LECTURE 8. Hydraulic machines and systems II 2002 MIT PSDAM LAB

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1 LECTURE 8 Hydraulic machines and systems II

2 Basic hydraulic machines & components Graphical Nomenclature Arrows show direction of flow Control Volume Pipe or hose with fluid flow Pipe or hose without fluid flow Pay attention to flows in/out Power out Power in port Motor Pump port Cylinder Valve Pressure Gauge Reservoir

3 Example I Pump & cylinder Solve for the the velocity of piston and the force exerted by piston Note where power crosses into and out of the system boundary T p = 10 in-lbf ω p = 1000 rpm 2π π Power in Power out p 4 = 14 psi x p 3 Pump Valve p 3 = 1014 psi Cylinder F cyl A cyl = 10 in 2 Reservoir

4 Example I Pump & cylinder cont. Force exerted by piston: F cyl = A cyl p cyl p cyl = p 3 -p 4 = 1014 psi 14 psi = 1000 psi F cyl = 10 in lbf / in 2 = lbf If we know F and v, we know the power output of the cylinder At the boundary of the hydraulic system we see one inflow & one outflow of power From the power balance: Σ P in = Σ P out + Σ P loss + Σ (de stored /dt); If P loss & (de stored /dt) are small compared to P out : Σ P in T p ω p ~ Σ P out ~ F cyl v cyl v cyl ~ (T p ω p )/ F cyl = (10 in-lbf) (1000/2π rev / min ) (2π rad / rev ) (1/60 min / s ) / ( lbf) = in/s Always check and list your units!!!

5 Example II Pump, motor, & cylinder Given the diagram, solve for T m and ω m Note where power crosses into and out of the system boundary D p = 0.5 in 3 / rev D m = 1 in 3 / rev T p = 7.16 in-lbf ω p = 1000 rpm 2π T m =? ω m =? p 3 p 4 = 33.2 psi x Pump Motor Valve p 3 = 44 psi F cyl A cyl = 5 in 2 Reservoir

6 Example II Pump, motor, & cylinder cont. Motor speed: We know that the mass flow rate through the pump and motor has to be the same. As we assume the liquid is incompressible, this means the volumetric flow rate is the same: Q p = ω p D p = Q m = ω m D m ω m = ω p (D p /D m ) = [1000/(2π) rev / min ] [( 0.5 in 3 / rev ) / ( 1 in 3 / rev ) ] = 500/(2π) rev / min Motor torque: Σ P in = Σ P out + Σ P loss + Σ (de stored /dt) ; If P loss & de stored /dt are small compared to P in : Σ P in ~ Σ P out T p ω p ~ T m ω m + F cyl v cyl ~ T m ω m + ( p cyl A cyl ) v cyl T m ~ [ T p ω p -( p cyl A cyl ) v cyl ] / ω m We can not solve as we don t know v cyl, we find v cyl via volumetric flow rate Volume cyl = A cyl x cyl ; Q cyl = d(volume cyl )/dt; Q cyl = d(a cyl x cyl )/dt = A cyl v cyl Q cyl = Q p = Q m therefore ω m D m = ω p D p = A cyl v cyl v cyl = ω p D p / A cyl T m ~ [ T p ω p -( p cyl A cyl ) v cyl ] / ω m ~ [ T p ω p -( p cyl A cyl ) (ω p D p / A cyl ) ] / ω m The numerical plug and chug is left to you, T m = 8.91 in lbf

7 Example III Pump, motor, & cylinder cont. Given the diagram, solve for T p, T m, and ω m Note where power crosses into and out of the system boundary D p = 0.5 in 3 / rev T p =? ω p = 1000 rpm 2π D m = 1 in 3 / rev T m =? ω m =? p 5 = 33.2 psi x p 1 p 2 p 3 p 3 p 1 = 10 psi Pump p 2 = 100 psi Motor p3 = 44 psi Valve p 4 = 44 psi F cyl A cyl = 5 in 2 Reservoir

8 Example III Pump, motor, & cylinder cont. Use a power balance on the pump to determine the pump torque: Σ P in = Σ P out + Σ P loss + Σ (de stored /dt); If P loss & (de stored /dt) are small compared to P in : Σ P in = Σ P out T p ω p = p p Q p T p = p p ( Q p ) / ω p = p p (D p ω p ) / ω p = p p (D p ) (100psi 10 psi) 0.5 in 3 / rev ( 1 / 2π ) rev / rad = 7.16 in lbf The solution to the rest of the problem is the solution to Example II

9 PROJECT I AND HWK 6

10 PLANETARY GEAR TRAINS

11 Planetary relationships (ala Patrick Petri) Say the arm is grounded. Planet gears = idler gears ' ω ri = N s ' ω si N r Now say the arm spins. we can say ' ω ri = ω ri ω a ω 2 ' ω si = ω si ω a N si = ω ri ω ai N ri ω si ω ai Finding the train ratio: Say the ring is grounded, sun = input, arm = output N si = 0 ω ai ω ai N s = N ri ω si ω ai ω si N R + N s

12 Planetary gear systems: Arm as output

13 THREADED MECHANISMS

14 Threaded mechanisms: Geometry Threaded mechanisms are used in applications such as: Bolts Lead screws (i.e. mills and lathes) General threaded mechanism geometry Control volume v s Force exert s Lead, l Torque applied ω Nut with interior threads Usually, either the nut or the screw is grounded Figure above shows the nut grounded Shaft with exterior threads

15 Threaded mechanisms: Modeling power flow From power balance for our control volume: ΣP in = [ΣP out ]+ Σ d (E stored ) P applied = [P exert + P loss ]+ d (E stretch ) dt dt Power in via work by applied Torque : Power out via work done by exerted Force : Power loss due to friction Torque : P applied = T applied ( ω ) ω P exert = F exert ( v) v P loss = T friction ( ω ) ω Rate of energy storage in stretched "cylinder": P stretch = F stretch ( v s ) v s From geometry: v = ( ω ) l 2π Lead = l