hydraulic system analysis

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MEL334: Low Cost Automation 1

Hydraulic Circuit Analysis

Dr. Sunil Jha


Introduction

Energy Losses in Fluid Power Systems.Bernoulli‘s equation & Continuity equation used to perform analysis of Fluid Power SystemCalculating Pressure drops, flow rates, HP losses for all components.

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Laminar or Turbulent Flow

Reynolds number<2000 : LAMINAR>4000 : TURBULENTCRITICAL ZONEGreater losses in Turbulent flowFluid power system designed to operate in LAMINAR flow region.


Friction Losses

Friction – Main cause of lossesLoss in Pressure headHead Loss

Losses in PipesLosses in Fittings

Head loss in Pipes – DARCY’s Equation

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DARCY’s Equation

HL = f (L/D)(v2/2g)f = friction factor (dimensionless)

L = length of pipeD = pipe inside diameterv = avg. fluid velocityg = Acceleration due to gravity

Used for Laminar as well as Turbulent flowDifference lies in evaluation of friction factor.


Frictional Losses in Laminar Flow

Friction Factor f = 64/(Reynolds Number)Hagen Poiseuille Equation

HL = (64/NR)(L/D)(v2/2g)

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Frictional Losses in Turbulent Flow

Relative roughness = ε/Dε = absolute roughnessD = Pipe inside diameter

Typical values of Roughness (µm)Drawn tubing = 1.524Commercial steel = 45.72Cast Iron = 121.92Galvanized Iron = 152.4Riveted Steel = 1828.8

Moody diagram used to calculate friction factor


Moody Diagram

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Losses in Valves & FittingsEnergy losses in valves and fittings such as Tees, Elbows, and BendsNature of flow through Valves and Fittings is very complex.Head Loss HL = Kv2/2gK factor for Valves and Fittings

Gate Valve (wide open) = 0.19¾ open = 0.90, ½ open = 4.5, ¼ open = 24.0


K-Factor

Return Bend = 2.2Standard Tee = 1.890 deg Elbow = 0.7545 deg Elbow = 0.42Ball Check valve = 4.0

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Equivalent Length TechniqueDarcy’s Equation

Friction head loss in pipes proportional to Square of Fluid VelocityLength of Pipe

Head Loss in valves & fittingProportional to square of fluid velocity

Possible to find a length of pipe that for the same flow rate would produce same head loss as a valve or fitting.


Equivalent Length TechniqueHL(valve or Fitting) = HL(Pipe)

K(v2/2g) = f(L/D)(v2/2g)Since velocities are equalEquivalent Length Le = KD/fConvenient method for analyzing hydraulic circuits where frictional energy losses are to be taken in to account.

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ProblemFor the Hydraulic System shown, following data are given:

Pump is adding 5 hp (3730 W) to fluidPump flow is 0.001896 m3/sPipe has 0.0254 m inside diaSp. Gravity of oil = 0.9Kinematic viscosity of oil is 100 cSElevation difference between station 1 & 2 is 6.096 mPipe lengths: 1 ft = 0.305 m, 4 ft = 1.22 m, 16 ft = 4.88 m

Find pressure available at inlet to hydraulic motor. The pressure at the oil top surface level in the hydraulic tank is atmospheric (0 MPa). The head loss HL due to friction between stations 1 & 2 is not given.


Problem

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Solution

Bernoulli’s equation between stations 1 & 2Z1+P1/ρ+v1

2/2g+Hp-Hm-HL = Z2+P2/ρ +v22/2g

Hp = Head added by pumpHm = Head removed by Hydraulic MotorHL = Head lost in friction

Hm = 0 , No Hyd motor between 1 & 2V1 = 0, P1/ρ = 0Z2 – Z1 = ?


Solution

Solve for v2

Evaluate velocity head at station 2Find Reynolds Number

NR = ρvD/µ = vD/(Kinematic Viscosity)SI unit Kinematic Viscosity = m2/sStokes = cm2/s

Laminar or Turbulent ?Find friction factor

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SolutionDetermine Head loss due to friction

Darcy’s equationCalculate LUse Leq for standard elbow, K = 0.9

Substitute in Bernoulli’s equation to solve for Pressure head at 2.Calculate Pump Head

Hp = 0.762 (HP)/Q(m3/s).Sg

Solve for P2 , use ρwater = 9797 N/m2


Problem for Exam

Determine the External load F that the hydraulic cylinder can sustain while moving in The extending direction.

hydraulic system analysis

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