SP9-1211
APPLICATION MANUALFor Central Air Conditioning
and Heating Systems
2
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5 P
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6 E
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7 C
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8 F
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9 V
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10 P
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11 A
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12 S
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Lin
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13 S
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x D
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14 H
eat G
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Coo
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Load
(Li
ne 1
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1.3)
15 E
quip
men
t Sel
ecte
d (F
ig.1
.7)
300
1200 1.3
19 S
elec
ted
No.
of T
erm
inat
ors
(Cho
ose
the
larg
er n
umbe
r fr
om L
ine
18)
18 R
ecom
men
ded
No.
of T
erm
inat
ors
(Lin
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÷ 1
7)
17 B
ase
Load
Fac
tors
16 A
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ted
Load
s: L
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(F
ig. 2
.3)
x Li
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21 A
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IK-W
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FIG
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.1:
KW
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3
SpacePak provides the Kwik-Way Heating & AirConditioning sizing sheet (see Figure 1.1) to help dealerscalculate the heat gain and/or heat loss of a structure toassure maximum comfort for the occupants. The form iseasy to use; provides an accurate analysis; and is essen-tial for sizing the structure and selecting the appropriateSpacePak equipment.
The Kwik-Way form also provides a room-by-room analy-sis for selecting the appropriate number of air outlets andbalancing orifices for each room. Proper system balanc-ing is the key to proper system operation.
The small amount of time that is spent in designing theSpacePak system will result in a satisfied customerevery time. We have provided a number of extremelyimportant guidelines in this application manual involvingthe air distribution system design, such as static pres-sures, plenum duct and supply tubing runs, and locationof room outlets.
We recommend reviewing these guidelines carefully, asthey are intended to save time on the job, help obtain thebest possible installation, and provide continuous, trou-ble-free operation.
SECTION 1: EQUIPMENT SELECTION JOB ESTIMATING & SYSTEM DESIGN
Before equipment can be selected for an installation, it isimperative that the heat gain (for cooling) and/or the heatloss (for heating) be calculated for the home or businessin which the Space Pak system will be installed to assuremaximum comfort for the occupants.
Prior to performing your calculations, complete a floorplan of the structure, such as the one shown in Figure1.2, which contains the following measurements (all ofwhich can be rounded off to the nearest foot):
1. Square footage (length x width) of each room.
2. Linear feet of exposed (outside) wall in each room. Ifmore than one exposed wall in each room, add thelengths together. If the wall has two different kinds ofconstruction (part brick, part frame), measure each sec-tion as though it were a separate wall. If the wall is par-tially above grade and more than 3-feet below grade,measure each section as though it were a separate wall.If there is an open stairway alongside an exposed wall,the entire wall area (all the way to the ceiling of the sec-ond floor) is to be added to the first floor.
3. Ceiling height in each room.
DINING ROOM306 Sq. Ft.
FAMILYROOM
216 Sq. Ft.
KITCHEN198 Sq. Ft.
BED. 4216 Sq. Ft.
BED. 3180 Sq. Ft.
BED. 2210 Sq. Ft.
BED. 1357 Sq. Ft.
LIVING ROOM580 Sq. Ft.
GARAGEBATH
126 Sq. Ft.
65'
38'
3'x7' 3'x4' 3'x4'
3'x4'3'x4'3'x4'
3'x4
'3'
x4'
3'x4
'
3'x4' 5'x4'
5'x4
'3'
x7'
3'x7' 8'x4'
18'
12'
11'
12'
15'
14'
15' 21'
17'
7'
18'
18' 17'
18'
12'
18'
20'
29'
7' P
arti
tio
n
18' P
arti
tio
n
N
S
W
E
FIGURE 1.2: FLOOR PLAN EXAMPLE
4
4. Square footage of partitions — the walls between aconditioned and unconditioned area (such as a livingroom and attached garage).
5. Square footage of all exposed doors and windows.
6. Square footage of all exposed sliding glass doors.These are considered “windows” for cooling and “doors”for heating.
7. Square footage of exposed ceiling in each room. Allrooms in a one-story structure have exposed ceilings. Allsecond floor rooms in a two-story structure haveexposed ceilings.
8. Square footage of exposed floors or floors over anunconditioned area in each room.
9. Indicate North-South and East-West directions.
Now, you’re ready to perform the heat gain/lost calcula-tions using the SpacePak Kwik-Way Heating & AirConditioning Sizing Sheet (Form PR108).
KWIK-WAY EXAMPLEFor our Kwik-Way example, we will use the floor plan andmeasurements as shown in Figure 1.2. This will be a
cooling-only installation with the indoor fan coil unit to belocated in a vented attic. Construction consists of one-story frame over a basement (R-11 insulation betweenfloor and basement); 3-1/2-inch insulation in the wallsand ceiling; 8-foot ceiling height; double-hung windowswith blinds; storm doors; masonry partition with no insu-lation; and infiltration (no mechanical ventilation).
There are three occupants, the house faces South andthe summer outdoor temperature is 90°F.
TOTAL STRUCTURE HEAT GAIN (LOSS)To determine an accurate whole-house heat gain (loss)for the structure for estimating purposes, you can com-plete the TOTAL STRUCTURE column first on the Kwik-Way sheet. However, this does not preclude performinga final room-by-room analysis.
L ft exp wall: At the top of the Kwik-Way sheet, on thisline, fill in the total linear feet of exposed wall for thestructure (see Figure 1.3).
Ceiling Ht: On this line, fill in the ceiling height of thestructure (see Figure 1.3). NOTE: If a room(s) has a dif-ferent height than the others, such as one with a cathe-dral-type ceiling, then you will have to factor in the room’sheight and linear exposed feet separately.
RECOMMENDED NO. OF OU
ROOM:
X =
AREASQ FT
HEATBTU
COOLBTU
HEAT COOL
TOTAL STRUCTURE
AREASQ FT
HEATBTU
COOLBTU
HEAT COOL
FACTORS
HEAT COOL
Lg x Wd
L ft exp wall
Ceiling Ht
Total Exp wall
ITEM OF CONSTRUCTION
Shade Exposure
N
NE & NW
E & W
SE & SW
S
1 WINDOWS (cooling) Including Glass Doors
2 Glass Doors (heating)
3 Windows (heating)
4 Standard Doors (non-glass)
5 Partitions (less doors & windows)
6 Exposed Walls (less 1, 2, 3, 4 & 5)
7 Ceiling/Roof
8 Floor
9 Ventilation (Mechanical)/Infiltration
10 People
11 Appliances
12 Sub-Total Lines 1 - 11
13 Sensible {Line 12 x Duct Factor (Fig.1.6)}
14 Heat Gain for Cooling Load (Line 13 x 1.3)
15 Equipment Selected (Fig.1.7)
300
1200
1.3
16 Adjusted Loads: LAF (Fig. 2.3) x Line 13 and Line 14
OUTDOOR UNIT :
ROOM:
X =
AREASQ FT
HEATBTU
COOLBTU
HEAT COOL HEAT COOL HEAT COOL
INDOOR UNIT :
FROM FLOOR PLANHouse Width = 65'House Depth = 38'Ceiling Height = 8'
2068
1648
20
50
25
92.32.0
2.5
.5
1.2
FIGURE 1.3: TOTAL STRUCTURE COOLING EXAMPLE
5
Total Exp wall: On this line, fill in the total squarefootage of exposed wall. Simply multiply linear feet ofexposed wall by ceiling height in each room.
FACTORS: For this column, select and enter the applic-able factors, as shown in Figure 1.4, which correspond tothe construction of the structure and the application:cooling-only or heating (see Figure 1.3). These factorsare also available in Table 1 on the back of the Kwik-Waysheet.
For cooling-only installation, select factors for Lines 1, 4,5, 6, 7, 8 and 9. For heating installations, select factorsfor Lines 2, 3, 4, 5, 6, 7, 8 and 9.
Example (cooling-only): Based on example house con-struction, the following factors would be selected:
DESCRIPTION FACTORSDouble-Hung Windows With Blinds
North . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20East & West. . . . . . . . . . . . . . . . . . . . . . . . 50South . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
Storm Doors (at 90°F) . . . . . . . . . . . . . . . . . . . . 9Masonry Partition w/o Insulation (at 90°F) . . . . 2.3Frame Walls w/3-1/2” Insulation (at 90°F) . . . . 2.0Ceiling w/3-1/2” Insulation (at 90°F) . . . . . . . . . 2.5Floors Over Basement, R-11 (at 90°F) . . . . . . . .5Infiltration (at 90°F) . . . . . . . . . . . . . . . . . . . . . . 1.2
FIGURE 1.4: HEAT GAIN (LOSS) FACTORS
6
Line 1 Windows (cooling): In the column AREA SQ FT, fillin the total square footage of all exposed windows andglass doors, based on the direction they are facing. Multiplythe square footages by the appropriate factors and enterresults in the column COOL BTU (see Figure 1.5).
Line 2 Glass Doors (heating): For heating installations,in the column AREA SQ FT, fill in the total square footageof all exposed glass doors. Multiply the square footageby the factor and enter result in the column HEAT BTU.
Line 3 Windows (heating): For heating installations, inthe column AREA SQ FT, fill in the total square footageof all exposed windows. Multiply the square footage bythe factor and enter result in the column HEAT BTU.
Line 4 Standard Doors: In the column AREA SQ FT, fillin the total square footage of all exposed standard doors(non-glass). Multiply the square footage by the factor andenter result in the column COOL BTU(HEAT BTU).
Line 5 Partitions: In the column AREA SQ FT, fill in thetotal square footage of all partitions (less doors and win-dows). Multiply the square footage by the factor andenter result in the column COOL BTU(HEAT BTU).
Line 6 Exposed Wall: In the column AREA SQ FT, sub-tract the square footages in lines 1, 2, 3, 4, & 5 from thetotal exposed wall square footage and enter the result.
Multiply the square footage by the factor and enter resultin the column COOL BTU(HEAT BTU).
Line 7 Ceiling/Roof and Line 8 Floor: In the columnAREA SQ FT, fill in the total square footage of the ceil-ings and floors. Multiply the square footage by the fac-tors and enter the results in the column COOL BTU(HEAT BTU).
Line 9 Ventilation/lnfiltration: Incoming outside air,from either mechanical ventilation or infiltration, must beaccounted for:
A. For infiltration (no mechanical ventilation avail-able, as in our example), in the column AREA SQFT, fill in the square footage of the total exposedwalls. Multiply the square footage by the factorand enter result in the column COOL BTU (HEATBTU).
B. If mechanical ventilation is available, select theappropriate factor from Table 1 (Kwik-Way sheet).Multiply the factor by the CFM of the mechanicalventilation and enter result in the column COOLBTU (HEAT BTU).
Line 10 People: In the column AREA SQ FT, fill in thetotal number of people living in the home. Multiply thenumber by the factor of 300 (constant) and enter result inthe column COOL BTU.
RECOMMENDED NO. OF OU
ROOM:
X =
AREASQ FT
HEATBTU
COOLBTU
HEAT COOL
TOTAL STRUCTURE
AREASQ FT
HEATBTU
COOLBTU
HEAT COOL
FACTORS
HEAT COOL
Lg x Wd
L ft exp wall
Ceiling Ht
Total Exp wall
ITEM OF CONSTRUCTION
Shade Exposure
N
NE & NW
E & W
SE & SW
S
1 WINDOWS (cooling) Including Glass Doors
2 Glass Doors (heating)
3 Windows (heating)
4 Standard Doors (non-glass)
5 Partitions (less doors & windows)
6 Exposed Walls (less 1, 2, 3, 4 & 5)
7 Ceiling/Roof
8 Floor
9 Ventilation (Mechanical)/Infiltration
10 People
11 Appliances
12 Sub-Total Lines 1 - 11
13 Sensible {Line 12 x Duct Factor (Fig.1.6)}
14 Heat Gain for Cooling Load (Line 13 x 1.3)
15 Equipment Selected (Fig.1.7)
300
1200
1.3
16 Adjusted Loads: LAF (Fig. 2.3) x Line 13 and Line 14
OUTDOOR UNIT : 3-Ton
ROOM:
X =
AREASQ FT
HEAT COOL HEAT COOL HEAT COOL
INDOOR UNIT : ESP-3642
FROM FLOOR PLANHouse Width = 65'House Depth = 38'Ceiling Height = 8'Windows-N = 3 (3' x 4') = 36'Windows-N = 1 (4' x 5') = 20'Windows-W =3 (3' x 4') = 36'Windows-E = 1 (4' x 5') = 20'Windows-S = 3 (3' x 4') = 36'Windows-S = 1 (4' x 8') = 32'Doors = 3 (3' x 7') = 63'Partitions = 1 (8' x 25' - 21') = 179'People = 3
20
50
25
92.32.0
2.5
.5
1.2
1.15
2068
1648
56
56
68
63179
1226
2470
2470
16483
1120
2800
1700
5674122452
6175
1235
1978900
1200
205392362030706
FIGURE 1.5: TOTAL STRUCTURE COOLING EXAMPLE (CONTINUED)
DUCT FACTOR
COOLING HEATING
Attic Unvented
Attic Vented
Crawl Space (Vented)
LOCATION
Basement
1.15
1.15
1.05
1.00
1.15
1.15
1.15
1.10
7
Line 11 Appliances: For appliances, the factor is a con-stant 1200. Enter this in the column COOL BTU. NOTE:For the room-by-room analysis, this 1200 factor would beincluded in the kitchen heat gain.
Line 12 Sub-Total: In the column COOL BTU (HEATBTU), enter the subtotal of Lines 1 through 11.
Line 13 Total Sensible: In the FACTORS column, entera factor for Duct Gain Cooling, based on where theplenum duct will be run, such as through an attic or base-ment (see Figure 1.6). Multiply the subtotal on Line 12 bythe factor and enter result in the column COOL BTU onLine 13. These duct factors are also available in Table 2on the back of the Kwik-Way sheet.
Line 14 Heat Gain For Cooling Load: Multiply theCOOL BTU on Line 13 by the factor of 1.3 and enterresult in the column COOL BTU on Line 14. This calcu-lation accounts for the moisture introduced into thedwelling by people, cooling or bathing. Experience hasshown this latent heat load is an additional 30% of thetotal sensible load.
EQUIPMENT SELECTIONLine 15 Equipment Selection: Based on the heat gainin Line 14 (and/or heat loss in Line 13), select the appro-priate cooling-only Model “ESP” system from the applic-able product specification sheet and enter the informa-tion on Line 15.
Example: For our 30,706 heat gain, you would select aModel ESP- 3642 (3-ton) system, as it has the closestcooling capacity to the 2.6 ton overall Heat Gain.
ROOM TERMINATORS (OUTLETS)Line 15 Equipment Selection: Based on the equipmentselected, determine the recommended number fullyopen outlets from Figure 1.7 and enter the informationon Line 15. This information is also available in Table 3on the back of the Kwik-Way sheet.
Example: 3-ton system = 17 terminators (fully open)
However, this is for estimating purposes only and doesnot preclude performing a room-by-room analysis whichis completely necessary to assure a balanced systemConsidering each room outlet (fully open or partially ori-ficed) requires a terminator/sound attenuating tubinginstallation kit, we recommend completing the room-by-room analysis before pricing the job.
FIGURE 1.6: DUCT GAIN (LOSS) FACTORS
NOMINALTONNAGE MODEL
ESP-2430
ESP-2430
ESP-3642
ESP-3642
ESP-4860
ESP-4860
2
2 1/2
3
3 1/2
4
5
RECOMMENDED
12
14
17
19
26
26
MINIMUM RECOMMENDEDNUMBER OF FULLY OPEN OUTLETS *
* The minimum or recommended number of outlets means fully open outlets. Anyoutlet having an orifice would be only a percentage of an outlet.
14
18
21
25
28
35
FIGURE 1.7: TERMINATOR SELECTION
8
RO
OM
: L
IVIN
GR
OO
M:
BE
D 1
RO
OM
: B
ED
2R
OO
M:
BE
D 3
RO
OM
: B
ED
4R
OO
M:
KIT
CH
EN
RO
OM
: D
ININ
GR
OO
M:
FAM
ILY
OU
TD
OO
R U
NIT
:
3 -
Ton
IND
OO
R U
NIT
: E
SP
-364
2R
EC
OM
ME
ND
ED
NO
. OF
TE
RM
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S:
17
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AT:
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=Ll
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erm
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ors
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ling
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l Exp
wal
l
ITE
M O
F C
ON
ST
RU
CT
ION
Sha
deE
xpos
ure
N
NE
& N
W
E &
W
SE
& S
W
S
1 W
IND
OW
S
(co
olin
g)
Incl
udin
g
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ss D
oors
2 G
lass
Doo
rs (
heat
ing)
3 W
indo
ws
(hea
ting)
4 S
tand
ard
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rs (
non-
glas
s)
5 P
artit
ions
(le
ss d
oors
& w
indo
ws)
6 E
xpos
ed W
alls
(le
ss 1
, 2, 3
, 4 &
5)
7 C
eilin
g/R
oof
8 F
loor
9 V
entil
atio
n (M
echa
nica
l)/In
filtr
atio
n
10 P
eopl
e
11 A
pplia
nces
12 S
ub-T
otal
Lin
es 1
- 1
1
13 S
ensi
ble
{Lin
e12
x D
uct F
acto
r (F
ig.1
.6)}
14 H
eat G
ain
for
Coo
ling
Load
(Li
ne 1
3 x
1.3)
15 E
quip
men
t Sel
ecte
d (F
ig.1
.7)
300
1200 1.3
19 S
elec
ted
No.
of T
erm
inat
ors
(Cho
ose
the
larg
er n
umbe
r fr
om L
ine
18)
18 R
ecom
men
ded
No.
of T
erm
inat
ors
(Lin
e 16
÷ 1
7)
17 B
ase
Load
Fac
tors
16 A
djus
ted
Load
s: L
AF
(F
ig. 2
.3
)
x Li
ne 1
3 an
d Li
ne 1
4
20 O
rific
es R
equi
red
21 A
djus
tabl
e D
ampe
rs R
equi
red
BA
SE
LO
AD
FA
CTO
R H
EAT
ING
=
Llin
e 13
Tot
al S
truc
ture
Hea
ting
Rec
omm
ende
d N
o. o
f Ter
min
ator
s
AR
EA
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ATB
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CO
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ATC
OO
L
NO
NE
.15
.35
.50
AR
EA
SQ
FT
HE
ATB
TU
CO
OL
BT
U
HE
ATC
OO
L
NO
NE
.15
.35
.50
KW
IK-W
AY
HE
AT
ING
& A
IR C
ON
DIT
ION
ING
SIZ
ING
SH
EE
T
20 50 25 9 2.3
2.0
2.5
.5 1.2
1.15
8 1648 56 56 68 63
179
1226
2470
2470
1648 3
1120
2800
1700
567
412
2452
6175
1235
1978
900
1200
2053
9
2362
030
706
21 x
17
= 3
7512
x 1
5 =
180
12 x
18
= 2
1611
x 1
8 =
198
17 x
18
= 3
0620
x 2
9 =
580
21 8 168
29 8 232
12 8 96
30 8 240
18 8 144
17 8 136
30 8 240
49 8 392
357
357
16824
893
202
600
12 12 208
210
210
232
600
300
416
525
105
278
12 84 180
180
96
600
168
450
90 115
12 216
216
216
240
600
432
540
108
288
1224
0
111
198
198
144
222
495
99 173
1224
0
2118
9
116
306
306
136
232
765
153
163
2040
0
84 216
216
240
168
540
108
288
1224
0
123
283
2118
9
263
580
580
526
1450
290
470
5612
9
2118
9
3280
0
2010
00
2162
2486
3232
2224
2558
3325
1423
1636
2127
2208
2539
3301
2618
3011
3914
2013
2315
3010
2116
2433
3163
5154
5927
7705
1.1
3555
1.97
1.97
1.1
3658
2.03
2.03
0.90
1914
1.06
1.06
0.92
3037
1.68
1.68
0.89
3483
1.93
1.93
0.90
2709
1.50
1.50
0.90
2847
1.58
1.58
1.1
8476
4.69
4.69
= 18
06
206
14 x
15
= 2
1012
x 1
8 =
216
144
288
179
392
130
01
300
130
0
1200
=
FIG
UR
E 2
.1:
KW
IK-W
AY
HE
AT
ING
AN
D A
IR C
ON
DIT
ION
ING
SIZ
ING
SH
EE
T
9
SECTION 2: ROOM-BY-ROOM ANALYSIS
Complete the heat gain (heat loss) for each room on theKwik-Way sheet (see Figure 2.1), following the sameprocedures you used for calculating the Total StructureHeat Gain (Loss). Complete each appropriate calculationthrough Line 14.
To assure proper balancing of the SpacePak system,room by room, the next concern is providing each roomin the house with the proper number of air outlets or roomterminators.
On your floor plan, now “rough in” the location of theindoor fan coil unit and the plenum duct run. The plenumduct is normally located in the attic or basement (seeFigure 2.2). Then, estimate the average length (perroom) of supply tubing runs to the outside corners ofeach room on your floor plan (see Figure 2.2).
Line 16 Adjusted Loads: Based on the average supplytubing run for each room, select the appropriate “LengthAdjustment Factor” from Figure 2.3 and enter the factorsin the AREA SQ FT columns for each room. This infor-mation is also available in Table 4 on the back of theKwik-Way sheet. Multiply COOL BTU on Line 14 (HEATBTU on Line 13) by the factors and enter results in col-umn COOL BTU (HEAT BTU) on Line 16 for each room.
Line 17 Base Load Factors: To obtain the base loadfactors, divide the Total Structure COOL BTU on Line(HEAT BTU on Line 13) by the recommended numberterminators on Line 15 and enter result on Line 17.
Line 18 Recommended No. of Terminators (Outlets):Divide the COOL BTU (HEAT BTU) for each room or Line16 by the cooling (heating) base load factor on Line 17and enter results on Line 18 for each room. DO NOTROUND OFF TO THE NEAREST WHOLE NUMBER - ifless or more than a whole number, leave the fraction.
Line 19 Selected No. of Terminators (Outlets): Forheating installations, select the larger of the two numberson Line 18 for each room and enter the results or Line 19for each room.
2" SUPPLY TUBING LENGTH ADJUSTMENT FACTOR CHART
FACTOR
RUN
.85 .88 .90 .94 1.0 1.1 1.25 1.50
6' 8' 10' 12' 15' 20' 25' 30'
FIGURE 2.3: LENGTH ADJUSTMENT FACTORS
DINING ROOM306 Sq. Ft.
FAMILYROOM
216 Sq. Ft.
KITCHEN198 Sq. Ft.BED. 4
216 Sq. Ft.
BED. 3180 Sq. Ft.
BED. 2210 Sq. Ft.
BED. 1357 Sq. Ft.
LIVING ROOM580 Sq. Ft.
GARAGEBATH126 Sq. Ft.
65'
38'
3'x7' 3'x4' 3'x4'
3'x4'3'x4'3'x4'
3'x4
'3'
x4'
3'x4
'
3'x4' 5'x4'
5'x4
'3'
x7'
3'x7' 8'x4'
18'
12'
12'
15'
14'
15' 21'
17'
7'
18'
18' 17'
18'
12'
18'
20'
29'
7' P
arti
tio
n
18' P
arti
tio
nN
S
W
E18' average run
10' average run11' average run9' average run
19' average run 19' average run
10' averagerun
9' averagerun10' average
run
11'
9'
10'
9'
25'
25'
25'
11'
10'10'
13'
10'
11'
12'
9'
24'
10'
ES
P
ReturnAir
FIGURE 2.2: FLOOR PLAN EXAMPLE
10
RO
OM
: L
IVIN
GR
OO
M:
BE
D 1
RO
OM
: B
ED
2R
OO
M:
BE
D 3
RO
OM
: B
ED
4R
OO
M:
KIT
CH
EN
RO
OM
: D
ININ
GR
OO
M:
FAM
ILY
OU
TD
OO
R U
NIT
:
3 -
Ton
IND
OO
R U
NIT
: E
SP
-364
2R
EC
OM
ME
ND
ED
NO
. OF
OU
TLE
TS
: 1
7S
UP
PLE
ME
NTA
L H
EAT
:
BA
SE
LO
AD
FA
CTO
R C
OO
LIN
G
=Ll
ine
14 T
otal
Str
uctu
re H
eatin
gR
ecom
men
ded
No.
of O
utle
ts
HE
ATC
OO
L
NO
NE
.15
.35
.50
TOTA
L S
TR
UC
TU
RE
HE
ATC
OO
L
FAC
TOR
S
HE
ATC
OO
L
Lg x
Wd
L ft
exp
wal
l
Cei
ling
Ht
Tota
l Exp
wal
l
ITE
M O
F C
ON
ST
RU
CT
ION
Sha
deE
xpos
ure
N
NE
& N
W
E &
W
SE
& S
W
S
1 W
IND
OW
S
(co
olin
g)
Incl
udin
g
Gla
ss D
oors
2 G
lass
Doo
rs (
heat
ing)
3 W
indo
ws
(hea
ting)
4 S
tand
ard
Doo
rs (
non-
glas
s)
5 P
artit
ions
(le
ss d
oors
& w
indo
ws)
6 E
xpos
ed W
alls
(le
ss 1
, 2, 3
, 4 &
5)
7 C
eilin
g/R
oof
8 F
loor
9 V
entil
atio
n (M
echa
nica
l)/In
filtr
atio
n
10 P
eopl
e
11 A
pplia
nces
12 S
ub-T
otal
Lin
es 1
- 1
1
13 S
ensi
ble
{Lin
e12
x D
uct F
acto
r (F
ig.1
.6)}
14 H
eat G
ain
for
Coo
ling
Load
(Li
ne 1
3 x
1.3)
15 E
quip
men
t Sel
ecte
d (F
ig.1
.7)
300
1200 1.3
19 S
elec
ted
No.
of O
utle
ts (
Cho
ose
the
larg
er n
umbe
r fr
om L
ine
18)
18 R
ecom
men
ded
No.
of O
utle
ts (
Line
16
÷ 17
)
17 B
ase
Load
Fac
tors
16 A
djus
ted
Load
s: L
AF
(F
ig. 2
.3
)
x Li
ne 1
3 an
d Li
ne 1
4
20 O
rific
es R
equi
red
21 A
djus
tabl
e D
ampe
rs R
equi
red
BA
SE
LO
AD
FA
CTO
R H
EAT
ING
=
Llin
e 13
Tot
al S
truc
ture
Hea
ting
Rec
omm
ende
d N
o. o
f Out
lets
HE
ATC
OO
L
NO
NE
.15
.35
.50
HE
ATC
OO
L
NO
NE
.15
.35
.50
HE
ATC
OO
L
NO
NE
.15
.35
.50
HE
ATC
OO
L
NO
NE
.15
.35
.50
HE
ATC
OO
L
NO
NE
.15
.35
.50
HE
ATC
OO
L
NO
NE
.15
.35
.50
HE
ATC
OO
L
NO
NE
.15
.35
.50
KW
IK-W
AY
HE
AT
ING
& A
IR C
ON
DIT
ION
ING
SIZ
ING
SH
EE
T
20 50 25 9 2.3
2.0
2.5
.5 1.2
1.15
8 1648 56 56 68 63
179
1226
2470
2470
1648 3
1210
2800
1700
567
412
2452
6175
1235
1978
900
1200
2053
9
2362
0
3070
6
21 x
17
= 3
7512
x 1
5 =
180
12 x
18
= 2
1611
x 1
8 =
198
17 x
18
= 3
0620
x 2
9 =
580
21 8 168
29 8 232
12 8 96
30 8 240
18 8 144
17 8 136
30 8 240
49 8 392
357
357
16824
893
202
600
12 12 208
210
210
232
600
300
416
525
105
278
12 84 180
180
96
600
168
450
90 115
12 216
216
216
240
600
432
540
108
288
1224
0
111
198
198
144
222
495
99 173
1224
0
2118
9
116
306
306
136
232
765
153
163
2040
0
84 216
216
240
168
540
108
288
1224
0
123
283
2118
9
263
580
580
526
1450
290
470
5612
9
2118
9
3280
0
2010
00
2162
2486
3232
2224
2558
3325
1423
1636
2127
2208
2539
3301
2618
3011
3914
2013
2315
3010
2116
2433
3163
5154
5927
7705
1.1
3555
1.97
1.97
1.1
3658
2.03
2.03
0.90
1914
1.06
1.06
0.92
3037
1.68
1.68
0.89
3483
1.93
1.93
0.90
2709
1.50
1.50
0.90
2847
1.58
1.58
1.1
8476
4.69
4.69
= 18
06
22
12
11
31
12
206
14 x
15
= 2
1012
x 1
8 =
216
144
288
179
392
130
01
300
130
0
1200
=
AR
EA
SQ
FT
HE
ATB
TU
CO
OL
BT
UA
RE
AS
Q F
TH
EAT
BT
UC
OO
LB
TU
AR
EA
SQ
FT
HE
ATB
TU
CO
OL
BT
UA
RE
AS
Q F
TH
EAT
BT
UC
OO
LB
TU
AR
EA
SQ
FT
HE
ATB
TU
CO
OL
BT
UA
RE
AS
Q F
TH
EAT
BT
UC
OO
LB
TU
AR
EA
SQ
FT
HE
ATB
TU
CO
OL
BT
UA
RE
AS
Q F
TH
EAT
BT
UC
OO
LB
TU
AR
EA
SQ
FT
HE
ATB
TU
CO
OL
BT
U
11
FIG
UR
E 2
.4:
KW
IK-W
AY
HE
AT
ING
AN
D A
IR C
ON
DIT
ION
ING
SIZ
ING
SH
EE
T
11
Line 20 Orifices Required: You must select the termi-nal-orifice combination from Figure 2.5 which producesclose to the number of terminators required on Line 19for each room. This information is also available in Table5 on the back of the Kwik-Way sheet.
Example (see Figure 2.4): Bedroom 4 requires 1.68 termi-nators, which is equal to 168% capacity. To assure equalair distribution, we selected the combination of two roomterminators, one fully open and one with a 35% orifice.
Line 21 Adjustable Dampers Required: For heatinginstallations, if there is a significant difference betweenthe number of cooling and heating terminators on Line18, then adjustable dampers will have to be used tochange the air flow from the cooling season to the heat-ing season. Enter the number of adjustable dampersrequired for each room.
DESIRED NUMBEROF TERMINALS *
TERMINAL - ORIFICECOMBINATION
.5
.65
.85
1.00
1.15
1.30
1.50
1.65
1.70
1.80
1.85
1.95
(1) .5
(1) .35
(1) .15
(1)
(1) .5 + (1) .35
(2) .35
(1) .35 + (1) .15 or (1) + (1) .5 or (3) .5
(1) + (1) .35 or (2) .5 + (1) .35
(2) .15
(2) .35 + (1) .5
(1) + (1) .15
(3) .35
2.00 (2)
* For a room with more than two (2) terminals, combinations of the above may be used to achieve the desired fractional number.
FIGURE 2.5: TERMINAL-ORIFICE COMBINATION
SECTION 3: SYSTEM DESIGN CONSIDERATIONS
PLENUM DUCTThe plenum duct can be run in practically any locationaccessible for the attachment of the supply tubing. Theplenum is normally located in the attic or basement, andit is usually more economical to run the plenum where itwill appreciably shorten the lengths of two or more sup-ply runs. In some two-story split level homes, it may beadvantageous to go from one level to another with theplenum duct. Whenever necessary, either between floorsor along the ceiling, the small size of the plenum makesit easy to box in.
The fan coil coil unit is designed to operate with a totalexternal static pressure of 1.5 inches of water column(minimum 1.2 — maximum 1.5). Excessive static pres-sure increases the air flow in individual runs and maycause some or all terminators to be noisy.
For systems with a tee installed as on Unit No. 1 (Figure3.1), the best results are obtained if not more than 60%of the total number of system outlets are attached to anyone branch of the tee. For systems with a tee installed ason Unit No. 2 (Figure 3.1), not more than 30% of the totalnumber of system outlets should be attached to the per-pendicular branch of the tee.
The larger system capacities (ESP-4860) are affectedmore by higher system static pressure than the smaller
UNIT NO. 1
UNIT NO. 2
FAN COIL UNIT
FAN COIL UNIT
NO MORE THAN60 % OF CAPACITYON ONE SIDE
30 % MAX. OFCAPACITY
MINIMUM 18"
MINIMUM 18"
FIGURE 3.1: ESP-4860 INSTALLATION
systems. The four and five ton system should be consid-ered and handled as two separate smaller units. Thisnecessitates the installation of the plenum tee a mini-mum of 18" from the unit (see Figure 3.1). No supplyruns should be installed between unit outlet and tee.
All tees and elbows must be a minimum of 18" from thefan coil unit or any other tee or elbow. Keep all tees andelbows to a minimum.
SUPPLY TUBINGIn the case of two-story or split-level applications, supplytubing may run from one story to another. It is smallenough to go in stud spaces, but this is often difficult inolder homes because of hidden obstructions in studspaces. It is more common to run the supply tubing fromthe attic down through second story closets to the firststory terminators. Supply tubing runs in the corners ofthe second story rooms can be boxed in and are hardlynoticeable since overall diameter is only 3-1/4".
At the plenum, all supply tubing connections must be aminimum of 18" from any plenum tee, plenum elbow orthe fan coil unit.
Individual supply tubing runs must be a minimum of 6feet, even if the distance between the sound attenuatingtubing and plenum is less than 6 feet.
ROOM TERMINATORSTerminators should be located in the ceiling or floor forvertical discharge. However, ceiling locations are not rec-ommended for heating where ceiling height is 10 feet ormore due to possible stratification and short circuiting ofair flow.
Horizontal discharge is acceptable for cooling-only sys-tems, but is sometimes more difficult to install. Twoexcellent spots for horizontal discharge are in the soffitarea above kitchen cabinets and in the top portion ofclosets. Horizontal discharge is not recommended forheating systems, as it will not maintain a proper floor-to-ceiling temperature difference.
Terminators should always be out of normal traffic pat-terns to prevent discharge air from blowing directly onoccupants. And they should not be located directly aboveshelves or large pieces of furniture. Outside wall or cornerlocations are recommended if the room has more thanone outside wall. Locating terminators away from interiordoors prevents short cycling of air to the return air box.
NOTE:The Kwik-Way method is appropriate for calculating thecooling loads of most buildings or normal structures butit might not be appropriate when calculating loads forsun rooms and buildings that do not comply with dataand factors in this application manual and Kwik-Wayform. If in doubt, calculate the actual cooling load usingthe long form manual “J”.
Some factors may vary state to state and your designtemps may be greater. SpacePak assumes no liabilityfor incorrect interpretation of this manual.
260 NORTH ELM STREET • WESTFIELD, MA 01085 • TEL: (413) 564-5530 • FAX (413) 564-58187555 TRANMERE DRIVE • MISSISSAUGA, ONTARIO L5S 1L4 CANADA • TEL: (905) 672-2991 • FAX (905) 672-2883
