Transcript HEAT LOAD CALCULATION 2

HEAT LOAD
CALCULATION 2
HEAT LOAD 2
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TOPICS COVERED

LOAD CALCULATIONS
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FORMS
FORMULAS
TABLES
FACTORS
AIR SIDE
PSYCHROMETRY
HEAT LOAD 2
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HEAT LOAD 2
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HEAT LOAD FORM
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SEE TRANSPARENCY
HEAT LOAD 2
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HEAT LOAD ESTIMATE
PROBLEM
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HEAT LOAD 2
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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
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DESIGN CONDITIONS
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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
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DESIGN CONDITIONS
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Indoor design conditions
– Base on requirement and standards
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– 75°F db / 55% rh
– 72°F db / 55% rh
– 71°F db / 50% rh
Peak time at
– 4.00 pm
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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
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23-26
22+/-2
4-6 (2)
1.85-4.65
1.4
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0.8-2
1.4
25-50*
40-150
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20
20
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*-Carrier Handbook
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QUESTION
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Fill in the table other information
for the following conditions
– Outdoor air 92db/80wb
– Indoor 75db/55%rh
HEAT LOAD 2
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SOLAR GAIN THROUGH
GLASS
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Cooling Load = Window Area x
Peak solar heat gain (Table 4) x
Storage Factor (Table 5) x Shade
Factor (Table 6)
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Refer to Table 4, Table 5 and
Table 6
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SOLAR AND TRANSMISSION
GAIN
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Heat Gain Through Walls &
Roofs = Area x Equivalent
Temp. Difference (Table 7 for wall &
Table 8 for roof) x Transmission
Coefficient (U) (Table 9)
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TRANSMISSION GAIN (EXCEPT
WALLS AND ROOFS)
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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)
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TRANSMISSION GAIN (EXCEPT
WALLS AND ROOFS) (2)
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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)
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INFILTRATION
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When ventilation exceeds
infiltration (+ve pressure), then
infiltration = 0
When rooms are design at -ve
pressure, have to consider
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INTERNAL LOADS
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People
– No. of people from no. of chairs or
base on per floor area
– Table 10 - Heat gain from people
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No. of people x Table 10
Activity, Sensible & Latent
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INTERNAL LOADS
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Power
– Table 53 - Heat gain from electric
motors
– Electric motors contribute only
sensible heat to space
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INTERNAL LOADS
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Lights - Estimate (w/ft2)
Lights - Type
– Incandescent
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rated lamp watt x 3.413
– Fluorescent lamp
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rated lamp watt x 1.25 x 3.413
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INTERNAL LOADS
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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
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SAFETY FACTOR
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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%
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ROOM SENSIBLE HEAT (RSH)
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Now all the load components
contributing to sensible load can be
added
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SUPPLY AIR DUCT LOSSES
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In transferring air from system
cooling coil to space, four losses
must be considered;
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Supply duct heat gain
supply duct leakage loss
fan heat
bypassed outdoor air
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SUPPLY DUCT HEAT GAIN
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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
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SUPPLY DUCT LEAKGE LOSS
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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%
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FAN HEAT
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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%
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BYPASS OUTDOOR AIR
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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
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EFFECTIVE ROOM SENSIBLE
HEAT (ERSH)
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This load determine the cfm
required across the cooling coil
ERSH = RSH + (SUPPLY DUCT
GAIN + SUPPLY DUCT LEAK
LOSS + FAN) + BYPASS
OUTDOOR AIR
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LATENT LOAD
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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.
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ROOM LATENT HEAT (RLH)
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HEAT LOAD 2
SUM OF ALL ROOM LATENT
LOAD
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INFILTRATION
PEOPLE
STEAM
APPLIANCES
ADDITIONAL HEAT GAINS
VAPOUR TRANSMISSION
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EFFECTIVE ROOM LATENT
LOAD (ERLH)
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ROOM LATENT HEAT (RLH) +
SUPPLY DUCT LEAKAGE LOSS +
BYPASS OUTDOOR AIR
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OUTDOOR AIR / VENTILATION
RATES
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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
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OUTDOOR AIR HEAT
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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
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QUESTION
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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
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RETURN DUCT LOSSES
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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
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RETURN AIR or BLOW THRU’
FAN
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Fan and motor heat appear on the
upstream side of the coil.
…..hp x 2545
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GRAND TOTAL HEAT
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Total heat load the coil must
remove from the air passing over it.
Also known as dehumidifier load
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REFRIGERATION LOAD
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Introducing two additional loads not
experience by the coil
– Piping sensible heat gain
– Pumping heat gain
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APPARATUS DEW POINT (ADP)
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Effective Room Sensible Heat (ERSH)
Effective Room Total Heat (ERTH)
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ADP obtained from plot ESHF line
or Table 65
Indicated ADP & Selected ADP
ADP must be > 48°F
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HEAT LOAD 2
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QUESTION
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ERSH = 115 000
ERLH = 15 000
Calculate the ESHF
Find ADP if room is 75°F/55%rh
from table
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DEHUMIDIFIED AIR QUANTITY
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Dehumidified rise = (trm-tadp)x(1-BF)
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Cfmda =
ERSH
1.09 x (trm-tadp)x(1-BF)
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SUPPLY AIR QUANTITY (cfmsa)
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Outlet temp. diff = RSH
=Fdes.dif
(trm - tgrille)
1.09 x cfmda
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Supply cfm =
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Bypass cfm = cfmsa - cfmda
RSH
= cfmsa
1.09 x Fdes. dif
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RESULTING ENTERING AND
LEAVING CONDITIONS AT
APPARATUS
 tedb= trm + cfmoa x (toa - trm)
cfmsa
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tldb = tadp + bf x (tedb - tadp)
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tewb & tlwb read from psych. chart
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SOME EX. OF COMPUTERISED
HEAT LOAD CALCULATIONS
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See Lotus files
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HEAT LOAD 2
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GO TO PSYCHRO 2
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