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Hydraulic Oil Cooler

Standard Oil Cooler 1-2 Pass Models :

Model	Size	Capacity kcal/hr	Oil Flow LPM	A	B	C	D	E	N1/N2	N3/N4	NS DRAIN
FOC 2	4 "x 18 "	1500	10-15	456	78	300	370	100	G 3/4 "	G 1 "	G 1/4 "
FOC 3	4 " x 20 "	2400	15-30	472	81	310	386	100	G 3/4 "	G 1 "	G 1/4 "
FOC 5	4 " x 24 "	3000	20-35	606	88	430	520	100	G 1 "	G 1 "	G 1/4 "
FOC 7.5	4 " x 30 "	5000	30-35	756	93	570	670	100	G 1 "	G 1 "	G 1/4 "
FOC 10	6 " x 18 "	6000	35-75	456	78	300	350	150	G 1 "	G 1 1/2 "	G 1/4 "
FOC 15	6 " x 20 "	10500	60-100	472	78	315	366	150	G 1 "	G 1 1/2 "	G 1/4 "
FOC 20	6 " x 24 "	13500	80-150	606	98	410	500	150	40 NB Flange	G 1 1/2 "	G 1/4 "
FOC 25	6 " x 30 "	18000	100-190	756	103	550	650	150	40 NB Flange	G 1 1/2 "	G 1/4 "
FOC 30	8 " x 24 "	24000	140-270	606	98	410	500	200	40 NB BSP	G 2 "	G 1/2 "
FOC 40	8 " x 24 "	32000	180-370	606	98	410	500	200	40 NB BSP	G 2 "	G 1/2 "
FOC 50	8 " x 30 "	34000	220-415	756	118	520	650	200	50 NB BSP	G 2 "	G 1/2 "
FOC 60	8 " x 36 "	38000	240-456	906	98	710	800	200	50 NB BSP	G 2 "	G 1/2 "

Construction Material

Model	Tubes	Shell	Tubessheet	End Cover	Baffies
Standard Models	Copper	IS 1239 ERW	IS 2062	CI	IS 2062
Marine Application	Cu Ni	SS 316 L	Brass/AB-2	LG-4	Brass

Max. Operating Pressure (Shell)	10 bar
Max. Operating Pressure (Tube)	5 bar
Max. Operating Temperature	100 0C
Water Flow Rate	1.3 Ã— Oil Flow Rate

Heat Exchange Selection :

Heat Load Q (kW) = m x Cp x Δ T
Where, m = Mass flow rate of oil = Oil flow rate ( LPM ) x p
m (kg/s) = ( LPM x p / 60,000 )
Heat load Q (kW) = m x Cp x (To-Ti)
Heat load Q (k cal /hr) = Q (kW) x 860.4

Sample Calculation
Heat load Q (kW) = m x Cp x Δ T
Where, m (kg/s) = 40 x 864 / 60,000 = 0.576 kg/s
Heat load Q (kW) = 0.576 x 1.965 x (50-45) = 5.66 kW
Heat load Q (k cal /hr) = 5.66 kW x 860.4 = 4870 k cal/hr
Thus selecting FOC 7.5 Model from Table for above heat load and
oil flow rate. Water flow rate = 1.3 times Oil flow rate
Apart from our standard models, we undertake design of shell and tube
heat exchangers as per customer requirement.

Required Parameters

Oil Grade
Density of Oil p (kg/m3)
Specific Heat of Oil Cp (kJ/kg °C)
Oil Inlet Temperature Ti (°C)
Oil Outlet Temp Required To (°C)
Oil Flow Rate LPM

Assumed Parameters

Oil Grade= ISO VG 68
Density of Oil p (kg/m3) = 864
Specific Heat of Oil Cp (kJ/kg °C) =1.965
Oil Inlet Temperature Ti ( °C) = 50
Oil Outlet Temp Required To ( °C) = 45
Oil Flow Rate LPM = 40

Flowtex Oil Cooling Package Models (AHOC1 - AHOC9)

Cooler Model	Oil Flow LPM	Heat Loan KW	A	B	C	D	E	F	G	H	I	J
AHOC1	10-30	2-5	400	369	198	250	200	DIA 12	175	110	DIA 8.5	G 1/2 "
AHOC2	30-70	4-10	527	466	420	350	300	DIA 12	225	110	DIA 11	G 1/2 "
AHOC3	50-90	8-15	540	466	453	350	300	DIA 12	225	110	DIA 11	G1 "
AHOC4	70-130	12-20	657	591	440	350	300	DIA 12	300	150	DIA 11	G 1 1/4 "
AHOC5	100-180	18-26	690	591	473	350	300	DIA 12	300	150	DUA 11	G 1 1/4 "
AHOC6	150-220	25-35	787	760	545	425	375	DIA 14	350	150	DIA 13	G 1 1/4 "
AHOC7	200-270	32-42	820	760	577	425	375	DIA 14	350	150	DIA 13	G 1 1/4 "
AHOC8	250-320	40-50	887	868	565	450	400	DIA 14	415	165	DIA 13	G 1 1/4 "
AHOC9	300-420	45-65	920	868	623	450	400	DIA 14	415	165	DIA 13	G1 1/2 "

Flowtex Oil Cooling Package Models (AHOC10 - AHOC13)

Cooler Model	Oil Flow LPM	Heat Loan KW	A	B	C	D	E	F	G	H	I	J
AHOC10	350-480	60-75	1241	1079	681	450	380	829	1029	18 Ø X 36	G 1/2 "	G 1/2 "
AHOC11	400-500	72-85	1241	1079	730	450	380	829	1029	18 Ø X 36 SLOT	G2 "	G2 "
AHOC12	450-500	85-93	1365	1208	715	450	380	958	1158	18 Ø X 36 SLOT	G2 "	G2 "
AHOC13	480-520	90-100	1365	1208	731	450	380	958	1158	18 Ø X 36 SLOT	G2 "	G2 "

Materials :

Core : Brazed aluminium bar & plate.
Connection : Aluminium female BSP.
Fan Cowl : Steel with powder coating finish.
Fan Guard : Steel
Fan : PAG
Motor Bracket : Steel with enamel finish.
Motor : Electric motor (Make CGl)

Cooler Type	Fan Dia (mm)	Fan Speed (RPM)	Noise Level dB (A)	Motor Voltage / Frequency (+ / 10%)	Motor Power HP Three Phase	Max. Working Pressure (Bar)
*AHOC1	250	1410	68	380/50	0.5	15
*AHOC2	350	1410	79	415/50	0.75	15
AHOC3	350	1410	79	415/50	0.75	15
AHOC4	450	1410	80	415/50	1	15
AHOC5	450	1410	80	415/50	1	15
AHOC6	550	1410	83	415/50	1.5	15
AHOC7	530	1410	83	415/50	1.5	15
AHOC8	630	1410	85	415/50	1.5	15
AHOC9	630	1410	85	415/50	2	15
AHOC10	750	1440	85	415/50	2	15
AHOC11	750	1440	95	415/50	3	15
AHOC12	900	1440	95	415/50	3	15
AHOC13	900	1440	95	415/50	3	15

Selection Procedure

Design Condition :

Oil gRADE : iso vg 68
Oil Flow Rate : 10 to 520 LPM
Oil Inlet Temperature = 70 0 C
Oil Outlet Tempeture = 650C
Ambient Air Temperature = 40 0C

Determine the heat load to be rejected in kW.

Heat Load to be rejected (Q) in kW.
Q = 0.03 x Oil Flow Rate in LPM x (Inlet Oil Temperature - Outlet Oil Temperature)
The Inlet oil temperature is generally the maximum system oil temperature.
The Outlet oil temperature is generally the desired oil temperature from the cooler

Determine Entering Temperature Difference. ( E.T.D)

E.T.D oC = Inlet Oil Temperature Inlet Air Temperature.
The Inlet Air Temperature is the highest Ambient Air temperature the application will see.

Determine the corrected heat dissipation to use the curves

Corrected Heat Load = Q ( kW ) x 30 oC / E.T.D

Select models from curves.

On the Graph of Heat Load ( kW) v/s Oil Flow Rate (LPM) plot the intersecting point of Corrected Heat Load and Oil Flow Rate. From the graph provided select appropriate model of AHOC. Select the nearest curve above the intersection point of the graph.

Determine oil pressure drop from table :

From the table of pressure drop (bar) and Oil Flow Rate (LPM), for the selected model of Hydraulic Oil Cooler the corresponding pressure drop indicates the maximum pressure drop for the given maximum oil flow rate.

Example :

Actual Conditions (Assumed):

Fluid : ISO VG 68
Oil Inlet Temperature : 70 oC
Oil Outlet Temperature : 51.8 oC
Ambient Temperature : 45oC
Flow Rate : 20 LPM.

Determine Heat Load in kW.

Q = 0.03 x Oil Flow Rate in LPM x (Inlet Oil Temperature - Outlet oil temperature)
= 0.03 x 20 ( 70 - 51.8 )
= 11 kW.

Calculate the Entering Temperature Difference. ( E.T.D)

E.T.D oC = Inlet Oil Temperature - Inlet Air Temperature.
= 70 - 45
= 25oC

Determine the corrected heat dissipation to use the curves.

Corrected Heat Load = Q ( kW ) x 30Â·C I E.T.D
= 11 x 301 25
= 13.2 kW.

Select models from curves :

Find the intersection point between the corrected heat load and flow rate on the performance curves. Any curve above this point will work for this application. In these case the intersection point on the graph indicates AHOC6.

Determine oil pressure drop from table :

The pressure drop should be found next. From the table of pressure drop ( bar) and max. oil flow rate I for AHOC6 the maximum pressure drop of oil will be 0.92 bar.

Similarly, you can try to select any AHOC models as per your requirement by following the above steps with your technical data.

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