HydraulicsHydraulics

HydraulicsHydraulics Pumps convert mechanical to hydraulic Pumps convert mechanical to hydraulic

power. power. Formulas:Formulas: Flow:Flow: QQapap = (D = (DppNNpp/231)e/231)eVpVp

where Qwhere Qapap = actual flow from pump (gph) = actual flow from pump (gph) DDpp = pump displacement (in. = pump displacement (in.33/rev)/rev) NNp p = pump speed (rpm)= pump speed (rpm) eeVpVp = volumetric efficiency of = volumetric efficiency of

pumppump 1 gal = 231 in1 gal = 231 in33

HydraulicsHydraulicsTorque: TTorque: Tspsp = = ΔΔp Dp Dpp/(2/(2ππ ×× 12 12 ×× e eTpTp))

where Twhere Tspsp = shaft torque of pump (lb∙ft) = shaft torque of pump (lb∙ft)

ΔΔp = pressure across pump (psi)p = pressure across pump (psi)

eeTpTp = torque efficiency of pump = torque efficiency of pump

Hydraulic power: PHydraulic power: Phyphyp = = ΔΔp Qp Qapap/1714 (hp)/1714 (hp)

Shaft power: PShaft power: Pspsp = P = Phyphyp /e /ePpPp = 2 = 2ππ T Tspsp N Npp/33000/33000

where ewhere ePpPp = power(overall) efficiency = power(overall) efficiency

= e= eVpVp × e × eTpTp

Hydraulic pump example:Hydraulic pump example: Problem # 114 in practice problems.Problem # 114 in practice problems.

Pump flow desired is 20 gpm with a motor speed Pump flow desired is 20 gpm with a motor speed between 1700 and 2200 rpm. Choose the smallest between 1700 and 2200 rpm. Choose the smallest pump.pump.

QQapap = (D = (DppNNpp/231)e/231)eVpVp

Solve for DSolve for Dpp using trial and error. using trial and error. For pump JT02: DFor pump JT02: Dpp = 231 in. = 231 in.33/gal×20 gpm/(2200 /gal×20 gpm/(2200

rpm×0.91) rpm×0.91) = 2.31 in.= 2.31 in.33/rev > 2.0 in./rev > 2.0 in.33/rev and /rev and won’t pump enough at won’t pump enough at fastest fastest speed.speed.

For pump JT03: DFor pump JT03: Dpp = 231×20/(1700×0.91) = 2.99 in. = 231×20/(1700×0.91) = 2.99 in.3 3

and okay since lowest speed won’t over pump and we and okay since lowest speed won’t over pump and we can set speed to deliver 20 gpm using the 3.0 in.can set speed to deliver 20 gpm using the 3.0 in.33/rev./rev.

Hydraulic pump practice problem:Hydraulic pump practice problem:

Calculate the shaft power to drive the Calculate the shaft power to drive the pump pump

for JT03 with data from table of Qfor JT03 with data from table of Qapap = = 20 gpm, 20 gpm, ΔΔp = 3000 psi system p = 3000 psi system pressure – 100 psi pump input pressure – 100 psi pump input

prressure, and eprressure, and ePpPp = 84%. = 84%.

Hydraulic pump practice problem:Hydraulic pump practice problem:

Calculate the shaft power to drive the pump Calculate the shaft power to drive the pump for JT03.for JT03.

Pump flow is 20 gpm, system pressure is Pump flow is 20 gpm, system pressure is 3000 psi, input pressure at pump is 100 psi, 3000 psi, input pressure at pump is 100 psi, and pump overall efficiency is 0.84.and pump overall efficiency is 0.84.

PPspsp = P = Phyphyp /e /ePpPp

= [20 gpm× = [20 gpm× (3000-100)psi]/(1714×0.84)(3000-100)psi]/(1714×0.84)

= 40.3 hp= 40.3 hp

Hydraulic CylinderHydraulic CylinderHydraulic cylinder converts hydraulic Hydraulic cylinder converts hydraulic

power to mechanical power.power to mechanical power.

HydraulicsHydraulics CylindersCylinders

FormulasFormulas

Force to extend: FForce to extend: Fextext = p = ppispisAApispis – p – prodrod(A(Apispis-A-Arodrod) ) (lb)(lb)

where pwhere ppispis = piston pressure (psi) = piston pressure (psi)

AApispis = piston area (in. = piston area (in.22))

pprodrod = rod pressure (psi) ≈ 0 = rod pressure (psi) ≈ 0

AArodrod = rod area (in. = rod area (in.22))

Force to retract: FForce to retract: Fretret = p = prodrod(A(Apispis-A-Arodrod) – p) – ppispisAApispis

where pwhere ppis pis ≈ 0≈ 0

HydraulicsHydraulics

Velocity extend: vVelocity extend: vextext = 231 Q = 231 Qapap/(A/(Apis pis ∙∙

12) (ft/min)12) (ft/min)

Velocity return: vVelocity return: vretret = 231∙ Q = 231∙ Qapap/ [(A/ [(Apispis--AArodrod)12])12]

Velocity can be defined in terms of Velocity can be defined in terms of extension distance and time.extension distance and time.

Hydraulic cylinder example:Hydraulic cylinder example:

Calculate the force and velocity to extend if system Calculate the force and velocity to extend if system pressure is 1800 psi, cylinder diameter is 3 in., rod pressure is 1800 psi, cylinder diameter is 3 in., rod diameter is 1 in., back pressure is 50 psi and pump diameter is 1 in., back pressure is 50 psi and pump flow to cylinder is 15 gpm.flow to cylinder is 15 gpm.

FFextext = p = ppispisAApispis – p – prodrod(A(Apispis-A-Arodrod) v) vextext = 231∙ Q = 231∙ Qapap/ (A/ (Apispis∙12)∙12)

= [1800 psi= [1800 psi××ππ××(3 in.)(3 in.)22/4]/4] -50 -50psi[psi[ππ××(3in.)(3in.)22/4 /4 - - ππ××1122/4)]/4)]

= 1800 = 1800 ×× 7.07 – 50(7.07- 0.79) 7.07 – 50(7.07- 0.79) = 12,412 lbs= 12,412 lbs

vvextext = 231 in. = 231 in.33/gal /gal ×× 15 gpm/ (7.07 in. 15 gpm/ (7.07 in.2 2 ×× 12 in./ft)12 in./ft) = 40.85 ft/min. /(60 s/min) = 0.68 ft/s= 40.85 ft/min. /(60 s/min) = 0.68 ft/s

Hydraulic cylinder practice Hydraulic cylinder practice example:example:

For the previous hydraulic cylinder For the previous hydraulic cylinder example, calculate the force to example, calculate the force to

retract if back pressure is assume retract if back pressure is assume zero. Also, calculate the velocity to zero. Also, calculate the velocity to

retract.retract.

Hydraulic cylinder practice Hydraulic cylinder practice example:example:

FFretret = p = prodrod(A(Apispis-A-Arodrod) – p) – ppispisAApispis where p where ppis pis ≈ 0≈ 0

vvretret = 231∙ Q = 231∙ Qapap/ [A/ [Apispis-A-Arodrod)12])12]

FFretret = 1800 psi ( 7.07 in. = 1800 psi ( 7.07 in.22 – 0.79 in. – 0.79 in.22))

= 11,304 lbs= 11,304 lbs

vvretret = 231 in. = 231 in.33/gal×15 gpm/[( 7.07 – 0.79 /gal×15 gpm/[( 7.07 – 0.79 in.in.22)12 in./ft])12 in./ft]

= 45.98 lb∙ft= 45.98 lb∙ft

HydraulicsHydraulics

Motors convert hydraulic to Motors convert hydraulic to mechanical power.mechanical power. FormulasFormulas

Speed: NSpeed: Nmm = 231 Q = 231 Qapap e eVmVm/D/Dm m (rpm)(rpm)

Torque: TTorque: Tmm = ( = (ΔΔp Dp Dmm/2/2π∙π∙12) e12) eTmTm (lb∙ft) (lb∙ft)

Power: PPower: Psmsm = = PPhymhym e ePm Pm (hp)(hp)

= (Q= (Qapap ΔΔp /1714) p /1714) eePmPm (hp)(hp)

Hydraulic motor example:Hydraulic motor example: Calculate the hydraulic motor torque efficiency Calculate the hydraulic motor torque efficiency

and motor output power from the following and motor output power from the following data: Flow is 10 gpm at 1000 psi, motor torque data: Flow is 10 gpm at 1000 psi, motor torque is 1545 in. lbs, motor speed is 185 rpm, motor is 1545 in. lbs, motor speed is 185 rpm, motor displacement is 11.5 in.displacement is 11.5 in.33 per revolution, and no per revolution, and no back pressure.back pressure.

TTmm = ( = (ΔΔp Dp Dmm/2/2ππ××12) e12) eTm Tm PPsmsm = (Q = (Qapap ΔΔp /1714) p /1714) eePmPm

= 2 = 2 ππTTmm N Nmm/33000/33000

eeTmTm = { = {[1545 in. lb/(12 in./ft)]×2 [1545 in. lb/(12 in./ft)]×2 ππ(12 in./ft)}(12 in./ft)} (1000 psi×11.5 in.(1000 psi×11.5 in.33/rev)/rev) = 84.4%= 84.4%

PPsmsm = = 2 2 ππ × [1545 lb∙in./(12 in./ft)] × 185 rpm × [1545 lb∙in./(12 in./ft)] × 185 rpm =4.54 hp =4.54 hp 33000 ft∙lb/min∙hp33000 ft∙lb/min∙hp

Hydraulic MotorHydraulic Motor

Hydraulic motor converts hydraulic Hydraulic motor converts hydraulic power to mechanical power.power to mechanical power.

Hydraulic motor practice example:Hydraulic motor practice example:

Calculate the overall efficiency of the Calculate the overall efficiency of the hydraulic motor in the previous hydraulic motor in the previous

exampleexample

Hydraulic motor practice example:Hydraulic motor practice example:

PPsmsm = = PPhyphyp e ePmPm

= (Q= (Qapap ΔΔp /1714) p /1714) eePmPm (hp)(hp)

eePmPm = 4.54 hp/ [(10 gpm×1000psi/1714) = 4.54 hp/ [(10 gpm×1000psi/1714) hp]hp]

= 77.8 %= 77.8 %