Work and Power Basics

1
Work and Power Basics
2
Work

• Linear motion Force x Stroke
• Inch pounds, foot lbs, or Newton meters
• Rotory motion work is Torque x Angular
  displacement. Units?

3
Power

• Work / Time
• Foot lbs/minute
• 33,000 foot lbs/min 1 HP
• Newton meters/second
• 1 Newton meter / second 1 Watt

4
Hydraulic Power

• UNITS ONLY
• Hydraulic power a function of flow x pressure,
  GPM x PSI
• GPM x in3/Gal x ft3/in3 ft3 / minute
• PSI lb / in2 x ft2 / in2 lbs / ft2
• Then GPM x PSI ft3/min x lbs / ft2
• Or ft lbs / min

5
Hydraulic Power

• Flow x Pressure actual English units
• Lets use 1 GPM and 1714 psi
• 1 Gallon/minute x 1714 psi
• 231 in3/gal x 1 GPM x 1714 lb/in2 x 144 in2/foot2
  
• (231 in3/min x 1 ft3/1728 in3) x (1714 x 144
  lb/foot2)
• 0.13368 ft3/min x 246,816 lb/ft2
• 32,994 ft lbs /min 33,000 ft lbs/min 1 HP
• HP (GPM x PSI)/1714
• REMEMBER THIS

6
Hydraulic Power

• Flow x Pressure
• SI units
• 1 Liter/ minute x 1 MPa
• 1 Liter 1/1000 m3/min 1 MegaPascal 106 N/m2
• (1/103 m3/min) x (106 Newtons/m2)
• 1000 Newton meters / minute
• 1000 N m /min x 1 min/60 sec
• 16.67 N m/sec 16.67 Watts 0.01667 kW
• kW (L/min x MPa) / 60 (Q ?P / 60)
• REMEMBER THIS

7
Example ProblemCar Crusher

• Need 125,000 lbs of force
• 8 foot stroke
• 10 seconds to extend?
• Target system max pressure of 1500 psi
• What is the cylinder size needed?
• 125,000 lbs/ A (area) 1500 psi
• Area 83.33 in2
• pr2 83.33 in2 r 5.15 inches (lets use
  5)

8
Car Crusher contd

• What will the system pressure be?
• Cylinder area 52 p 78.53 in2
• 125,000 lbs / 78.53 in2 1592 psi
• Set PRV to 1600?
• What is the Hydraulic power required?
• HP GPM x PSI / 1714 (we need GPM)
• Q cyl stroke x area /time
• Q 96 in x 78.53 in2/ 10 sec 754 in3/sec
• 754 in3/sec x 1 gal/231 in3 x 60 sec/min

9
Car Crusher contd

• Hydraulic Power contd
• Q 195.8 GPM
• (195.8 GPM x 1592 PSI) / 1714 181.9 HP
• What determines the actual ?P during the
  extension of the cylinder?
• What happens when we hit 1600 psi?
• Hmmmmmm

10
What about oil temp at the PRV?

• We have 195.8 GPM flowing over the valve
• This oil might weigh 8 lbs / gallon
• Thats 1566.4 lbs/minute
• One HP is 42.445 Btu/minute
• What is the specific heat of petroleum oil?
• I find (Google.. What else?) a value of
• c (0.388 0.00045t) / (sqrt spec grav)
• T deg F and spec grav for oil is approx 0.9

11
What about oil temp at the PRV?

• c 0.465 Btu/lb R
• 181.8 HP x 42.445 Btu/min/HP 7716.5 Btu/min
• 7716.5 Btu/min / 1566.4 lb/min 4.926 Btu/lb
• 4.926 Btu/lb /0.4659 Btu/lbR 10.57R
• Means 10.57 F temp rise across the valve as long
  as the full flow is passing there.