Mitsubishi Electric GUG-01SL-E Technical Manual page 41

II : Calculation of cooling load to determine the required capacity
Cooling Load Classifications
Class
(a) Indoor penetration heat
(b) Indoor generated heat
(c) Reheating load
(d) Ventilation load
(a) The heat penetrating into the room is often 30 to 40% of the total cooling load.
(b) It applies only when heat generated in the room.
(c) It applies only when reheating is necessary.
In this example, the outdoor and return air conditions are assumed as shown in table below.
Outdoor Air
Cooling
Return Air
(1) Calculation of ventilation load
Ventilation load = ρ x Q
= ρ [kg/m
= ρ [kg/m
ρ : Density of air
Q : Ventilation air volume
f
h
: Outdoor air enthalpy
O
h
: Return air enthalpy
R
S : Area of floor space
n : The number of people per unit area
V
: Required fresh air rates per person
f
Here, calculate ventilation load per unit area for convenience to summate with other loads.
Ventilation load per unit area [W/m
(2) Other cooling load
In this example, other cooling loads are regarded as a general value in Tokyo shown in the table below.
Calculate them in an appropriate way in the actual designing.
Typical load values during cooling
Indoor
Ventilation penetration
load 33.8% heat 30.3%
53.0 W/m 2 47.6 W/m
Indoor
generated heat
(persons, lighting
equipment) 35.9%
56.4 W/m
2
157.0 W/m
Cooling load per unit area
When the volume of ventilation air per persons is 25 m
approximately 157.0 W/m .
2
Required cooling capacity to make up for above cooling loads
= 157.0 [W/m2] × 200 [m2] = 31.4 [kW]
9
CHAPTER
Heat generated from walls
Heat generated from glass
Accumulated heat load in walls
Generated heat from occupants
Generated heat from electrical equipment
Sensible heat
Latent heat
Dry Bulb Temp. Relative Humidity
33 63%
˚C
26 50%
˚C
x Δh
f
] x Q [m /h ] x ( h - h
3 3
f O
] x S [m ] x n [person/m
3 2
] = ρ [kg/m
] x n [person/m
2 3
= 1.2 [kg/m ] x 0.2 [person/m
3
=190.8[kJ/h m
2
•
Ventilation Load
2
Indoor
Generated Heat
Indoor Penetration Heat
2 Conditions: Middle south-facing floor of a typical office building.
Model selection and capacity calculation
(qws)
from direct sunlight
from conduction and convection
(qss)
(q )
FS
(q )
FL
Wet Bulb Temp.
27˚C
18.7˚C
) [kJ/kg(DA)]
R
] x V [m /h person] x ( h
2 3
•
f
] x V [m /h·person] x (h
2 3
f
] x 25 [m /h·person] x ( 84.6 – 52.8 ) [kJ/kg(DA)]
2 3
] = 53.0[W/m ]
2
Load type
Persons
Lighting Equipment
Total
/h, and the number of persons per 1 m
3
Heat Load
(q )
GS
Sensible heat (q s)
H
Latent heat (q )
HL
Sensible heat (q
Latent heat (q
(q )
RL
Enthalpy
84.6 kJ/kg(DA)
52.8 kJ/kg(DA)
- h ) [kJ/kg(DA)]
O R
1.2 kg/m 3
1000 m /h
3
84.6 kJ/kg(DA)
52.8 kJ/kg(DA)
200 m
2
0.2 person/m
2
25 m /h/person
3
- h ) [kJ/kg(DA)]
O R
53.0 W/m 2
Load
53.0 W/m
26.4 W/m
30.0 W/m
47.6 W/m
157.0 W/m
is 0.2, the cooling load will be
2
(q s)
G
s)
E
)
EL
Enthalpy difference
31.8 kJ/kg(DA)
2
2
2
2
2
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