Sizing a hydraulic motor for the windlass

This is a follow-up to the post Sizing an anchor windlass where we calculated the mass that our anchor windlass must lift given the scope, chain and anchor sizes we are working with. The end result was that we needed to pull up 10,140 lbs.

The following calculations were worked out and sent to me by my Dad. You can catch a picture of him in another earlier post with the smudge on his nose, helping me install hull plates!

Required Power Output

Must lift (and lower) 10,000 pounds at 1 foot per second.

Thus required power output to chain = 10,000 lb-ft/sec.

Convert lb-ft/sec to hp:
From http://www.engineeringtoolbox.com/unit-converter-d_185.html#Power
1 hp (English horse power) = 745.7 W = 0.746 kW = 550 ft lb/s = 2,545 Btu/h = 33.000 ft lb/m = 1.0139 metric horse power ~= 1.0 KVA

Power output = 10140 lb-ft/sec * ( 1 hp/550 ft-lb/sec ) = 18.43 hp

Required Power Input

Assume hydraulic motor is 80% (mechanical and volumetric) efficient. Then the hydraulic motor must be receiving more input power from the hydraulic fluid, some of which is lost as heat in the motor:

Power input = 18.18 hp / .8 = 23.04 hp

Required Pump Power Output

Our system is comprised of a hydraulic pump that generates the pressure in the fluid and the hydraulic motor that generates the work. The hydraulic motor connected to the windlass is in the bow. The hydraulic pump is located at the other end of the ship in the engine room. The hydraulic pump producing the flow also must be delivering at least 23.04 hp. As some energy is lost to heat in the fluid line due to the drop in line pressure between the pump and motor, the pump must produce more than 23.04 hp to make up for line drop.

Required pressure and flow

Required input energy received from hydraulic fluid can be expressed in terms of Q*P where Q = flow (gallons/min), and P = pressure drop (psi) across the motor. Q*P is measured in units of energy.

Convert gal-psi/min to lb-ft/sec:
Hydraulic power delivered from 1 gallon/min at 1 psi
= Q*P = (1 gal/min * 1 psi) * ( 0.1337 Cu.Ft./gal ) * ( 1 min/60 sec ) * (1 lb/Sq.In / psi) * ( 144 Sq.In / Sq.Ft ) = 0.3209 lb-ft/sec

Convert gal-psi/min to hp:
Input Power (hp) = 0.3209 lb-ft/sec * ( 1 hp / 550 ft-lb/sec) = 5.834e-4 hp = (1 / 1714) hp

Thus power (hp) = Q (gal/min) * P (psi) / 1714

For example, to deliver 1 hp to the motor, Q*P = 1714

As calculated above, motor’s input power required is 23.04 hp. Either P or Q can be calculated based on the other. Lets assume P = 2000 psi. Then Q must be:

Q = power (hp) * 1714 / P (psi) = 23.04 * 1714 / 2000 = 19.75 gal/min

So now to find a pump and motor combination that match this flow and horsepower requirement….

Update on Sept 27, 2009:
I found a nice website – Ideal Windlass that has a nice listing of windlasses that might work for the tugboat.