Figure 1 - Winch Driven by Motor and Reduction Gearbox

TEP 4205 INDUSTRIAL HYDRAULICS

4Winch Application

Figure 1 - Winch driven by motor and reduction gearbox.

4.1Lifting the load

Winch drum torque

Motor torque

Motor pressure

motor mechanical efficiency

reduction gearbox mechanical efficiency

winch drum mechanical efficiency

4.2Lowering the load

4.3Numerical Values

Data

r = 0.25 m

M = 2.5 Tonne

D = 574 cm3 rev-1 = 9.1 x 10-5 m3 rad-1

R = 5:1 (reduction)

m = 0.78 (starting), 0.92 (running)

R = d = 0.94

Ideal pressure

Therefore, during start-up from rest:

T = 0.69

Pup = 193 bar & Pdown = 92 bar

And when operating at speed:

5 Hydraulic Motor for Driving a Winch

A winch is to be driven by a hydraulic motor and in order to provide a wide range of operating speeds at the maximum supply flow the motor displacement can be set at either a maximum or minimum value. These displacement values (Dmax and Dmin) are pre-set by mechanical stops in the motor but their level can be chosen from within the range shown in the data. The winch operator makes the selection of the two displacements by changing the position of a control valve in the hydraulic circuit.

The mechanical transmission has a speed-reducing gearbox and to design the system it is required to select an appropriate motor from the two sizes given in the data and determine the reduction ratio of the gearbox that will provide the specified performance.

1)Calculate the gearbox ratio that is required for each motor to give the maximum winch torque when the motor is operating at a selected pressure.

2)Determine the motor speeds from 1) that are required to give the maximum cable speed of 50m/min and select a suitable motor from the data.

3)Calculate the flow that is required to drive the selected motor when in maximum displacement at the speed that produces a cable speed of 15m/min.

4)Using the flow from 3) determine the required minimum motor displacement that will provide the maximum cable speed of 50m/min.

5)For this flow calculate the motor speed for an oil viscosity of 10cSt. This needs to account for the effect of oil viscosity on the volumetric efficiency.

Data

Motors (Maximum pressure for continuous operation = 300bar)

Motor displacement (Dmax/Dmin) cm3/rev / 55/11 / 80/17
Maximum speed at Dmax rev/min / 4200 / 3750
Maximum speed for D < Dmax rev/min / 6300 / 5600

Winch drum diameter0.5m

Gearbox mechanical efficiency ()92%

Maximum winch load200kN

Maximum cable speed at maximum load 15m/min

Maximum cable speed 50m/min

Motor mechanical efficiency at a) Dmax, b) D<Dmax () 95%, 90%

Motor volumetric efficiency for Dmax (35 cSt oil viscosity) () 95%

Motor volumetric efficiency for D < Dmax (35 cSt oil viscosity) () 90%

5.1Gearbox ratio

Choose the maximum operating pressure of 300bar.

Winch drum torque = Load force x drum radius =

Motor torque at 300bar:

Required reduction gearbox ratio (n)

This gives values of n required for the two types of motor of 218 and 150.

5.2Motor selection

Cable speed (Nm = motor speed)

Motor speed which gives motor speeds of 6939 and 4775

rev/min respectively.

Select the larger motor, as its speed is less than the maximum allowable value given in the data.

5.3Flow required

Motor speed for a cable speed of 15m/min

The flow, Q, required for this speed

5.4Minimum motor displacement

For the flow given in 5.3, the minimum motor displacement required to operate the winch at a cable speed of 50 m/min is given by:

5.5Maximum motor speed with oil having a viscosity of 10cSt

At minimum displacement the volumetric efficiency of 90% was quoted for an oil viscosity of 35cSt which represents a leakage flow of 10%. This leakage will increase as the oil viscosity reduces so that here it will be and the volumetric efficiency will therefore be 65%.

6Hydraulic system for gantry crane

A hydraulically powered gantry crane, shown in Figure.2, is to lift loads up to 300 Tons. It is required to design the hydraulic systems for driving both the winch and the crane wheels.

Details of the installation are given below:

a)Crane Winch (2 Drums)

Total load= 300 Tonne

Figure 2 Gantry crane

Two pulley blocks each having 4 pulleys to provide 8 lengths of cable for each of the two lifting hooks.

Maximum lift= 10m
Winch drum diameter= 600mm
Lift speed (variable) (max)= 0.02m/s

Each of the two winch drums are to be powered separately, and are required to be held stationary in any position.

b)Wheel Drive

The vehicle has eight wheels that are grouped together into units having two wheels each. In these units the two wheels are connected by a chain drive, with one of the wheels being driven by a hydraulic motor (i.e. four motors in total). There is no requirement for brakes on these wheels.
Supply pipe lengths are 30m on one side and 80m on the other.
As far as is possible the torque on each side must be within 10% of each other.
Braking will be carried out by reducing the pump supply and means must be provided to prevent both cavitation and excessive pressures.
Gear reduction ratio between wheels and motor = 2.5:1
Wheel drive system efficiency = 80%
Maximum wheel torque at low speed = 4250Nm
Maximum wheel torque at high speed = 2120Nm
Wheel diameter = 0.46m
Maximum speed = 30m/min
Minimum speed =1.2m/min
Maximum pressure = 250bar

CABLE INFORMATION

Cable diameter mm / Minimum breaking force (kN)
35 / 785
40 / 1000
48 / 1460

6.1Gantry crane

Load = 3000kN

8 Cables (falls) and 2 pulley blocks

Cable tension

Allow 10% for friction -

Choose cable of 40mm diameter

For one cable layer the winch drum torque

The winches can be driven by either low or high speed motors and the reduction gearbox ratio needs to be selected accordingly.

a)Low speed motor

Choose 10:1 reduction gearbox and allow 8% for mechanical losses

The motor torque

For 80% motor efficiency at 250bar the motor displacement is given by:

There are 4 pulleys and 8 lengths of cable on each side which gives a maximum cable speed of:

The winch drum speed and the motor speed will be 47.8 rev/min.

b)High-speed motor

For a high speed motor, a gear box ratio of 100:1 can be used which, using the same gear box efficiency as in a), will require a motor having a displacement of 225 cm3/rev operating at a speed of 478 rev/min.

c)Motor flow

For both the high and low speed motor drives, the motor flows will be the same. Thus, assuming a motor volumetric efficiency of 90% the flow required for each winch is:

The required maximum pump displacement for supplying both winches, assuming a volumetric efficiency of 95% and a drive speed of 1500 rev/min, will be:

For a mechanical efficiency of 95% for the pump the required input power is:

The total output power from the winches is:

Appropriate winch motors and supply pump can be selected from manufacturers’ literature and the assumed values of the mechanical and volumetric efficiencies used in the analysis can be checked.

6.2Wheel drive

For a torque at each wheel of 4250Nm, two wheels require 8500 Nm torque. Thus for a gearbox reduction ratio of 2.5:1 the required motor torque is given by:

(80% wheel drive mechanical efficiency)

At a vehicle speed of 1.2m/min (U) with 0.46m wheel diameter,

Hence

Table 1 Motor performance

U(m/min) / 1.2 / 30.0
Wheel Speed (rev/min) / 0.83 / 20.75
Motor Speed (rev/min) / 2.08 / 51.9
Motor Dm(cm3/rev) / 1476 / 738
Nominal Flow (L/min) / 3.06 / 38.3
Motor Torque (Nm) / 4250 / 2120
Pressure (bar) / 215 / 214

As for the winch drive, low or high-speed motors can be selected for the wheel drive. By way of example, a dual displacement low speed motor is considered here which gives the performance outlined in Table 1.

The maximum total nominal flow (two motors each side) = . Assuming that the volumetric efficiency of the motors is 90% and 95% for the pump gives a maximum pump flow of:

For 1500 rev/min pump speed, the required pump displacement is:

As for the winch drive, the pump and motors having displacements that are closest to those required can be selected from manufacturers’ literature.

6.3Pipe sizes

For motor flows of 80L/min to each side.

The pressure loss

f - Friction factor from Moody diagram (chapter 8)

For the pipe the value of the Reynolds No.

Taking a value of fluid viscosity of 40cSt () and length L of the longest side of80m gives:

Table 2 Pipe pressure loss

d mm / Re / / / U m/s
20 / 2122 / 0.03 / 9.4 / 4.2
25 / 1700 / 0.038 / 3.9 / 2.7
30 / 1415 / 0.043 / 1.8 / 1.9

The effect of the pipe diameter on the pressure loss can be seen from Table 2

A circuit that will provide the necessary functions is shown in Figure 3. This contains the following major functions:

Crane Winch

Pilot operated check valve to protect from hose failures
Cross line relief valves
Brake control by:

-Selection of the DCV

-High pressure in the circuit

Counterbalance valves for lowering the load

Wheel drive

Cross line relief valves
Brake valves
Purge valve to extract flow for cooling from the low pressure side of the circuit (drain flows may also be passed to the cooler inlet).
Motor displacement selection valve.
Boost pump to make up drain flows from the pump and motors.

9Simple actuator cushion

Using the analysis given in chapter 3, 4.2, the performance of an actuator cushion using a simple orifice or adjustable restrictor valve can be determined for the actuator application having the given data. The peak actuator pressure needs to be limited to 300 bar that occurs at the commencement of cushioning.

Data:

Actuator piston diameter140 mm

Cushion plug diameter 43 mm

Mass40 000 kg

Initial velocity 0.3 ms-1

From chapter 3, 4.2, the velocity variation is given by:

For a maximum pressure of 300 bar the restrictor size is given by:

and, taking a value for CD of 0.65 gives:

Now:

The velocity will reduce to 37% of its initial value (0.11 m/s) in 8.6 mm.

P Chapple February 2004.WORKED EXAMPLE SOLUTIONS FEBRUARY