HEAT LOAD CALCULATION 2

advertisement
HEAT LOAD
CALCULATION 2
HEAT LOAD 2
1
home
back
next
TOPICS COVERED

LOAD CALCULATIONS
–
–
–
–
–
–
FORMS
FORMULAS
TABLES
FACTORS
AIR SIDE
PSYCHROMETRY
HEAT LOAD 2
2
home
back
next
HEAT LOAD 2
3
HEAT LOAD FORM

SEE TRANSPARENCY
HEAT LOAD 2
4
home
back
next
HEAT LOAD ESTIMATE
PROBLEM






HEAT LOAD 2

General office
Ceiling 10’
Wall 4 ½” brick, plastered both side
Window- ordinary ¼” single glass
internal shade
Door- plywood sandwich air space
Occupants - 35 person
Lighting - Fluorescent
5
home
back
next
DESIGN CONDITIONS

Outdoor design conditions
– 92°F db / 80°F wb or 95db/83wb
– Overestimating effect
– Night time temp. & rh different
– 76°F db/ 75°F wb / 95% rh
HEAT LOAD 2
6
home
back
next
DESIGN CONDITIONS

Indoor design conditions
– Base on requirement and standards

– 75°F db / 55% rh
– 72°F db / 55% rh
– 71°F db / 50% rh
Peak time at
– 4.00 pm
HEAT LOAD 2
7
home
back
next
CURRENT AIR-CONDITIONING DESIGN DATA
COMPARISON
INDOOR
TEMP.(o C)
LIGHTING LOAD
(w/ft2)
OFFICE EQUIP.
(w/ft2)
OUTLET
VELOCITY
(ft/min.)
FRESH AIR
(cfm/person)
JKR
ASHRAE 95
CIBSE
24
23-26
22+/-2
4-6 (2)
1.85-4.65
1.4
-
0.8-2
1.4
25-50*
40-150
-
20
20
20
*-Carrier Handbook
HEAT LOAD 2
8
QUESTION

Fill in the table other information
for the following conditions
– Outdoor air 92db/80wb
– Indoor 75db/55%rh
HEAT LOAD 2
9
home
back
next
SOLAR GAIN THROUGH
GLASS

Cooling Load = Window Area x
Peak solar heat gain (Table 4) x
Storage Factor (Table 5) x Shade
Factor (Table 6)

Refer to Table 4, Table 5 and
Table 6
HEAT LOAD 2
10
home
back
next
SOLAR AND TRANSMISSION
GAIN

Heat Gain Through Walls &
Roofs = Area x Equivalent
Temp. Difference (Table 7 for wall &
Table 8 for roof) x Transmission
Coefficient (U) (Table 9)
HEAT LOAD 2
11
home
back
next
TRANSMISSION GAIN (EXCEPT
WALLS AND ROOFS)


Heat gain through all glass = Area x
Temp. Difference (OA - RA) x
Transmission coefficient (U) (Table 9)
Heat gain through shade wall,
partition = Area x Temp. Difference
(OA - RA - 5°F) x Transmission
coefficient (U) (Table 9)
HEAT LOAD 2
12
home
back
next
TRANSMISSION GAIN (EXCEPT
WALLS AND ROOFS) (2)

Heat gain through wall, partition
(adjacent to Kitchen,Boiler Room) =
Area x Temp. Difference (OA - RA +
15°F to 25°F) x Transmission
coefficient (U) (Table 9)
HEAT LOAD 2
13
home
back
next
INFILTRATION



When ventilation exceeds
infiltration (+ve pressure), then
infiltration = 0
When rooms are design at -ve
pressure, have to consider
HEAT LOAD 2
14
home
back
next
INTERNAL LOADS

People
– No. of people from no. of chairs or
base on per floor area
– Table 10 - Heat gain from people


No. of people x Table 10
Activity, Sensible & Latent
HEAT LOAD 2
15
home
back
next
INTERNAL LOADS

Power
– Table 53 - Heat gain from electric
motors
– Electric motors contribute only
sensible heat to space
HEAT LOAD 2
16
home
back
next
INTERNAL LOADS


Lights - Estimate (w/ft2)
Lights - Type
– Incandescent

rated lamp watt x 3.413
– Fluorescent lamp

rated lamp watt x 1.25 x 3.413
HEAT LOAD 2
17
home
back
next
INTERNAL LOADS

Appliances
– most appliances contributes both
sensible and latent heat load.
– They contribute latent heat by virtue of
their function ex. drying, cooking
– Hood to remove this is most effective
– Table 50 & 51- Heat Gain from
Restaurant
HEAT LOAD 2
18
home
back
next
SAFETY FACTOR



Safety factor added for possible
error in the survey
Over estimating safety factor will
cause oversized air conditioning
equipment - difficult to maintain
space conditions
< 5%
HEAT LOAD 2
19
home
back
next
ROOM SENSIBLE HEAT (RSH)

Now all the load components
contributing to sensible load can be
added
HEAT LOAD 2
20
home
back
next
SUPPLY AIR DUCT LOSSES

In transferring air from system
cooling coil to space, four losses
must be considered;
–
–
–
–
Supply duct heat gain
supply duct leakage loss
fan heat
bypassed outdoor air
HEAT LOAD 2
21
home
back
next
SUPPLY DUCT HEAT GAIN



Supply air in the duct at 50°F to
60°F passes through surrounding
environment above 90°F - potential
heat gain to supply air
Insulation reduces this gain
Typical figure < 2% of RSH
HEAT LOAD 2
22
home
back
next
SUPPLY DUCT LEAKGE LOSS




Lost capacity in the supply air duct
depends on duct shape, duct
pressure and workmanship.
Low pressure (0 - 2”s.p) : <5%
Medium pressure (2” - 6”s.p) : 2% 3%
High pressure (6” & above) : <1%
HEAT LOAD 2
23
home
back
next
FAN HEAT


Draw through (draw through the
cooling coil) fan add heat to air
supply. Electrical losses for motor
which lies in the air stream also add
heat
< 5%
HEAT LOAD 2
24
home
back
next
BYPASS OUTDOOR AIR




Some of the air passing through the
coil remain untreated.
Load equivalent to infiltration load
Depends on bypass factors use
Load = cfm x (toa-trm) x bf x 1.09
HEAT LOAD 2
25
home
back
next
EFFECTIVE ROOM SENSIBLE
HEAT (ERSH)


This load determine the cfm
required across the cooling coil
ERSH = RSH + (SUPPLY DUCT
GAIN + SUPPLY DUCT LEAK
LOSS + FAN) + BYPASS
OUTDOOR AIR
HEAT LOAD 2
26
home
back
next
LATENT LOAD


The latent counterpart of infiltration,
internal loads, and supply duct load
are also calculated to determine the
Room Latent Heat and Effective
Room Latent Heat
Vapor Transmission - only for low
or high dew point application.
HEAT LOAD 2
27
home
back
next
ROOM LATENT HEAT (RLH)

HEAT LOAD 2
SUM OF ALL ROOM LATENT
LOAD
–
–
–
–
–
–
INFILTRATION
PEOPLE
STEAM
APPLIANCES
ADDITIONAL HEAT GAINS
VAPOUR TRANSMISSION
28
home
back
next
EFFECTIVE ROOM LATENT
LOAD (ERLH)

ROOM LATENT HEAT (RLH) +
SUPPLY DUCT LEAKAGE LOSS +
BYPASS OUTDOOR AIR
HEAT LOAD 2
29
home
back
next
OUTDOOR AIR / VENTILATION
RATES




HEAT LOAD 2
Outdoor air/ventilation rates from
Table 11-Ventilation Std.
Outdoor air quantity can be
determined either by cfm/person or
cfm/ft2 or airchange rate (ach)
Air change is defined as the quantity
of changed air every hour
cfm = vol x ach
60
30
home
back
next
OUTDOOR AIR HEAT



Outdoor air heat comprised of both
sensible and latent load
Except for bypassed air, the load
appears on the upstream of the coil
Requirements of outdoor/fresh air
base on air change rate or cfm/sq.ft
or per person
HEAT LOAD 2
31
home
back
next
QUESTION

Calculate the outdoor/fresh air
requirement of the office space, if
the recommended fresh air change
is 2 ach.
Answer :
(ach/60) x Vol.
= (2/60) x 32000
= 1067 cfm
HEAT LOAD 2
32
home
back
next
RETURN DUCT LOSSES





Return duct are normally shorter
than supply duct
Temperature of air at about 75°F 80°F
Return duct slightly negative
1% for Return duct heat gain &
1% for return duct leakage loss
HEAT LOAD 2
33
home
back
next
RETURN AIR or BLOW THRU’
FAN


Fan and motor heat appear on the
upstream side of the coil.
…..hp x 2545
HEAT LOAD 2
34
home
back
next
GRAND TOTAL HEAT


Total heat load the coil must
remove from the air passing over it.
Also known as dehumidifier load
HEAT LOAD 2
35
home
back
next
REFRIGERATION LOAD

Introducing two additional loads not
experience by the coil
– Piping sensible heat gain
– Pumping heat gain
HEAT LOAD 2
36
home
back
next
APPARATUS DEW POINT (ADP)

Effective Room Sensible Heat (ERSH)
Effective Room Total Heat (ERTH)

ADP obtained from plot ESHF line
or Table 65
Indicated ADP & Selected ADP
ADP must be > 48°F


HEAT LOAD 2
37
home
back
next
QUESTION




ERSH = 115 000
ERLH = 15 000
Calculate the ESHF
Find ADP if room is 75°F/55%rh
from table
HEAT LOAD 2
38
home
back
next
DEHUMIDIFIED AIR QUANTITY

Dehumidified rise = (trm-tadp)x(1-BF)

Cfmda =
ERSH
1.09 x (trm-tadp)x(1-BF)
HEAT LOAD 2
39
home
back
next
SUPPLY AIR QUANTITY (cfmsa)

Outlet temp. diff = RSH
=Fdes.dif
(trm - tgrille)
1.09 x cfmda

Supply cfm =

Bypass cfm = cfmsa - cfmda
RSH
= cfmsa
1.09 x Fdes. dif
HEAT LOAD 2
40
home
back
next
RESULTING ENTERING AND
LEAVING CONDITIONS AT
APPARATUS
 tedb= trm + cfmoa x (toa - trm)
cfmsa

tldb = tadp + bf x (tedb - tadp)

tewb & tlwb read from psych. chart
HEAT LOAD 2
41
home
back
next
SOME EX. OF COMPUTERISED
HEAT LOAD CALCULATIONS

See Lotus files
HEAT LOAD 2
42
home
back
next
HEAT LOAD 2
43
GO TO PSYCHRO 2
HEAT LOAD 2
44
home
back
next
Download