Engineering

Power and Work

1. Define the terms “power” and “work”.
2. Describe the forms of power produced by a hydraulic
system.
3. Calculate the horsepower being supplied by a hydraulic
system.
4. Calculate the heat being generated by a hydraulic
system.
5. Explain how heat is generated in a hydraulic system.
6. Explain how work is performed in a hydraulic system.
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Directions
1. Read the text assignment given by your
instructor.
2. Read the Background Information section for
this experiment.
3. Perform the steps in the Experiment Procedure
section using the equipment identified in the
Equipment list.
4. Answer the Checkout Activities at the end of
this experiment.
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BACKGROUND INFORMATION
1. Power: Rate at which energy is being used
2. Most common form of power available in factories is electrical.
3. Electrical power:
• Most common and available form of power in factories
• Must be converted into mechanical power
• Must be transmitted to point of use in machines.
4. Hydraulic systems are power drives that transmit power through a
liquid to the point of use.
• This “fluid” power is then converted into mechanical power by an
actuator.
• Power for linear motion is given by a cylinder while rotary motion is
given by a motor
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Amount of Mechanical Power
1. Determined from speed and force or torque
2. Units are: horsepower (English), kW (SI)
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Generation of Fluid Power
1. Power in a hydraulic system:
• Exists in the form of pressurized fluid
• Gets transmitted by the fluid flow
2. Hydraulic Pump:
• Gets its power from the electric motor
• Motor converts electric power into mechanical power.
• Pump generates a flow of oil to the system
• Actuator converts potential energy of fluid into
mechanical
3. Resistance from the actuator or internal friction cause
buildup of pressure at the pump’s outlet.
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Calculation of Fluid Power
1. Fluid power is affected linearly by:
• The amount of pressure
• Fluid flow
2. Fluid power can be calculated as follows:
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Power Conversion
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Figure 1. Power conversion in a fluid
power system.
Dual Hi-Lo Hydraulic Systems
• Increasing either the flow rate or pressure,
increases proportionally the power
• Hydraulic press systems:
– Often have two pumps that operate together:
• A large pump extends the press at high speed and
low pressure
• A small pump provides a small flow but at high
pressure.
– This method is called a hi-low system and greatly
reduces the power required by the system.
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Power Use, Heat Generation
• Power Used — The amount of fluid power used by an
actuator can be found by using a variation of the
formulas for the fluid power generated by the pump. The
net fluid power used by an actuator depends on the
pressure difference between the ports of the actuator.
• Heat Generation — Unfortunately, the power generated
in a hydraulic system does not all get converted into
mechanical power at the actuator. Some of this power is
lost by frictional resistance of each of the components.
This power is converted into heat
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Amount of Heat Generated
• The amount of heat generated by any component can be determined by
finding the pressure drop and the flow rate across the component. As
illustrated in Figure 2, the power generated by the pump is partially
converted into heat by each component in the system. The rest of the power
is converted into mechanical power. It is as follows:
• Heat is an important consideration in the design and operation of hydraulic
systems. If the hydraulic system temperature becomes too high the oil will
lose its lubricity causing components to wear. High heat will also cause
varnish to develop on the internal surfaces of the components which will
cause problems such as valve sticking.
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Heat Generation in a Circuit
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Figure 2. Heat generation in a hydraulic system.
EQUIPMENT LIST
Item No. Quantity Description
HYDRAULIC INSTRUMENTATION MODULE:
1 2 Gages
2 1 Flowmeter
BASIC HYDRAULIC VALVE MODULE:
3 1 Directional Control Valve
7 1 Needle Valve
HYDRAULIC ACTUATOR MODULE:
8 1 Cylinder 1 1/2 Bore X 4” stroke
LOOSE COMPONENTS:
7 Hoses
ADDITIONAL COMPONENTS REQUIRED:
Stopwatch
Set of Allen Wrenches
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EXPERIMENT PROCEDURES
This experiment is divided into two
parts:
I. Effect of Pressure Drop on Heat
Generation
II. Performing Work
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Part I – Effect of Pressure Drop on Heat
Generation
In this part of the experiment, you will demonstrate that energy in the
form of fluid power is converted into heat when fluid flows through
an orifice and the greater the pressure drop across this orifice, the
greater the heat generated.
1. Set up the 850 basic hydraulic system in either the bench or satellite
configuration illustrated in Experiment 1-2.
2. Set up the circuit illustrated in Figures 3 and 4.
– In this circuit the needle valve is used to generate heat. By adjusting the
needle valve’s setting the pressure drop can be increased or decreased
in order to change the rate of heat generation.
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Diagram for Heat Generation
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Figure 3. Diagram of circuit for generating
heat.
Pictorial of Circuit for
Heat Generation
3. Perform the following checkout
procedures before starting the
power unit.
A. Check the oil level. Fill if
necessary.
B. Press the stop pushbutton on
the motor starter to make sure
the starter is in the off position.
C. Plug in the power cord to a
115VAC wall outlet.
D. Reduce the relief valve to its
minimum pressure setting
(turn CCW fully).
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Figure 4. Pictorial of circuit for generating heat.
Experimental Procedure, Part I (Cont’d)
4. Close the needle valve (CW fully).
5. Put your hand on the body of the needle valve. It should be at room
temperature. Record the oil temperature below.
Oil Temperature __________°F/°C
6. Start the power unit.
7. Set the relief valve at 500 psi/3447 k
8. Open the isolation valve and then open the needle valve until the
pressure drop across the needle valve is 400 psi/2760 kPa and the
flow meter reads at least 2.5 gpm/9.5 lpm.
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Part I (Cont’d)
9. Allow the system to run for about five minutes. Then put your hand
on the body of the needle valve. Does it feel hotter than before?
Record your findings below.
10. Put your hand on other parts of the system. Do they feel hot also?
Do they feel as hot as the needle valve? Record your findings below
along with the oil temperature.
– What you feel with your hand at the hot component will eventually occur
at the reservoir if operated long enough.
– Oil Temperature __________°F/°C
– In the next steps you will now generate heat using the relief valve.
11. Reduce the relief valve setting to minimum, close the isolation valve
and turn off the power unit.
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Part II – Performing Work
Demonstrate that a pressure drop occurs across the
actuator when it is performing work
Show that no heat is generated by this
1. Continuing from Part I, set up the circuit shown
– Use a directional control valve to reciprocate a cylinder.
– Connect two pressure gages to the cylinder ports for reading the
pressure drop across the actuator.
– A load device should now be attached as shown to allow you to
put a load in the actuator’s rod.
– By turning the socket head screws on the load device CW you
will increase the load on the cylinder.
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Circuit for Performing Work
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Figure 5. Diagram of circuit for performing work.
Part II (Cont’d)
• 2. Turn on the power unit.
• 3. Increase the relief valve’s setting to 500 psi/3447 kPa and
open the isolation valve.
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Figure 6. Adjusting the cylinder load device.
• 4. Using a socket head
wrench, turn both the
screws on the load
device CCW as
illustrated in Figure 6
until they are both loose.
This will remove all load
on the cylinder.
Part II (Cont’d)
• 5. Extend and retract the cylinder several times using the direct control
valve to make sure that the system is working.
– Observe the readings of pressure gages A, B, and S.
– Record their readings in row no. 1 of the chart below when the cylinder is
extending.
– Also, record the time required to extend the cylinder. Leave the cylinder in
retracted position.
– Use the needle valve to lengthen stroke time in Order to observe gage readings
better. (An extend time of 1.5 to 2.0 seconds will provide sufficient time.)
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Part II (Cont’d)
• 6. With the cylinder stopped and retracted, put a light load on the cylinder by turning
the screws on the load device CW about 1/4 turn.
• 7. Extend the cylinder and observe the pressure gages A and B. Readjust the screws
on the load device until the pressure difference is about 150 psi/1035 kPa. This is a
light load.
• 8. Observe the pressures readings of gages A, B, and S while the cylinder is
extending. Record them in row no. 2 of the chart on the previous page. Also, record
the time to extend.
• 9. Repeat steps 6 through 8 for a medium load of 250 psi/1725 kPa and a heavy load
of 350 psi/2415 kPa.
• 10. With a heavy load on the cylinder record below the readings of pressure gages A
and B when the cylinder is extending and then when it is retracting. Notice whether
the delta p is higher when the cylinder is retracting or extending.
• 11. Reduce the relief valve’s pressure setting to a minimum, close the isolation valve
and turn off the power unit.
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Checkout Activities
1. What form of power does a hydraulic actuator produce?
2. What factors determine the amount of power generated by the hydraulic system?
3. What form of power does a hydraulic system use to transmit power?
4. What causes energy loss in a hydraulic system? What form does this energy take?
5. Using the data from the experiment Part H, calculate the total amount of power being
generated by the power unit for each load. Assume pump flow rate to be 2.5 gpm.
Show your work.
Load Power (HP / kW)
No-Load /
Light /
Medium /
Heavy /
7. In the experiment Part H, step 10, what were the delta p’s required to extend the load
and retract load. Explain why one was higher.
Extend psi/kPa
Retract psi/kPa
8. What is the hydraulic equivalent of force in a mechanical system?
9. What is the hydraulic equivalent of speed in a mechanical system?
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