Selection procedure, Example – Thermal Transfer Systems OCA Series User Manual
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Selection Procedure
Performance Curves are based on 50SSU oil entering the cooler 100°F higher
than the ambient air temperature used for cooling. This is also referred to as
a 100°F Entering Temperature Difference (ETD).
step 1
determine the Heat load. This will vary with different systems,
but typically coolers are sized to remove 25 to 50% of the input
nameplate horsepower.
(Example: 100 HP Power Unit x .33 = 33 HP Heat load.)
Convert HP to BTU/MIN: HP x 42.41 = BTU/MIN
step 2
determine entering temperature difference (etd).
Desired oil entering cooler °F – Ambient air temp. °F = Actual ETD
step 3
determine Curve Horsepower Heat load.
Enter the information from above:
E.T.D. Temperature Correction Factor:
Btu/Mincorrected = Input Btu/Min x
100 x Cv
Desired E.T.D.
step 4
enter curves at oil flow through cooler and curve horsepower.
Any curve above the intersecting point will work.
step 5
determine oil pressure drop from Curves:
l
= 5 PSI; n = 10 PSI; s = 20 PSI; ; = 30 PSI; H = 50 PSI.
Multiply pressure drop from curve by correction factor found in
oil
P correction curve.
Example
FLUID = SAE 20 OIL
SYSTEM ELECTRIC NAMEPLATE HORSEPOWER = 300HP
ENTERING TEMPERATURE = 200°F
AMBIENT TEMPERATURE = 75°F
FLOW RATE = 200GPM
100
2.04
2.44
4.44
6.44
8.84
1.11
1.57
1.86
2.58
4.23
6.48
9.42
13.60
1.30
3.04
3.54
0.770
110
1.74
2.14
3.64
5.14
6.74
1.08
1.49
1.76
2.39
3.77
5.74
8.37
11.67
1.24
2.44
2.94
0.760
120
1.54
1.84
3.04
4.24
5.64
1.06
1.42
1.64
2.19
3.30
5.95
7.27
9.77
1.18
2.14
2.54
0.749
130
1.44
1.64
2.64
3.44
4.54
1.03
1.34
1.53
1.98
2.84
4.18
6.23
7.84
1.12
1.94
2.24
0.738
140
1.34
1.54
2.27
2.94
3.74
1.01
1.27
1.42
1.79
2.42
3.51
5.24
6.15
1.07
1.94
2.04
0.726
150
1.24
1.34
1.94
2.54
3.14
0.99
1.21
1.34
1.65
2.08
2.94
4.39
4.81
1.02
1.74
1.94
0.716
200
0.97
1.00
1.24
1.44
1.64
0.93
1.03
1.12
1.22
1.37
2.63
1.78
1.99
0.94
1.24
1.34
0.675
250
0.85
0.86
0.96
1.01
1.09
0.89
0.97
1.00
1.07
1.15
1.25
1.26
1.27
0.87
1.04
1.09
0.596
mil-l
ester
sae 5
sae 10
sae 20
sae 30
sae 40
iso 22
iso 32
iso 46
iso 68
iso 100
iso 150
iso 220
iso 320
7808
polyglycol phosphate
50%eg
entering
liquid
temp
C
p
pressure drop CorreCtion FaCtors
100
1.12
1.16
1.26
1.39
1.46
1.09
1.15
1.19
1.27
1.38
1.44
1.57
1.85
1.20
0.93
0.84
0.86
110
1.10
1.13
1.21
1.33
1.41
1.07
1.14
1.17
1.26
1.32
1.40
1.49
1.68
1.15
0.90
0.81
0.85
120
1.07
1.11
1.18
1.28
1.36
1.05
1.12
1.15
1.21
1.28
1.36
1.41
1.54
1.10
0.89
0.80
0.85
130
1.05
1.09
1.14
1.25
1.30
1.04
1.10
1.14
1.18
1.25
1.31
1.35
1.45
1.06
0.86
0.78
0.84
140
1.04
1.06
1.12
1.20
1.26
1.03
1.09
1.11
1.17
1.21
1.27
1.31
1.40
1.04
0.85
0.77
0.83
150
1.02
1.05
1.10
1.17
1.23
1.03
1.07
1.10
1.14
1.18
1.23
1.28
1.34
1.02
0.84
0.75
0.83
200
0.99
1.00
1.02
1.05
1.08
0.99
1.00
1.01
1.02
1.03
1.09
1.10
1.15
0.99
0.80
0.72
0.81
250
0.96
0.97
0.98
0.99
1.00
0.96
0.97
0.97
0.97
0.98
1.00
1.02
1.03
0.98
0.77
0.70
0.80
mil-l
ester
sae 5
sae 10
sae 20
sae 30
sae 40
iso 22
iso 32
iso 46
iso 68
iso 100
iso 150
iso 220
iso 320
7808
polyglycol phosphate
50%eg
entering
liquid
temp
C
V
VisCosity CorreCtion FaCtors
Determine heat load.
Generally, about 25% to 33% of the system horsepower is removed.
300hp x 0.33 = 99hp
Since the graphs have the heat load in terms of BTU/min, the units must
be converted.
99hp x 42.4167 = 4,199 BTU/min
Calculate the entering temperature difference (E.T.D.). The E.T.D. is the
inlet oil temperature minus the entering air temperature.
ETD=200-75 =125
Calculate the corrected curve heat load.
Corrected curve heat load = actual heat load x (100/ETD) x Cv (viscosity
correction factor obtained from the Cv table).
4,199 BTU/min x (100/125) x 1.02= 3,426 BTU/min
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. Usually the smallest cooler is most desired. In this case the
intersecting point on the single pass graph indicates that the OCA-450
will suffice.
The pressure drop should be found next. Find the point on the curve that is
directly above the intersecting point. This point on the curve indicates the
pressure drop.
P
≈ 6psi
These curves are made for SAE 10 oil entering at 200°F. Therefore,
the pressure drop needs to be corrected. The 1.24 is the pressure drop
correction factor obtained in the Cp table.
P
CORRECTED
= 6 x 1.24= 7.44 psi