Dave Gerr - Boat Mechanical Systems Handbook-How to Design, Install, and Recognize Proper Systems in Boats

Finding Required Pump Power

The power to drive a pump is determined

by the head and flow rate (in gpm or lpm) and

the motor and pump efficiency.

Formula 19-3. Required Pump Power

or

Where

Apparent Head =total apparent head, ft.

or m

gpm =gallons per minute

lpm =liters per minute

SG =specific gravity of the liquid being

pumped

Pump e =pump efficiency—between

55% and 75%, use 60% as average*

Motor e =motor efficiency—for electric

motors approximately 75%

NOTE: If you set SG and pump and motor

efficiency all to 1, you get what is termed pure

water horsepower, the theoretical power

required to lift fresh water to the head speci-

fied at the given flow rate.

*Pump efficiency is actually a function of

flow rate and specific pump speed. You can

read exact efficiency from manufacturers’

pump-performance curves. At flow rates

from 100 to 200 gpm (380 to 560 lpm), for

instance, efficiency varies from around

50 percent to 75 percent.

Example:For our deck washdown, we

find the following:

Referring to Table 6-2 for specific gravi-

ties, we see that seawater has an SG of 1. 028 :

To ensure the motor does not run overloaded,

add about 15 percent.

0 .63 hp × 1. 15 = 0 .72 hp; use a^3 / 4 hp motor

hp × 745. 7 =watts, so 745. 7 × 0 .75 hp = 559

watts; use a 560-watt DC motor

amps =watts ÷volts, so 560 watts ÷12 volt

DC =47 amps

or

To ensure the motor does not run overloaded,

add about 15 percent.

470 × 1. 15 =540 watts; use about a 550 watt

DC motor

amps =watts ÷volts, so 550 watts ÷12 volt

DC =46 amps

The maximum draw will be about 46 to

47 amps. Assume about 40 to 42 amps con-

tinuous at 12 volts. This will take a lot of juice

out of the battery bank. A large yacht like this

will probably have two generators, and it

would make more sense to run this motor off

120 volts AC (240 volts in Europe).

Pump Type

Since this pump has a positive suction head

and won’t be passing solids, we could use a

wa tts

Apparent Head, m lpm SG

6 .12 Pump e Mo

=

××

××ttor e

hp

Apparent Head, ft. gpm SG

3,956 Pump e Mo

=

××

××ttor e

PART SIX:PLUMBING SYSTEMS WITH NOTES ON FIRE SUPPRESSION

TABLE 19-5.FLOW SPEED VERSUS PIPE DIAMETER

Nominal
Diameter Min.Flow Max.Flow
Schedule 40 Pipe gpm for lpm for gpm for lpm for
in. DN, mm 2 ft./sec. 0.6 m/sec. 9 ft./sec. 3 m/sec.

(^1) / 8 6 0.4 1.3 1.6 6.0
(^1) / 4 8 0.4 1.7 2.0 7.6
(^3) / 8 10 1.2 4.5 5.4 20.3
(^1) / 2 15 1.9 7.2 8.5 32.3
(^3) / 4 20 3.3 12.6 15.0 56.7
1 25 5.4 20.4 24.3 91.8
11 / 4 32 9.3 35.3 42.0 159.0
11 / 2 40 12.7 48.1 57.2 216.4
2 50 20.9 79.2 94.2 356.6
21 / 2 65 29.9 113.1 134.4 508.8
3 80 46.1 174.6 207.5 785.7
31 / 2 90 61.7 233.5 277.6 1,050.7
4 100 79.4 300.6 357.4 1,352.9
hp
91 .5ft. 2gpm 1.028 SG
3,956 0.60 Pump

=

××

×

1

ee0.7 5 Motor e

0 .63 hp

×

=

wa tts

28 m 45 lpm 1.028 SG

6 .12 0.60 Pump e

=

××

×× 00 .7 5 Motor e

=470 watts

Formula 19-3.