Reluctance Sensors (VRS) applications are in pump RPM, plant floor machinery, electric motor speed, wheel speed sensing, and engine transmission. c) Most Suitable Sensing System For the rubber wheels of trolley running on a steel surface, there are certain sensors that applicable. For detecting steel surface, the two magnetic technologies applicable are hall- effect sensors and variable reluctance sensors (Corp 2019, 1). Based on the design parameters and advantages of each sensor, it is possible to narrow down to the most suitable sensing system. A combination of different techniques can also be applied to come with the best sensor. The best sensor for this task is Light Detection and Ranging system. The LIDAR sensor measures the speed of trolley on the steel track accurately as well as the curvature and distance. It has numerous advantages and its range is 0.5 to 100 mph ( Ahmadian, Craft, and Stuart 2014, 1 ). This sensor eliminates need for frequent calibration and mechanical failures. Task 2 a) Actuator Requirements Based on the given parameters, the actuator can be designed as shown below. Fig. 1: Design For components to move, a force is required to overcome the force created by the spring force and the components. The maximum forces created by the actuators are calculated below. The maximum sinusoidal paths are calculated using the formulas below: Frequency f = 1/T, Location on Y-axis y = A sin (2πft),
Sensors and Actuators Design Velocity v = 2πfA cos (2πft), Acceleration a = - (2πf) 2 [A sin(2πft) = -4π 2 f 2 y (Webassign.net 2019, 1) For f = 0.15 Hz and A = 0.5M , we have get maximums as; Distance = 0.5 sin (90) = 0.5 m Velocity = 2π × 0.15 × 0.5 cos 0= 0. 471 m/s Acceleration = - (2π × 0.15) 2 0.5 sin 90 = -0.444 m 2 /s. For f = 0.15 M and A = -0.5 Hz, we get: Distance = -0.5 sin (90) = -0.5 m Velocity = 2π × 0.15 ×- 0.5 cos0= -0. 471 m/s Acceleration = - (2π × 0.15) 2 (-0.5) sin 90 = 0.444 m 2 /s For f = 1 Hz and A = 0.05 M, the values obtained are; Distance = 0.05 sin (90) = 0.05 m Velocity = 2π × 1 × 0.05 cos0= 0. 314 m/s Acceleration = - (2π × 1) 2 0.05 sin 90 = -1.973 m 2 /s For f = 1 Hz and A = -0.05 M, the values obtained are; Distance = -0.05 sin (90) = -0.05 m Velocity = 2π × 1 × (-0.05) cos0= -0. 314 m/s Acceleration = - (2π × 1) 2 (-0.05) sin 90 = 1.973 m 2 /s From above results, the value of acceleration are a = 1.973 m 2 /s and distance is 0.5 m. From Newton’s second law f = ma= -ky Total force = ma + (-ky) = (500 ×1.973) + (2000 × 0.5) = 1986.5 N Velocity = 0.471 m/s b) Outline Design and Parameters Assumption made: 210 bar supplied and 25% excess flow allowed.
Sensors and Actuators Design Area , A = 1.25 Famx 0.67 Ps = 1.25 □ 1986.5 0.67 □ 210□ 10S = 0.00017 m 2 Maximum Flow From the valve = Q = AV max = 0.00017 × 0.471 = 0.00008 m 3 Peak hydraulic power = 210 × 10 5 × 0.00008 = 1680W =1.68 kW c) Block Diagram Model The block diagram model with mass, actuator, and system is as shown and it relates to actuator position and voltage: Fig. 2 Block Diagram Model
Sensors and Actuators Design From the model above, the input is current and voltage which are both electrical which goes to the hydraulic power supply. The hydraulic instigates flow rate and constant amount of supply pressure which are then controlled by the servo valve that produces flow rate and pressure which are fed to the actuator. The actuator outputs force at a specific mass and some velocity.
- Fall '19
- Miles per hour, Actuators Design