Where
MLWF =most likely water flow,
cu. ft./min. or m^3 /min.
1CA=one channel cross-section area,
sq. ft. or m^2
no. channels =the number of channels
from the keel-cooler inlet to the keel-
cooler exitNOTE: The number of channels are the
tubes, pipes, or other shapes that run from
the keel-cooler inlet to the outlet and then
back to the engine. Two tubes running side by
side with water entering aft and exiting for-
ward on each are two channels. The same
two tubes running side by side but with water
entering aft, turning around at a U, and re-
turning back through the other tube aft,com-
prise just one channel for these calculations.
Example: For our example 350 bhp
(261 kW) engine, we can try using copper-
nickel tubing 1-in. (25 mm) diameter. We’ll try
5 tubes on each side of the keel, port and star-
board (10 tubes or “channels” total), with
flow running from the stern forward in each.
We then getmax. flow =97 gpm × 0. 134
= 12 .98 cu. ft./min.
min. flow =61 gpm × 0. 134 = 8 .17 cu. ft./min.
MLWF =(0. 67 ×[12.98 cu. ft./min. – 8.17 cu.
ft./min.]) + 8.17 cu. ft./min. = 11 .4 cu. ft./min.
Inside diameter of 1-in. tube = 0. 877 in. ID
Inside section area =π×(0. 877 in. ID ÷2)^2
= 0 .60 sq. in.
0 .60 sq. in. ÷ 144 = 0 .0042 sq. ft. inside
section area
FV= 11 .4 cu. ft./min. ÷(0.0042 sq. ft. × 10
channels×60 sec./min.) = 4 .5 ft./sec.This falls in the desired range between 2 and
8 ft./sec. If the flow velocity (FV) had been
over 8 ft./sec., you would have had to in-
crease tube diameter, the number of tubes, or
both. Had the flow velocity been less than 2
ft./sec., you would have had to reduce the
tube diameter, the number of tubes, or both.
Reducing diameter would require longer tube
length exposed to the sea.
The required 1-in. copper-nickel tube
length (for 350 bhp, over 8 knots) is 4. 58 in./
bhp, 1,603 in., or 133.6 ft., which divided by
10 tubes = 13 .2 ft. length for each tube.Or
max. flow =367 L/min. × 0. 001
= 0 .37 m^3 /min.
min. flow =231 L/min. × 0. 001 = 0 .23 m^3 /min.
MLWF =(0. 67 ×[0.37 m^3 /min.- 0.23 m^3 /min.]) + 0.23 m^3 /min.
= 0 .32 m^3 /min.
Inside diameter of 25 mm tube =21 mm ID
Inside section area =π×(21 mm ID ÷2)^2
÷ 100 = 3 .46 cm^2
3 .46 cm^2 ÷10,000 = 0 .00035 m^2 inside section
area
FV= 0 .32 m^3 /min. ÷(0.00035 m^2
×10 channels ×60 sec./min.) = 1 .52 m/sec.
This falls in the desired range between
0 .6 m/sec. and 2.45 m/sec. If the flow velocity
(FV) had been over 2.45 m/sec., you would
have had to either increase tube diameter, the
number of tubes, or both. Had the flow
velocity been less than 0.6 m/sec. you would
have had to reduce the tube diameter, the
number of tubes, or both. Reducing diameter
would require longer tube length exposed to
the sea.
The required 25 mm copper-nickel tube
length (for 261 kW, over 8 knots) is 11.8 cm/kW,
4,130 cm, or 41.3 m, which divided by 10 tubes
= 4 .1 m length for each tube.REDUCINGKEEL-COOLERLENGTH It is often
desirable to reduce keel-cooler length. For
our previous example, we could do this by
doubling the number of tubes, returning the
flow with a U at the end of each pair of tubes.
With the flow rate under the maximum of
8ft./sec. or 2.45 m/sec., we’re not likely to
have erosion problems at the U connections.
Note that we have “doubled” the number of
tubes but not the number of “channels” for
flow, so the flow rate remains the same. Doing
this will halve the length of the keel cooler
from 13.2 to 6.6 feet (from 4.1 to 2.05 m).
As long as the tubes are arranged so that
water can flow all around their exteriors,
they can be packed fairly closely together to
make a compact unit. Thus, a shorter, com-
pact keel cooler can more readily be set flush
into a recess built into the hull to reduce ap-
pendage drag—or, for the configuration in
our example, into two recesses on either side
of the hull.PART THREE: EXHAUST SYSTEMS