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TSM 363 Fluid Power Systems

Hydrostatic Transmission

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Hydrostatic Transmission is popularin ORE

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Mechanical Drivetrain

A mechanical drivetrain systemconsists of an engine, a clutch,a transmission, a universal drive-line, a differential, and the wheels.

Transmission is used to increaseor decrease speed

A differential is needed in

mechanical drivetrain, becauseit allows the outside wheel torotate faster than in the insidewheel when the vehicle turns.

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Hydrostatic Transmission DrivetrainWith a hydrostatic transmission, itconsists of an engine, a clutch,a hydrostatic transmission, auniversal driveline, a differential,and the wheels.Change the displacement of thepump (or motor) can change thevehicle speed.

HT improves maneuverability:> wide range of T/n ratio> compacted, less mass inertia> dynamic braking (inst. reverse)> able to stall & undamaged> no interruption of power

HT has lower efficiency

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Basic Concept ofHydrostatic Transmission

A hydrostatic transmission is simply a pump and motor(s)connected in a circuit.

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Basic Types of HST

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(a) Fixed pump-fixed motor

(d) Variable pump-variable motor

(b) Variable pump-fixed motor

(c) Fixed pump-variable motor

Typical Closed-Circuit HST Configurations

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Typical P erformance of VP-FM HST

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Typical P erformance of FP-VM HST

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Typical P erformance of VP-VM HST

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Typical HST Releasing & Recharging

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Typical Controls of Closed-Circuit HSP

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Typical Open-Circuit HST Configurations

(b) Four-way valve controlledfixed pump-fixed motor

(a) Needle valve controlledfixed pump-fixed motor

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Typical E/H Controlled Open-Circuit HSP

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Typical HST Arrangement - 1

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Typical HST Arrangement - 2

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Typical HST Arrangement - 3

TSM 363 L 10

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Example:Typical Calculations of HST

Suppose both the pump and the motor in a hydrostatictransmission have disp. D M = 1.925 in3/rev, and the volumetricefficiency are 0.90 for both the pump and motor. The pumpis driven by an engine at 2400 rpm, what is the motor speed?

Theoretical flowQ t = (DPnE)/231

= (1.925x2400)/231 = 20 gpmPump flow

QP = Qt Eff P = 20x0.90 = 18 gpmMotor speednM = (QP231)/ D M Eff M

= (18X231)/1.925x0.90 = 1944 rpm

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A Special HST Application

TSM 363 L t 10

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Lecture Summary

Discussed the basic concepts of HST: Closed-circuit HST: FP-FM; VP-FM; FP-VM; VP-VM

Open-circuit HSTApplications of HST:

In-line designSprit designV-controlled designSprit-torque design