor, CLF =1 for 24-hour light usage) (Table A28-38)
Power:
Qp = P Ef CLF
Where P = power rating from electrical plans or manufacturer’s data Ef = efficiency factors and arrangements to suit circumstances CLF = cooling load factor (a schedule factor) (Table A28-37)
Appliances:
Qsensible = Qis Fua Fra (CLF)/Ff|
Qlatent — Qil Fua
Where QiS; Qii = sensible and latent heat gain from appliances
Fua, Fra, Ffi = use factors, radiation factors, flue loss factors
CLF = cooling load factor (a schedule factor) (Tables A28-37, A28-39)
Ventilation and infiltration air:
Qsensible _ p Vflow (To — Ti)
Qlatent = 2500 p Vflow (W0 — Wi)
Qtotal — P Vflow (Ho ” Htooni)
Where p = air density
Vflow = ventilation or infiltration flow rate (Thermal comfort & IAQ requirements) T0, Ti = outside, inside air temperature W0, Wi = outside, inside air humidity ratio Ho, Hi = outside, inside air enthalpy
Example 8-1
A PSU classroom is 6 m long, 6 m wide and 3 m high. There is a 2.5 m x 4 m window in the east wall. Only the east wall/window is exterior. Assume the thermal conditions in adjacent spaces (west, south, north, above and below) are the same as those of the classroom. Determine the
Cooling load at 9:00 am, 12:00 noon on July 21.
Other known conditions include:
Latitude = 40° N
Ground reflectance = 0.2
Clear sky with a clearness number =1.0
Overall window heat transmission coefficient = 7.0 W/m2K
Room dry-bulb temperature = 25.5 °C
Permissible temperature exceeded = 2.5%
Schedule of occupancy: 20 people enter at 8:00 am and stay for 8 hours
Lighting schedule: 300 W on at 8:00 am for 8 hours Exterior wall structure:
Outside surface, A0 Face brick (100 mm), A2 Insulation (50 mm), B3 Concrete block (100 mm), C3 Inside surface, E0 Exterior window:
Single glazing, 3 mm No exterior shading, SC = 1.0
Solution:
(1) Exterior wall:
|
Laver_______ Unit resistance fm2 K/W)
|
|
U=l/R = 0.643 W/m2 K |
From Table A28-33A, find wall type 13.
From Table A28-32, CLTD9:oo = 9 and CLTDi2:oo = 14 for east wall.
CLTDcorrected = CLTD + (25.5 — T,) + (Tm — 29.4)
Ti = inside temperature
Tm = mean outdoor temperature
Tm = (maximum outdoor temperature) — (daily range)/2
At 9:00 am CLTDcor = 9 + (25.5 — 25.5) + (31 -9/2 — 29.4) = 6.1 K at 12:00 am CLTDcor = 14 + (25.5 — 25.5) + (31 -9/2 — 29.4) = 11.1 K
Q = U A (CLTD)
= 0.643 (W/m2 K) x (6 x 3 — 4 x 2.5) m2 x 6.1 = 34 W (at 9 am)
Q = 0.643 (W/m2 K) x (6 x 3 — 4 x 2.5) m2 x 11.1 = 62 W (at 12 noon)
(2) Window conduction:
From Table A28-34, CLTD9:0o = 1 and CLTD^oo = 5.
CLTDcor = CLTD + (25.5 — T{) + (Tm — 29.4)
= 1 + 0 + (31 — 9/2-29.4) =-1.9 K (at 9 am)
CLTDcor = 5 + 0 + (31 — 9/2 — 29.4) = 2.1 K (at 12 noon)
Q = U A (CLTD)
= 7.0 W/m2 K x 4 x 2.5 m2 x (-1.9 K) = -133 W (at 9 am)
Q = 7.0 W/m2 K x 4 x 2.5 m2 x 2.1 K = 147 W (at 12 noon)
(3) Solar load through glass:
From Table A28-35B, find zone type is A
From Table A28-36, find CLF = 576 at 9 am and CLF = 211 at 12 noon
Q = A (SC) (SCL)
= (2.5 x 4 m2) x 1.0×576 = 5760 W (at 9 am)
Q = (2.5 x 4 m2) x 1.0 x 211 =2110 W (at 12 noon)
(4) Cooling load from partitions, ceiling, floors:
Q = 0
(5) People:
From Table A28-35B, find zone type is B
From Table A28-3, find sensible/latent heat gain = 70 W.
From Table A28-37, find CLF9:oo (1) = 0.65 and CLF^oo (3) =0.85
Qsensibie = N (sensible heat gain) CLF
= 20 x 70 W x 0.65 = 910 W (at 9 am)
Qsensibie = 20 x 70 x 0.85 = 1190 W (at 12 noon)
Qiatent = N (latent heat gain)
= 20 x 45 = 900 W (at 9 am and 12 noon)
(6) Lighting:
From Table A28-35B, find zone type is B
From Table A28-38, find CLF9:0o (1) = 0.75 and CLF^oo (3) = 0.93 Qei = W F„i Fsa (CLF)
= 300 Wxlx 1 x0.75 = 225 W (at 9 am)
Qd = 300 x 1 x 1 x 0.93 = 279 W (at 12 noon)
(7) Appliances:
Qsensible 0 Qiatent — 0
(8) Infiltration:
Since room air pressure is positive, we have:
Qsensible 0 Qiatent — 0
Total cooling load:
|
Component |
9:00 am |
12:00 noon |
|
Wall |
34 |
62 |
|
Window conduction |
-133 |
147 |
|
Window solar transmission |
5,760 |
2,110 |
|
Partitions |
0 |
0 |
|
People |
910 |
1,190 |
|
Lights |
225 |
279 |
|
Appliance |
0 |
0 |
|
Infiltration |
0 |
0 |
|
Total 6,796 W 3,788 W |
Posted in Fundamentals of Heating. Ventilating, and Air-Conditioning