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)

A0

0.059

A2

0.076

B3

1.173

C3

0.125

E0

0.121

Total

1.554

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


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