Keeprite KSH71, KSH User Manual

Steam Coils

Bulletin K70-KSH-PDI-11

1064628

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T ype KSH

Features

RIPPLED FINS produce a rippled air flow pattern for

maximum heat transfer. These ripples also assure
permanent “fin-tube bond” through greater flexibility under
expansion and contraction.

ST AGGERED TUBES create air turbulence to give
maximum air side heat transfer.

MECHANICAL EXP ANSION BOND ensures permanent
metal to metal contact. (No low conductivity materials
used as a bonding agent).

FIN COLLARS are drawn wide and smooth to provide
maximum contact area.

NOMENCLATURE

K SH 8 2 24 x 48

KEEPRITE
REFRIGERATION

COIL TYPE
STEAM HEAT

FINS PER INCH
(STD. 8 FINS/INCH)

ROWS DEEP

“W” DIMENSION
(SEE P 4)

NOMINAL TUBE
LENGTH

STEAM BAFFLE disperses entering steam thereby
preventing blow through or short circuiting and
ensuring equal steam pressure throughout the supply
header.

THE IMPORT ANT DIFFERENCE

KEEPRITE REFRIGERATION

SURFACE

RIPPLED FINS STAGGERED TUBES

Highly Efficient result

Rippled Air Flow assure intimate

and prolonger contact between air

and the cooling surface.

COIL SELECTION

KSH Coils are a general purpose coil for reheat applications or when
outdoor air temperature is

above 35°F.

For applications where
modulating control is required and entering air is below freezing
“Type DT” coils should be used.

KSH Coils are not recommended for use with steam pressures
above 25 PSIG.

The following example outlines the procedure for determining the coil
size, fin spacing, rows deep, etc.

SPECIFIED:

(a) Air Volume (Std. Air)........................................................8,000 CFM

(b) Design Face Velocity (Max,)...............................................700 FPM

(c) Steam Pressure....................................................................10 PSIG

(d) Entering Air Temperature..........................................................40°F.

(e) Leaving Air Temperature.........................................................125°F.

(f) Heating Load............................................................741,000 BTU/Hr.

REQUIRED:

(a) Coil Size and Model (d) BTU/Hr. Capacity.
(b) Coil Nomenclature (e) Lbs. of Condensate/Hr.
(c) Leaving Air Temperature (f) Air Side Friction Loss

PROCEDURE:

A. Determine Coil Face Dimensions

1. Coil Face Area Req’d =
Design Face Vel. 700

SpecifiedCFM*

=

8000

11.4 sq. ft.

=

*(Specified CFM at Std. Air)

2. From Table 3, select a “24 x 72” coil with 12 sq. ft. face area as
having face dimensions most suitable for this job.
B. Determine Coil Model Number

1. Temperature Rise Required = 125° - 40° = 85° F .

2. From Table 2, the Conversion Factor for 10 psig steam and 40° F.
entering air is .878.

3. Then,
Conv. Factor .878

4. Actual Coil Face V el. =
Coil Face Area 12

Req’d Temp. Rise

850

=

= 96.8° F. (Req’d Base Rise)

Specified CFM = 8000

= 667 FPM

5. From Figure 1 (or Table 1 by interpolation), find “Model 82” coil has
Base Temp. Rise of 100° F. at 667 FPM. Hence, select a “Model
82” coil.
C. Determine Coil Nomenclature ‡

From Coil Designation Chart, below, determine coil nomenclature as
follows: KSH82 - 24x72.

TABLE 1 BASE TEMPERATURE RISE AND STATIC PRESSURE

Temp. Rise based on 5 psig. steam and 0° F Ent. Air Temp. --- Static Pressure based on Std. Air (70°F, and 29.92” Hg.)

ROWS FACE VELOCITY - FEET PER MINUTE - STANDARD AIR

DEEP MODEL 300 400 500 600 700 800 1000 1200

141

ROW 71

272116.4 .047 106.8 .076 98.8 .112 92.0 .153 86.3 .199 81.5 .251 73. 8 .368 67.6 .503

ROW 82 130.7 .053 120 .3 .087 112.0 .128 104.4 .175 98.0 .228 93.0 .287 84.2 .421 77.4 .576

T.R. S.P. T.R . S . P. T.R. S.P. T.R. S . P. T.R . S.P. T.R . S.P. T.R . S.P. T.R . S.P.

51.5 .020 45.4 .034 41.2 .049 37.9 .067 35.3 .087 33.1 .110 30.0 . 162 27.4 .221

72.8 .027 64.4 .044 58.5 . 065 53.9 .088 50.3 .115 47.1 .145 42.6 .213 38.8 .291

81 83.8 .031 74.4 .051 67.6 .074 62.2 .101 57.9 .132 54.7 .166 49.1 .243 44.9 .333

TABLE 2 STEAM CONVERSION FACTORS

ENT. STEAM PRESSUR E PS IG - STEAM TEMPERATURE ° F - LATENT HE AT BTU PER LB.

AIR 0 2 5 10 15 25

TEMP. 212° 218° 227 ° 239° 250° 267°

°F. 970 966 961 953 946 934

-30 1.065 1.094 1.132 1. 186 1.232 1.307

-20

-10

0 .933 .9 62 1.000 1.054 1.099 1.175

10
20
30 .8 01 .830 .868 .9 22 .967 1.042
40
50
60 .669 .698 .736 .790 .835 .910
70
80 .581 .610 .648 .702 .747 .822
90 .537 .566 .604 .658 .703 .772

100
125 .383 .412 .450 .504 .549 .624
150 .273 .302 .340 .394 .439 .514

1.021 1.050 1.088 1.142 1. 188 1.263
.977 1.006 1.004 1.098 1.144 1.219

.889 .9 18 .956 1.010 1.056 1.131
.845 .8 74 .912 .966 1.011 1.087

.757 .786 .824 .878 .923 .998
.713 .742 .780 .834 .879 .954

.625 .654 .692 .746 .791 .866

.493 .522 .560 .614 .659 .734

D. Determine Leaving Air Temperature

1. Actual Temp. Rise = Base Rise x Conv. Factor

2. Actual Temp. Rise @ 10 psig and 40°F. EA = 100 x .878 = 87.8°F

3. Leaving Air Temperature = 40° + 87.8° = 127.8°F.
E. Determine BTU/Hr. Capacity

1. BTU/Hr. Capacity = 1.09 x Temp. Rise x CFM
= 1.09 x 87.8 x 8000 = 766,000
F. Determine Lbs. of Condensate/Hr.

1 .Lbs. of Cond./Hr. =
Latent heat @ 10 PSIG 953

Total BTU/Hr. Capacity = 766.000

= 804

G. Determine Air Side Friction

1. Air Friction = .210 inches of water from Fig. 1
(or from Table 1 by interpolation).

COILS IN SERIES

Occasionally, it may be necessary to use two or more coils in series in
order to heat the air to the required final temperature. Likewise, if row
control is required, it would be necessary to furnish individual coils.
Suppose, in the preceding example, that it had been desired to have
a final air temperature considerably higher than 125°F. It would then
be necessary to select an additional coil or coils to place after the
first coil. In calculating the temperature rise through these additional
coils, the leaving air temperature of the first coil is used as the
entering air temperature to the second coil, etc. The method of
computation is identical to that previously shown.

DETERMINA TION OF MIXTURE AIR TEMPERA TURE

Air entering the coil is usually a mixture of both return air and fresh air.
Determine mixture air temperature per following example:

SPECIFIED:

Return Air Temperature...............................................................70°F.

Fresh Air Temperature.............................................................-200 F.

CFM (Return Air).........................................................................2000

CFM (Fresh Air)...........................................................................1000

REQUIRED:

Mixture Air Temperature (° F.)

SOLUTION:

MixtureAirTemp.=

(2000) x (70) + (1000) x (-20)

3000

= 140,000 + (-20,000) = 120,000 = 40°F.

3000 3000

To Calculate Conversion
Factors not given in above
table, use following formula:
Conversion factor =
Steam Temp. - Ent. Air T emp.
227