TUTORIL QUESTIONS FOR COURSE TEP 4195 Data: Hydraulic Oil Density 870 kg/m 3 bsolute viscosity 0.03 Ns/m 2 Spool valve discharge coefficient 0.62. 1) hydrostatic transmission has a variable displacement axial-piston pump, with a maximum displacement of 12 cm 3 /rev, and a fixed capacity motor of 120cm 3 /rev displacement. The motor speed is 138 rev/min when driving a load torque of 200Nm and when the pump is set to maximum displacement and driven at a speed of 1450 rev/min. Note This information enables the value of the total leakage and volumetric efficiency to be determined by calculating the ideal pump flow and motor speed. The total leakage is assumed to be proportional to the pressure. If the load is increased to 400 Nm and the pump control set to 50% maximum capacity, estimate: (a) the steady state speed of the motor (58.5 rev/min) (b) the input power to the pump and the overall efficiency (3.6 kw, 69%) The pump mechanical efficiency is 95% at full stroke and 90% at half stroke. The motor mechanical efficiency is 95%. 2) a) For the circuit shown in Figure 1, it is required to determine the size of the restrictor ( as a rated flow) that is to be used during extension of the actuator. For the data given calculate the rated flow of the restrictor valve (L/min) at a rated pressure drop of 10bar that will provide the specified actuator velocity. (19L/min) P Chapple pril 2006. Tutorial Questions 1
P P Q P P S B P S Q S P T P smax Data Figure 1 ctuator Circuit Opposing force on the actuator rod = 20 kn ctuator velocity = 0.5 ms -1 ctuator piston diameter = 50 mm ctuator rod diameter = 25 mm Relief valve set pressure (P smax ) = 200 bar Pump flow = 80 L/min The pressure in the return line from the actuator can be assumed to be zero. b) Show how the circuit can be modified in order that the extending actuator velocity will be controlled when the force is negative (pulling). c) In order to prevent the actuator velocity changing with changes in the load force a pressure compensated flow control valve can be used in place of the restrictor valve. Draw a sketch of this type of valve and briefly describe its method of operation. 3) 1 2 Velocity U E Force F Q 1 Q 2 P 1 P 2 Rod for equal area actuator B α = 1 / 2 P S P T P s P Chapple pril 2006. Tutorial Questions 2
i) four-way spool valve is used to position a hydraulic actuator with a supply pressure of 250 bar. The valve spool is a fully annular design of 6 mm diameter, with a maximum movement of 0.5 mm and C q = 0.62. For a double ended actuator (equal area) having areas of 12.5 cm 2 calculate the maximum power which can be transmitted to the actuator and its speed and thrust at this condition. (20.8 x 10 3 N, 0.46m/s, 9.6 kw) ii) For an unequal area actuator with zero force conditions show that the ratio of extension velocity (u E ) to retraction velocity (u R ) is given by: u u E R = 1 2 iii) For an unequal area actuator having areas of 12.5 and 25 cm 2 respectively, and using the valve and supply pressure in i) calculate the thrust and actuator pressures when extending at 0.49 ms -1. (58bar, 48bar, 8.5kN) 4) Question based on Chapter 6 a) Show that the volume that can be discharged from a gas type accumulator is given by the following equation: P2 P 0 V0 = V 1 P2 γ 1 P 1 p 0 is the precharge pressure V 0 is the stored volume at the precharge pressure P 1 is the minimum pressure required for operation of the system P 2 is the maximum required pressure ssume that the gas is compressed isothermally from P 0 to P 2 and expands adiabatically from P 2 to P 1 with the adiabatic index γ. b) hydraulic actuator is required to perform 60 cycles per hour, each cycle requiring 0.9 litres of oil to operate the actuator in a time of 8s. fixed displacement pump supplies 1 litre/min to the system the outlet pressure of which is controlled by a relief valve at a pressure of 100 bar. Determine the volumetric capacity of the accumulator that is required to operate the system at a minimum pressure of 75 bar. The precharge pressure is set at 90% of the minimum pressure and the adiabatic index to be used is 1.6. (5.6L) P Chapple pril 2006. Tutorial Questions 3
5) closed loop hydrostatic transmission system is to be used to drive the tracks of an excavator for which there are two motors, each one being used to drive a track as shown in the Figure 1. Figure 1 Excavator track drive a) Using the given data calculate the motor torque required to start moving the excavator up an incline of 20 0 and the motor speed required to drive the excavator on level ground at maximum speed. Use these values to select a suitable motor type from the table that will provide the required performance. (288rev/min, B) b) Calculate: i. The maximum flow from each pump that is required to drive the selected motors. (25L/min) ii. The maximum displacement of each pump. (14.6 cm 3 /rev) iii. The pump pressure and the total input power to the pump that are required to drive the excavator at the maximum speed on level ground. (144bar, 6.65kW) c) Calculate the maximum speed when using a hydraulic fluid having a viscosity of 20 cst. (268rev/min) d) Sketch a circuit for the hydrostatic system, which includes: Relief valves for limiting the maximum circuit pressure Brake control valves The method for supplying boost flow into the circuit. e) It is intended to use variable displacement motors for the excavator track drive system. Calculate the minimum value of motor displacement that can be used to drive the excavator on level ground at the maximum motor flow from b) i) at the maximum pressure of 210bar. (67.8cm 3 /rev) P Chapple pril 2006. Tutorial Questions 4
Data Total weight 5400N Maximum speed (U max ) 6m/s Static friction force to start moving the excavator 450N Friction force at speeds > 0 250 + 250U (U = velocity m/s) N Track drive wheel diameter 0.4m Motor mechanical efficiency 85% for starting at zero speed ( η MS ) 92% for motor speeds greater than zero ( η MD ). Pump and motor volumetric efficiencies 95% (with a hydraulic fluid viscosity of 32 centistoke cst) ( ηpv, η MV ) Pump mechanical efficiency Pump speed Maximum pressure 95% ( η PM ) 1800 rev/min (N) 210 bar ( P max ) Motor Data Motor type Theoretical displacement (cm 3 /rev) 102 250 B 83 300 C 67 350 Maximum speed (rev/min) 6) Figure 1 Pump and motor system Data Torque required at maximum motor speed 175Nm Motor displacement 82cm 3 /rev (D M ) Motor mechanical efficiency 92% for motor speeds greater than zero ( η MM ). Motor volumetric efficiency 93% ( η MV ) Pump mechanical efficiency 95% ( η PM ) Pump volumetric efficiency 96% ( η PV ) P Chapple pril 2006. Tutorial Questions 5
Pump speed 1800rev/min (N P ) Pump displacement 15cm 3 /rev (D P ) Fluid specific heat 2100J/kg/ 0 C (C P ) Heat dissipated in the cooler for a water inlet temperature of 20 0 TOW C 3 kw 40 ( TOW is the difference between the cooler oil inlet and water inlet temperatures). Pump external drain leakage flow = 50% of the total pump leakage a) For operating the motor against the stated torque that is required at the maximum motor speed calculate: 1) The pressure required at the motor inlet. 2) The pump outlet flow. 3) The motor speed at this operating condition. 4) The flow at the cooler inlet (point in Figure 2). b) ssuming that all of the volumetric and mechanical losses are dissipated into the hydraulic system fluid and that there is no heat transfer from the fluid to the environment through the pipes calculate: 1) The total heat generated by the losses. 2) The temperature increase in the hydraulic fluid between the pump inlet and the cooler inlet assuming that there is perfect mixing of the flows at point. 3) The temperature of the hydraulic fluid at the cooler inlet (point ) that is required to dissipate the heat that is generated in 1). (Note that the heat extracted from the fluid in the cooler is dependent on the difference Tow between the temperatures of the hydraulic fluid at the cooler inlet and the cooling water inlet). 4) The reduction in the fluid temperature through the cooler and the temperature in the reservoir. P Chapple pril 2006. Tutorial Questions 6