Basic Circuits

„ Most people have an intuitive concept of Brief Overview of Fluid Power what at basic cylinder circuit or motor circuit would look like. Chapter 1 „ Two mechanical parameters, torque (T) and shaft speed (N), are converted to two (Material taken from Fluid Power Circuits and Controls, Cundiff, 2001) different fluid parameters, pressure (P) and flow (Q), using a pump. And then converted back to mechanical power.

Fluid Power Concept Gear Pump

Kurtz, 1999, Machine Design for Mobile Applications

Pump Systems

„ Open loop: provides cooling medium

„ Closed Loop: temperature build up

Vickers, Mobile Hydraulics Manual

Lab-Volt, Hydraulic Trainer Manual

1 Review of Engineering Concepts Brief Review

„ Work done in a 1-min interval is: „ 1 horsepower is 33,000 lbf-in / min, „ Work = Force * Distance therefore:

„ Work = F * 2πrN = 2πTN (lbf-in) „ hp = (P / 12) / 33,000

„ Since power is the rate of doing work, the „ If torque is expressed in lbf-in and N is work done in 1 min is: shaft speed in rpm, then power in hp is given by: „ P = Work / t = 2πTN / 1 (lbf-in/min) 2π(T /12)N „ hp = 33,000 „ hp = TN / 63,025

Basic Concept of Hydraulic Basic Concept of Hydraulic Cylinder Cylinder con’t

„ The distance moved is „ The force is related to relative to the fluid the pressure developed volume delivered to the at cap end.

cylinder. „ F = pA

„ x = V / A „ Where F = force (lbf) 2 „ Where x = distance (in) „ p = pressure (lbf /in ) 3 „ V = fluid volume (in ) „ A = area (in2) 2 „ A = area (in )

Basic Concept of Hydraulic Double-Acting Hydraulic Cylinder Cylinder con’t

„ Work done is given by:

„ Work = Fx „ Work = (pA) (V / A) = pV „ Power is work per unit time

„ Power = pV / t

Goering, 2003, Off-Road Vehicle Engineering Principles

2 Basic Concept of Hydraulic Basic Concept of Hydraulic Cylinder con’t Cylinder con’t

„ The units used for pressure are typically „ Flow is defined as volume per unit time, 2 Q = V / t; therefore, lbf /in , or psi, and the units of flow are gal/min, or GPM. „ Power = pQ „ To obtain hydraulic power with units of „ Mechanical power= T * N lbf-ft/min: „ Hydraulic power = p * Q „ Phyd= 231pQ / 12

„ Where Phyd = hydraulic power (lbf-ft/min) „ p = pressure (psi)

„ Q = flow (GPM)

Basic Concept of Hydraulic Basic Concept of a Hydraulic Cylinder con’t Motor

„ To obtain hydraulic horsepower: „ A flow of fluid is delivered to a hydraulic motor having displacement V . „ Phyd = (231pQ/ 12) / 33,000 m „ P = pQ / 1714 „ The displacement of a hydraulic motor is hyd the volume of fluid required to produce one revolution. „ Typical units are in3/rev. „ When a flow Q is delivered to this motor, it rotates at N rpm.

Basic Concept of a Hydraulic Basic Concept of a Hydraulic Motor con’t Motor con’t

„ N = Q / Vm „ Mechanical horsepower:

„ Where N = rotational „ Pmech = 2πTN / 33,000 speed (rpm) (Mobile Hydraulics Manual, Vickers 1998) „ Where Pmech = mechanical power (hp) „ Q = flow (in3/min) „ T = torque (lbf-ft) „ N = rotational speed (rpm) „ Vm = displacement (in3/rev) „ Substituting for N:

„ Pmech = 2πT(Q / Vm) / 33,000 3 „ Where Vm = displacement (in / rev)

(Design Engineering Handbook, Parker 2001)

3 Basic Concept of a Hydraulic Basic Concept of a Hydraulic Motor con’t Motor con’t

„ Hydraulic horsepower is proportional to the product „ If the units for flow are GPM, and the units of pressure drop and flow.

for torque are lb -in: „ ΔpQ 2πTQ /Vm f hp = = „ Pmech = 2πT /12(231Q / Vm) 1714 1714 33,000 „ Solving for torque: „ T = ΔpVm „ Pmech = 2πTQ / Vm 1714 2π „ Where T = torque (lbf-in) „ Δp = pressure drop across motor (psi) 3 „ Vm = displacement (in / rev)

Basic Circuit Analysis Basic Circuit Analysis con’t

„ The fluid power circuit has four 3. Directional control valve (DCV). The components. DCV directs the flow of fluid based on its 1. Pump. The pump develops a flow of fluid position. through the circuit. 4. Cylinder. The cylinder converts hydraulic 2. Relief Valve. The relief valve protects the energy into a force acting over some circuit. If the pressure rises high enough distance, known as the stroke. to offset the spring force keeping the valve closed, the valve opens, and flow returns to the reservoir, thus limiting the maximum circuit pressure.

Figure 1.3 Basic Circuit Analysis con’t

„ Pressure drop occurs as fluid flows through a section of hose, fitting, valve, or actuator. „ These individual pressure drops must be summed to calculate the total pressure required to achieve the functional objective.

4 Efficiency Efficiency con’t

„ Some energy is “lost” in hydraulic „ If this transfer could be done with a systems; the input mechanical energy is gearbox, typical efficiencies would be: not delivered as output mechanical „ Single reduction gearbox: 98 - 99% energy. „ Double reduction gearbox: 96 – 97% „ Triple reduction gearbox: 95% „ Mechanical energy at one location is delivered to a second location. „ Typical efficiency for a hydraulic pump to convert mechanical energy to hydraulic energy is 85%.

Efficiency con’t Summary

„ Overall efficiency of circuit „ Key advantages of fluid power are: in Figure 1.1, ignoring pressure drops, is then: 1. High power density (high power output per unit mass of system) „ 0.85 * 0.85 = 0.72 2. Control (speed of actuators easily „ This means that only 72% of the input mechanical controlled) energy is delivered as 3. Not damaged when overloaded (relief valve output mechanical energy. opens to protect system). 4. Flexible power delivery very attractive in mobile applications.

Summary con’t

„ The key disadvantage is the inefficiency. „ A fluid power option should not be used unless the advantages offset the inefficiency.

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