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Section 1.3 - Hydraulic System Power Requirements

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Typical Problem

Suppose you were asked to determine the mechanical horsepower (HP) required to retract a landing gear in a required time period. How would you do the calculations?

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Given- Force Requirements = 5000 lb  (this is the force that has to be moved)
           Distance moved = 2 ft   (this is the distance you must move the force)
           Time required = 30 s (this is the time required to move that distance)

Power is given as Force times velocity for a constant force (P=Fv).  If the force is not constant, then you can use the average force over the time required.  The velocity in this case is the average velocity, namely, the distance traveled over the time required.  Therefore,
Power=Force x distance / time

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We convert to horsepower (HP) using the conversion factor 550eq004.gif (181 bytes) equals 1 HP.  Therefore, by multiplying 334 eq004.gif (181 bytes) by [1 HP/ (550 eq004.gif (181 bytes))], we find that we need 0.61 HP. 

Thus, an actuating cylinder must then be mounted which can deliver 0.61 HP. The actuating cylinder for the retractable landing gear is mounted so that it can move in order that the piston rod in the actuating cylinder won't bend.  A flexible hose to the oil pressure lines is put at the cylinder attachment so that it won't break during movement.

Selection of an Actuating Cylinder

The selection of the actuating cylinder depends upon two parameters:

  1. Piston stroke - the distance that it must travel to do the job.
  2. Piston head area which must be large enough to develop the proper force with the pressure available.

Flow Requirements to Accomplish Task

The hydraulic system oil flow rate, Q, may be measured in gallons per minute (gpm).  The flow rate required can be related to the volume of fluid required to be moved (in cubic inches-cu in) and the time required for the job (in minutes).

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The volume of fluid required to be moved is given by the input force times the piston stroke (in inches) divided by the system oil pressure.  Remember that force divided by pressure is an area, and, multiplied by the piston stroke defines the volume moved.  Therefore,

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If the pressure in the system = 2000 psi, find Q of previous problem.

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Hydraulic Horsepower

The hydraulic horsepower is the power provided by the hydraulic system. It is directly proportional to the rate of flow, the pressure, a constant and inversely proportional to the efficiency of the system.  The coefficient equals 0.000583 and is the conversion factor between gallon-lbs/(minute-square inches) and horsepower.  Therefore:

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Find the hydraulic HP of the previous problem if the system has an efficiency of 1.

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F-111 sweep back problem

Let's look at a typical problem.  Find the hydraulic and mechanical HP required to vary the sweep back of an experimental  F-111 wing, given the following data:

Force required = 160,000 lb;
Cross-sectional area of the actuating cylinder piston A = 32 square in
Fluid Pressure P = 5000 lb/sq. in. = 5000 psi
Piston stroke D=30 inches
Time required for sweeping the wing T=75 seconds=1.25 minutes

Hydraulic HP is found by getting the flow rate, Q, in gpm, FIRST

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Now having found Q, we can now find the Hydraulic HP, assuming an efficiency of 1, using

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The mechanical HP is found using

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By comparing both results, we can see that the hydraulic system will meet the requirements of the mechanical system HP.

Note: In reality, the actual value is about 9.7 HP, but because systems that deliver the horsepower are rated in 1/4 HP increments, the results would be listed to the next 1/4 HP increment, namely 9.75 HP

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Updated: September 28, 2005