Chapter_8

8. The Cooling Load

8.1 Describe a situation where the heat gain to the space is (a) greater than the

cooling load at a given time; (b) less than the cooling load at a given time; (c)

equal to the cooling load at a given time.

Answer:

Based on Fig. 8.2

(a) Heat gain to the space is greater than the cooling load at a given time at the

first half of the usage or operation.

(b) Heat gain to the space is less than the cooling load at a given time at the

second half of the usage or operation.

(c) Heat gain to the space is equal to the cooling load at a given time at the middle

of the usage or operation.

8.2 Southern coastal regions of the United States experience periods of very high

humidity. Explain how this might influence selection of design conditions.

Answer:

The very high humidity outside conditions influence the selection of inside design

condition through selection of desired room relative humidity not at the lowest side of

range that cause higher latent cooling load or a higher moisture content than indoor

design conditions but at the highest side of humidity range.

8.3 Determine the ASHRAE Standard 90.1 design conditions for the following

locations. Include the maximum outdoor temperature, the outdoor mean

coincident wet bulb temperature, the indoor dry bulb temperature, the relative

humidity, the elevation, and the latitude. (a) Washington, D.C., (b) San

Francisco, California, (c) Denver, Colorado, (d) Dallas, Texas.

Answer:

(a) Washington, D.C.

Maximum outdoor temperature = 95 F (35 C)

Outdoor mean coincident wet bulb temperature = 76 F (25 C)

Indoor dry bulb temperature = 75 F (24 C)

Relative humidity = 50 %

Elevation = 66 ft (20 m)

Latitude = 38.85 deg

(b) San Francisco, California







8. The Cooling Load

(c) Denver, Colorado







(d) Dallas, Texas







8.4 Determine the wall conduction transfer function coefficients for a wall

composed of 4 in. brick [k = 7 (Btu-in.)/(hr-ft2-F)], 1/2 in. plywood, 3 1/2 in.

mineral fiber insulation (R-11), and 1/2 in. gypsum board.

Solution:

Wall Layers (Table 5-4)

Layer Thickness, in Density, lb/ft3

Conductivity,

(Btu-in)/(hr-

ft2-F)

Specific Heat,

Brick 4 130 7 0.19

Plywood 0.5 34 0.8 0.29

Mineral fiber

insulation

(R-11)

3.5 1.2 0.3185 0.17

Gypsum board 0.5 50 1.11 0.26

CTF Coefficients (From PRF/RTF Generator)

n nX ,

Btu/(h-ft2-F)

nY ,

Btu/(h-ft2-F)

nZ ,

Btu/(h-ft2-F)

nΦ

0 4.033132292 0.001517739 0.623505633 0

1 -4.890102268 0.017652429 -0.943189491 0.643204868

2 0.893455637 0.010001789 0.358387909 -0.017086206

3 -0.006945024 0.000370798 -0.009166607 1.3166E-05

4 2.65956E-06 6.21048E-07 5.90198E-06 -1.31644E-09

5 -1.59894E-10 4.86714E-11 -2.45933E-10 1.12031E-14

8.5 Change the insulation in Problem 8-4 to R-19, and determine the conduction

transfer function coefficients.

8. The Cooling Load

Solution:

Wall Layers (Table 5-4)


Conductivity,

(Btu-in)/(hr-

ft2-F)

Specific Heat,

Brick 4 130 7 0.19

Plywood 0.5 34 0.8 0.29

Mineral fiber

insulation

(R-19)

6 1.2 0.3158 0.17

Gypsum board 0.5 50 1.11 0.26


n nX ,

Btu/(h-ft2-F)

nY ,

Btu/(h-ft2-F)

nZ ,

Btu/(h-ft2-F)

nΦ

0 4.033132964 0.000387523 0.611908143 0

1 -4.978563054 0.008315152 -0.968315133 0.66508466

2 0.982856227 0.008091923 0.388562266 -0.02728007

3 -0.019978162 0.000729205 -0.014736957 0.000228003

4 8.14087E-05 5.69043E-06 0.000111178 -9.35628E-08

5 -1.44779E-08 2.93347E-09 -3.98891E-08 4.13516E-12

6 4.33848E-13 8.69125E-14 6.47835E-13 -2.32896E-17

8.6 A roof is composed of asphalt roll roofing, 1/2 in. plywood, 3 1/2 in. mineral

fiber insulation (R-11), and 1/2 in. gypsum board. Determine the conduction

transfer function coefficients.

Solution:

Roof layers (Table 5-4)


Conductivity,

(Btu-in)/(hr-

ft2-F)

Specific Heat,

Asphalt roll

roofing, C =

6.50 Btu/(hr-

ft2-F)

-- 70 -- 0.36

Plywood 0.5 34 0.8 0.29

Mineral fiber

insulation

(R-11)

3.5 1.2 0.3185 0.17

Gypsum board 0.5 50 1.11 0.26

8. The Cooling Load


n nX ,

Btu/(h-ft2-F)

nY ,

Btu/(h-ft2-F)

nZ ,

Btu/(h-ft2-F)

nΦ

0 0.479438185 0.046513387 0.623539389 0

1 -0.398071691 0.03488033 -0.543490526 0.003066927

2 0.000215969 0.000188726 0.00153361 -2.70423E-07

3 -1.13694E-08 4.98289E-09 -3.36675E-08 1.75447E-14

4 5.74319E-16 2.39706E-17 1.43501E-15 5.26759E-28

8.7 The roof of Problem 8-6 is changed to have a suspended ceiling with a 12 in.

air space above it. Determine the conduction transfer function coefficients.

Solution:



Conductivity,

(Btu-in)/(hr-

ft2-F)

Specific Heat,

Asphalt roll

roofing, C =

6.50 Btu/(hr-

ft2-F)

-- 70 -- 0.36

Air Space

R=0.93 (F-ft2-

hr)/Btu

12 >3.5

Plywood 0.5 34 0.8 0.29

Mineral fiber

insulation

(R-11)

3.5 1.2 0.3185 0.17

Gypsum board 0.5 50 1.11 0.26


n nX ,

Btu/(h-ft2-F)

nY ,

Btu/(h-ft2-F)

nZ ,

Btu/(h-ft2-F)

nΦ

0 0.373706533 0.026229879 0.623038586 0

1 -0.308505531 0.038174505 -0.630432109 0.136813194

2 0.000527467 0.001323929 0.073124072 -1.60845E-05

3 -4.88864E-08 8.3104E-08 -2.15052E-06 1.88841E-12

4 4.50557E-15 1.14735E-15 1.54748E-13 4.57357E-21

8.8 A roof is composed of asphalt roll roofing, 4 in. of 120 lb/ft3 limestone

concrete, 2 in. of expanded polystyrene, and 0.5 in. of acoustical tile.

Determine the conduction transfer function coefficients.

8. The Cooling Load

Solution:


Layer Thickness,

in

Density,

lb/ft3

Conductivity,

(Btu-in)/(hr-

ft2-F)

Specific

Heat,

Resistance,

F-ft2-

hr/Btu

Asphalt roll

roofing, C =

6.50 Btu/(hr-

ft2-F)

-- 70 -- 0.36 0.153846

Limestone

concrete 4 120 7.9 -- 0.50633

Expanded

Polystyrene 2 1.0 0.36 0.29 --

Acoustical

Tiles 0.5 -- 0.40 0.31 1.25


n nX ,

Btu/(h-ft2-F)

nY ,

Btu/(h-ft2-F)

nZ ,

Btu/(h-ft2-F)

nΦ

0 0.150349016 0.124281292 0.146532186 0

1 -0.016417239 0.009650483 -0.012600407 7.66193E-10

2 3.30335E-12 1.13173E-12 1.81747E-12 4.87108E-35

8.9 A wall has an incident solar radiation of 300 Btu/(hr-ft2), an outside air

temperature of 95 F, and an outside wind speed of 15 mph. The wall has a

solar absorptivity of 0.8, a thermal emissivity of 0.9, negligible thermal mass,

an outside-surface-to-inside-surface U-factor of 0.1 Btu/(hr-ft2-F), and an

inside surface temperature of 72 F. Determine the conduction heat flux for

each hour.

Given:

Incident solar radiation, t

G = 300 Btu/(hr-ft2)

Outside air temperature, o

t = 95 F

Inside surface temperature, θ,, jist = 72 F

Solar absorptivity = α = 0.8

Thermal emissivity = ε = 0.9

Outside-surface-to-inside-surface U-factor , U = 0.1 Btu/(hr-ft2-F)

Outside wind speed = 15 mph

Required:

Conduction heat flux for each hour = θ,,, joutconductionq ′′

Solution:

Eq. 8-24

8. The Cooling Load

skyrgrc

skyskyrggroctjis

joshhhU

thththGUtt

−−

−−

+++

++++=

αθ

θ

,,

,,

0tt

sky= - 10.8 F = 95 F – 10.8 F = 84.2 F

( )[ ] ( )[ ]0, 2

cos12

cos tttskysky

ααα −+=

90=α

( )[ ]( ) ( )[ ]( ) 36.87952

90cos12.842

90cos,

=−+=αskyt F

Assume θ,, jost = 150 F

Eq. 8-13. ( )[ ] [ ]223

1 b

otcaVtCh +∆=

tC = 0.096 Btu/(hr-ft

2-F

4/3)

a = 0.203 Btu/(hr-ft2-F-mph)

b = 0.89

( )[ ] ( )[ ] 28985723.215203.095150096.0289.0

23

1

=+−=c

h Btu/(hr-ft2-F)

Eq. 8-17 and 8-18.

( )

−

−=

−

−

θ

θεσ,,

4

,,

4

josg

josggs

grtt

ttFh

( )

−

−=

−

−

θ

θεσ,,

4

,,

4

jossky

josskyskys

skyrtt

ttFh

5.0==−− skysgs

FF 8101714.0 −×=σ Btu/(hr-ft

2-R

4)

67.55467.45995 =+=g

t

03.54767.45936.87 =+=sky

t

67.60967.459150,,

=+=θjost

( )( ) ( )

−

−×= −

−67.60967.554

67.60967.5545.0101714.09.0

44

8

grh

61009964.0=−gr

h Btu/(hr-ft2-F)

( )( ) ( )

−

−×= −

−67.60903.547

67.60903.5475.0101714.09.0

44

8

skyrh

59858705.0=−skyr

h Btu/(hr-ft2-F)

skyrgrc

skyskyrggroctjis

joshhhU

thththGUtt

−−

−−

+++

++++=

αθ

θ

,,

,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )59858705.061009964.028985723.21.0

36.8759858705.09561009964.09528985723.23008.0721.0,,

+++

++++=

θjost

783646.159,,

=θjost F

By further trial and error.

163625.159,,

=θjost F

8. The Cooling Load

( )[ ] ( )[ ] 293004.215203.095163625.159096.0289.0

23

1

=+−=c

h Btu/(hr-ft2-F)

833625.61867.459163625.159,,

=+=θjost

( )( ) ( )

−

−×= −

−833625.61867.554

833625.61867.5545.0101714.09.0

44

8

grh

625091.0=−gr

h Btu/(hr-ft2-F)

( )( ) ( )

−

−×= −

−833625.61803.547

833625.61803.5475.0101714.09.0

44

8

skyrh

613452.0=−skyr

h Btu/(hr-ft2-F)

skyrgrc

skyskyrggroctjis

joshhhU

thththGUtt

−−

−−

+++

++++=

αθ

θ

,,

,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )613452.0625091.0293004.21.0

36.87613452.095625091.095293004.23008.0721.0,,

+++

++++=

θjost

163625.159,,

=θjost F

( ) ( ) 716363.872163625.1591.0,,,,,,,

=−=−=′′θθθ jisjosjoutconduction

ttUq Btu/(hr-ft2)

( ) ( ) 127450.1471636625.1599529286725.2,,,,

−=−=−=′′θθ osocjoutconvection

tthq Btu/(hr-ft2)

( ) 2403008.0,,,

===′′tjoutsolar

Gq αθ Btu/(hr-ft2)

( ) ( )θθθ ,,,,,, josskyskyrjosggrjradiation

tthtthq −+−=′′−−

( ) ( )163625.15936.87613452.0163625.15995625091.0,,

−+−=′′θjradiation

q

156187.84,,

−=′′θjradiation

q Btu/(hr-ft2)

To check:

θθθθ ,,,,,,,,,,, jradiationjoutconvectionjoutsolarjoutconductionqqqq ′′+′′+′′=′′

156187.84127450.147240716363.8 −−=

716363.8716363.8 =

Therefore:

Conduction heat flux for each hour = 716363.8,,,

=′′θjoutconduction

q Btu/(hr-ft2)

8.10 Compute the solar irradiation for a west-facing wall in Amarillo, Texas, for

each hour of the day on July 21. Assume 0.4 percent outdoor design

conditions. The wall has a solar absorptivity of 0.8, a thermal emissivity of

0.9, negligible thermal mass, an outside-surface-to-inside-surface U-factor of

0.1 Btu/(hr-ft2-F), and an inside surface temperature of 72 F. Determine the

conduction heat flux for each hour.

Solution: Table B-1a, Amarillo, Texas, Latitude = 35.23 deg, Longitude = 101.70 deg

Dry bulb at 0.4 percent = 96 F, DR = 23.3 F, wind speed = 15 mph

From Table 6-1, July 21, d = 20.6 deg

A = 346.6 Btu/hr-ft2, B = 0.186, C = 0.138

l = 35.23 deg

8. The Cooling Load

For west-facing wall, ψ = 90 deg

Afternoon hours, walls facing west of south, || ψφγ −=

Morning hours, walls facing west of south, || ψφγ +=

Since ψ = 90 deg, γ > 90, it is in shade.

Therefore, from sunrise (6:00 A.M. LST) – noon (12:00 NOON LST)

DG = 0

wsNDddFCGGG == θ

and from noon (12:00 NOON LST) – sunset (6:00 P.M. LST)

θcosNDD

GG =

wsNDddFCGGG == θ

l = 35.23 deg, d = 20.6 deg

Then,

( )6.20sin23.35sin6.20coscos23.35cossin 1 += −hβ

−=

−= −−

23.35coscos

6.20sin23.35sinsincos

coscos

sinsinsincos 11

β

β

β

βφ

l

dl

|90||| −=−= φψφγ

For vertical surface

( )γβθ coscoscos 1−=

=

=

ββ sin186.0exp

6.346

sinexp B

AG

ND

Direct normal solar radiation

θcosNDD

GG =

Diffuse solar radiation

wsNDdFCGG =

2

cos1 ∑+=

wsF

α=∑ = 90 deg

2

90cos1+=

wsF = 0.5

( ) ( )NDNDwsNDd

GGFCGG 069.05.0138.0 ===

h :

h , morning = ( )[ ]( )1500:12 TIME−

h , afternoon = ( )( )15TIME

8. The Cooling Load

Spreadsheet (Solar Irradiation), neglecting energy reflected onto the surface.

TIME h β φ γ θ GND GD Gd Gt

6:00 A.M. 90 11.71 107.07 IN SHADE IN SHADE






12:00 NOON 0 75.37 0.00 90.00 90.00 285.98 0.00 19.73 19.73

1:00 P.M. 15 70.31 45.97 44.03 75.98 284.47 68.92 19.63 88.55

2:00 P.M. 30 59.90 68.94 21.06 62.09 279.55 130.84 19.29 150.13

3:00 P.M. 45 48.04 81.88 8.12 48.56 269.90 178.64 18.62 197.27

4:00 P.M. 60 35.82 91.23 1.23 35.84 252.24 204.48 17.40 221.88

5:00 P.M. 75 23.63 99.27 9.27 25.29 217.93 197.04 15.04 212.08

6:00 P.M. 90 11.71 107.07 17.07 20.60 138.62 129.76 9.56 139.32

For conduction heat flux for each hour = θ,,, joutconductionq ′′

( )XDRttdo

−=

Tabulation of Outside Temperature (Equation 8.2, and Table 81)

TIME X ot

6:00 A.M. 0.98 73.17

7:00 A.M. 0.93 74.33

8:00 A.M. 0.84 76.43

9:00 A.M. 0.71 79.46

10:00 A.M. 0.56 82.95

11:00 A.M. 0.39 86.91

12:00 NOON 0.23 90.64

1:00 P.M. 0.11 93.44

2:00 P.M. 0.03 95.30

3:00 P.M. 0.00 96.00

4:00 P.M. 0.03 95.30

5:00 P.M. 0.10 93.67

6:00 P.M. 0.21 91.11

For 4:00 P.M., 88.221=t

G Btu/(hr-ft2)

Eq. 8-24

skyrgrc

skyskyrggroctjis

joshhhU

thththGUtt

−−

−−

+++

++++=

αθ

θ

,,

,,

0tt

sky= - 10.8 F = 95.3 F – 10.8 F = 84.5 F

( )[ ] ( )[ ]0, 2

cos12

cos tttskysky

ααα −+=

90=α

( )[ ]( ) ( )[ ]( ) 66325.873.952

90cos15.842

90cos,

=−+=αskyt F


8. The Cooling Load

Eq. 8-13. ( )[ ] [ ]223

1 b

otcaVtCh +∆=


2-F

4/3)


b = 0.89

( )[ ] ( )[ ] 2897513.215203.03.95150096.0289.0

23

1

=+−=c

h Btu/(hr-ft2-F)

Eq. 8-17 and 8-18.

( )

−

−=

−

−

θ

θεσ,,

4

,,

4

josg

josggs

grtt

ttFh

( )

−

−=

−

−

θ

θεσ,,

4

,,

4

jossky

josskyskys

skyrtt

ttFh

5.0==−− skysgs

FF 8101714.0 −×=σ Btu/(hr-ft

2-R

4)

97.55467.4593.95 =+=g

t

33325.54767.45966325.87 =+=sky

t

67.60967.459150,,

=+=θjost

( )( ) ( )

−

−×= −

−67.60997.554

67.60997.5545.0101714.09.0

44

8

grh

61055587.0=−gr

h Btu/(hr-ft2-F)

( )( ) ( )

−

−×= −

−67.60933325.547

67.60933325.5475.0101714.09.0

44

8

skyrh

59904013.0=−skyr

h Btu/(hr-ft2-F)

skyrgrc

skyskyrggroctjis

joshhhU

thththGUtt

−−

−−

+++

++++=

αθ

θ

,,

,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )59904013.061055587.02897513.21.0

66325.8759904013.03.9561055587.03.952897513.288.2218.0721.0,,

+++

++++=

θjost

697282.142,,

=θjost F


04919.143,,

=θjost F

( )[ ] ( )[ ] 28723809.215203.03.9504919.143096.0289.0

23

1

=+−=c

h Btu/(hr-ft2-F)

71919.60267.45904919.143,,

=+=θjost

( )( ) ( )

−

−×= −

−71919.60297.554

71919.60297.5545.0101714.09.0

44

8

grh

59938718.0=−gr

h Btu/(hr-ft2-F)

( )( ) ( )

−

−×= −

−71919.60233325.547

71919.60233325.5475.0101714.09.0

44

8

skyrh

58796622.0=−skyr

h Btu/(hr-ft2-F)

8. The Cooling Load

skyrgrc

skyskyrggroctjis

joshhhU

thththGUtt

−−

−−

+++

++++=

αθ

θ

,,

,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )58796622.059938718.028723809.21.0

66325.8758796622.03.9559938718.03.9528723809.288.2218.0721.0,,

+++

++++=

θjost

04919.143,,

=θjost F

( ) ( ) 104919.77204919.1431.0,,,,,,,


ttUq Btu/(hr-ft2)

( ) ( ) 213766.10904919.1433.9528723809.2,,,,


tthq Btu/(hr-ft2)

( ) 504.17788.2218.0,,,

===′′tjoutsolar




( ) ( )04919.14366325.8758796622.004919.1433.9559938718.0,,


q

185316.61,,


q Btu/(hr-ft2)

To check:


185316.61213766.109504.177104919.7 −−=

104919.7104919.7 =

Therefore:



q Btu/(hr-ft2)

Tabulation of conduction heat flux for each hour:

TIME Gt θ,, jost

θ,,, joutsolarq ′′

θ,,, joutconvectionq ′′ θ,, jradiation

q ′′ θ,,, joutconductionq ′′

6:00 A.M. 72.08 0 2.477187 -2.469665 0.007522

7:00 A.M. 73.20 0 2.568034 -2.448524 0.119510

8:00 A.M. 75.22 0 2.731785 -2.409507 0.322278

9:00 A.M. 78.15 0 2.966431 -2.351515 0.615916

10:00 A.M. 81.52 0 3.234314 -2.282226 0.952088

11:00 A.M. 85.15 0 3.535165 -2.200360 1.334805

12:00 NOON 19.73 93.60 15.784 -6.707806 -6.916013 2.160181

1:00 P.M. 88.55 112.10 70.84 -42.438980 -24.391456 4.009564

2:00 P.M. 150.13 127.53 120.104 -73.530403 -41.020163 5.553434

3:00 P.M. 197.27 138.41 157.816 -96.91230 -54.262842 6.640828

4:00 P.M. 221.88 143.05 177.504 -109.213766 -61.185316 7.104919

5:00 P.M. 212.08 139.50 169.664 -104.799653 -58.113987 6.750360

6:00 P.M. 139.32 121.31 111.456 -68.872965 -37.651621 4.931413

8.11 Compute the solar irradiation for a south-facing wall in Billings, Montana, for


conditions. The wall has a solar absorptivity of 0.9, a thermal emissivity of

0.9, negligible thermal mass, and outside-surface-to-inside-surface U-factor of

0.1 Btu/(hr-ft2-F), and an inside surface temperature of 72 F. Determine the

conduction heat flux for each hour.

8. The Cooling Load

Solution: Table B-1a, Billings, Montana, Latitude = 45.80 deg, Longitude = 108.53

deg

Dry bulb at 0.4 percent = 93 F, , DR = 25.8 F, wind speed = 10 mph


A = 346.6 Btu/hr-ft2, B = 0.186, C = 0.138

l = 45.80 deg

For south-facing wall, ψ = 0 deg

Afternoon hours, walls facing west of south, φγ =

Morning hours, walls facing west of south, φγ =

Therefore,

θcosNDD

GG = = 0

wsNDddFCGGG == θ

l = 45.80 deg, d = 20.6 deg

Then,


−=

−= −−

80.45coscos

6.20sin80.45sinsincos

coscos

sinsinsincos 11

β

β

β

βφ

l

dl

φγ =

For vertical surface

( )γβθ coscoscos 1−=

=

=

ββ sin186.0exp

6.346

sinexp B

AG

ND

Direct normal solar radiation

θcosNDD

GG =

Diffuse solar radiation

wsNDdFCGG =

2

cos1 ∑+=

wsF

α=∑ = 90 deg

2

90cos1+=

wsF = 0.5

( ) ( )NDNDwsNDd

GGFCGG 069.05.0138.0 ===

h :

h , morning = ( )[ ]( )1500:12 TIME−

h , afternoon = ( )( )15TIME

8. The Cooling Load

Spreadsheet (Solar Irradiation), neglecting energy reflected onto the surface.

TIME h β φ γ θ GND GD Gd Gt

6:00 A.M. 90 14.61 104.68 14.68 20.60 165.80 155.20 11.44 166.64

7:00 A.M. 75 24.91 94.53 4.53 25.29 222.85 201.50 15.38 216.87

8:00 A.M. 60 35.35 83.65 6.35 35.84 251.31 203.72 17.34 221.06

9:00 A.M. 45 45.54 70.90 19.10 48.56 267.08 176.78 18.43 195.21

10:00 A.M. 30 54.83 54.34 35.66 62.09 276.06 129.20 19.05 148.25

11:00 A.M. 15 61.96 31.02 58.98 75.98 280.74 68.01 19.37 87.39

12:00 NOON 0 64.80 0.00 90.00 90.00 282.20 0.00 19.47 19.47

1:00 P.M. 15 61.96 31.02 58.98 75.98 280.74 68.01 19.37 87.39

2:00 P.M. 30 54.83 54.34 35.66 62.09 276.06 129.20 19.05 148.25

3:00 P.M. 45 45.54 70.90 19.10 48.56 267.08 176.78 18.43 195.21

4:00 P.M. 60 35.35 83.65 6.35 35.84 251.31 203.72 17.34 221.06

5:00 P.M. 75 24.91 94.53 4.53 25.29 222.85 201.50 15.38 216.87

6:00 P.M. 90 14.61 104.68 14.68 20.60 165.80 155.20 11.44 166.64

For conduction heat flux for each hour = θ,,, joutconductionq ′′

( )XDRttdo

−=

Tabulation of Outside Temperature (Equation 8.2, and Table 81

TIME X ot

6:00 A.M. 0.98 67.72

7:00 A.M. 0.93 69.01

8:00 A.M. 0.84 71.33

9:00 A.M. 0.71 74.68

10:00 A.M. 0.56 78.55

11:00 A.M. 0.39 82.94

12:00 NOON 0.23 87.07

1:00 P.M. 0.11 90.16

2:00 P.M. 0.03 92.23

3:00 P.M. 0.00 93.00

4:00 P.M. 0.03 92.23

5:00 P.M. 0.10 90.42

6:00 P.M. 0.21 87.58

.

For 4:00 P.M., 06.221=t

G Btu/(hr-ft2)

Eq. 8-24

skyrgrc

skyskyrggroctjis

joshhhU

thththGUtt

−−

−−

+++

++++=

αθ

θ

,,

,,

0tt

sky= - 10.8 F = 92.23 F – 10.8 F = 81.43 F

( )[ ] ( )[ ]0, 2

cos12

cos tttskysky

ααα −+=

90=α

8. The Cooling Load

( )[ ]( ) ( )[ ]( ) 59325.8423.922

90cos143.812

90cos,

=−+=αskyt F


Eq. 8-13. ( )[ ] [ ]223

1 b

otcaVtCh +∆=


2-F

4/3)


b = 0.89

( )[ ] ( )[ ] 61889226.110203.023.92150096.0289.0

23

1

=+−=c

h Btu/(hr-ft2-F)

Eq. 8-17 and 8-18.

( )

−

−=

−

−

θ

θεσ,,

4

,,

4

josg

josggs

grtt

ttFh

( )

−

−=

−

−

θ

θεσ,,

4

,,

4

jossky

josskyskys

skyrtt

ttFh

5.0==−− skysgs

FF

8101714.0 −×=σ Btu/(hr-ft2-R

4)

9.55167.45923.92 =+=g

t

26325.54467.45959325.84 =+=sky

t

67.60967.459150,,

=+=θjost

( )( ) ( )

−

−×= −

−67.6099.551

67.6099.5515.0101714.09.0

44

8

grh

60590202.0=−gr

h Btu/(hr-ft2-F)

( )( ) ( )

−

−×= −

−67.60926325.544

67.60926325.5445.0101714.09.0

44

8

skyrh

59446797.0=−skyr

h Btu/(hr-ft2-F)

skyrgrc

skyskyrggroctjis

joshhhU

thththGUtt

−−

−−

+++

++++=

αθ

θ

,,

,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )59446797.060590202.061889226.11.0

59325.8459446797.023.9260590202.023.9261889226.106.2219.0721.0,,

+++

++++=

θjost

134046.158,,

=θjost F


480121.157,,

=θjost F

( )[ ] ( )[ ] 62248507.110203.023.92480121.157096.0289.0

23

1

=+−=c

h Btu/(hr-ft2-F)

150121.61767.459480121.157,,

=+=θjost

8. The Cooling Load

( )( ) ( )

−

−×= −

−150121.6179.551

150121.6179.5515.0101714.09.0

44

8

grh

61807841.0=−gr

h Btu/(hr-ft2-F)

( )( ) ( )

−

−×= −

−150121.61726325.544

150121.61726325.5445.0101714.09.0

44

8

skyrh

60654201.0=−skyr

h Btu/(hr-ft2-F)

skyrgrc

skyskyrggroctjis

joshhhU

thththGUtt

−−

−−

+++

++++=

αθ

θ

,,

,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )60654201.061807841.062248507.11.0

59325.8460654201.023.9261807841.023.9262248507.106.2219.0721.0,,

+++

++++=

θjost

480121.157,,

=θjost F

( ) ( ) 548012.872480121.1571.0,,,,,,,


ttUq Btu/(hr-ft2)

( ) ( ) 867347.105480121.15723.9262248507.1,,,,


tthq Btu/(hr-

ft2)

( ) 954.19806.2219.0,,,

===′′tjoutsolar




( ) ( )480121.15759325.8460654201.0480121.15723.9261807841.0,,


q

538642.84,,


q Btu/(hr-ft2)

To check:


538642.84867345.105954.198548012.8 −−=

548012.8548012.8 =

Therefore:



q Btu/(hr-ft2)

Tabulation of conduction heat flux for each hour:

TIME Gt θ,, jost

θ,,, joutsolarq ′′

θ,,, joutconvectionq ′′ θ,, jradiation

q ′′ θ,,, joutconductionq ′′

6:00 A.M. 166.64 120.84 149.976 -85.868752 -59.223479 4.883770

7:00 A.M. 216.87 136.96 195.183 -110.328147 -78.359179 6.495674

8:00 A.M. 221.06 140.11 198.954 -111.704987 -80.438172 6.810840

9:00 A.M. 195.21 135.33 175.689 -98.265984 -71.090310 6.332707

10:00 A.M. 148.25 124.59 133.425 -74.261154 -53.904671 5.259174

11:00 A.M. 87.39 109.68 78.651 -42.827019 -32.056161 3.767820

12:00 NOON 19.47 91.60 17.523 -7.170045 -8.393234 1.959722

1:00 P.M. 87.39 116.21 78.651 -41.715251 -32.514620 4.421129

2:00 P.M. 148.25 136.50 133.425 -71.354780 -55.620633 6.449588

3:00 P.M. 195.21 150.80 175.689 -93.580228 -74.228323 7.880448

4:00 P.M. 221.06 157.48 198.954 -105.867345 -84.538642 8.548012

5:00 P.M. 216.87 154.79 195.183 -104.412162 -82.491867 8.278971

6:00 P.M. 166.64 137.89 149.976 -81.265660 -62.120863 6.589477

8. The Cooling Load

8.12 For the wall described in Problem 8-4, with an outside surface temperature

profile given by Table 8-4 and a constant inside surface temperature of 70 F,

determine the inside conduction heat flux for each hour.

Solution:

∑ ′′Φ+∑++∑−−=′′=

−=

−=

−

qN

nnjinconductionn

yN

nnjosnjos

zN

nnjisnjisjinconduction

qtYtYtZtZq1

,,,1

,,,,01

,,,,0,,, δθδθθδθθθ

Table 8-4 Exterior Surface Temperatures for Example 8-2

Hour θ,, jost , F Hour θ,, jos

t , F

1 79.55 13 106.15

2 77.80 14 108.95

3 76.40 15 110.00

4 75.35 16 108.95

5 75.00 17 106.50

6 75.70 18 102.65

7 77.45 19 98.10

8 80.60 20 93.55

9 85.15 21 89.70

10 90.40 22 86.20

11 96.35 23 83.40

12 101.95 24 81.30


n nX ,

Btu/(h-ft2-F)

nY ,

Btu/(h-ft2-F)

nZ ,

Btu/(h-ft2-F)

nΦ

0 4.033132292 0.001517739 0.623505633 0

1 -4.890102268 0.017652429 -0.943189491 0.643204868

2 0.893455637 0.010001789 0.358387909 -0.017086206

3 -0.006945024 0.000370798 -0.009166607 1.3166E-05

4 2.65956E-06 6.21048E-07 5.90198E-06 -1.31644E-09

5 -1.59894E-10 4.86714E-11 -2.45933E-10 1.12031E-14

For the first hour,

( )( )

( )20,,,521,,,422,,,323,,,224,,,1

20,,521,,422,,323,,224,,11,,0

20,,521,,422,,323,,224,,11,,01,,,

jinconductionjinconductionjinconductionjinconductionjinconduction

josjosjosjosjosjos

jisjisjisjisjisjisjinconduction

qqqqq

tYtYtYtYtYtY

tZtZtZtZtZtZq

′′Φ+′′Φ+′′Φ+′′Φ+′′Φ+

++++++

++++−−=′′

8. The Cooling Load

( )( )( )( ) ( )( )

( )( ) ( )( )( )( )

( )( )( )( ) ( )( )

( )( ) ( )( )( )( )

( )( ) ( )( ) ( )( )( )( ) ( )( )

354012.001012031.101031644.1

0103166.10017086206.00643204868.0

55.931086714.4

70.891021048.620.86000370798.0

4.83010001789.030.81017652429.0

55.79001517739.0

701045933.2

701090198.570009166607.0

70358387909.070943189491.0

70623505633.0

149

5

11

7

10

6

1,,,

=

×+×−+

×+−++

×+

×++

+

++

×+

×+−+

+−

−−=′′

−−

−

−

−

−

−

jinconductionq

Repeating the process for four days. See Table.

Table. Interior Surface Heat Flux for Exercise 8.12.

Heat Flux, (Btu/(h-ft2)

Hours Day 1 Day 2 Day 3 Day 4

1 0.354012 1.327263 1.327271 1.327271

2 0.526122 1.125464 1.12547 1.12547

3 0.579476 0.948367 0.94837 0.94837

4 0.566396 0.793441 0.793443 0.793443

5 0.523356 0.663097 0.663098 0.663098

6 0.479758 0.565766 0.565766 0.565766

7 0.463573 0.516509 0.516509 0.516509

8 0.49645 0.529031 0.529031 0.529031

9 0.598146 0.618199 0.618199 0.618199

10 0.783437 0.79578 0.79578 0.79578

11 1.049264 1.05686 1.05686 1.05686

12 1.384809 1.389485 1.389485 1.389485

13 1.762783 1.76566 1.76566 1.76566

14 2.136777 2.138548 2.138548 2.138548

15 2.465986 2.467076 2.467076 2.467076

16 2.717857 2.718528 2.718528 2.718528

17 2.863531 2.863944 2.863944 2.863944

18 2.893727 2.893982 2.893982 2.893982

19 2.810904 2.81106 2.81106 2.81106

20 2.630477 2.630573 2.630573 2.630573

21 2.382741 2.382801 2.382801 2.382801

22 2.106007 2.106044 2.106044 2.106044

23 1.826011 1.826033 1.826033 1.826033

24 1.56159 1.561604 1.561604 1.561604

8.13 For the wall described in Problem 8-5, with an outside surface temperature

profile given by Table 8-4 and a constant inside surface temperature of 70 F,

determine the inside conduction heat flux for each hour.

Solution:

8. The Cooling Load

∑ ′′Φ+∑++∑−−=′′=

−=

−=

−

qN

nnjinconductionn

yN

nnjosnjos

zN

nnjisnjisjinconduction

qtYtYtZtZq1

,,,1

,,,,01

,,,,0,,, δθδθθδθθθ

Table 8-4 Exterior Surface Temperatures for Example 8-2

Hour θ,, jost , F Hour θ,, jos

t , F

1 79.55 13 106.15

2 77.80 14 108.95

3 76.40 15 110.00

4 75.35 16 108.95

5 75.00 17 106.50

6 75.70 18 102.65

7 77.45 19 98.10

8 80.60 20 93.55

9 85.15 21 89.70

10 90.40 22 86.20

11 96.35 23 83.40

12 101.95 24 81.30


n nX ,

Btu/(h-ft2-F)

nY ,

Btu/(h-ft2-F)

nZ ,

Btu/(h-ft2-F)

nΦ

0 4.033132964 0.000387523 0.611908143 0

1 -4.978563054 0.008315152 -0.968315133 0.66508466

2 0.982856227 0.008091923 0.388562266 -0.02728007

3 -0.019978162 0.000729205 -0.014736957 0.000228003

4 8.14087E-05 5.69043E-06 0.000111178 -9.35628E-08

5 -1.44779E-08 2.93347E-09 -3.98891E-08 4.13516E-12

6 4.33848E-13 8.69125E-14 6.47835E-13 -2.32896E-17

For the first hour,

( )( )

′′Φ+′′Φ+

′′Φ+′′Φ+′′Φ+′′Φ+

+++++++

+++++−−=′′

19,,,620,,,5

21,,,422,,,323,,,224,,,1

19,,620,,521,,422,,323,,224,,11,,0

19,,620,,521,,422,,323,,224,,11,,01,,,

jinconductionjinconduction

jinconductionjinconductionjinconductionjinconduction

josjosjosjosjosjosjos

jisjisjisjisjisjisjisjinconduction

qq

qqqq

tYtYtYtYtYtYtY

tZtZtZtZtZtZtZq

( )( )( )( ) ( )( )

( )( ) ( )( )( )( ) ( )( )

( )( )( )( ) ( )( )

( )( ) ( )( )( )( ) ( )( )

( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )

218022.001032896.201013516.401035628.9

0000228003.0002728007.0066508466.0

10.981069125.855.931093347.2

70.891069043.520.86000729205.0

4.83008091923.030.81008315152.0

55.79000387523.0

701047835.6701098891.3

70000111178.070014736957.0

70388562266.070968315133.0

70611908143.0

17128

149

6

138

1,,,

=

×−+×+×−+

+−++

×+×+

×++

+

++

×+×−+

+−+

+−

−−=′′

−−−

−−

−

−−

jinconductionq

8. The Cooling Load

Repeating the process for four days. See Table.

Table. Interior Surface Heat Flux for Exercise 8.12.

Heat Flux, (Btu/(h-ft2)

Hours Day 1 Day 2 Day 3 Day 4

1 0.218022 0.851033 0.85104 0.85104

2 0.32874 0.722758 0.722762 0.722762

3 0.365628 0.610642 0.610645 0.610645

4 0.359693 0.512044 0.512046 0.512046

5 0.333284 0.428016 0.428017 0.428017

6 0.303724 0.362628 0.362629 0.362629

7 0.287675 0.324302 0.324303 0.324303

8 0.298978 0.321752 0.321753 0.321753

9 0.349551 0.363712 0.363712 0.363712

10 0.449513 0.458318 0.458318 0.458318

11 0.599704 0.605179 0.605179 0.605179

12 0.794342 0.797747 0.797747 0.797747

13 1.019912 1.022029 1.022029 1.022029

14 1.250352 1.251669 1.251669 1.251669

15 1.459298 1.460116 1.460116 1.460116

16 1.626105 1.626614 1.626614 1.626614

17 1.732281 1.732597 1.732597 1.732597

18 1.768815 1.769011 1.769011 1.769011

19 1.735893 1.736015 1.736015 1.736015

20 1.640481 1.640557 1.640557 1.640557

21 1.498965 1.499012 1.499012 1.499012

22 1.333912 1.333942 1.333942 1.333942

23 1.163292 1.16331 1.16331 1.16331

24 0.999033 0.999045 0.999045 0.999045

8.14 On a warm sunny day, the metal surface of the roof of a car can become quite

hot. If the roof of the car has 310 Btu/(hr-ft2) total solar radiation incident on

it, the outdoor air temperature is 90 F, and the windspeed is 6 mph, estimate

the maximum possible surface temperature. Assume the solar absorptivity and

thermal emissivity are both 0.9.

Solution:

tG = 310 Btu/(hr-ft

2)

oV = 6 mph

αε = = 0.90


For metal surface, θθ ,,,, jisjostt =

Then

( ) 0,,,,,,,

=−=′′θθθ jisjosjoutconduction

ttUq

( ) 2793109.0,,,

===′′tjoutsolar




8. The Cooling Load

0=−gr

h , because the horizontal roof has no view to the ground.

( )θθ ,,,, josskyskyrjradiation

tthq −=′′−

By trial and error, try =θ,, jost 150 F

0tt

sky= - 10.8 F = 90 F – 10.8 F = 79.2 F

Eq. 8-13. ( )[ ] [ ]223

1 b

otcaVtCh +∆=


2-F

4/3)


b = 0.89

( )[ ] ( )[ ] 06840041.16203.090150096.0289.0

23

1

=+−=c

h Btu/(hr-ft2-F)

( ) ( ) 1040246.641509006840041.1,,,,


tthq Btu/(hr-ft2)

( )

−

−=

−

−

θ

θεσ,,

4

,,

4

jossky

josskyskys

skyrtt

ttFh

1=−skys

F

( )( ) ( )

−

−×= −

−67.60987.538

67.60987.5380.1101714.09.0

44

8

skyrh

17302931.1=−skyr

h Btu/(hr-ft2-F)

( ) ( ) 05047482.831502.7917302931.1,,,,

−=−=−=′′− θθ josskyskyrjradiation

tthq Btu/(hr-ft2)


8455006.13105047482.831040246.64279,,,

=−−=′′θjoutconduction

q > 0

Try again, so that 0,,,


q

=θ,, jost 198.588261 F

0tt

sky= - 10.8 F = 90 F – 10.8 F = 79.2 F

( )[ ] ( )[ ] 09999879.16203.090588261.198096.0289.0

23

1

=+−=c

h Btu/(hr-ft2-F)

( ) ( ) 4469552.119588261.1989009999879.1,,,,


tthq Btu/(hr-ft2)

( )

−

−=

−

−

θ

θεσ,,

4

,,

4

jossky

josskyskys

skyrtt

ttFh

( )( ) ( )

−

−×= −

−258261.65887.538

258261.65887.5380.1101714.09.0

44

8

skyrh

33642155.1=−skyr

h Btu/(hr-ft2-F)

( ) ( ) 553045.159588261.1982.7933642155.1,,,,

−=−=−=′′− θθ josskyskyrjradiation

tthq Btu/(hr-

ft2)


000000.0553045.1594469552.119279,,,

=−−=′′θjoutconduction

q

Therefore

8. The Cooling Load

=θ,, jost 198.59 F

8.15 Determine the transmitted direct and diffuse solar radiation through a 100 ft2

double-pane window with shading coefficient 0.8 for 3 P.M. on July 21 in

Amarillo, Texas.

Solution:

For Amarillo, Texas, 3 P.M. on July 21, refer to Prob. 8-10.

64.178=D

G Btu/(hr-ft2), 62.18=

dG Btu/(hr-ft

2), 56.48=θ deg

Use Equation with a shading coefficient.

( ) [ ]j

jjDDirect

tGSCTSHG ∑==

5

0

cosθ

or ( ) [ ]j

jjDSLDirect

tGSCAq ∑==

5

0

cosθ&

( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCTSHG

or ( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCAq&

Table 6-2: j

jt

0 -0.00885

1 2.71235

2 -0.62062

3 -7.07329

4 9.75995

5 -3.89922

Transmitted direct solar radiation:

( )( )( )

( ) ( ) ( )

−+−

−+−=

543

2

56.48cos89922.356.48cos75995.956.48cos07329.7

56.48cos62062.056.48cos71235.200885.064.1788.0100

Directq&

87.023,12=Direct

q& Btu/hr

Transmitted diffuse solar radiation:

( )( )( )

−+−−+−=

7

89922.3

6

75995.9

5

07329.7

4

62062.0

3

71235.2

2

00885.062.1828.0100

diffuseq&

207.1190=diffuse

q& Btu/hr



Billings, Montana .

Solution:

For Billings, Montana, 3 P.M. on July 21, refer to Prob. 8-11.

78.176=D

G Btu/(hr-ft2), 43.18=

dG Btu/(hr-ft

2), 56.48=θ deg

8. The Cooling Load


( ) [ ]j

jjDDirect

tGSCTSHG ∑==

5

0

cosθ

or ( ) [ ]j

jjDSLDirect

tGSCAq ∑==

5

0

cosθ&

( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCTSHG

or ( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCAq&

Table 6-2: j

jt

0 -0.00885

1 2.71235

2 -0.62062

3 -7.07329

4 9.75995

5 -3.89922


( )( )( )

( ) ( ) ( )

−+−

−+−=

543

2

56.48cos89922.356.48cos75995.956.48cos07329.7

56.48cos62062.056.48cos71235.200885.078.1768.0100

Directq&

68.898,11=Direct

q& Btu/hr


( )( )( )

−+−−+−=

7

89922.3

6

75995.9

5

07329.7

4

62062.0

3

71235.2

2

00885.043.1828.0100

diffuseq&

062.1178=diffuse

q& Btu/hr

8.17 A large office space has an average occupancy of 20 people from 8:00 A.M. to

5:00 P.M. Lighting is 2.5 W/ft2 of recessed, unvented fluorescent fixtures on

from 8:00 A.M. to 6:00 P.M. Computers, photocopiers, fax machines, etc.

create a heat gain of 1 W/ft2. Compute the sensible and latent heat gain at 4:00

P.M. for the space, assuming a floor area of 4000 ft2. For the sensible heat

gain, estimate the radiative and convective portions.

Solution:

For 20 people occupant:

Sensible Heat = 20(73) = 1460 W

Latent Heat = 20(59) = 1180 W

Lighting Load:

suFWFq =&

W = 2.5(4000) = 10,000 W

0.1=u

F

8. The Cooling Load

2.1=s

F

( )( )( ) 000,122.10.1000,10 ==q& W

Miscellaneous Equipment:

ulFCPFq =&

C = 1.0 W/W

0.1=l

F

0.1=u

F

)4000)(0.1(=P = 4000 W

( )( )( )( )0.10.140000.1=q& = 4000 W

Tabulation: (Sensible and latent heat gain)

Sensible Heat Gain, W Latent Heat Gain, W

Occupants 1460 1180

Lighting 12,000 0

Equipment 4,000 0

Total 17,460 1180

Tabulation: (Radiative and convective portions of sensible heat gain) Ref. to Table 8-

20.

Radiative Convective

Percentage Watts Percentage Watts

Occupants 70 1022 30 438

Lighting 59 7080 41 4920

Equipment 70 2800 30 1200

Total 10,902 6,558

8.18 A space has an occupancy of 40 people engaged in sedentary activity form

8:00 A.M. to 5:00 P.M. The average light level is 28 W/m2 of vented

fluorescent fixtures with a ceiling plenum return. Office equipment amounts to

5 kW. Estimate the sensible and latent heat gain to the space for a floor area of

750 m2 at 4:00 P.M. For the sensible heat gain, estimate the radiative and

convective portions.

Solution:



Latent Heat = 40(59) = 2360 W

Lighting Load:

suFWFq =&

W = 28(750) = 21,000 W

0.1=u

F

5.1=s

F

( )( )( ) 500,315.10.1000,21 ==q& W

8. The Cooling Load


ulFCPFq =&

C = 1.0 W/W

0.1=l

F

0.1=u

F

P = 5000 W

( )( )( )( )0.10.150000.1=q& = 5000 W



Occupants 2920 2360

Lighting 21,000 0

Equipment 5,000 0

Total 28,920 2360


20.



Occupants 70 2044 30 876

Lighting 59 12,390 41 8610

Equipment 70 3500 30 1500

Total 17,934 10,986

8.19 A room has 5000 W of vented fluorescent light fixtures on form 6:00 A.M. to

6:00 P.M. The air flows from the lights through a ducted return. Compute the

heat gain to space at 5:00 P.M. assuming that 20 percent of heat from the

lights is convected to the return air.

Solution:

Lighting Load:

suFWFq =&

W = 5,000 W

0.1=u

F

2.1=s

F

( )( )( ) 60002.10.15000 ==q& W

8.20 A large office complex has a variable occupancy pattern. Twenty people arrive

at 8:00 A.M. and leave at 4:00 P.M. Forty people arrive at 10:00 A.M. and

leave at 4:00 P.M. Ten people arrive at 1:00 P.M. and leave at 5:00 P.M.

Assume seated, light activity, and compute the sensible and latent heat gains at

4:00 P.M. and 6:00 P.M.

Solution:

At 4:00 P.M. Total Number of Occupants = 20 + 40 +10 = 70

8. The Cooling Load



Latent Heat = 70(59) = 4130 W

At 6:00 P.M. Total Number of Occupants = 0

Sensible Heat = 0 W

Latent Heat = 0 W

8.21 The attic space shown in Fig. 8-10 has H = 3 ft, W = 24 ft, L = 36 ft, and all

interior surfaces have emissivities of 0.9. For a time when the inside surface

temperatures are Ft 1221

= , Ft 1412

= , Ft 1003

= , Ft 924

= and Ft 955

= ,

estimate the net thermal radiation incident on each surface using the

MRT/balance method.

Solution:

( ) 3.44512336 22

1=+=A ft

2

( ) 3.44512336 22

2=+=A ft

2

( )( )( ) 362432

13

==A ft2

( )( )( ) 362432

14

==A ft2

( )( ) 86436245

==A ft2

Zone Surface Description

Surface Name Area, ft2 T, F

1 North-facing

pitched roof 445.3 122

2 South-facing


3 West-facing end

wall 36 100

4 East-facing end

wall 36 92

5 Floor 864 95

8. The Cooling Load

( )∑ −==

N

iijijf

AA1

,1 δ

9.0,

===ijjf

εεε

( )

( )

( )

( )∑ −

∑ −=

∑ −

∑ −=

=

=

=

=

N

iiji

N

iijii

N

iijii

N

iijiii

jf

A

tA

A

tAt

1

1

1

1

,

1

1

1

1

δ

δ

δε

δε

Surface 1. 3.445=j

A ft2, 3.138186436363.445 =+++=

fA ft

2

( )( ) ( )( ) ( )( ) ( )( )86436363.445

958649236100361413.445,

+++

+++=

jft

88.109,

=jf

t F

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

9.0

9.01

3.1381

3.4451

9.0

9.01

1, −

++

−=

fjF

8719.0,

=fj

F

( )( ) 61.57567.45988.1091222

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 1394.161.5758719.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)

( )jfjjrjinsurfradiation

tthq,,,,,

−=′′− θ

( )( ) 8095.1388.1091221394.1,1,,,,,

=−=′′=′′−− θθ insurfradiationjinsurfradiation

qq Btu/(hr-ft2)

Surface 2. 3.445=j

A ft2, 3.138186436363.445 =+++=

fA ft

2

( )( ) ( )( ) ( )( ) ( )( )86436363.445

958649236100361223.445,

+++

+++=

jft

76.103,

=jf

t F

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

9.0

9.01

3.1381

3.4451

9.0

9.01

1, −

++

−=

fjF

8719.0,

=fj

F

( )( ) 05.58267.45976.1031412

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 1781.105.5828719.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)


tthq,,,,,

−=′′− θ

8. The Cooling Load

( )( ) 8724.4376.1031411781.1,2,,,,,


qq Btu/(hr-ft2)

Surface 3. 36=j

A ft2, 6.1790864363.4453.445 =+++=

fA ft

2

( )( ) ( )( ) ( )( ) ( )( )864363.4453.445

9586492361413.4451223.445,

+++

+++=

jft

09.113,

=jf

t F

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

9.0

9.01

6.1790

361

9.0

9.01

1, −

++

−=

fjF

8982.0,

=fj

F

( )( ) 22.56667.45909.1131002

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 1172.122.5668982.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)


tthq,,,,,

−=′′− θ

( )( ) 6241.1409.1131001172.1,3,,,,,

−=−=′′=′′−− θθ insurfradiationjinsurfradiation

qq Btu/(hr-ft2)

Surface 4. 36=j

A ft2, 6.1790864363.4453.445 =+++=

fA ft

2

( )( ) ( )( ) ( )( ) ( )( )864363.4453.445

95864100361413.4451223.445,

+++

+++=

jft

25.113,

=jf

t F

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

9.0

9.01

6.1790

361

9.0

9.01

1, −

++

−=

fjF

8982.0,

=fj

F

( )( ) 30.56267.45925.113922

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 0942.130.5628982.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)


tthq,,,,,

−=′′− θ

( )( ) 2518.2325.113920942.1,4,,,,,


qq Btu/(hr-ft2)

Surface 5. 864=j

A ft2, 6.96236363.4453.445 =+++=

fA ft

2

8. The Cooling Load

( )( ) ( )( ) ( )( ) ( )( )36363.4453.445

9236100361413.4451223.445,

+++

+++=

jft

84.128,

=jf

t F

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

9.0

9.01

6.962

8641

9.0

9.01

1, −

++

−=

fjF

8259.0,

=fj

F

( )( ) 59.57167.45984.128952

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 0568.159.5718259.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)


tthq,,,,,

−=′′− θ

( )( ) 7621.3584.128950568.1,5,,,,,


qq Btu/(hr-ft2)

Intermediate Variable for MRT/Balance Calculation

Surface fA , ft

2

fε

ft , F

fjF

,

avgjt

,, R

jrh

,, Btu/(hr-ft

2-F)

1 445.3 0.9 109.88 0.8719 575.61 1.1394

2 445.3 0.9 103.76 0.8719 582.05 1.1781

3 36 0.9 113.09 0.8982 566.22 1.1172

4 36 0.9 113.25 0.8982 562.30 1.0942

5 864 0.9 128.84 0.8259 571.59 1.0568

( )

∑

∑ −=′′

=

=

N

jj

N

jifjjrj

balance

A

tthA

q

1

1,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )86436363.4453.445

8647621.35362518.23366241.143.4458724.433.4458095.13

++++

−+−+−++=′′

balanceq

60026.36.1826

24.6576−=

−=′′

balanceq

MRT/Balance Calculation:

Surface

Initial

Estimate,

θ,,, jinsurfradiationq

−′′ ,

Btu/(hr-ft2)

Initial

Estimate,


−& ,

Btu/hr


−′′

with Balance,

Btu/(hr-ft2)


−&

with Balance,

Btu/hr

1 13.8095 6149.37 17.4098 7752.57

2 43.8724 19536.38 47.4727 21139.58

3 -14.6241 -526.47 -11.0238 -396.86

4 -23.2518 -837.06 -19.6515 -707.46

5 -35.7621 -30898.50 -32.1618 -27787.80

8. The Cooling Load

Sum -6576.24 0.00

8.22 The attic space shown in Fig. 8-10 has H = 2 m, W = 10 m, L = 20 m, and all

interior surfaces have emissivities of 0.9. For a time when the inside surface

temperatures are Ct 401

= , Ct 482

= , Ct 353

= , Ct 384

= and Ft 325

= ,

estimate the net thermal radiation incident on each surface using the

MRT/balance method.

Solution:

( ) 96.20310220 22

1=+=A m

2

( ) 96.20310220 22

2=+=A m

2

( )( )( ) 101022

13

==A m2

( )( )( ) 101022

14

==A m2

( )( ) 20020105

==A m2


Surface Name Area, m2 T, C

1 North-facing


2 South-facing


3 West-facing end

wall 10 35

4 East-facing end

wall 10 38

5 Floor 200 32

( )∑ −==

N

iijijf

AA1

,1 δ

9.0,

===ijjf

εεε

( )

( )

( )

( )∑ −

∑ −=

∑ −

∑ −=

=

=

=

=

N

iiji

N

iijii

N

iijii

N

iijiii

jf

A

tA

A

tAt

1

1

1

1

,

1

1

1

1

δ

δ

δε

δε

Surface 1. 96.203=j

A m2, 96.423200101096.203 =+++=

fA m

2

( )( ) ( )( ) ( )( ) ( )( )200101096.203

32200381035104896.203,

+++

+++=

jft

91.39,

=jf

t C

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

8. The Cooling Load

9.0

9.01

96.423

96.2031

9.0

9.01

1, −

++

−=

fjF

8587.0,

=fj

F

( )( ) 11.31315.27391.39402

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 9783.511.3138587.01067.5438

,=×= −

jrh W/(m

2-C)


tthq,,,,,

−=′′− θ

( )( ) 5380.091.39409783.5,1,,,,,


qq W/m2.

Surface 2. 96.203=j

A m2, 96.423200101096.203 =+++=

fA m

2

( )( ) ( )( ) ( )( ) ( )( )200101096.203

32200381035104096.203,

+++

+++=

jft

06.36,

=jf

t C

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

9.0

9.01

96.423

96.2031

9.0

9.01

1, −

++

−=

fjF

8587.0,

=fj

F

( )( ) 18.31515.27306.36482

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 0976.618.3158587.01067.5438

,=×= −

jrh W/(m

2-C)


tthq,,,,,

−=′′− θ

( )( ) 8054.7206.36480976.6,2,,,,,


qq W/m2.

Surface 3. 10=j

A m2, 92.6172001096.20396.203 =+++=

fA m

2

( )( ) ( )( ) ( )( ) ( )( )2001096.20396.203

3220038104896.2034096.203,

+++

+++=

jft

02.40,

=jf

t C

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

9.0

9.01

92.617

101

9.0

9.01

1, −

++

−=

fjF

8985.0,

=fj

F

8. The Cooling Load

( )( ) 66.31015.27302.40352

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 1097.666.3108985.01067.5438

,=×= −

jrh W/(m

2-C)


tthq,,,,,

−=′′− θ

( )( ) 6707.3002.40351097.6,3,,,,,


qq W/m2.

Surface 4. 10=j

A m2, 92.6172001096.20396.203 =+++=

fA m

2

( )( ) ( )( ) ( )( ) ( )( )2001096.20396.203

3220035104896.2034096.203,

+++

+++=

jft

97.39,

=jf

t C

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

9.0

9.01

92.617

101

9.0

9.01

1, −

++

−=

fjF

8985.0,

=fj

F

( )( ) 14.31215.27397.39382

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 1974.614.3128985.01067.5438

,=×= −

jrh W/(m

2-C)


tthq,,,,,

−=′′− θ

( )( ) 2089.1297.39381974.6,4,,,,,


qq W/m2.

Surface 5. 200=j

A m2, 92.427101096.20396.203 =+++=

fA m

2

( )( ) ( )( ) ( )( ) ( )( )101096.20396.203

381035104896.2034096.203,

+++

+++=

jft

65.43,

=jf

t C

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

9.0

9.01

92.427

2001

9.0

9.01

1, −

++

−=

fjF

8598.0,

=fj

F

( )( ) 98.31015.27365.43322

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 8646.598.3108598.01067.5438

,=×= −

jrh W/(m

2-C)

8. The Cooling Load


tthq,,,,,

−=′′− θ

( )( ) 3226.6865.43328646.5,5,,,,,


qq W/m2.


Surface fA , m

2

fε

ft , C

fjF

,

avgjt

,, K

jrh

,, W/(m

2-C)

1 203.96 0.9 39.91 0.8587 313.11 5.9783

2 203.96 0.9 36.06 0.8587 315.18 6.0976

3 10 0.9 40.02 0.8985 310.66 6.1097

4 10 0.9 39.97 0.8985 312.14 6.1974

5 200 0.9 43.65 0.8598 310.98 5.8648

( )

∑

∑ −=′′

=

=

N

jj

N

jifjjrj

balance

A

tthA

q

1

1,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )200101096.20396.203

2003226.68102089.12106707.3096.2038054.7296.203538.0

++++

−+−+−++=′′

balanceq

378844.192.627

804.865==′′

balanceq


Surface

Initial

Estimate,


−′′ ,

W/m2

Initial

Estimate,


−& ,

W


−′′

with Balance,

W/m2


−&

with Balance,

W

1 0.538 109.731 -0.8408 -171.50

2 72.8054 14849.39 71.4266 14568.16

3 -30.6707 -306.707 -32.0495 -320.50

4 -12.2089 -122.089 -13.5877 -135.88

5 -68.3226 -13664.5 -69.7014 -13940.3

Sum 865.804 0.0

8.23 One approach to reducing attic heat transfer is to install a radiant barrier, e.g.

aluminum foil on one or more surfaces. If we were to line the inside of the

pitched roof surfaces of Problem 8-21 with aluminum foil ( 1.0=ε ), and

everything else were to remain the same, how would the radiation flux

incident on the attic floor change? Please answer quantitatively.

Solution:

( ) 3.44512336 22

1=+=A ft

2

( ) 3.44512336 22

2=+=A ft

2

( )( )( ) 362432

13

==A ft2

( )( )( ) 362432

14

==A ft2

8. The Cooling Load

( )( ) 86436245

==A ft2

1.021

== εε , 9.0543

=== εεε


Surface Name Area, ft2 T, F

1 North-facing


2 South-facing


3 West-facing end

wall 36 100

4 East-facing end

wall 36 92

5 Floor 864 95

( )∑ −==

N

iijijf

AA1

,1 δ

( )

( )∑ −

∑ −=

=

=

N

iijii

N

iijiii

jf

A

tAt

1

1

,

1

1

δε

δε

Surface 1. 3.445=j

A ft2, 3.138186436363.445 =+++=

fA ft

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.08649.0369.0361.03.445

959.0864929.0361009.0361411.03.445,

+++

+++=

jft

38.97,

=jf

t F

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )6421.0

86436363.445

9.08649.0369.0361.03.445,

=+++

+++=

jfε

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

6421.0

6421.01

3.1381

3.4451

6421.0

6421.01

1, −

++

−=

fjF

5757.0,

=fj

F

( )( ) 36.56967.45938.971222

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 7281.036.5695757.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)


tthq,,,,,

−=′′− θ

( )( ) 9258.1738.971227281.0,1,,,,,


qq Btu/(hr-ft2)

8. The Cooling Load

Surface 2. 3.445=j

A ft2, 3.138186436363.445 =+++=

fA ft

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.08649.0369.0361.03.445

959.0864929.0361009.0361221.03.445,

+++

+++=

jft

43.96,

=jf

t F

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )6421.0

86436363.445

9.08649.0369.0361.03.445,

=+++

+++=

jfε

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

6421.0

6421.01

3.1381

3.4451

6421.0

6421.01

1, −

++

−=

fjF

5757.0,

=fj

F

( )( ) 38.57867.45943.961412

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 7632.038.5785757.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)


tthq,,,,,

−=′′− θ

( )( ) 0158.3443.961417632.0,2,,,,,


qq Btu/(hr-ft2)

Surface 3. 36=j

A ft2, 6.1790864363.4453.445 =+++=

fA ft

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.08649.0361.03.4451.03.445

959.0864929.0361411.03.4451221.03.445,

+++

+++=

jft

51.98,

=jf

t F

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )5021.0

864363.4453.445

9.08649.0361.03.4451.03.445,

=+++

+++=

jfε

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

5021.0

5021.01

6.1790

361

5021.0

5021.01

1, −

++

−=

fjF

4971.0,

=fj

F

8. The Cooling Load

( )( ) 93.55867.45951.981002

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 5947.093.5584971.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)


tthq,,,,,

−=′′− θ

( )( ) 8861.051.981005947.0,3,,,,,


qq Btu/(hr-ft2)

Surface 4. 36=j

A ft2, 6.1790864363.4453.445 =+++=

fA ft

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.08649.0361.03.4451.03.445

959.08641009.0361411.03.4451221.03.445,

+++

+++=

jft

80.98,

=jf

t F

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )5021.0

864363.4453.445

9.08649.0361.03.4451.03.445,

=+++

+++=

jfε

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

5021.0

5021.01

6.1790

361

5021.0

5021.01

1, −

++

−=

fjF

4971.0,

=fj

F

( )( ) 07.55567.45980.98922

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 5825.007.5554971.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)


tthq,,,,,

−=′′− θ

( )( ) 9610.380.98925825.0,4,,,,,


qq Btu/(hr-ft2)

Surface 5. 864=j

A ft2, 6.96236363.4453.445 =+++=

fA ft

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.0369.0361.03.4451.03.445

929.0361009.0361411.03.4451221.03.445,

+++

+++=

jft

55.116,

=jf

t F

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )1598.0

36363.4453.445

9.0369.0361.03.4451.03.445,

=+++

+++=

jfε

8. The Cooling Load

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

1598.0

1598.01

6.962

8641

1598.0

1598.01

1, −

++

−=

fjF

0911.0,

=fj

F

( )( ) 45.56567.45955.116952

1,

=++=avgj

t R

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 1129.045.5650911.0101713.0438

,=×= −

jrh Btu/(hr-ft

2-F)


tthq,,,,,

−=′′− θ

( )( ) 4330.255.116951129.0,5,,,,,


qq Btu/(hr-ft2)


Surface fA , ft

2

fε

ft , F

fjF

,

avgjt

,, R

jrh

,, Btu/(hr-ft

2-F)

1 445.3 0.1 97.38 0.5757 569.36 0.7281

2 445.3 0.1 96.43 0.5757 578.38 0.7632

3 36 0.9 98.51 0.4971 558.93 0.5947

4 36 0.9 98.80 0.4971 555.07 0.5825

5 864 0.9 116.55 0.0911 565.45 0.1129

( )

∑

∑ −=′′

=

=

N

jj

N

jifjjrj

balance

A

tthA

q

1

1,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )86436363.4453.445

864433.236961.3368861.03.4450158.343.4459258.17

++++

−+−+−++=′′

balanceq

4163.116.1826

99.20852==′′

balanceq


Surface

Initial

Estimate,


−′′ ,

Btu/(hr-ft2)

Initial

Estimate,


−& ,

Btu/hr


−′′

with Balance,

Btu/(hr-ft2)


−&

with Balance,

Btu/hr

1 17.9258 7982.36 6.5095 2898.69

2 34.0158 15147.24 22.5995 10063.56

3 -0.8861 -31.90 -12.3024 -442.89

4 -3.961 -142.60 -15.3773 -553.58

5 -2.433 -2102.11 -13.8493 -11965.8

Sum 20852.99 0.00

Therefore the radiation flux incident on the attic floor change or reduced by

( ) %57%10080.27787

80.1196580.27787=

−.

8. The Cooling Load

8.24 If we were to line the inside of the pitched roof surfaces of Problem 8-22 with

aluminum foil ( 1.0=ε ), and everything else were to remain the same, how

would the radiation flux incident on the attic floor change? Please answer

quantitatively.

Solution:

( ) 96.20310220 22

1=+=A m

2

( ) 96.20310220 22

2=+=A m

2

( )( )( ) 101022

13

==A m2

( )( )( ) 101022

14

==A m2

( )( ) 20020105

==A m2


Surface Name Area, m2 T, C

1 North-facing


2 South-facing


3 West-facing end

wall 10 35

4 East-facing end

wall 10 38

5 Floor 200 32

( )∑ −==

N

iijijf

AA1

,1 δ

9.0,

===ijjf

εεε

( )

( )∑ −

∑ −=

=

=

N

iijii

N

iijiii

jf

A

tAt

1

1

,

1

1

δε

δε

Surface 1. 96.203=j

A m2, 96.423200101096.203 =+++=

fA m

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.02009.0109.0101.096.203

329.0200389.010359.010481.096.203,

+++

+++=

jft

87.33,

=jf

t C

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )5151.0

200101096.203

9.02009.0109.0101.096.203,

=+++

+++=

jfε

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

8. The Cooling Load

5151.0

5151.01

96.423

96.2031

5151.0

5151.01

1, −

++

−=

fjF

4177.0,

=fj

F

( )( ) 08.31015.27387.33402

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 8244.208.3104177.01067.5438

,=×= −

jrh W/(m

2-C)


tthq,,,,,

−=′′− θ

( )( ) 3136.1787.33408244.2,1,,,,,


qq W/m2.

Surface 2. 96.203=j

A m2, 96.423200101096.203 =+++=

fA m

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.02009.0109.0101.096.203

329.0200389.010359.010401.096.203,

+++

+++=

jft

12.33,

=jf

t C

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )5151.0

200101096.203

9.02009.0109.0101.096.203,

=+++

+++=

jfε

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

5151.0

5151.01

96.423

96.2031

5151.0

5151.01

1, −

++

−=

fjF

4177.0,

=fj

F

( )( ) 71.31315.27312.33482

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 9248.271.3134177.01067.5438

,=×= −

jrh W/(m

2-C)


tthq,,,,,

−=′′− θ

( )( ) 5210.4312.33489248.2,2,,,,,


qq W/m2.

Surface 3. 10=j

A m2, 92.6172001096.20396.203 =+++=

fA m

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.02009.0101.096.2031.096.203

329.0200389.010481.096.203401.096.203,

+++

+++=

jft

37.34,

=jf

t C

8. The Cooling Load

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )3719.0

2001096.20396.203

9.02009.0101.096.2031.096.203,

=+++

+++=

jfε

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

3719.0

3719.01

92.617

101

3719.0

3719.01

1, −

++

−=

fjF

3682.0,

=fj

F

( )( ) 28.30915.27337.34352

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 4361.284.3073682.01067.5438

,=×= −

jrh W/(m

2-C)


tthq,,,,,

−=′′− θ

( )( ) 5347.137.34354361.2,3,,,,,


qq W/m2.

Surface 4. 10=j

A m2, 92.6172001096.20396.203 =+++=

fA m

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.02009.0101.096.2031.096.203

329.0200359.010481.096.203401.096.203,

+++

+++=

jft

25.34,

=jf

t C

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )3719.0

2001096.20396.203

9.02009.0101.096.2031.096.203,

=+++

+++=

jfε

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

3719.0

3719.01

92.617

101

3719.0

3719.01

1, −

++

−=

fjF

3682.0,

=fj

F

( )( ) 28.30915.27325.34382

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 4705.228.3093682.01067.5438

,=×= −

jrh W/(m

2-C)


tthq,,,,,

−=′′− θ

8. The Cooling Load

( )( ) 2644.925.34384705.2,4,,,,,


qq W/m2.

Surface 5. 200=j

A m2, 92.427101096.20396.203 =+++=

fA m

2

( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( ) ( ) ( )9.0109.0101.096.2031.096.203

389.010359.010481.096.203401.096.203,

+++

+++=

jft

70.41,

=jf

t C

( )

( )∑ −

∑ −=

=

=

N

iiji

N

iijii

jf

A

A

1

1

,

1

1

δ

δεε

( )( ) ( )( ) ( )( ) ( )( )1374.0

101096.20396.203

9.0109.0101.096.2031.096.203,

=+++

+++=

jfε

j

j

f

j

j

j

fj

A

AF

ε

ε

ε

ε −

++

−=

11

1

1,

1374.0

1374.01

92.427

2001

1374.0

1374.01

1, −

++

−=

fjF

0979.0,

=fj

F

( )( ) 31015.27370.41322

1,

=++=avgj

t K

( )3

,,,4

avgjfjjrtFh σ≈

( )( )( ) 6615.03100979.01067.5438

,=×= −

jrh W/(m

2-C)


tthq,,,,,

−=′′− θ

( )( ) 4166.670.41326615.0,5,,,,,


qq W/m2.


Surface fA , m

2

fε

ft , C

fjF

,

avgjt

,, K

jrh

,, W/(m

2-C)

1 203.96 0.1 33.87 0.4177 310.08 2.8244

2 203.96 0.1 33.12 0.4177 313.71 2.9248

3 10 0.9 34.37 0.3682 307.84 2.4361

4 10 0.9 34.25 0.3682 309.28 2.4705

5 200 0.9 41.70 0.0979 310.00 0.6615

( )

∑

∑ −=′′

=

=

N

jj

N

jifjjrj

balance

A

tthA

q

1

1,,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )200101096.20396.203

2004166.6102644.9105347.196.203521.4396.2033136.17

++++

−++++=′′

balanceq

88842.1792.627

5.11232==′′

balanceq

8. The Cooling Load


Surface

Initial

Estimate,


−′′ ,

W/m2

Initial

Estimate,


−& ,

W


−′′

with Balance,

W/m2


−&

with Balance,

W

1 17.3136 3531.28 -0.5748 -117.24

2 43.521 8876.54 25.6326 5228.02

3 1.5347 15.35 -16.3537 -163.54

4 9.2644 92.64 -8.6240 -86.24

5 -6.4166 -1283.32 -24.3050 -4861.00

Sum 11232.5 0.00

Therefore the radiation flux incident on the attic floor change or reduced by

( ) %65%1003.13940

48613.13940=

−.

8.25 If the attic air temperature in Problem 8-21 is 85 F, estimate the convective

heat flux from each interior surface.

Solution: 85=i

t F

( )ijisjicjinconvection

tthq −=′′θθ ,,,,,,,

Convective Heat Flux, using Table 8-8

Surface Area,

jA , ft

2

jich

,,,

Btu/(hr-ft2-

F)

θ,, jist , F θ,,, jinconvection

q ′′ ,

Btu/(hr-ft2)

θ,,, jinconvectionjqA ′′ ,

Btu/hr

1 445.3 0.42 122 15.54 6919.96

2 445.3 0.42 141 23.52 10473.46

3 36 0.56 100 8.40 302.40

4 36 0.56 92 3.92 141.12

5 864 0.18 95 1.80 1555.20

θ,,, jinconvectionq& 19392.14

8.26 If the attic air temperature in Problem 8-22 is 29 C, estimate the convective

heat flux from each interior surface.

Solution: 29=i

t C

( )ijisjicjinconvection

tthq −=′′θθ ,,,,,,,

Convective Heat Flux, using Table 8-8

Surface Area,

jA , m

2

jich

,,,

W/(m2-C)

θ,, jist , C θ,,, jinconvection

q ′′ ,

W/m2

θ,,, jinconvectionjqA ′′ ,

W

1 203.96 2.39 40 26.29 5362.11

2 203.96 2.39 48 45.41 9261.82

3 10 3.18 35 19.08 190.80

4 10 3.18 38 28.62 286.20

5 200 1.02 32 3.06 612.00

θ,,, jinconvectionq& 15712.93

8. The Cooling Load

8.27 Using the detailed model presented in Section 8-9, estimate the solar radiation

absorbed by each pane of a double-pane window with 1/8-in sheet glass for 3

P.M. on July 21 in Amarillo, Texas. You may neglect the solar radiation

incident from the inside of the window.

Solution:


64.178=D

G Btu/(hr-ft2), 62.18=

dG Btu/(hr-ft

2), 56.48=θ deg

6618.056.48coscos ==θ

Table 8-12. Polynomial Coefficient for a Double Pane Window with 1/8-in. Sheet

Glass (36)

outerja

,

innerja

,

jt

0.01407 0.00228 -0.00401

1.06226 0.34559 0.7405

-5.59131 -1.19908 7.2035

12.15034 2.22366 -20.1176

-11.78092 -2.05287 19.68824

4.2007 0.72376 -6.74585

Equation 8-50.

Neglect the solar radiation incident from the inside of the window.

douterdiffuseDouterDirectjouterabsorbedGGq

,,,,,ααθ +=′′

dinnerdiffuseDinnerDirectjinnerabsorbedGGq

,,,,,ααθ +=′′

( ) ( ) ( )

( ) ( ) ( ) 063365.06618.02007.46618.078092.116618.0150134.12

6618.059131.56618.006226.16618.001407.0

543

210

,

=+−+

−+=outerDirect

α

059951.07

2007.4

6

78092.11

5

15034.12

4

59131.5

3

06226.1

2

01407.02

,=

+−+−+=

outerdiffuseα

( ) ( ) ( )

( ) ( ) ( ) 048446.06618.072376.06618.005287.26618.022366.2

6618.019908.16618.034559.06618.000228.0

543

210

,

=+−+

−+=innerDirect

α

045096.07

72376.0

6

05287.2

5

22366.2

4

19908.1

3

34559.0

2

00228.02

,=

+−+−+=

innerdiffuseα


,,,,,ααθ +=′′

( ) ( ) 44.1262.18059951.064.178063365.0,,,

=+=′′θjouterabsorbed

q Btu/(hr-ft2)

( ) ( ) 50.962.18045096.064.178048446.0,,,

=+=′′θjinnerabsorbed

q Btu/(hr-ft2)

8.28 Using the detailed model presented in Section 8-9, estimate the solar radiation

absorbed by each pane of a double-pane window witn 1/8-in sheet glass for 3

P.M. on July 21 in Billings, Montana. You may neglect the solar radiation

incident from the inside of the window.

Solution:

8. The Cooling Load


78.176=D

G Btu/(hr-ft2), 43.18=

dG Btu/(hr-ft

2), 56.48=θ deg

6618.056.48coscos ==θ

Table 8-12. Polynomial Coefficient for a Double Pane Window with 1/8-in. Sheet

Glass (36)

outerja

,

innerja

,

jt

0.01407 0.00228 -0.00401

1.06226 0.34559 0.7405

-5.59131 -1.19908 7.2035

12.15034 2.22366 -20.1176

-11.78092 -2.05287 19.68824

4.2007 0.72376 -6.74585

Equation 8-50.

Neglect the solar radiation incident from the inside of the window.


,,,,,ααθ +=′′

dinnerdiffuseDinnerDirectjinnerabsorbedGGq

,,,,,ααθ +=′′

( ) ( ) ( )

( ) ( ) ( ) 063365.06618.02007.46618.078092.116618.0150134.12

6618.059131.56618.006226.16618.001407.0

543

210

,

=+−+

−+=outerDirect

α

059951.07

2007.4

6

78092.11

5

15034.12

4

59131.5

3

06226.1

2

01407.02

,=

+−+−+=

outerdiffuseα

( ) ( ) ( )

( ) ( ) ( ) 048446.06618.072376.06618.005287.26618.022366.2

6618.019908.16618.034559.06618.000228.0

543

210

,

=+−+

−+=innerDirect

α

045096.07

72376.0

6

05287.2

5

22366.2

4

19908.1

3

34559.0

2

00228.02

,=

+−+−+=

innerdiffuseα


,,,,,ααθ +=′′

( ) ( ) 31.1243.18059951.078.176063365.0,,,

=+=′′θjouterabsorbed

q Btu/(hr-ft2)

( ) ( ) 40.943.18045096.078.176048446.0,,,

=+=′′θjinnerabsorbed

q Btu/(hr-ft2)

8. The Cooling Load

8.29 Compute the hourly cooling loads for Example 8-16, using the heat balance

method.

Solution: See Figure 8-6.

Table 8-4, Exterior Surface Temperatures

Hour θ,, jost , F Hour θ,, jost , F

1 79.55 13 106.15

2 77.80 14 108.95

3 76.40 15 110.00

4 75.35 16 108.95

5 75.00 17 106.50

6 75.70 18 102.65

7 77.45 19 98.10

8 80.60 20 93.55

9 85.15 21 89.70

10 90.40 22 86.20

11 96.35 23 83.40

12 101.95 24 81.30

Same as Table 8-2. Wall 1 Layers, Listed from outside to inside. (Refer to Table 8-

17a)

Layer Thickness, in. Density, lb/ft3

Conductivity,

(Btu-in.)/(hr-

ft2-F)

Specific Heat,

Btu/(lb-F)

1 in. Stucco 1.0 116.0 4.8 0.20

5 in. insulation 5.0 5.7 0.3 0.20

¾ in. plaster or

gypsum 0.75 100.0 5.04 0.20

Same as Table 8-2. Roof 1 Layers, Listed from outside to inside. (Refer to Table 8-

17b)


Conductivity,

(Btu-in.)/(hr-

ft2-F)

Specific Heat,

Btu/(lb-F)

1/2 in. slag or

stone 0.5 55.0 9.96 0.40

3/8 in. felt and

membrane 0.375 70.0 1.32 0.40

2 in.

heavyweight

concrete

2.0 140.0 12.0 0.20

Ceiling air

space, R = 1.0

(F-ft2-hr)/Btu

Acoustic tile 0.75 30.0 0.42 0.20

8. The Cooling Load

Same as Table 8-3. Wall CTF Coefficients

n nX , Btu/(h-ft2-

F)

nY , Btu/(h-ft2-

F)

nZ , Btu/(h-ft2-

F) nΦ

0 2.068783191 0.0036061 1.419234575

1 -2.649608242 0.026045615 -1.80125067 0.328930497

2 0.628395811 0.009821804 0.428280141 -0.005290438

3 -0.007851015 0.000250326 -0.00655245 1.04454E-05

4 4.38178 x 10-6

3.69865 x10-7

1.26801 x 10-5

-1.21994 x 10-9

5 -3.70418x10-10

2.38394 x 10-11

-1.35736 x 10-9

5.01896 x 10-15

Same as Table 8-3. Roof CTF Coefficients

n nX , Btu/(h-ft2-

F)

nY , Btu/(h-ft2-

F)

nZ , Btu/(h-ft2-

F) nΦ

0 3.249652077 0.029969794 0.530411955

1 -3.703823801 0.094934868 -0.494218555 0.553537309

2 0.590586137 0.011045673 0.09978853 -0.000433935

3 -0.000454717 9.15139 x 10-6

-2.24332 x 10-5

3.93274 x 10-9

4 3.49402 x 10-9

2.04672 x 10-11

1.79613 x 10-10

9.09325 x 10-19

Table 8-9, Zone Surface Temperature

Surface Name Area, ft2 ist , F

1 North Wall 360 72

2 East Wall 360 73

3 South Wall 280 77

4 South Window 80 85

5 West Wall 360 76

6 Roof 900 78

7 Floor 900 72

Table 8-10, Intermediate Variables for MRT/Balance Calculation.

Surface fA fε ft , F fjF , avgjt , , R jrh , ,

Btu/(hr-

ft2-F)

1 2880 0.9 75.35 0.8889 533.34 0.924

2 2880 0.9 75.22 0.8889 533.78 0.926

3 2960 0.9 74.78 0.8916 535.56 0.938

4 3160 0.9 74.72 0.8977 539.53 0.966

5 2880 0.9 74.85 0.8889 535.09 0.933

6 2340 0.9 73.81 0.8667 535.58 0.912

7 2340 0.9 76.12 0.8667 533.73 0.903

8. The Cooling Load

Table 8-11 MRT/Balance Calculation

Surface

Initial Estimate

θ,,, jinsurfradiationq −′′ ,

Btu/(hr-ft2)

Initial Estimate

θ,,, jinsurfradiationq −& ,

Btu/hr

θ,,, jinsurfradiationq −′′

with Balance,

Btu/(hr-ft2)

θ,,, jinsurfradiationq −&

with Balance,

Btu/hr

1 -3.0929 -1113.46 -3.0931 -1113.51

2 -2.0585 -741.04 -2.0586 -741.10

3 2.0798 582.33 2.0796 582.29

4 9.9298 794.38 9.9296 794.37

5 1.0757 387.26 1.0756 387.21

6 3.8207 3438.63 3.8206 3438.50

7 -3.7196 -3347.63 -3.7197 -3347.76

Sum 0.47 0.00

( )2

2/000145381.0

3240

/47.0fthrBtu

ft

hrBtuqbalance −==′′

Table 8-15 Surface Information

Surface Name Area, ft2 ist , F ich ,

1 North Wall 360 72 0.56

2 East Wall 360 73 0.56

3 South Wall 280 77 0.56

4 South Window 80 85 0.56

5 West Wall 360 76 0.56

6 Roof 900 78 0.18

7 Floor 900 72 0.18

Table 8-23 (Part) Incident Solar Radiation

Incident Solar Radiation, Btu/(hr-ft2)

Hour oT South Wall Roof

1 75.86 0.00 0.00

2 74.64 0.00 0.00

3 73.67 0.00 0.00

4 72.94 0.00 0.00

5 72.70 0.00 0.00

6 73.19 0.00 0.00

7 74.40 0.00 0.00

8 76.59 10.30 33.69

9 79.75 23.97 101.85

10 83.39 34.91 166.53

11 87.52 70.25 223.05

12 91.41 106.19 267.81

13 94.33 131.29 297.90

14 96.27 142.69 311.32

15 97.00 139.17 307.19

16 96.27 121.11 285.78

17 94.57 90.46 248.51

8. The Cooling Load

18 91.90 50.96 197.83

19 88.74 30.01 137.00

20 85.58 17.72 69.92

21 82.91 2.35 5.98

22 80.48 0.00 0.00

23 78.53 0.00 0.00

24 77.07 0.00 0.00

Table 8-24 Solar Heat Gain Factors and Solar Heat Gain for Window in Example 8-

16

Hour θcos DG ,

Btu/(hr-

ft2)

dG ,

Btu/(hr-

ft2)

sunlitTSHGF ,

Btu/(hr-ft2)

sunlitASHGF ,

Btu/(hr-ft2)

TSHGq& ,

Btu/(hr-

ft2)

ASHGq& ,

Btu/(hr-

ft2)

1 -0.82026 0.00 0.00 0.00 0.00 0.00 0.00 2 -0.85007 0.00 0.00 0.00 0.00 0.00 0.00 3 -0.84089 0.00 0.00 0.00 0.00 0.00 0.00 4 -0.79337 0.00 0.00 0.00 0.00 0.00 0.00 5 -0.71073 0.00 0.00 0.00 0.00 0.00 0.00 6 -0.59861 0.00 0.00 0.00 0.00 0.00 0.00 7 -0.46465 0.00 0.00 0.00 0.00 0.00 0.00 8 -0.31798 0.00 10.30 8.23 0.56 579.2 12.8

9 -0.16859 0.00 23.97 19.15 1.30 1348.1 29.9

10 -0.02667 0.00 34.91 27.89 1.90 1963.5 43.5

11 0.098118 26.08 44.17 41.69 3.89 2935.1 89.2

12 0.197265 54.63 51.56 66.40 6.37 4674.8 146.0

13 0.264016 74.68 56.61 88.27 7.75 6214.0 177.7

14 0.293821 83.79 58.89 98.95 8.30 6966.1 190.2

15 0.284648 80.98 58.19 95.61 8.13 6731.3 186.4

16 0.237124 66.54 54.57 79.07 7.22 5566.4 165.6

17 0.154487 42.10 48.36 54.41 5.34 3830.2 122.5

18 0.042368 10.93 40.04 33.13 2.59 2332.4 59.4

19 -0.09159 0.00 30.01 23.98 1.63 1688.1 37.4

20 -0.23826 0.00 17.72 14.15 0.96 996.5 22.1

21 -0.38765 0.00 2.35 1.88 0.13 132.2 2.9

22 -0.52957 0.00 0.00 0.00 0.00 0.00 0.0

23 -0.65436 0.00 0.00 0.00 0.00 0.00 0.0

24 -0.75351 0.00 0.00 0.00 0.00 0.00 0.0

Table 8-26 Internal Heat Gains and Infiltration Heat Gains

Heat Gain, Btu/hr

Hour People,

Latent

People,

Sensible Lights Equipment Infiltration

1 0 0 921.8 614.5 391.3

2 0 0 921.8 614.5 268.1

3 0 0 921.8 614.5 169.6

4 0 0 921.8 614.5 95.6

5 0 0 921.8 614.5 71.0

6 0 0 921.8 614.5 120.3

7 0 0 921.8 614.5 243.5

8 0 0 921.8 614.5 465.3

9 2000 2500 4608.9 3072.6 785.6

10 2000 2500 4608.9 3072.6 1155.2

11 2000 2500 4608.9 3072.6 1574.2

12 2000 2500 4608.9 3072.6 1968.4

8. The Cooling Load

13 2000 2500 4608.9 3072.6 2264.1

14 2000 2500 4608.9 3072.6 2461.3

15 2000 2500 4608.9 3072.6 2535.2

16 2000 2500 4608.9 3072.6 2461.3

17 2000 2500 4608.9 3072.6 2288.8

18 0 0 921.8 614.5 2017.7

19 0 0 921.8 614.5 1697.4

20 0 0 921.8 614.5 1377.0

21 0 0 921.8 614.5 1106.0

22 0 0 921.8 614.5 859.5

23 0 0 921.8 614.5 662.4

24 0 0 921.8 614.5 514.5

For wall and roof

( ) ( ) θθθ ,,,

1

,,,,, convernalintiopiltrationinfa

N

j

ijisjicjsystem qttcmtthAq &&& −−−−−= ∑=

( ) ( ) θθθ ,,,

1


N

j

ijisjicjsystem qttcmtthAq &&& +−+−=− ∑=

SOUTH WALL AND ROOF COOLING LOAD

Note: Only the south wall and the roof are exposed to the outside

skyrgrco

skyskyrggroctjoutjiso

joshhhX

thththGHtYt

−−

−−

+++

++++−=

αθθ

θ

,,,,

,,

∑∑∑=

−

=

−

=

− +′′Φ++−=qxy N

n

njoutconductionn

N

n

njosn

N

n

njisnjout qtXtYH1

,,,

1

,,

1

,,,, δθδθδθθ

∑∑∑=

−

=

−

=

−′′Φ+++−−=′′

qxy N

n

njoutconductionn

N

n

njosnjoso

N

n

njisnjisojoutconduction qtXtXtYtYq1

,,,

1

,,,,

1

,,,,,, δθδθθδθθ

jrco

jinihgradiationbalancejfjricjinjosojinsolar

jishhZ

qqththHtYqt

,

,,,,,,,,,,,,

,,++

′′+′′+++++′′=

− θθθθ

θ

∑∑∑=

−

=

−

=

− +′′Φ++−=qyz

N

n

njinconductionn

N

n

njosn

N

n

njisnjin qtYtZH1

,,,

1

,,

1

,,,, δθδθδθθ

∑∑∑=

−

=

−

=

−′′Φ+++−−=′′

qyzN

n

njinconductionn

N

n

njosnjoso

N

n

njisnjisojinconduction qtYtYtZtZq1

,,,

1

,,,,

1

,,,,,, δθδθθδθθ

9.0=α , neglect ground reflectivity

( )[ ] [ ]22

, 31 b

otoc aVtCh +∆= , Eq. 8-13, assuming mphVo 12= wind and the surface

facing windward.

Table 8-6:

096.0=tC Btu/(h-ft2-F

4/3)

203.0=a (h-ft2-F-mph)

89.0=b

8. The Cooling Load

( )

−

−=

−

θ

θεσ

,,

4

,,

4

,

josg

josggs

grtt

ttFh , Eq. 8-17, assuming a surface emissivity 90.0=ε

( )

−

−=

−

θ

θεσ

,,

4

,,

4

,

jossky

josskyskys

skyrtt


( )428101714.0 RfthrBtu −−×= −σ

oskysky ttt

−+

=

2cos1

2cos,

ααα

90=α deg

5.0== −− skysgs FF

Surface 3 (South Wall)

( )2

,,, 16.1 fthrBtuq jinsolar −=′′θ

( )2/000145381.0 fthrBtuqbalance −=′′

2,,,3240 ft

portionRadiativeq jinihgradiation =′′

− θ

Ftis 77= initial

56.0, =ich

938.0, =jrh

Ft jf 78.74, =

068783191.2=oX

0036061.0=oY

419234575.1=oZ

2280 ftAj =

Hour ot θ,, jost

Radiative

portion of

internal heat

gain, Btu/hr

θ,,, jinihgradiationq −′′ ,

Btu/(hr-ft2)

1 75.86 79.55 740.5 0.228551

2 74.64 77.80 740.5 0.228551

3 73.67 76.40 740.5 0.228551

4 72.94 75.35 740.5 0.228551

5 72.70 75.00 740.5 0.228551

6 73.19 75.70 740.5 0.228551

7 74.40 77.45 740.5 0.228551

8 76.59 80.60 5452.5 0.228551

9 79.75 85.15 5452.5 1.682865

10 83.39 90.40 5452.5 1.682865

11 87.52 96.35 5452.5 1.682865

12 91.41 101.95 5452.5 1.682865

13 94.33 106.15 5452.5 1.682865

14 96.27 108.95 5452.5 1.682865

15 97.00 110.00 5452.5 1.682865

8. The Cooling Load

16 96.27 108.95 5452.5 1.682865

17 94.57 106.50 740.5 1.682865

18 91.90 102.65 740.5 0.228551

19 88.74 98.10 740.5 0.228551

20 85.58 93.55 740.5 0.228551

21 82.91 89.70 740.5 0.228551

22 80.48 86.20 740.5 0.228551

23 78.53 83.40 740.5 0.228551

24 77.07 81.30 740.5 0.228551

Using the equations and starting inside surface temperature from 77 F, The result for

the inside surface temperature are as follows after some iterations until nearly

converged. After 13 iterations.

Hour θ,, jost θ,, jist θ,, joutH θ,, jinH ( )ijisicj tthA −θ,,,

1 75.8674 75.0059 -160.24 106.6839 471.33

2 74.3818 74.8741 -156.41 106.3048 450.66

3 73.2878 74.8060 -153.78 106.1100 439.98

4 72.4215 74.6225 -151.59 105.5779 411.21

5 72.0227 74.8920 -150.32 106.3655 453.47

6 71.7099 74.3385 -147.45 104.7519 366.68

7 72.7907 74.8132 -149.34 106.1329 441.11

8 76.2976 74.4707 -151.04 105.1210 387.40

9 81.4499 75.2247 -155.10 105.8477 505.63

10 88.5659 75.4013 -170.03 106.3371 533.32

11 99.3424 75.7867 -180.22 107.4227 593.75

12 112.6382 76.0371 -204.00 108.1051 633.01

13 123.6443 76.4191 -229.59 109.1801 692.92

14 131.0799 76.8179 -252.39 110.3164 755.44

15 133.6412 77.1884 -266.87 111.3880 813.54

16 131.2950 77.4541 -272.93 112.1717 855.21

17 124.0862 77.6075 -267.93 112.6451 879.25

18 113.0573 77.0661 -254.41 112.5599 794.37

19 103.0188 76.7954 -231.43 111.8064 751.92

20 95.1573 76.0809 -212.08 109.7502 639.88

21 87.8040 76.2769 -196.92 110.3488 670.63

22 82.3582 75.4244 -179.08 107.8813 536.94

23 79.8183 75.4643 -172.08 108.0068 543.20

24 77.2393 75.1348 -163.56 107.0550 491.54

Surface 6 (Roof)

( )[ ] [ ]22

, 31 b

otoc aVtCh +∆= , Eq. 8-13, assuming mphVo 12= wind

Table 8-6:


4/3)

203.0=a (h-ft2-F-mph)

89.0=b

8. The Cooling Load

( )

−

−=

−

θ

θεσ

,,

4

,,

4

,

josg

josggs

grtt


( )

−

−=

−

θ

θεσ

,,

4

,,

4

,

jossky

josskyskys

skyrtt


( )428101714.0 RfthrBtu −−×= −σ

8.10−= osky tt

0=α deg

0.1

0.0

=

=

−

−

skys

gs

F

F

( )2



2,,,3240 ft


− θ

Ftis 78= initial

18.0, =ich downward heat flow

912.0, =jrh

Ft jf 81.73, =

249652077.3=oX

029969794.0=oY

530411955.0=oZ

2900 ftAj =

Hour ot θ,, jost

Radiative

portion of

internal heat

gain, Btu/hr


Btu/(hr-ft2)

1 75.86 79.55 740.5 0.228551

2 74.64 77.80 740.5 0.228551

3 73.67 76.40 740.5 0.228551

4 72.94 75.35 740.5 0.228551

5 72.70 75.00 740.5 0.228551

6 73.19 75.70 740.5 0.228551

7 74.40 77.45 740.5 0.228551

8 76.59 80.60 5452.5 0.228551

9 79.75 85.15 5452.5 1.682865

10 83.39 90.40 5452.5 1.682865

11 87.52 96.35 5452.5 1.682865

12 91.41 101.95 5452.5 1.682865

13 94.33 106.15 5452.5 1.682865

14 96.27 108.95 5452.5 1.682865

15 97.00 110.00 5452.5 1.682865

16 96.27 108.95 5452.5 1.682865

8. The Cooling Load

17 94.57 106.50 740.5 1.682865

18 91.90 102.65 740.5 0.228551

19 88.74 98.10 740.5 0.228551

20 85.58 93.55 740.5 0.228551

21 82.91 89.70 740.5 0.228551

22 80.48 86.20 740.5 0.228551

23 78.53 83.40 740.5 0.228551

24 77.07 81.30 740.5 0.228551





1 79.83 77.64 -278.0914 41.9007 912.94

2 74.54 76.63 -249.5663 40.4245 749.71

3 73.74 75.49 -247.4426 38.6037 565.51

4 75.01 75.24 -256.9090 38.1651 525.51

5 69.32 75.05 -223.4381 38.0160 493.58

6 72.11 74.07 -238.8704 36.3427 334.87

7 71.43 74.34 -231.4721 36.8002 378.51

8 73.78 74.18 -230.3845 36.4694 352.52

9 77.76 75.22 -233.4380 36.5969 522.38

10 83.37 75.85 -247.5313 37.4416 623.51

11 91.66 76.78 -254.9153 38.7028 774.25

12 102.60 77.96 -279.1179 40.2936 965.82

13 111.67 79.69 -304.8093 42.8260 1245.81

14 118.59 81.45 -332.4318 45.4792 1531.46

15 120.61 83.05 -346.4516 48.0072 1789.93

16 120.47 83.92 -363.9490 49.4201 1930.60

17 117.81 84.86 -379.0489 51.0254 2082.91

18 106.45 83.93 -351.4262 51.3085 1931.97

19 101.65 82.60 -349.2826 49.3072 1717.79

20 95.92 81.99 -333.7814 48.4796 1617.98

21 91.06 81.07 -325.0945 47.1369 1469.38

22 86.47 80.14 -305.8842 45.7667 1318.83

23 83.12 79.03 -290.9112 44.0572 1138.12

24 81.72 78.10 -286.3728 42.5985 988.28

For the window: South window , 280 ftAj = . Double Pane Window.l

skyrgrocairspace

skyskyrggroocjisairspacejouterabsorbed

joshhhU

thththtUqt

−−

−−

+++

++++′′=

,

,,,,,,

,,

θθ

θ

jricairspace

jinihgradiationbalancejgjriicjosairspacejinnerabsorbed

jishhU

qqththtUqt

,,

,,,,,,,,,,,

,,++

′′+′′++++′′=

− θθθ

θ

( )FfthrBtuh ic −−= 2

, 56.0 from Table 8-8

( )FfthrBtuUairspace −−= 219.1 from Table 5-3a

8. The Cooling Load

( )2/000145381.0 fthrBtuqbalance −=′′ assumed the same to simplify calculation but

must be recalculated in actual.

( )[ ] [ ]22

, 31 b

otoc aVtCh +∆= , Eq. 8-13, assuming mphVo 12= wind and the surface

facing windward.

Table 8-6:


4/3)

203.0=a (h-ft2-F-mph)

89.0=b

( )

−

−=

−

θ

θεσ

,,

4

,,

4

,

josg

josggs

grtt


( )

−

−=

−

θ

θεσ

,,

4

,,

4

,

jossky

josskyskys

skyrtt


( )428101714.0 RfthrBtu −−×= −σ

oskysky ttt

−+

=

2cos1

2cos,

ααα

90=α deg

5.0== −− skysgs FF

Other assumed constant.

966.0, =jrh Btu/(hr-ft2-F)

72.74, =jft F

θθ ααα ,,,,,,,,, jinsolarinnerdiffusedouterdiffuseDouterDirectjouterabsorbed qGGq ′′++=′′

θθ ααα ,,,,,,,,, jinsolarouterdiffusedinnerdiffuseDinnerDirectjinnerabsorbed qGGq ′′++=′′

2,,,3240 ft

qq

diffuse

jinsolar

&=′′

θ

diffuseddiffuse AGq τ=&

Using Table 8-12

6797.02

25

0

=+

= ∑=j

j

diffusej

tτ

[ ]j

j

outerjouterDirect a∑=

=5

0

,, cosθα

0600.02

25

0

,

, =+

= ∑=j

outerj

outerdiffusej

aα

[ ]j

j

innerjinnerDirect a∑=

=5

0

,, cosθα

0451.02

25

0

,

, =+

= ∑=j

innerj

innerdiffusej

aα

Tabulation for2,,,

3240 ft

qq

diffuse

jinsolar

&=′′

θ

8. The Cooling Load

Hour DG dG θ,,, jinsolarq ′′ outerDirect ,α innerDirect ,α θ,,, jouterabsorbedq ′′ θ,,, jinnerabsorbedq ′′

1 0.00 0.00 0.0000 0.0000 0.0000

2 0.00 0.00 0.0000 0.0000 0.0000

3 0.00 0.00 0.0000 0.0000 0.0000

4 0.00 0.00 0.0000 0.0000 0.0000

5 0.00 0.00 0.0000 0.0000 0.0000

6 0.00 0.00 0.0000 0.0000 0.0000

7 0.00 0.00 0.0000 0.0000 0.0000

8 0.00 10.30 0.1729 0.6253 0.4749

9 0.00 23.97 0.4023 1.4552 1.1051

10 0.00 34.91 0.5859 2.1193 1.6094

11 26.08 44.17 0.7413 0.0749 0.0266 4.6346 2.7291

12 54.63 51.56 0.8654 0.0827 0.0380 7.6493 4.4513

13 74.68 56.61 0.9501 0.0765 0.0418 9.1524 5.7327

14 83.79 58.89 0.9884 0.0731 0.0430 9.6985 6.3174

15 80.98 58.19 0.9766 0.0741 0.0427 9.5368 6.1371

16 66.54 54.57 0.9159 0.0795 0.0405 8.6010 5.2111

17 42.10 48.36 0.8116 0.0832 0.0341 6.4381 3.6670

18 10.93 40.04 0.6720 0.0499 0.0149 2.9764 2.0091

19 0.00 30.01 0.5037 1.8218 1.3835

20 0.00 17.72 0.2974 1.0757 0.8169

21 0.00 2.35 0.0394 0.1427 0.1083

22 0.00 0.00 0.0000 0.0000 0.0000

23 0.00 0.00 0.0000 0.0000 0.0000

24 0.00 0.00 0.0000 0.0000 0.0000

skyrgrocairspace

skyskyrggroocjisairspacejouterabsorbed

joshhhU

thththtUqt

−−

−−

+++

++++′′=

,

,,,,,,

,,

θθ

θ

jricairspace

jinihgradiationbalancejgjriicjosairspacejinnerabsorbed

jishhU

qqththtUqt

,,

,,,,,,,,,,,

,,++

′′+′′++++′′=

− θθθ

θ

Initial Final Hour

θ,, jost θ,, jist θ,, jost θ,, jist

( )ijisicj tthA −θ,,,

1 77.80 85 74.4558 74.1276 95.32

2 76.40 85 73.4759 73.6983 76.08

3 75.35 85 72.6973 73.3572 60.80

4 75.00 85 72.1118 73.1006 49.31

5 75.70 85 71.9193 73.0163 45.53

6 77.45 85 72.3123 73.1884 53.24

7 80.60 85 73.2832 73.6138 72.30

8 85.15 85 75.2826 74.6647 119.38

9 90.40 85 78.3247 76.7651 213.48

10 96.35 85 81.5099 78.3464 284.32

8. The Cooling Load

11 101.95 85 85.6971 80.5932 384.98

12 106.15 85 89.9042 83.0706 495.96

13 108.95 85 92.8451 84.8310 574.83

14 110.00 85 94.6398 85.8326 619.70

15 108.95 85 95.1639 85.9958 627.01

16 106.50 85 94.1976 85.2315 592.77

17 102.65 85 92.0248 83.7110 524.65

18 98.10 85 88.5035 81.0223 404.20

19 93.55 85 85.5392 79.4932 335.69

20 89.70 85 82.7004 78.0407 270.62

21 86.20 85 80.1889 76.6794 209.64

22 83.40 85 78.1741 75.7568 168.30

23 81.30 85 76.6033 75.0685 137.47

24 77.80 85 75.4285 74.5538 114.41

Cooling Loads

( ) ( ) θθθ ,,,

1


N

j

ijisjicjsystem qttcmtthAq &&& −−−−−= ∑=

( ) ( ) θθθ ,,,

1


N

j

ijisjicjsystem qttcmtthAq &&& +−+−=− ∑=

Cooling Load, Btu/hr

Hour Wall Roof Window Infiltration People Lights Equipment Total

1 471.33 912.94 95.32 391.3 0 921.8 614.5 3407.19

2 450.66 749.71 76.08 268.1 0 921.8 614.5 3080.85

3 439.98 565.51 60.8 169.6 0 921.8 614.5 2772.19

4 411.21 525.51 49.31 95.6 0 921.8 614.5 2617.93

5 453.47 493.58 45.53 71 0 921.8 614.5 2599.88

6 366.68 334.87 53.24 120.3 0 921.8 614.5 2411.39

7 441.11 378.51 72.3 243.5 0 921.8 614.5 2671.72

8 387.4 352.52 119.38 465.3 0 921.8 614.5 2860.9

9 505.63 522.38 213.48 785.6 4500 4608.9 3072.6 14208.59

10 533.32 623.51 284.32 1155.2 4500 4608.9 3072.6 14777.85

11 593.75 774.25 384.98 1574.2 4500 4608.9 3072.6 15508.68

12 633.01 965.82 495.96 1968.4 4500 4608.9 3072.6 16244.69

13 692.92 1245.81 574.83 2264.1 4500 4608.9 3072.6 16959.16

14 755.44 1531.46 619.7 2461.3 4500 4608.9 3072.6 17549.4

15 813.54 1789.93 627.01 2535.2 4500 4608.9 3072.6 17947.18

16 855.21 1930.60 592.77 2461.3 4500 4608.9 3072.6 18021.38

17 879.25 2082.91 524.65 2288.8 4500 4608.9 3072.6 17957.11

18 794.37 1931.97 404.2 2017.7 0 921.8 614.5 6684.54

19 751.92 1717.79 335.69 1697.4 0 921.8 614.5 6039.1

20 639.88 1617.98 270.62 1377 0 921.8 614.5 5441.78

21 670.63 1469.38 209.64 1106 0 921.8 614.5 4991.95

22 536.94 1318.83 168.3 859.5 0 921.8 614.5 4419.87

23 543.2 1138.12 137.47 662.4 0 921.8 614.5 4017.49

24 491.54 988.28 114.41 514.5 0 921.8 614.5 3645.03

8. The Cooling Load

8.30 A common retrofit to older buildings is to add fibreglass insulation above the

acoustic tile. For the roof in the Example 8-16, add 6 in. of fibreglass

insulation above the acoustic tiles and compute the hourly cooling loads, using

the heat balance method.

Solution: Adding 6 in fibreglass insulation above acoustic tiles

Same as Table 8-2. Roof 1 Layers, Listed from outside to inside. (Refer to Table 8-

17b)


Conductivity,

(Btu-in.)/(hr-

ft2-F)

Specific Heat,

Btu/(lb-F)

1/2 in. slag or

stone 0.5 55.0 9.96 0.40

3/8 in. felt and

membrane 0.375 70.0 1.32 0.40

2 in.

heavyweight

concrete

2.0 140.0 12.0 0.20

Ceiling air

space, R = 1.0

(F-ft2-hr)/Btu

Fiberglass

Insulation 6 1.2 0.25 0.23

Acoustic tile 0.75 30.0 0.42 0.20

Same as Table 8-3. Roof CTF Coefficients (from

n nX , Btu/(h-ft2-F) nY , Btu/(h-ft

2-F) nZ , Btu/(h-ft

2-F) nΦ

0 18.421565774 0.001891421 2.311785784

1 -24.442342884 0.035304525 -3.675905395 0.730323546

2 6.611505751 0.032925868 1.540893799 -0.082391241

3 -0.523690051 0.002854181 -0.104148989 0.000987299

4 0.005952081 1.44369 x 10-5

0.000365249 -4.673217 x 10-8

5 -2.67113 x 10-7

1.39615 x 10-9

-1.56572 x 10-8

3.978621 x 10-13

6 2.014513 x 10-12

2.51190 x 10-15

1.273613 x 10-13

-1.031689 x10-20

All other data same as above (Problem 8-29). Surface 6 (Roof)

( )[ ] [ ]22

, 31 b

otoc aVtCh +∆= , Eq. 8-13, assuming mphVo 12= wind

Table 8-6:


4/3)

203.0=a (h-ft2-F-mph)

89.0=b

( )

−

−=

−

θ

θεσ

,,

4

,,

4

,

josg

josggs

grtt


8. The Cooling Load

( )

−

−=

−

θ

θεσ

,,

4

,,

4

,

jossky

josskyskys

skyrtt


( )428101714.0 RfthrBtu −−×= −σ

8.10−= osky tt

0=α deg

0.1

0.0

=

=

−

−

skys

gs

F

F

( )2



2,,,3240 ft


− θ

Ftis 78= initial

18.0, =ich downward heat flow

912.0, =jrh

Ft jf 81.73, =

421565774.18=oX

001891421.0=oY

311785784.2=oZ

2900 ftAj =

Hour ot θ,, jost

Radiative

portion of

internal heat

gain, Btu/hr


Btu/(hr-ft2)

1 75.86 79.55 740.5 0.228551

2 74.64 77.80 740.5 0.228551

3 73.67 76.40 740.5 0.228551

4 72.94 75.35 740.5 0.228551

5 72.70 75.00 740.5 0.228551

6 73.19 75.70 740.5 0.228551

7 74.40 77.45 740.5 0.228551

8 76.59 80.60 5452.5 0.228551

9 79.75 85.15 5452.5 1.682865

10 83.39 90.40 5452.5 1.682865

11 87.52 96.35 5452.5 1.682865

12 91.41 101.95 5452.5 1.682865

13 94.33 106.15 5452.5 1.682865

14 96.27 108.95 5452.5 1.682865

15 97.00 110.00 5452.5 1.682865

16 96.27 108.95 5452.5 1.682865

17 94.57 106.50 740.5 1.682865

18 91.90 102.65 740.5 0.228551

19 88.74 98.10 740.5 0.228551

8. The Cooling Load

20 85.58 93.55 740.5 0.228551

21 82.91 89.70 740.5 0.228551

22 80.48 86.20 740.5 0.228551

23 78.53 83.40 740.5 0.228551

24 77.07 81.30 740.5 0.228551





1 85.9623 76.6359 -1621.75 179.0263 751.02

2 80.5683 76.3246 -1510.56 177.9768 700.59

3 78.7324 76.0565 -1474.23 177.0677 657.15

4 77.7634 75.7671 -1455.64 176.0844 610.26

5 77.1969 75.6352 -1444.20 175.6365 588.90

6 74.1636 75.5253 -1378.56 175.2683 571.10

7 74.0934 75.3422 -1373.73 174.6450 541.43

8 72.5218 75.2051 -1325.13 174.1815 519.23

9 72.5451 75.5439 -1304.60 173.8802 574.11

10 74.5344 75.7470 -1326.82 174.5679 607.02

11 77.8622 75.9252 -1354.11 175.1681 635.88

12 83.7972 77.4119 -1437.02 180.2173 876.73

13 91.8507 76.4487 -1577.70 176.9236 720.69

14 99.6776 77.7050 -1729.11 181.1850 924.22

15 109.6422 78.3952 -1943.42 183.5154 1036.02

16 115.9635 79.2971 -2097.10 186.5732 1182.13

17 116.1266 80.0221 -2133.12 189.0407 1299.58

18 116.4963 79.9122 -2183.93 190.1202 1281.77

19 114.6622 79.7368 -2171.93 189.5267 1253.36

20 107.4077 79.5591 -2036.68 188.9357 1224.58

21 103.3641 77.8678 -1971.29 183.1864 950.58

22 97.9290 78.7250 -1864.04 186.1145 1089.46

23 92.7770 77.6236 -1759.48 182.3752 911.02

24 85.7356 77.2037 -1613.84 180.9592 843.00

Then,

Cooling Load, Btu/hr

Hour Wall Roof Window Infiltration People Lights Equipment Total

1 471.33 751.02 95.32 391.30 0 921.8 614.5 3245.27

2 450.66 700.59 76.08 268.10 0 921.8 614.5 3031.73

3 439.98 657.15 60.80 169.60 0 921.8 614.5 2863.83

4 411.21 610.26 49.31 95.60 0 921.8 614.5 2702.68

5 453.47 588.90 45.53 71.00 0 921.8 614.5 2695.20

6 366.68 571.10 53.24 120.30 0 921.8 614.5 2647.62

7 441.11 541.43 72.30 243.50 0 921.8 614.5 2834.64

8 387.40 519.23 119.38 465.30 0 921.8 614.5 3027.61

8. The Cooling Load

9 505.63 574.11 213.48 785.60 4500 4608.9 3072.6 14260.32

10 533.32 607.02 284.32 1155.20 4500 4608.9 3072.6 14761.36

11 593.75 635.88 384.98 1574.20 4500 4608.9 3072.6 15370.31

12 633.01 876.73 495.96 1968.40 4500 4608.9 3072.6 16155.60

13 692.92 720.69 574.83 2264.10 4500 4608.9 3072.6 16434.04

14 755.44 924.22 619.70 2461.30 4500 4608.9 3072.6 16942.16

15 813.54 1036.02 627.01 2535.20 4500 4608.9 3072.6 17193.27

16 855.21 1182.13 592.77 2461.30 4500 4608.9 3072.6 17272.91

17 879.25 1299.58 524.65 2288.80 4500 4608.9 3072.6 17173.78

18 794.37 1281.77 404.20 2017.70 0 921.8 614.5 6034.34

19 751.92 1253.36 335.69 1697.40 0 921.8 614.5 5574.67

20 639.88 1224.58 270.62 1377.00 0 921.8 614.5 5048.38

21 670.63 950.58 209.64 1106.00 0 921.8 614.5 4473.15

22 536.94 1089.46 168.30 859.50 0 921.8 614.5 4190.50

23 543.20 911.02 137.47 662.40 0 921.8 614.5 3790.39

24 491.54 843.00 114.41 514.50 0 921.8 614.5 3499.75

8.31 Compute the total hourly cooling loads for the building described by the plans

and specifications furnished by your instructor, using the heat balance method.

8.32 Compute the sol-air temperatures for a west-facing wall in Amarillo, Texas for


conditions. The wall has a solar absorption of 0.8, a thermal emissivity of 0.9,

and an exterior surface conductance of 3.0 Btu/(hr-ft2-F).

Solution:


64.178=D

G Btu/(hr-ft2), 62.18=

dG Btu/(hr-ft

2), 56.48=θ deg

6618.056.48coscos ==θ , 96=o

t F

ootoehRhGtt δεα −+=

0=o

hRδε F for vertical surfaces.

26.19762.1864.178 =+=+=dDt

GGG Btu/(hr-ft2)

8.0=α

0.3=o

h Btu/(hr-ft2-F)

( )( ) 60.14800.326.1678.096 =−+=e

t F.

8.33 Compute the sol-air temperatures for a south-facing wall in Billings, Montana

for each hour of the day on July 21. Assume 0.4 percent outdoor design

conditions. The wall has a solar absorption of 0.8, a thermal emissivity of 0.9,

and an exterior surface conductance of 3.0 Btu/(hr-ft2-F).

Solution:


78.176=D

G Btu/(hr-ft2), 43.18=

dG Btu/(hr-ft

2), 56.48=θ deg

6618.056.48coscos ==θ , 93=o

t F

8. The Cooling Load


0=o


21.19543.1878.176 =+=+=dDt

GGG Btu/(hr-ft2)

8.0=α

0.3=o

h Btu/(hr-ft2-F)

( )( ) 06.14500.321.1958.093 =−+=e

t F.



Amarillo, Texas.

Solution:


64.178=D

G Btu/(hr-ft2), 62.18=

dG Btu/(hr-ft

2), 56.48=θ deg


( ) [ ]j

jjDDirect

tGSCTSHG ∑==

5

0

cosθ

or ( ) [ ]j

jjDSLDirect

tGSCAq ∑==

5

0

cosθ&

( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCTSHG

or ( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCAq&

Table 6-2: j

jt

0 -0.00885

1 2.71235

2 -0.62062

3 -7.07329

4 9.75995

5 -3.89922


( )( )( )

( ) ( ) ( )

−+−

−+−=

543

2

56.48cos89922.356.48cos75995.956.48cos07329.7

56.48cos62062.056.48cos71235.200885.064.1788.0100

Directq&

87.023,12=Direct

q& Btu/hr


( )( )( )

−+−−+−=

7

89922.3

6

75995.9

5

07329.7

4

62062.0

3

71235.2

2

00885.062.1828.0100

diffuseq&

207.1190=diffuse

q& Btu/hr

8. The Cooling Load



Billings, Montana.

Solution:


78.176=D

G Btu/(hr-ft2), 43.18=

dG Btu/(hr-ft

2), 56.48=θ deg


( ) [ ]j

jjDDirect

tGSCTSHG ∑==

5

0

cosθ

or ( ) [ ]j

jjDSLDirect

tGSCAq ∑==

5

0

cosθ&

( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCTSHG

or ( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCAq&

Table 6-2: j

jt

0 -0.00885

1 2.71235

2 -0.62062

3 -7.07329

4 9.75995

5 -3.89922


( )( )( )

( ) ( ) ( )

−+−

−+−=

543

2

56.48cos89922.356.48cos75995.956.48cos07329.7

56.48cos62062.056.48cos71235.200885.078.1768.0100

Directq&

68.898,11=Direct

q& Btu/hr


( )( )( )

−+−−+−=

7

89922.3

6

75995.9

5

07329.7

4

62062.0

3

71235.2

2

00885.043.1828.0100

diffuseq&

062.1178=diffuse

q& Btu/hr

8.36 Compute the solar irradiation and sol-air temperatures for a flat roof for the

conditions of Problem 8-32.

Solution: Table B-1a, Amarillo, Texas, Latitude = 35.23 deg, Longitude = 101.70 deg

Dry bulb at 0.4 percent = 96 F, DR = 23.3 F, wind speed = 15 mph


A = 346.6 Btu/hr-ft2, B = 0.186, C = 0.138

l = 35.23 deg

8. The Cooling Load

For horizontal surface

βθ sincos =

( ) NDN

N

tGC

C

CG

+=

3cosθ

0.1=N

C

l = 35.23 deg, d = 20.6 deg

At 3:00 P.M.

h , afternoon = ( )( ) ( )( ) 4515315 ==TIME deg

Then,


( )6.20sin23.35sin6.20cos45cos23.35cossin 1 += −β

04.48=β

( ) 9.269

04.48sin186.0exp

6.346

sinexp

==

=

βB

AG

NDBtu/(hr-ft

2)

( )( ) 2389.269138.004.48sin =+=t

G

96=o

t F


0=o


8.0=α

0.3=o

h Btu/(hr-ft2-F)

( )( ) 47.15900.32388.096 =−+=e

t F.

8.37 If wall 2 from Table 8-17a is exposed to the sol-air temperature profile shown

in Table 8-16, compute the conduction heat flux for hour 17. The room air

temperature is 74 F. Use periodic response factors.

Solution:

Table 8-18, wall 2, Periodic Response Factor

Y Wall 2

0PY 0.000520

1PY 0.001441

2PY 0.006448

3PY 0.012194

4PY 0.015366

5PY 0.016223

6PY 0.015652

7PY 0.014326

8PY 0.012675

9PY 0.010957

8. The Cooling Load

10PY 0.009313

11PY 0.007816

12PY 0.006497

13PY 0.005360

14PY 0.004395

15PY 0.003587

16PY 0.002915

17PY 0.002362

18PY 0.001909

19PY 0.001539

20PY 0.001239

21PY 0.000996

22PY 0.000799

23PY 0.000641

Table 8-16. Sol-air Temperature

Hour et , F

1 80.7

2 79.7

3 78.8

4 78.2

5 78.0

6 78.4

7 80.7

8 87.1

9 92.9

10 98.3

11 103.6

12 108.9

13 120.3

14 138.1

15 151.2

16 157.8

17 157.3

18 148.6

19 131.0

20 101.4

21 86.8

22 84.7

23 83.0

24 81.8

Equation 8-73.

8. The Cooling Load

( )∑ −=′′=

−

23

0,,,,,

nrcnjePnjinconduction

ttYq δθθ

74=rc

t F

Then. For hour 17.

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )7474747474

7474747474

7474747474

7474747474

74747474

19,,2320,,2221,,2122,,2022,,19

23,,1824,,171,,162,,153,,14

4,,135,,126,,117,,108,,9

9,,810,,711,,612,,513,,4

14,,315,,216,,117,,017,,,

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−=′′

jePjePjePjePjeP

jePjePjePjePjeP

jePjePjePjePjeP

jePjePjePjePjeP

jePjePjePjePjinconduction

tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYq

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )746.148000641.074131000799.0

744.101000996.0748.86001239.0747.84001539.07483001909.0

748.81002362.0747.80002915.0747.79003587.0748.78004395.0

742.78005360.07478006497.0744.78007816.0747.80009313.0

741.87010957.0749.92012675.0743.98014326.0746.103015652.0

749.108016223.0743.120015366.0741.138012194.0

742.151006448.0748.157001441.0743.157000520.017,,,

−+−+

−+−+−+−+

−+−+−+−+

−+−+−+−+

−+−+−+−+

−+−+−+

−+−+−=′′jinconduction

q

31058.417,,,

=′′jinconduction

q Btu/(hr-ft2)

8.38 If wall 3 from Table 8-17a is exposed to the sol-air temperature profile shown

in Table 8-16, compute the conduction heat flux for each hour of the day. The

room air temperature is 72 F. Use periodic response factors.

Solution:


Y Wall 3

0PY 0.000530

1PY 0.000454

2PY 0.000446

3PY 0.000727

4PY 0.001332

5PY 0.002005

6PY 0.002544

7PY 0.002884

8PY 0.003039

9PY 0.003046

10PY 0.002949

11PY 0.002783

12PY 0.002576

8. The Cooling Load

13PY 0.002349

14PY 0.002116

15PY 0.001889

16PY 0.001672

17PY 0.001471

18PY 0.001286

19PY 0.001119

20PY 0.000970

21PY 0.000838

22PY 0.000721

23PY 0.000619


Hour et , F

1 80.7

2 79.7

3 78.8

4 78.2

5 78.0

6 78.4

7 80.7

8 87.1

9 92.9

10 98.3

11 103.6

12 108.9

13 120.3

14 138.1

15 151.2

16 157.8

17 157.3

18 148.6

19 131.0

20 101.4

21 86.8

22 84.7

23 83.0

24 81.8

Equation 8-73.

( )∑ −=′′=

−

23

0,,,,,


ttYq δθθ

72=rc

t F

Then. For hour 17.

8. The Cooling Load

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )7272727272

7272727272

7272727272

7272727272

72727272

19,,2320,,2221,,2122,,2022,,19

23,,1824,,171,,162,,153,,14

4,,135,,126,,117,,108,,9

9,,810,,711,,612,,513,,4

14,,315,,216,,117,,017,,,

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−=′′

jePjePjePjePjeP

jePjePjePjePjeP

jePjePjePjePjeP

jePjePjePjePjeP


tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYq

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )726.148000619.072131000721.0

724.101000838.0728.86000970.0727.84001119.07283001286.0

728.81001471.0727.80001672.0727.79001889.0728.78002116.0

722.78002349.07278002576.0724.78002783.0727.80002949.0

721.87003046.0729.92003039.0723.98002884.0726.103002544.0

729.108002005.0723.120001332.0721.138000727.0

722.151000446.0728.157000454.0723.157000530.017,,,

−+−+

−+−+−+−+

−+−+−+−+

−+−+−+−+

−+−+−+−+

−+−+−+


q

860298.017,,,


q Btu/(hr-ft2)

Tabulation; Conduction Heat Fluxes

Hour q ′′ , Btu/(hr-ft2) Hour q ′′ , Btu/(hr-ft

2)

1 1.834797 13 0.827812

2 1.823278 14 0.797059

3 1.772074 15 0.790339

4 1.692400 16 0.809799

5 1.594165 17 0.860298

6 1.485286 18 0.946962

7 1.371827 19 1.070471

8 1.258399 20 1.224457

9 1.148681 21 1.395286

10 1.046254 22 1.562934

11 0.955537 23 1.703281

12 0.881414 24 1.796333

8.39 If roof 2 from Table 8-17b is exposed to the sol-air temperature profile shown

in the last column of Table 8-23, compute the conduction heat flux for hour

15. The room air temperature is 73 F. Use periodic response factors.

Solution:

Table 8-18, roof 2, Periodic Response Factor

Y Roof 2

0PY 0.000004

1PY 0.000658

2PY 0.004270

3PY 0.007757

4PY 0.008259

8. The Cooling Load

5PY 0.006915

6PY 0.005116

7PY 0.003527

8PY 0.002330

9PY 0.001498

10PY 0.000946

11PY 0.000591

12PY 0.000366

13PY 0.000225

14PY 0.000138

15PY 0.000085

16PY 0.000052

17PY 0.000032

18PY 0.000019

19PY 0.000012

20PY 0.000007

21PY 0.000004

22PY 0.000003

23PY 0.000002


Hour et , F

1 68.68

2 67.64

3 66.67

4 65.94

5 65.70

6 66.19

7 67.40

8 79.70

9 103.30

10 126.35

11 147.44

12 164.75

13 176.70

14 182.67

15 182.16

16 175.01

17 162.12

18 144.25

19 122.84

20 99.56

21 77.70

8. The Cooling Load

22 73.48

23 71.53

24 70.07

Equation 8-73.

( )∑ −=′′=

−

23

0,,,,,


ttYq δθθ

73=rc

t F

Then. For hour 15.

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )7373737373

7373737373

7373737373

7373737373

73737373

16,,2317,,2218,,2119,,2020,,19

21,,1822,,1723,,1624,,151,,14

2,,133,,124,,115,,106,,9

7,,88,,79,,610,,511,,4

12,,313,,214,,115,,015,,,

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−=′′

jePjePjePjePjeP

jePjePjePjePjeP

jePjePjePjePjeP

jePjePjePjePjeP


tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYq

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )7301.175000002.07312.162000003.0

7325.144000004.07384.122000007.07356.99000012.0737.77000019.0

7348.73000032.07353.71000052.07307.70000085.07368.68000138.0

7364.67000225.07367.66000366.07394.65000591.0737.65000946.0

7319.66001498.0734.67002330.0737.79003527.0733.103005116.0

7335.126006915.07344.147008259.07375.164007757.0

737.176004270.07367.1820000658.07316.182000004.015,,,

−+−+

−+−+−+−+

−+−+−+−+

−+−+−+−+

−+−+−+−+

−+−+−+


q

352221.215,,,


q Btu/(hr-ft2)

8.40 If roof 3 from Table 8-17b is exposed to the sol-air temperature profile shown

in the last column of Table 8-23, compute the conduction heat flux for each

hour of the day. The room air temperature is 72 F. Use periodic response

factors.

Solution:


Y Roof 3

0PY 0.001590

1PY 0.002817

2PY 0.006883

3PY 0.009367

4PY 0.009723

5PY 0.009224

6PY 0.008501

7PY 0.007766

8PY 0.007076

8. The Cooling Load

9PY 0.006443

10PY 0.005865

11PY 0.005338

12PY 0.004859

13PY 0.004422

14PY 0.004025

15PY 0.003664

16PY 0.003335

17PY 0.003035

18PY 0.002763

19PY 0.002515

20PY 0.002289

21PY 0.002083

22PY 0.001896

23PY 0.001726


Hour et , F

1 68.68

2 67.64

3 66.67

4 65.94

5 65.70

6 66.19

7 67.40

8 79.70

9 103.30

10 126.35

11 147.44

12 164.75

13 176.70

14 182.67

15 182.16

16 175.01

17 162.12

18 144.25

19 122.84

20 99.56

21 77.70

22 73.48

23 71.53

24 70.07

Equation 8-73.

8. The Cooling Load

( )∑ −=′′=

−

23

0,,,,,


ttYq δθθ

72=rc

t F

Then. For hour 15.

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )7272727272

7272727272

7272727272

7272727272

72727272

16,,2317,,2218,,2119,,2020,,19

21,,1822,,1723,,1624,,151,,14

2,,133,,124,,115,,106,,9

7,,88,,79,,610,,511,,4

12,,313,,214,,115,,015,,,

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−+−+

−+−+−+−=′′

jePjePjePjePjeP

jePjePjePjePjeP

jePjePjePjePjeP

jePjePjePjePjeP


tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYtY

tYtYtYtYq

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )7201.175001726.07212.162001896.0

7225.144002083.07284.122002289.07256.99002515.0727.77002763.0

7248.73003035.07253.71003335.07207.70003664.07268.68004025.0

7264.67004422.07267.66004859.07294.65005338.0727.65005865.0

7219.66006443.0724.67007076.0727.79007766.0723.103008501.0

7235.126009224.07244.147009723.07275.164009367.0

727.176006883.07267.182002817.07216.182001590.015,,,

−+−+

−+−+−+−+

−+−+−+−+

−+−+−+−+

−+−+−+−+

−+−+−+


q

135683.415,,,


q Btu/(hr-ft2)


Hour q ′′ , Btu/(hr-ft2) Hour q ′′ , Btu/(hr-ft

2)

1 5.318479 13 2.833505

2 4.835486 14 3.436577

3 4.376813 15 4.135683

4 3.945417 16 4.865839

5 3.541579 17 5.560573

6 3.165454 18 6.156739

7 2.818726 19 6.600222

8 2.504999 20 6.849346

9 2.242781 21 6.878035

10 2.094136 22 6.680250

11 2.132466 23 6.293914

12 2.384542 24 5.815357

8.41 If wall 2 from Table 8-17a is exposed to the sol-air temperature profile

calculated in Problem 8-31, compute the conduction heat flux for each hour of

the day. The room air temperature is 74 F. Use periodic response factors.

8.42 Determine the transmitted and absorbed solar radiation through a 100 ft2


Amarillo, Texas.

Solution:


64.178=D

G Btu/(hr-ft2), 62.18=

dG Btu/(hr-ft

2), 56.48=θ deg

8. The Cooling Load


( ) [ ]j

jjDDirect

tGSCTSHG ∑==

5

0

cosθ

or ( ) [ ]j

jjDSLDirect

tGSCAq ∑==

5

0

cosθ&

( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCTSHG

or ( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCAq&

Table 6-2: j

jt

0 -0.00885

1 2.71235

2 -0.62062

3 -7.07329

4 9.75995

5 -3.89922


( )( )

( )

( ) ( )

( )

−

+−

−+−

=5

43

2

56.48cos89922.3

56.48cos75995.956.48cos07329.7

56.48cos62062.056.48cos71235.200885.0

64.1788.0100Direct

q&

87.023,12=Direct

q& Btu/hr


( )( )( )

−

+−−+−

=

7

89922.3

6

75995.9

5

07329.7

4

62062.0

3

71235.2

2

00885.0

62.1828.0100diffuse

q&

207.1190=diffuse

q& Btu/hr

Transmitted solar heat gain

( ) ( )shadeshadesunlitSLTSHG

TSHGFSCATSHGFSCAq +=&

[ ] ∑+

+∑===

5

0

5

0 22cos

j

j

d

j

jjDsunlit

j

tGtGTSHGF θ

∑+

==

5

0 22

j

j

dshadej

tGTSHGF

08.13214207.119087.12023 =+=+=diffuseDirectTSHG

qqq &&& Btu/hr

Absorbed solar heat gain

( ) ( )[ ]ishadeshadesunlitSLASHG

NASHGFSCAASHGFSCAq +=&

[ ] ∑+

+∑===

5

0

5

0 22cos

j

j

d

j

jjDsunlit

j

aGaGASHGF θ

∑+

==

5

0 22

j

j

dshadej

aGASHGF

8. The Cooling Load

267.0=+

=oi

i

ihh

hN from page 180, Chapter 6.

Table 6-2: j

ja

0 0.01154

1 0.77674

2 -3.94657

3 8.57811

4 -8.38135

5 3.01188

[ ] ( ) ( )

( ) ( )54

325

0

56.48cos01188.356.48cos38135.8

56.48cos57811.856.48cos94657.356.48cos77674.001154.0cos

+−

+−+=∑=

j

jj

a θ

[ ] 0581.0cos5

0

=∑=

j

jj

a θ

+−+−+=∑

+= 7

01188.3

6

38135.8

5

57811.8

4

94657.3

3

77674.0

2

01154.02

22

5

0j

j

j

a

0541.02

25

0

=∑+=j

j

j

a

0=shade

A

[ ]i

j

j

d

j

jjDSLASHG

Nj

aGaGAq

∑

++∑=

==

5

0

5

0 22cosθ&

( )( ) ( )( ) ( )( )[ ]( )267.00541.062.180581.064.1788.0100 +=ASHG

q&

2.243=ASHG

q& Btu/hr

8.43 Determine the transmitted and absorbed solar radiation through a 100 ft2


Billings, Montana.

Solution:


78.176=D

G Btu/(hr-ft2), 43.18=

dG Btu/(hr-ft

2), 56.48=θ deg


( ) [ ]j

jjDDirect

tGSCTSHG ∑==

5

0

cosθ

or ( ) [ ]j

jjDSLDirect

tGSCAq ∑==

5

0

cosθ&

( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCTSHG

or ( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCAq&

8. The Cooling Load

Table 6-2: j

jt

0 -0.00885

1 2.71235

2 -0.62062

3 -7.07329

4 9.75995

5 -3.89922


( )( )( )

( ) ( ) ( )

−+−

−+−=

543

2

56.48cos89922.356.48cos75995.956.48cos07329.7

56.48cos62062.056.48cos71235.200885.078.1768.0100

Directq&

68.898,11=Direct

q& Btu/hr


( )( )( )

−+−−+−=

7

89922.3

6

75995.9

5

07329.7

4

62062.0

3

71235.2

2

00885.043.1828.0100

diffuseq&

062.1178=diffuse

q& Btu/hr




[ ] ∑+

+∑===

5

0

5

0 22cos

j

j

d

j

jjDsunlit

j

tGtGTSHGF θ

∑+

==

5

0 22

j

j

dshadej

tGTSHGF


qqq &&& Btu/hr




[ ] ∑+

+∑===

5

0

5

0 22cos

j

j

d

j

jjDsunlit

j

aGaGASHGF θ

∑+

==

5

0 22

j

j

dshadej

aGASHGF

267.0=+

=oi

i

ihh


Table 6-2: j

ja

0 0.01154

1 0.77674

2 -3.94657

3 8.57811

4 -8.38135

5 3.01188

8. The Cooling Load

[ ] ( ) ( )

( ) ( )54

325

0

56.48cos01188.356.48cos38135.8

56.48cos57811.856.48cos94657.356.48cos77674.001154.0cos

+−

+−+=∑=

j

jj

a θ

[ ] 0581.0cos5

0

=∑=

j

jj

a θ

+−+−+=∑

+= 7

01188.3

6

38135.8

5

57811.8

4

94657.3

3

77674.0

2

01154.02

22

5

0j

j

j

a

0541.02

25

0

=∑+=j

j

j

a

0=shade

A

[ ]i

j

j

d

j

jjDSLASHG

Nj

aGaGAq

∑

++∑=

==

5

0

5

0 22cosθ&

( )( ) ( )( ) ( )( )[ ]( )267.00541.043.180581.078.1768.0100 +=ASHG

q&

7.240=ASHG

q& Btu/hr

8.44 For the conduction heat fluxes determined in Problem 8-37, determine the

hourly conduction heat gains if the wall area is 800 ft2, and determine the

hourly cooling loads if the zone matches the MW 1 zone from Table 8-21. Plot

and compare the hourly heat gains vs. the hourly cooling loads.

Solution:


Y Wall 2

0PY 0.000520

1PY 0.001441

2PY 0.006448

3PY 0.012194

4PY 0.015366

5PY 0.016223

6PY 0.015652

7PY 0.014326

8PY 0.012675

9PY 0.010957

10PY 0.009313

11PY 0.007816

12PY 0.006497

13PY 0.005360

14PY 0.004395

15PY 0.003587

16PY 0.002915

17PY 0.002362

8. The Cooling Load

18PY 0.001909

19PY 0.001539

20PY 0.001239

21PY 0.000996

22PY 0.000799

23PY 0.000641


Hour et , F

1 80.7

2 79.7

3 78.8

4 78.2

5 78.0

6 78.4

7 80.7

8 87.1

9 92.9

10 98.3

11 103.6

12 108.9

13 120.3

14 138.1

15 151.2

16 157.8

17 157.3

18 148.6

19 131.0

20 101.4

21 86.8

22 84.7

23 83.0

24 81.8

Equation 8-73.

( )∑ −=′′=

−

23

0,,,,,


ttYq δθθ

74=rc

t F

Cooling load at the current hour

δθδθδθδθθθ 2323332210, −−−−+++++= qrqrqrqrqrq

CL&L&&&&&

Radiant Time Factors for MW1 Zone (Nonsolar for wall)

r MW1

0r 0.5166863

1r 0.2083323

8. The Cooling Load

2r 0.1084617

3r 0.0623217

4r 0.0378535

5r 0.0237341

6r 0.0151477

7r 0.0097658

8r 0.0063351

9r 0.0041280

10r 0.0027008

11r 0.0017737

12r 0.0011722

13r 0.0007799

14r 0.0005248

15r 0.0003575

16r 0.0002476

17r 0.0001764

18r 0.0001313

19r 0.0001009

20r 0.0000790

21r 0.0000673

22r 0.0000581

23r 0.0000518

Heat Gain Type Recommended Radiative

Fraction

Recommended Convective

Fraction

Conduction heat gain

through walls 0.63 0.37


Hour θ,,, jinconductionq ′′ ,

Btu/(hr-ft2)

θ,,, jinconductionq& ,

Btu/hr

Radiative Heat

Gain,

Btuhr

Convective

Heat Gain,

Btuhr

1 6.826660 5461.328 3440.637 2020.691

2 6.123124 4898.499 3086.054 1812.445

3 5.416716 4333.373 2730.025 1603.348

4 4.744078 3795.262 2391.015 1404.247

5 4.124142 3299.314 2078.568 1220.746

6 3.565991 2852.793 1797.259 1055.533

7 3.074145 2459.316 1549.369 909.9469

8 2.653095 2122.476 1337.16 785.3161

9 2.316047 1852.838 1167.288 685.5499

10 2.090090 1672.072 1053.405 618.6666

8. The Cooling Load

11 1.995356 1596.285 1005.659 590.6254

12 2.033981 1627.185 1025.126 602.0584

13 2.196728 1757.382 1107.151 650.2315

14 2.476091 1980.873 1247.95 732.9229

15 2.894856 2315.885 1459.007 856.8774

16 3.501937 2801.550 1764.976 1036.573

17 4.310580 3448.464 2172.532 1275.932

18 5.267715 4214.172 2654.928 1559.244

19 6.268377 5014.702 3159.262 1855.44

20 7.175417 5740.334 3616.41 2123.923

21 7.832174 6265.739 3947.416 2318.324

22 8.091384 6473.107 4078.058 2395.05

23 7.930616 6344.493 3997.03 2347.462

24 7.463121 5970.497 3761.413 2209.084

Hour CL

q,θ

& , Radiative Cooling

Load

Btuhr

Cooling Load

Btu/hr

1 3589.827 5610.518

2 3339.806 5152.251

3 3049.102 4652.45

4 2743.216 4147.463

5 2439.781 3660.527

6 2150.43 3205.963

7 1882.753 2792.7

8 1642.257 2427.573

9 1435.515 2121.065

10 1272.891 1891.557

11 1164.409 1755.035

12 1115.414 1717.473

13 1126.43 1776.662

14 1196.284 1929.207

15 1330.703 2187.58

16 1544.805 2581.378

17 1848.453 3124.385

18 2232.64 3791.884

19 2667.183 4522.623

20 3104.091 5228.014

21 3481.797 5800.121

22 3734.36 6129.41

23 3825.4 6172.862

24 3765.087 5974.171

Plot and comparison the hourly heat gains vs. the hourly cooling loads.

8. The Cooling Load



hourly cooling loads if the zone matches the MW 2 zone from Table 8-21. Plot


Solution:


Y Roof 2

0PY 0.000004

1PY 0.000658

2PY 0.004270

3PY 0.007757

4PY 0.008259

5PY 0.006915

6PY 0.005116

7PY 0.003527

8PY 0.002330

9PY 0.001498

10PY 0.000946

11PY 0.000591

12PY 0.000366

13PY 0.000225

14PY 0.000138

15PY 0.000085

16PY 0.000052

17PY 0.000032

18PY 0.000019

19PY 0.000012

8. The Cooling Load

20PY 0.000007

21PY 0.000004

22PY 0.000003

23PY 0.000002


Hour et , F

1 68.68

2 67.64

3 66.67

4 65.94

5 65.70

6 66.19

7 67.40

8 79.70

9 103.30

10 126.35

11 147.44

12 164.75

13 176.70

14 182.67

15 182.16

16 175.01

17 162.12

18 144.25

19 122.84

20 99.56

21 77.70

22 73.48

23 71.53

24 70.07

Equation 8-73.

( )∑ −=′′=

−

23

0,,,,,


ttYq δθθ

73=rc

t F



CL&L&&&&&

Radiant Time Factors for MW2 Zone (Nonsolar for roof)

r MW2

0r 0.2550941

1r 0.1139586

2r 0.0695853

3r 0.0513341

4r 0.0425855

8. The Cooling Load

5r 0.0377131

6r 0.0346113

7r 0.0324116

8r 0.0307118

9r 0.0293093

10r 0.0280908

11r 0.0269960

12r 0.0259840

13r 0.0250392

14r 0.0241403

15r 0.0232829

16r 0.0224620

17r 0.0216740

18r 0.0209139

19r 0.0201816

20r 0.0194781

21r 0.0188000

22r 0.0181458

23r 0.0175139


Fraction


Fraction


through roof 0.84 0.16



Btu/(hr-ft2)


Btu/hr

Radiative Heat

Gain,

Btuhr

Convective

Heat Gain,

Btuhr

1 1.327703 1327.703 1115.271 212.432

2 0.868037 868.037 729.1511 138.886

3 0.524499 524.499 440.5792 83.920

4 0.275061 275.061 231.0512 44.010

5 0.095494 95.494 80.21496 15.279

6 -0.034007 -34.007 -28.5659 -5.441

7 -0.126443 -126.443 -106.212 -20.231

8 -0.188270 -188.270 -158.147 -30.123

9 -0.214827 -214.827 -180.455 -34.372

10 -0.161902 -161.902 -135.998 -25.904

11 0.044532 44.532 37.40688 7.125

12 0.440807 440.807 370.2779 70.529

13 1.001723 1001.723 841.4473 160.276

14 1.664517 1664.517 1398.194 266.323

8. The Cooling Load

15 2.352221 2352.221 1975.866 376.355

16 2.989244 2989.244 2510.965 478.279

17 3.510027 3510.027 2948.423 561.604

18 3.862843 3862.843 3244.788 618.055

19 4.013206 4013.206 3371.093 642.113

20 3.945164 3945.164 3313.938 631.226

21 3.661131 3661.131 3075.35 585.781

22 3.182315 3182.315 2673.145 509.170

23 2.560590 2560.590 2150.896 409.694

24 1.906732 1906.732 1601.655 305.077

Hour CL

q,θ


Load

Btuhr

Cooling Load

Btu/hr

1 1534.366 1746.799

2 1366.341 1505.227

3 1219.922 1303.842

4 1096.499 1140.509

5 993.5208 1008.8

6 907.2695 901.8284

7 834.4896 814.2587

8 773.1012 742.978

9 723.5337 689.1614

10 695.0685 669.1641

11 706.636 713.7611

12 774.1008 844.6299

13 900.6262 1060.902

14 1077.049 1343.372

15 1285.924 1662.279

16 1505.818 1984.097

17 1714.674 2276.278

18 1892.026 2510.081

19 2020.79 2662.903

20 2088.472 2719.698

21 2087.881 2673.662

22 2017.739 2526.909

23 1885.527 2295.222

24 1714.504 2019.581

8. The Cooling Load




hourly cooling loads if the zone matches the HW zone from Table 8-21. Plot


Solution:


Y Roof 3

0PY 0.001590

1PY 0.002817

2PY 0.006883

3PY 0.009367

4PY 0.009723

5PY 0.009224

6PY 0.008501

7PY 0.007766

8PY 0.007076

9PY 0.006443

10PY 0.005865

11PY 0.005338

12PY 0.004859

13PY 0.004422

14PY 0.004025

15PY 0.003664

16PY 0.003335

17PY 0.003035

18PY 0.002763

19PY 0.002515

8. The Cooling Load

20PY 0.002289

21PY 0.002083

22PY 0.001896

23PY 0.001726


Hour et , F

1 68.68

2 67.64

3 66.67

4 65.94

5 65.70

6 66.19

7 67.40

8 79.70

9 103.30

10 126.35

11 147.44

12 164.75

13 176.70

14 182.67

15 182.16

16 175.01

17 162.12

18 144.25

19 122.84

20 99.56

21 77.70

22 73.48

23 71.53

24 70.07

Equation 8-73.

( )∑ −=′′=

−

23

0,,,,,


ttYq δθθ

72=rc

t F



CL&L&&&&&

Radiant Time Factors for HW Zone (Nonsolar for roof)

r HW

0r 0.2241915

1r 0.0768588

2r 0.057783

3r 0.0501948

4r 0.0456539

8. The Cooling Load

5r 0.0424286

6r 0.0398994

7r 0.0377894

8r 0.03596

9r 0.0343321

10r 0.0328598

11r 0.0315095

12r 0.0302635

13r 0.0291043

14r 0.0280197

15r 0.0270197

16r 0.0260402

17r 0.0251315

18r 0.0242692

19r 0.0234499

20r 0.0226689

21r 0.0219226

22r 0.0212093

23r 0.0205246


Fraction


Fraction


through roof 0.84 0.16



Btu/(hr-ft2)


Btu/hr

Radiative Heat

Gain,

Btuhr

Convective

Heat Gain,

Btuhr

1 5.318479 6382.175 5361.027 1021.148

2 4.835486 5802.583 4874.17 928.413

3 4.376813 5252.176 4411.828 840.348

4 3.945417 4734.500 3976.98 757.520

5 3.541579 4249.895 3569.912 679.983

6 3.165454 3798.545 3190.778 607.767

7 2.818726 3382.471 2841.276 541.195

8 2.504999 3005.999 2525.039 480.960

9 2.242781 2691.337 2260.723 430.614

10 2.094136 2512.963 2110.889 402.074

11 2.132466 2558.959 2149.526 409.433

12 2.384542 2861.450 2403.618 457.832

13 2.833505 3400.206 2856.173 544.033

14 3.436577 4123.892 3464.07 659.823

8. The Cooling Load

15 4.135683 4962.820 4168.768 794.051

16 4.865839 5839.007 4904.766 934.241

17 5.560573 6672.688 5605.058 1067.630

18 6.156739 7388.087 6205.993 1182.094

19 6.600222 7920.266 6653.024 1267.243

20 6.849346 8219.215 6904.141 1315.074

21 6.878035 8253.642 6933.059 1320.583

22 6.680250 8016.300 6733.692 1282.608

23 6.293914 7552.697 6344.265 1208.431

24 5.815357 6978.428 5861.88 1116.549

Hour CL

q,θ


Load

Btuhr

Cooling Load

Btu/hr

1 4984.0208 6005.1688

2 4877.1210 5805.5340

3 4758.3912 5598.7392

4 4631.4482 5388.9682

5 4498.7533 5178.7363

6 4362.2897 4970.0567

7 4224.0738 4765.2688

8 4086.4113 4567.3713

9 3954.7026 4385.3166

10 3844.1950 4246.2690

11 3775.5998 4185.0328

12 3762.9000 4220.7320

13 3810.6849 4354.7179

14 3915.8395 4575.6625

15 4069.1773 4863.2283

16 4257.1152 5191.3562

17 4463.0434 5530.6734

18 4668.6910 5850.7850

19 4855.7853 6123.0283

20 5007.4101 6322.4841

21 5109.3355 6429.9185

22 5151.6479 6434.2559

23 5134.5171 6342.9481

24 5073.2483 6189.7973

8. The Cooling Load


8.47 For the solar heat gains determined in Problem 8-41, determine the hourly

cooling loads if the zone matches the MW 1 zone from Table 8-21. Plot and

compare the hourly heat gains vs. the hourly cooling loads.

8.48 For the solar heat gains determined in Problem 8-42, determine the hourly

cooling loads if the zone matches the MW 2 zone from Table 8-21. Plot and

compare the hourly heat gains vs. the hourly cooling loads.

Solution:


64.178=D

G Btu/(hr-ft2), 62.18=

dG Btu/(hr-ft

2), 56.48=θ deg


( ) [ ]j

jjDDirect

tGSCTSHG ∑==

5

0

cosθ

or ( ) [ ]j

jjDSLDirect

tGSCAq ∑==

5

0

cosθ&

( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCTSHG

or ( ) ∑+

==

5

0 22

j

j

ddiffusej

tGSCAq&

Table 6-2: j

jt

0 -0.00885

1 2.71235

2 -0.62062

3 -7.07329

4 9.75995

5 -3.89922


8. The Cooling Load

( )( )

( )

( ) ( )

( )

−

+−

−+−

=5

43

2

56.48cos89922.3

56.48cos75995.956.48cos07329.7

56.48cos62062.056.48cos71235.200885.0

64.1788.0100Direct

q&

87.023,12=Direct

q& Btu/hr


( )( )( )

−

+−−+−

=

7

89922.3

6

75995.9

5

07329.7

4

62062.0

3

71235.2

2

00885.0

62.1828.0100diffuse

q&

207.1190=diffuse

q& Btu/hr




[ ] ∑+

+∑===

5

0

5

0 22cos

j

j

d

j

jjDsunlit

j

tGtGTSHGF θ

∑+

==

5

0 22

j

j

dshadej

tGTSHGF


qqq &&& Btu/hr




[ ] ∑+

+∑===

5

0

5

0 22cos

j

j

d

j

jjDsunlit

j

aGaGASHGF θ

∑+

==

5

0 22

j

j

dshadej

aGASHGF

267.0=+

=oi

i

ihh


Table 6-2: j

ja

0 0.01154

1 0.77674

2 -3.94657

3 8.57811

4 -8.38135

5 3.01188

[ ] ( ) ( )

( ) ( )54

325

0

56.48cos01188.356.48cos38135.8

56.48cos57811.856.48cos94657.356.48cos77674.001154.0cos

+−

+−+=∑=

j

jj

a θ

[ ] 0581.0cos5

0

=∑=

j

jj

a θ

+−+−+=∑

+= 7

01188.3

6

38135.8

5

57811.8

4

94657.3

3

77674.0

2

01154.02

22

5

0j

j

j

a

8. The Cooling Load

0541.02

25

0

=∑+=j

j

j

a

0=shade

A

[ ]i

j

j

d

j

jjDSLASHG

Nj

aGaGAq

∑

++∑=

==

5

0

5

0 22cosθ&

( )( ) ( )( ) ( )( )[ ]( )267.00541.062.180581.064.1788.0100 +=ASHG

q&

2.243=ASHG

q& Btu/hr

Tabulation for other time.

TIME θ DG , Btu/(hr-ft

2-F) d

G , Btu/(hr-ft2-F) TSHG

q& ,

Btu/hr

ASHGq& ,

Btu/hr

Total

Heat

Gain,

Btu/hr

6:00 A.M. IN SHADE 0 0 0

7:00 A.M. IN SHADE 0 0 0

8:00 A.M. IN SHADE 0 0 0

9:00 A.M. IN SHADE 0 0 0

10:00 A.M. IN SHADE 0 0 0

11:00 A.M. IN SHADE 0 0 0

12:00 NOON 90.00 0.00 19.73 1261.16 22.79 1283.95

1:00 P.M. 75.98 68.92 19.63 4241.02 116.45 4357.47

2:00 P.M. 62.09 130.84 19.29 9400.27 177.38 9577.65

3:00 P.M. 48.56 178.64 18.62 13,214.08243.11 13457.19

4:00 P.M. 35.84 204.48 17.40 15,242.35248.76 15491.11

5:00 P.M. 25.29 197.04 15.04 14,746.12213.68 14959.8

6:00 P.M. 20.60 129.76 9.56 9,702.76 135.56 9838.32

8. The Cooling Load


Fraction


Fraction

Transmitted solar radiation 1 0

Absorbed solar radiation 0.63 0.37

Radiant Time Factors for MW2 Zone (Solar for window (transmitted), nonsolar for

window (absorbed))

r MW2

Solar

MW2

Nonsolar

0r 0.1845232 0.2550941

1r 0.0965271 0.1139586

2r 0.0678909 0.0695853

3r 0.0544979 0.0513341

4r 0.0471169 0.0425855

5r 0.0425672 0.0377131

6r 0.0394930 0.0346113

7r 0.0372431 0.0324116

8r 0.0354716 0.0307118

9r 0.0339895 0.0293093

10r 0.0326910 0.0280908

11r 0.0315102 0.0269960

12r 0.0304145 0.0259840

13r 0.0293807 0.0250392

14r 0.0283964 0.0241403

15r 0.0274512 0.0232829

16r 0.0265404 0.0224620

17r 0.0256644 0.0216740

18r 0.0248185 0.0209139

19r 0.0239991 0.0201816

20r 0.0232084 0.0194781

21r 0.0224435 0.0188000

22r 0.0217037 0.0181458

23r 0.0209893 0.0175139

Transmitted Solar

Heat Gain Absorbed Solar Heat Gain

Hours

Radiative Portion Radiative Portion Convective Portion

1 0.00 0.000 0.000

2 0.00 0.000 0.000

3 0.00 0.000 0.000

4 0.00 0.000 0.000

5 0.00 0.000 0.000

8. The Cooling Load

6 0.00 0.000 0.000

7 0.00 0.000 0.000

8 0.00 0.000 0.000

9 0.00 0.000 0.000

10 0.00 0.000 0.000

11 0.00 0.000 0.000

12 1261.16 14.358 8.432

13 4241.02 73.364 43.087

14 9400.27 111.749 65.631

15 13,214.08 153.159 89.951

16 15,242.35 156.719 92.041

17 14,746.12 134.618 79.062

18 9,702.76 85.403 50.157

19 0.00 0.000 0.000

20 0.00 0.000 0.000

21 0.00 0.000 0.000

22 0.00 0.000 0.000

23 0.00 0.000 0.000

24 0.00 0.000 0.000

Transmitted Solar

Heat Gain

Absorbed Solar

Heat Gain Hours

Radiative Portion

Cooling Load

Radiative Portion

Cooling Load

Total Cooling Load

1 2296.737 21.081 2317.818

2 2206.373 20.193 2226.566

3 2125.283 19.399 2144.682

4 2050.5 18.669 2069.169

5 1980.269 17.986 1998.255

6 1913.556 17.340 1930.896

7 1849.713 16.723 1866.436

8 1788.342 16.133 1804.475

9 1729.19 15.566 1744.756

10 1672.104 15.021 1687.125

11 1616.964 14.496 1631.46

12 1770.772 25.842 1805.046

13 2305.585 75.466 2424.138

14 3393.872 113.217 3572.72

15 4560.078 154.391 4804.42

16 5536.922 165.534 5794.497

17 6058.874 152.817 6290.753

18 5641.988 114.074 5806.219

19 4005.592 40.886 4046.478

20 3316.949 31.990 3348.939

21 2937.656 27.624 2965.28

22 2697.786 25.088 2722.874

23 2529.678 23.387 2553.065

24 2401.49 22.114 2423.604

8. The Cooling Load


8.49 A room has an internal heat gain of 2000 W, 50 percent radiative and 50

percent convective, from 8 A.M. to 6 P.M., and 200 W with the same

radiative-convective split the rest of the day. If the room matches the LW zone

from Table 8-21, determine the hourly cooling loads. Plot and compute the

hourly heat gains vs. the hourly cooling loads.

Solution:



CL&L&&&&&

Radiant Time Factors for LW Zone (Nonsolar)

r LW

0r 0.5061886

1r 0.2296165

2r 0.1186367

3r 0.0639036

4r 0.0353306

5r 0.0198884

6r 0.0113417

7r 0.0065336

8r 0.0037952

9r 0.0022230

10r 0.0013148

11r 0.0007872

12r 0.0004812

13r 0.0003009

14r 0.0001962

15r 0.0001352

8. The Cooling Load

16r 0.0000989

17r 0.0000781

18r 0.0000666

19r 0.0000598

20r 0.0000552

21r 0.0000525

22r 0.0000510

23r 0.0000509

Hours

Radiative

Portion

Heat Gain,

W

Convective

Portion of

Heat Gain,

W

Heat Gain,

W

Radiative

Portion of

Cooling

Load, W

Cooling

Load, W

1 100 100 200 114.3982 214.3982

2 100 100 200 108.5883 208.5883

3 100 100 200 105.2325 205.2325

4 100 100 200 103.2857 203.2857

5 100 100 200 102.1520 202.1520

6 100 100 200 101.4908 201.4908

7 100 100 200 101.1036 201.1036

8 100 100 200 100.8786 200.8786

9 1000 1000 2000 556.2718 1556.2718

10 1000 1000 2000 762.8049 1762.8049

11 1000 1000 2000 869.4890 1869.4890

12 1000 1000 2000 926.9319 1926.9319

13 1000 1000 2000 958.6695 1958.6695

14 1000 1000 2000 976.5152 1976.5152

15 1000 1000 2000 986.6731 1986.6731

16 1000 1000 2000 992.5061 1992.5061

17 1000 1000 2000 995.8759 1995.8759

18 1000 1000 2000 997.8308 1997.8308

19 100 100 200 543.4443 643.4443

20 100 100 200 337.4980 437.4980

21 100 100 200 231.1580 331.1580

22 100 100 200 173.9156 273.9156

23 100 100 200 142.2946 242.2946

24 100 100 200 124.5167 224.5167

8. The Cooling Load


8.50 A room has an internal heat gain of 2000 W, 50 percent radiative and 50

percent convective, from 8 A.M. to 6 P.M., and 200 W with the same

radiative-convective split the rest of the day. If the room matches the MW 2

zone from Table 8-21, determine the hourly cooling loads. Plot and compute

the hourly heat gains vs. the hourly cooling loads.

Solution:



CL&L&&&&&

Radiant Time Factors for MW2 Zone (Nonsolar)

r MW2

0r 0.2550941

1r 0.1139586

2r 0.0695853

3r 0.0513341

4r 0.0425855

5r 0.0377131

6r 0.0346113

7r 0.0324116

8r 0.0307118

9r 0.0293093

10r 0.0280908

11r 0.0269960

12r 0.0259840

13r 0.0250392

14r 0.0241403

8. The Cooling Load

15r 0.0232829

16r 0.0224620

17r 0.0216740

18r 0.0209139

19r 0.0201816

20r 0.0194781

21r 0.0188000

22r 0.0181458

23r 0.0175139

Hours

Radiative

Portion

Heat Gain,

W

Convective

Portion of

Heat Gain,

W

Heat Gain,

W

Radiative

Portion of

Cooling

Load, W

Cooling

Load, W

1 100 100 200 342.5868 442.5868

2 100 100 200 332.9230 432.9230

3 100 100 200 324.1049 424.1049

4 100 100 200 315.8900 415.8900

5 100 100 200 308.1385 408.1385

6 100 100 200 300.7621 400.7621

7 100 100 200 293.7077 393.7077

8 100 100 200 286.9350 386.9350

9 1000 1000 2000 494.7934 1494.7934

10 1000 1000 2000 576.4015 1576.4015

11 1000 1000 2000 618.8125 1618.8125

12 1000 1000 2000 645.5066 1645.5066

13 1000 1000 2000 665.0110 1665.0110

14 1000 1000 2000 680.7894 1680.7894

15 1000 1000 2000 694.4093 1694.4093

16 1000 1000 2000 706.6597 1706.6597

17 1000 1000 2000 717.9691 1717.9691

18 1000 1000 2000 728.5850 1728.5850

19 100 100 200 524.2820 624.2820

20 100 100 200 446.0156 546.0156

21 100 100 200 406.7745 506.7745

22 100 100 200 383.1091 483.1091

23 100 100 200 366.5084 466.5084

24 100 100 200 353.5212 453.5212

8. The Cooling Load


8.51 Compare the results from Problems 8-49 and 8-50. How do the damping and

time delay effects of the two zones compare?

Solution:

Problem 8-49.

Dampening Effect = 2000 W - 1997.8308 W = 2.17 W, At 6:00 P.M. (Peak)

Problem 8-50.

Dampening Effect = 2000 W – 1728.5850 W = 271.4 W, At 6:00 P.M. (Peak)

Therefore same time delay but Problem 8-50 has better dampening effect.

8.52 For the heat gains specified in Problem 8-17, determine the hourly sensible

and latent cooling loads if the zone is the LW zone from Table 8-21.

Solution:



Latent Heat = 20(59) = 1180 W

Lighting Load:

suFWFq =&

W = 2.5(4000) = 10,000 W

0.1=u

F

2.1=s

F

( )( )( ) 000,122.10.1000,10 ==q& W


ulFCPFq =&

C = 1.0 W/W

0.1=l

F

0.1=u

F

8. The Cooling Load

)4000)(0.1(=P = 4000 W

( )( )( )( )0.10.140000.1=q& = 4000 W



Occupants 1460 1180

Lighting 12,000 0

Equipment 4,000 0

Total 17,460 1180


20.



Occupants 70 1022 30 438

Lighting 59 7080 41 4920

Equipment 70 2800 30 1200

Total 10,902 6,558



CL&L&&&&&

Radiant Time Factors for LW Zone (Nonsolar)

r LW

0r 0.5061886

1r 0.2296165

2r 0.1186367

3r 0.0639036

4r 0.0353306

5r 0.0198884

6r 0.0113417

7r 0.0065336

8r 0.0037952

9r 0.0022230

10r 0.0013148

11r 0.0007872

12r 0.0004812

13r 0.0003009

14r 0.0001962

15r 0.0001352

16r 0.0000989

17r 0.0000781

18r 0.0000666

8. The Cooling Load

19r 0.0000598

20r 0.0000552

21r 0.0000525

22r 0.0000510

23r 0.0000509

Hours

Radiative

Portion

of

Sensible

Heat

Gain, W

Convective

Portion of

Sensible

Heat Gain,

W

Sensible

Heat

Gain, W

Radiative

Portion of

Sensible

Cooling

Load, W

Sensible

Cooling

Load, W

Latent

Cooling

Load, W

1 0 0 0 148.0019 148.0019 0

2 0 0 0 88.0902 88.0902 0

3 0 0 0 53.4499 53.4499 0

4 0 0 0 33.3379 33.3379 0

5 0 0 0 21.6222 21.6222 0

6 0 0 0 14.7820 14.7820 0

7 0 0 0 10.7811 10.7811 0

8 0 0 0 8.4558 8.4558 0

9 10902 6558 17460 5525.0181 12083.0181 1180

10 10902 6558 17460 8026.9619 14584.9619 1180

11 10902 6558 17460 9319.3405 15877.3405 1180

12 10902 6558 17460 10015.2101 16573.2101 1180

13 10902 6558 17460 10399.6842 16957.6842 1180

14 10902 6558 17460 10615.8732 17173.8732 1180

15 10902 6558 17460 10738.9289 17296.9289 1180

16 10902 6558 17460 10809.5916 17367.5916 1180

17 10902 6558 17460 10850.4113 17408.4113 1180

18 7080 4920 12000 8939.6332 13859.6332 0

19 0 0 0 4492.5576 4492.5576 0

20 0 0 0 2422.0254 2422.0254 0

21 0 0 0 1343.0840 1343.0840 0

22 0 0 0 758.8934 758.8934 0

23 0 0 0 434.8782 434.8782 0

24 0 0 0 252.1943 252.1943 0

8.53 For the heat gains specified in Problems 8-18, determine the hourly sensible

and latent cooling loads if the zone is the MW 1 zone from Table 8-21.

Solution:



Latent Heat = 40(59) = 2360 W

8. The Cooling Load

Lighting Load:

suFWFq =&

W = 28(750) = 21,000 W

0.1=u

F

5.1=s

F

( )( )( ) 500,315.10.1000,21 ==q& W


ulFCPFq =&

C = 1.0 W/W

0.1=l

F

0.1=u

F

P = 5000 W

( )( )( )( )0.10.150000.1=q& = 5000 W



Occupants 2920 2360

Lighting 21,000 0

Equipment 5,000 0

Total 28,920 2360


20.



Occupants 70 2044 30 876

Lighting 59 12,390 41 8610

Equipment 70 3500 30 1500

Total 17,934 10,986



CL&L&&&&&


r MW1

0r 0.5166863

1r 0.2083323

2r 0.1084617

3r 0.0623217

4r 0.0378535

5r 0.0237341

6r 0.0151477

7r 0.0097658

8. The Cooling Load

8r 0.0063351

9r 0.0041280

10r 0.0027008

11r 0.0017737

12r 0.0011722

13r 0.0007799

14r 0.0005248

15r 0.0003575

16r 0.0002476

17r 0.0001764

18r 0.0001313

19r 0.0001009

20r 0.0000790

21r 0.0000673

22r 0.0000581

23r 0.0000518

Hours

Radiative

Portion

of

Sensible

Heat

Gain, W

Convective

Portion of

Sensible

Heat Gain,

W

Sensible

Heat

Gain, W

Radiative

Portion of

Sensible

Cooling

Load, W

Sensible

Cooling

Load, W

Latent

Cooling

Load, W

1 0 0 0 323.1635 323.1635 0

2 0 0 0 212.7134 212.7134 0

3 0 0 0 141.0366 141.0366 0

4 0 0 0 94.4100 94.4100 0

5 0 0 0 64.0172 64.0172 0

6 0 0 0 44.2019 44.2019 0

7 0 0 0 31.2572 31.2572 0

8 0 0 0 22.7744 22.7744 0

9 17,934 10,986 28,920 9282.6151 20268.6151 2360

10 17,934 10,986 28,920 13014.4061 24000.4061 2360

11 17,934 10,986 28,920 14956.3947 25942.3947 2360

12 17,934 10,986 28,920 16071.7173 27057.7173 2360

13 17,934 10,986 28,920 16748.7724 27734.7724 2360

14 17,934 10,986 28,920 17173.0030 28159.0030 2360

15 17,934 10,986 28,920 17443.4549 28429.4549 2360

16 17,934 10,986 28,920 17617.5528 28603.5528 2360

17 17,934 10,986 28,920 17730.2375 28716.2375 2360

18 0 0 0 8538.0169 8538.0169 0

19 0 0 0 4850.2216 4850.2216 0

20 0 0 0 2936.8790 2936.8790 0

8. The Cooling Load

21 0 0 0 1840.2239 1840.2239 0

22 0 0 0 1175.3459 1175.3459 0

23 0 0 0 759.1104 759.1104 0

24 0 0 0 493.8629 493.8629 0


and latent cooling loads if the zone is the MW 2 zone from Table 8-21.

Solution:

Lighting Load:

suFWFq =&

W = 5,000 W

0.1=u

F

2.1=s

F

( )( )( ) 60002.10.15000 ==q& W



CL&L&&&&&


r MW2

0r 0.2550941

1r 0.1139586

2r 0.0695853

3r 0.0513341

4r 0.0425855

5r 0.0377131

6r 0.0346113

7r 0.0324116

8r 0.0307118

9r 0.0293093

10r 0.0280908

11r 0.0269960

12r 0.0259840

13r 0.0250392

14r 0.0241403

15r 0.0232829

16r 0.0224620

17r 0.0216740

18r 0.0209139

19r 0.0201816

20r 0.0194781

21r 0.0188000

22r 0.0181458

23r 0.0175139

8. The Cooling Load


Fraction


Fraction

Lighting: Recessed

fluorescent, vented to

return air

0.59 0.41

Hours

Radiative

Portion

of

Sensible

Heat

Gain, W

Convective

Portion of

Sensible

Heat Gain,

W

Sensible

Heat

Gain, W

Radiative

Portion of

Sensible

Cooling

Load, W

Sensible

Cooling

Load, W

Latent

Cooling

Load, W

1 0 0 0 1100.9959 1100.9959 0

2 0 0 0 1057.7017 1057.7017 0

3 0 0 0 1017.9344 1017.9344 0

4 0 0 0 980.7315 980.7315 0

5 0 0 0 945.5262 945.5262 0

6 0 0 0 911.9596 911.9596 0

7 3540 2460 6000 1723.0093 4183.0093 0

8 3540 2460 6000 2037.7840 4497.7840 0

9 3540 2460 6000 2198.6593 4658.6593 0

10 3540 2460 6000 2297.9606 4757.9606 0

11 3540 2460 6000 2369.1977 4829.1977 0

12 3540 2460 6000 2425.9762 4885.9762 0

13 3540 2460 6000 2474.4650 4934.4650 0

14 3540 2460 6000 2517.7592 4977.7592 0

15 3540 2460 6000 2557.5265 5017.5265 0

16 3540 2460 6000 2594.7294 5054.7294 0

17 3540 2460 6000 2629.9347 5089.9347 0

18 3540 2460 6000 2663.5013 5123.5013 0

19 0 0 0 1852.4516 1852.4516 0

20 0 0 0 1537.6769 1537.6769 0

21 0 0 0 1376.8016 1376.8016 0

22 0 0 0 1277.5003 1277.5003 0

23 0 0 0 1206.2631 1206.2631 0

24 0 0 0 1149.4847 1149.4847 0


and latent cooling loads if the zone is the HW zone from Table 8-21.

Solution:

At 4:00 P.M. Total Number of Occupants = 20 + 40 +10 = 70



Latent Heat = 70(59) = 4130 W

8. The Cooling Load

At 6:00 P.M. Total Number of Occupants = 0

Sensible Heat = 0 W

Latent Heat = 0 W


Percentage Watts/occupant Percentage Watts/occupant

Occupants 70 51.1 30 21.9



CL&L&&&&&



CL&L&&&&&

Radiant Time Factors for HW Zone (Nonsolar for roof)

r HW

0r 0.2241915

1r 0.0768588

2r 0.057783

3r 0.0501948

4r 0.0456539

5r 0.0424286

6r 0.0398994

7r 0.0377894

8r 0.03596

9r 0.0343321

10r 0.0328598

11r 0.0315095

12r 0.0302635

13r 0.0291043

14r 0.0280197

15r 0.0270197

16r 0.0260402

17r 0.0251315

18r 0.0242692

19r 0.0234499

20r 0.0226689

21r 0.0219226

22r 0.0212093

23r 0.0205246

8. The Cooling Load

Hours

Radiative

Portion

of

Sensible

Heat

Gain, W

Convective

Portion of

Sensible

Heat Gain,

W

Sensible

Heat

Gain, W

Radiative

Portion of

Sensible

Cooling

Load, W

Sensible

Cooling

Load, W

Latent

Cooling

Load, W

1 0 0 0 693.5878 693.5878 0

2 0 0 0 666.3273 666.3273 0

3 0 0 0 640.9370 640.9370 0

4 0 0 0 617.1902 617.1902 0

5 0 0 0 594.8819 594.8819 0

6 0 0 0 573.8265 573.8265 0

7 0 0 0 553.8991 553.8991 0

8 0 0 0 534.9261 534.9261 0

9 1022 438 1460 725.7382 1163.7382 1180

10 1022 438 1460 767.4643 1205.4643 1180

11 3066 1314 4380 1208.7168 2522.7168 3540

12 3066 1314 4380 1343.6543 2657.6543 3540

13 3066 1314 4380 1437.4234 2751.4234 3540

14 3577 1533 5110 1619.1469 3152.1469 4130

15 3577 1533 5110 1716.2854 3249.2854 4130

16 3577 1533 5110 1797.3909 3330.3909 4130

17 511 219 730 1243.4868 1462.4868 590

18 0 0 0 1028.9339 1028.9339 0

19 0 0 0 941.2624 941.2624 0

20 0 0 0 880.6041 880.6041 0

21 0 0 0 832.3848 832.3848 0

22 0 0 0 791.4952 791.4952 0

23 0 0 0 755.5210 755.5210 0

24 0 0 0 723.1648 723.1648 0

8.56 Compute the total hourly cooling loads for the building described by the plans

and specifications furnished by your instructor, using the RTS method.

- end -

Chapter_8

Documents

c indoor dry bulb temperature

c relative humidity

cooling load c denver

c heat gain

given time c equal

ft2f specific heat

high humidity

humidity range