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APPLICATION MANUAL - mesteksa.commesteksa.com/fileuploads/Literature/SpacePak/SpacePak/SP9-1211.pdf · KWIK-WAY HEATING & AIR CONDITIONING SIZING SHEET = = FIGURE 1.1: KWIK-WAY HEATING

May 18, 2019

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buihuong

SP9-1211

APPLICATION MANUALFor Central Air Conditioning

and Heating Systems

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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 doorsfor 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, youre 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 90F.

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 roomsheight 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 90F) . . . . . . . . . . . . . . . . . . . . 9Masonry Partition w/o Insulation (at 90F) . . . . 2.3Frame Walls w/3-1/2 Insulation (at 90F) . . . . 2.0Ceiling w/3-1/2 Insulation (at 90F) . . . . . . . . . 2.5Floors Over Basement, R-11 (at 90F) . . . . . . . .5Infiltration (at 90F) . . . . . . . . . . . . . . . . . . . . . . 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

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TU

CO

OL

BT

U

HE

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

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' averagerun

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

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