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EOC Series selection

Step 1 Formula 1

APPLYING INTO A CASE DRAIN LINE In circumstances where the system is a closed loop or when return line flow is not available, the case drain flow can be utilized to help cool the system However, inmany instances, the case drain flow alone will not be enough to reject all of the heat generated by the system. Case drain lines should not be treated as a normal return lines since the pressure drop allowable usually can vary from 2-10 PSI max. Check with your pump manufacturer for the appropriate pressure drop tolerance before applying any cooler. To size the sys- tem for case flow or case flow plus any additional flushing loops, please use the following method. Formula: Tc exit = The corrected temperature of the oil exiting the cooler. Tc exit = { T - [ Q / (case flow gpm x 210) ]} Example: Tc exit = { 150 - [ 44,538 / (8 x 210) ]} = 123.5 Fs = Re-circulation Cooling Application (Kidney Loop) When applying any American Industrial air-cooled heat exchanger into a re-circulation (filtration loop) some important differences should be noted. The standard air-cooled heat transfer calculation can be used however some preliminary calculations must be done prior to using the formula. Before applying the standard air-cooled heat transfer formula, the air oil cooler exiting temperature must be derived from. Example Re-circulation Loop Application Fluid - Oil SAE 5w Flow - 15 GPM re-circulating Desired Reservoir Temp - 125 o F Q x Cv ____________ Tc exit - t ambient 44,538 Btu/hr x 1.13 Cv ________________ = 2,142 123.5 o F-100 o F

HP (to be removed) x 2545 Loop Flow (GPM)

 T =

Example

20HP x 2545 15gpm x 210

 T =

= 16.6 o F

Step 2 Formula 2

HP(to be removed) x 2545 x CV (T1-  T) - Ambient o F

Fs =

Example

220HP x 2545 x 1.06 (125-16.2) - 90 o F

Fs =

= 2,869.9 Fs

Step 3 Selection from the heat energy dissipation chart (page 172.) EOC-575-3-2P See example line 2pass curve. SELECTION To select a model, locate the flow rate (GPM) at the bot - tom of the flow vs Fs graph. Proceed upward until the GPM intersects with the calculated Fs. The curve closest above the intersection point will meet these conditions. Examples:

Return Line Fs = 1,258 GPM = 40 "return flow" Model = EOC-375-4

Case Line Fs = 2,142 GPM = 8 "case flow" Model = EOC-575-4-2P

Recirculation Loop Fs = 2,869.9 GPM = 15 "loop flow" Model = EOC-575-3-2P

PRESSURE DROP Determine theoil pressuredrop fromthecurvesas indicated. For viscositiesother than50ssu,multiply theactual indicated pressure drop (psi) for your GPM by the Cp value in the pressure differential curve for your viscosity value. Examples: EOC-375 EOC-575-2P @GPM = 40 @GPM = 8 Indicated pressure drop 7.8 PSI 4 PSI Cp correction factor (pg.173) 1.61 1.45 Corrected Pressure drop 12.56 PSI 5.8 PSI

Ambient Temp - 90 o F Input potential 60 HP Heat to be removed 1/3 x 60HP = 20HP Fan drive requirements 3/60/230-460 motor.

EOC - 337 - 1 - __ - S - R65 - ____ - ____

Model

Coating blank = Enamel (standard) G = Galvanize (cabinet) T = Heresite (core) X = Epoxy (cabinet) STS = Stainless Steel (cabinet)

EOC = standard EOCF = with washable filter

Tubing blank = Copper C = Carbon Steel Bypass *

Size

Motor

blank = no motor 1 = single phase 1EXP = explosion proof 3 = three phase 3EXP = explosion proof

blank = one pass 2P = two pass Passes

R65 = 65psi R30 = 30psi

Connection Type blank = NPT S = SAE

5 = 575 volt 4 = 12v DC 6 = 24v DC 9 = hydraulic

note: AIHTI reserves the right to make reasonable design changes without notice.

216

355 American Industrial Drive LaCrosse, VA 23950

tel: 434-757-1800

fax: 434-757-1810

email: sales@aihti.com

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