Download presentation

Presentation is loading. Please wait.

Published byCarlee Granger Modified over 2 years ago

1
x Disp. (m) x Vel. (m/s) x Acc. (m/s 2 ) m Mass (kg) k Spring cons. (N/m) c Damp. Cons. (Ns/m) f Force (N) θ Angular pos. (rad) θ Angular vel. (rad/s) θ Ang. acc. (rad/s 2 ) I G mass m. ineratia (kg-m 2 ) K r Rot. Spring const. (Nm/rad) C r Rot. damp. const. Nm/(rad/s) M Moment (Nm) q Charge (Coulomb) i Current (Amper) L Induc. (Henry) 1/C C:Capac. (Farad) R Resistance (Ohm) V Voltage (Volt) V f Volume (m 3 ) Q f Flow rate (m 3 /s) I f Fluid inertia (kg/m 4 ) 1/C f C f :Fluid capacitance R f Fluid resistance P Pressure (N/m 2 ) H t Heat (Joule) Q t Heat flow rate (J/s) 1/C t C t :Thermal capacitance R t Thermal resistor T Temp. ( o C).... Analogy in Mechanical (trans.-rot.), Electrical, Fluid, Thermal Systems

2
Fluid inertia in pipes: Fluid reservoir capacitance: C f : Fluid reservoir capacitance, A d :Cross section area of reservoir, g=9.81 m/s 2 I f :Fluid inertia, ρ: Density, L: Pipe length A: Cross section of pipe Fluid resistance of pipes in laminar flow: R f :Resistance, µ:Viscosity, L:Pipe length, d:Pipe diameter Reynolds number in laminar flow<2000 Reynolds Number= D.Rowell & D.N.Wormley, System Dynamics:An Introduction, Prentice Hall, 1997 V f Volume (m 3 ) Q f Flow Rate (m 3 /s) I f Fluid inertia (kg/m 4 ) 1/C f C f :Fluid capacitance R f Fluid resistance P Pressure (N/m 2 )

3
1 2 1 p k1 Q k1 3 4 Q3Q3 5 2 Q5Q5 papa Equivalent Electrical Circuit of Fluidic Systems In this lesson, we will learn and study the modeling fluidic systems by creating equivalent circuits. There is an analogy between the fluidic and electrical quantities. Let’s consider Example 5.1. A fluidic system is given here. We will create the equivalent circuit for the system by using the analogy. p n V n Fluid. Elec. R n I n L n QnQn C n Example 5.1

4
p n V n Fluid. Elec. R n I n L n QnQn C n Fluidic System Equivalent Electrical Circuit - VaVa + V k1 R1R1 L1L1 1 p k1 In the fluidic system the pumps is connected the pipe #1. The pump’s output pressure is Pk. The atmospheric pressure is Pa. The top of the tank #1 is open to the atmosphere. In the electrical circuit, a supply voltage operating at Vk-Va is connected to the resistor R1 and inductor L1. R1 represents the resistance to a flow in the pipe #1. L1 represents the inertia of a flow in the pipe #1. papa

5
Fluidic System Equivalent Electrical Circuit - VaVa + V k1 R1R1 L1L1 1 p k1 Q k1 papa 1 2 In the fluidic system the tank #1 is connected to both of the pipe #1 and the pipe#2. However, in the electrical system, the capacitor #1 is connected to both of the resistor R1, inductor L1 and the resistor R2, inductor L2. The other end of the capacitor #1 is connected to the voltage Va because the tank #1 is open to the atmospehere. The capacity of a tank in a fluidic system corresponds to a capacitor in an electrical circuit. There is an external flow with the flow rate Qk1 in the fluidic system. The current supply qk1_dot is placed in the electrical circuit. R2R2 L2L2 C1C1

6
Fluidic System Equivalent Electrical Circuit - VaVa + V k1 R1R1 L1L1 1 p k1 Q k1 papa 1 2 R2R2 L2L2 C1C1 3 4 Q3Q3 A R3R3 L3L3 R4R4 L4L4 - + V4V4 A In the fluidic system, the pipe #2 is connected to both of the pipe #3 and the pipe#4 at the point A. On the other hand, in the electrical system, the line with the resistor R2, inductor L2 is connected to the lines the resistor R3, inductor L3 and the resistor R4, inductor L4 at the point A. There is a pressure P4 in the pipe #4 due to the placement of pipe #4 in the vertical direction because the flow is opposite to the gravity in pipe #4. So, the voltage supply V4 is placed in the circuit due to the analogy. The positive end of the V4 is connected to the point A. The current produced from V4 flows in the opposite direction.

7
Fluidic System Equivalent Electrical Circuit - VaVa + V k1 R1R1 L1L1 1 p k1 Q k1 papa 1 2 R2R2 L2L2 C1C1 3 4 Q3Q3 A R3R3 L3L3 R4R4 L4L4 - + V4V4 A 5 2 Q5Q5 R5R5 L5L5 C2C2 In the fluidic system the tank #2 is connected to both of the pipe #1 and the pipe#2. On the other hand, in the electrical system, the capacitor #2 is connected both of the resistor R4, inductor L4 and the resistor R5, inductor L5.

8
Electrical systems can be analyzed instead of fluidic systems Nowadays computer aided engineering (CAD/CAE) is used. Dynamic (Transient) behaviour Steady-state behaviour By analyzing the equivalent circuit, transient or steady-state dynamic behavior at a desired point of a circuit can be calculated. Thus, the corresponding flow rates and pressures at a desired line of a fluidic system are found. Nowadays, circuits and fluidic systems can be modeled and analyzed by the computers. Before the developments in computer technology, in order to analyze fluidic systems engineers have used the equivalent circuits, which are produced easily and cheaply. Complex fluidic systems have been easily analyzed with the experiments of circuits.

Similar presentations

OK

H t Heat (Joule) Q t Heat flow rate (J/s) 1/C t C t :Thermal capacitance R t Thermal resistance T Temp. ( o C) Modeling of Thermal Systems In this lesson,

H t Heat (Joule) Q t Heat flow rate (J/s) 1/C t C t :Thermal capacitance R t Thermal resistance T Temp. ( o C) Modeling of Thermal Systems In this lesson,

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on environmental studies Ppt on construction of 3 phase induction motor Ppt on any one mathematician jobs A ppt on amelia earhart Water level indicator using seven segment display ppt online Ppt on needle stick injury policy Download ppt on reaching the age of adolescence for class 8 Product mix ppt on nestle jobs Ppt on rural entrepreneurship in india Ppt on 3 idiots movie in english