Week4 DM312 AJ.pdf

DM312 Mechatronic Design and Applications Lecture 4: Bond Graph Guided Functional Design Dr Aleksandar Josifovic Week 4 - DM312 In this lesson: Part 1 Two port elements Bond Graphs Multi-port elements Bond Graphs Bond graph – Causality Defining system equations Numerical Solution of Bond Graphs Part...

DM312 Mechatronic Design and Applications Lecture 4: Bond Graph Guided Functional Design Dr Aleksandar Josifovic Week 4 - DM312 In this lesson: Part 1 Two port elements Bond Graphs Multi-port elements Bond Graphs Bond graph – Causality Defining system equations Numerical Solution of Bond Graphs Part 2 Tutorial – Chenhao Ran - Defining Bond graph of Mechatronic systems DM312 – Class admin Completed No lecture on 20/02/2024 Self study and revise material from weeks 1-5 Recap from Week 3 - Bond graph mechanical system Recap from Week 3 – Hydro electric power station Flow P Q Reser voir P Q Pipe Hydraulic domain P Q Turbine τ ω Shaft τ ω Gener ator Mechanical rotation domain V i Transf ormer V i Electrical domain Grid Recap from Week 3 – Hydro electric power station Component Explain Energy variable Energy variable function (Se, Sf, Energy Domain Effort Flow R, I, C) Description Flow Source of flow Sf Hydraulic Q (volumetric flow) P (pressure) Reservoir Store (accumulate) water Transfer the flow through restriction Convert hydraulic power to mechanical Transfer of power (with inertia and friction losses) Convert mechanical power to electrical Transforms electrical to desired V and i Send the power to the network via resistance C Hydraulic Q (volumetric flow) P (pressure) R Hydraulic Q (volumetric flow) P (pressure) GY (Gyrator) Mechanical τ (Torque) ω (Rotational speed) I and R Mechanical τ (Torque) ω (Rotational speed) GY (Gyrator) Electrical V (voltage) i (Current) TF (Transformer) Electrical V (voltage) i (Current) R Electrical V (voltage) i (Current) Pipe Turbine Shaft Generator Transformer Grid Part 1: Two port elements Bond Graphs Two-port Elements - Transformer Energy entering transformer is equal to energy leaving transformer. Effort and flow may be different but proportional on either side. The energy domain remains the same on both sides of the transformer Transformer, symbol TF as in electrical transformer Gear box Lever Hydrostatic pump They conserve power and are reversible   Gear box   =   TF   Two-port Elements - Transformer Transformer examples in different energy domains v2 𝜏2 𝜔2 n2 V2 V1 V1 i1 TF 1/m V2 i2 m – modulus of the transformer F1 v1 TF b/a F2 v2 τ1 ω1 v P F A 𝜏1 𝜔1 v1 Q n1 TF n τ2 ω2 n – gearbox ratio n1 /n2 P Q TF 1/A A – piston area F V Recap from Week 3 – Hydro electric power station Component Explain Energy variable Energy variable function (Se, Sf, Energy Domain Effort Flow R, I, C) Description Flow Source of flow Sf Hydraulic Q (volumetric flow) P (pressure) Reservoir Store (accumulate) water Transfer the flow through restriction Convert hydraulic power to mechanical Transfer of power (with inertia and friction losses) Convert mechanical power to electrical Transforms electrical to desired V and i Send the power to the network via resistance C Hydraulic Q (volumetric flow) P (pressure) R Hydraulic Q (volumetric flow) P (pressure) GY (Gyrator) Mechanical τ (Torque) ω (Rotational speed) I and R Mechanical τ (Torque) ω (Rotational speed) GY (Gyrator) Electrical V (voltage) i (Current) TF (Transformer) Electrical V (voltage) i (Current) R Electrical V (voltage) i (Current) Pipe Turbine Shaft Generator Transformer Grid Two-port Elements - Gyrator Energy entering gyrator is equal to energy leaving gyrator. The energy domain is different on either side of the gyrator Gyrator, symbol GY Petrol engine Electric motor Piston Cylinder … They are often irreversible V Electric Motor i   = V i GY   Two-port Elements - Gyrator Gyrator examples in different energy domains Electric motor Domain: Electric -> Mechanical rotation Centrifugal pump Domain: Mechanical rotation -> Hydraulic Generator (diesel) Domain: Mechanical rotation -> Electric e1 f1 GY r e2 f2 r – gyrator ratio Wind turbine Domain: Hydraulic -> Mechanical rotation Two-port Elements - Gyrator What type of gyrator is this? Piston pump Domain: Mechanical translation -> Hydraulic What is the difference between Transformer and Gyrator? Q Q v P F A v F P F v P Q A Gyrator Transformer Power transformation in gyrator is irreversible Multi-port elements Bond Graphs Multi-port junctions Also called “Three-Port” but can use 3 or more. Two types 0 and 1. They conserve power and are reversible. Are therefore ideal routers of energy. Power is distributed among the various other elements. V V V i 0 i i 1 V i V i V i 0 junctions (common effort) Do not store, dissipate, generate or transform energy Equality of effort: e1 = e2 = e3 = e4 Flow sums to zero: f1 + f2 + f3 + f4 = 0 + / - sign depends on direction of arrow + means arrow pointing to 0 junction - means arrow pointing away from 0 junction 0 junctions (common effort) Example in electric domain V1 i1 0 V2 i2 V1= V2=V3 V3 i3 i1+ i2+i3=0 … i1= i2+i3 15A For example: V=12V 5A 10A Electrical domain 15L/s 5L/s 10L/s For example: P=2bar Hydraulic domain 1 junctions (common flow) Do not store, dissipate, generate or transform energy Equality of flow: f1 = f2 = f3 = f4 Effort sums to zero: e1 + e2 + e3 + e4 = 0 + / - sign depends on direction of arrow + means arrow pointing to port - means arrow pointing away from port 1 junctions (common flow) Example in mechanical domain 1 1 1 2 2 1= 2= 3 1+ 2+ 3=0 … 1= 2+  3  3 1 junctions (common flow) Words to symbols Try putting the symbols into the earlier example Engine   Gear box SE   TF     Shaft 1   C     GY Wheels F v I F v Road Losses in bond graphs What about including mechanical resistance? Engine   Gear box   Shaft     TF Road F v I F v Friction Friction SE Wheels F v 1 1     R C GY F v 1   R Hydro – electric powerplant example Flow P Q Reser voir Sf P Q Pipe P Q Turbine τ ω Shaft τ ω Gener ator V i Transf ormer V i Grid Car seat suspension example Ideal velocity is a source of flow Springs modelled by a C Masses modelled by an I Dampers are modelled by an R Driver mass, m2 Seat mass, m1 Seat suspension Car seat suspension example Driver mass Seat mass Car floor Sf Bond graph - Causality Causality Establishes cause and effect relationships between factors of power. The inputs are characterised by the causal stroke. Causal stroke indicates the direction in which the effort signal is directed. Flow directed in opposite direction to causal stroke “applies an Effort” e f “responds with a Flow” Causality - effort e System A f System B Effort is the output of A and flow is the output of B A e f B Effort pushes, vertical bar on the side of the half-arrow Se Causality - flow f System A e System B Flow is the output of A and effort is the output of B A e f B Flow points, vertical bar on the opposite side of the half-arrow Sf Causality for I, R, C elements I element (storage element) Receives effort (cause) and generates flow (effect) J C element (storage element) Receives flow and generates effort I C J Resistive (or dissipative) R elements can have any type of causal structure. Resistive causality J Conductive causality R J R Bond graph – System equations Equations defining Bond graph elements 𝑒 = 𝑅𝑓 1 𝑒 = න 𝑓𝑑𝑡 𝑐 𝑒 = 𝐼 𝑓ሶ Equations defining Bond graph elements 𝑒1 𝑓1 TF 𝑒2 𝑒1 𝑓2 𝑒2 = 𝑓1 = , 𝜆 𝜆 𝑓2 𝜆 - transformer ratio 𝑒1 𝑓1 𝑒2 TF 𝑓2 𝑓2 = 𝜆𝑓1 , 𝑒1 = 𝜆𝑒2 Equations defining Bond graph elements 𝑒1 𝑓1 GY 𝑒2 𝑓2 𝑒1 𝑒2 𝑓2 = , 𝑓1 = 𝑘 𝑘 𝑘 - gyrator ratio 𝑒1 𝑓1 𝑒2 GY 𝑓2 𝑒2 = 𝑘𝑓1 , 𝑒1 = 𝑘𝑓2 Numerical Solution of Bond Graphs Numerical Solution of Bond Graphs This model allows us to visualize how variables are related and to calculate how the system performs V I Battery Say we want wheel speed of 100 rad/s at 5 Nm torque   Motor Gear box     Wheel   I Friction SE V I GY    TF  1   R Numerical Solution of Bond Graphs Wheel speed of 100 rad/s at 5 Nm torque V I Battery Motor   Gear box   System variables: TF - 15:1 GY – 1:100 R – 0.1 Nms I – 1kg   Friction SE V I GY    TF  5Nm 1   I  100 rad/s R Wheel Numerical Solution of Bond Graphs Say we want wheel speed of 100 rad/s at 5 Nm torque Battery V I Motor   Gear box   System variables: TF - 15:1 GY – 1:100 R – 0.1 Nms I – 1kg   Friction SE V   TF GY I 1:100  15:1  1   R 0.1 5Nm 1 100 rad/s Wheel Numerical Solution of Bond Graphs Say we want wheel speed of 100 rad/s at 5 Nm torque Battery What is the Voltage required to spin the wheel at 100 rad/s and 5 Nm torque? V I Motor   Gear box     Friction V? SE V   TF GY I 1:100  15:1  1   R 0.1 5Nm 1 100 rad/s Wheel Numerical Solution of Bond Graphs Junctions will transmit common effort and sum flow (0) or transmit common flow and sum effort (1)   V Wheel Motor Gear box Battery   I   Friction SE V   TF GY I 1:100  15:1 100 1 100  R 0.1 5 100 1 Numerical Solution of Bond Graphs Junctions will transmit common effort and sum flow (0) or transmit common flow and sum effort (1)   V Wheel Motor Gear box Battery   I   Friction SE V   TF GY I 1:100  15:1 100 1 5 100 100  10Nm effort = resistance * flow R 0.1 𝜏 = 𝑅 ∗ 𝜔 =0.1*100 1 Numerical Solution of Bond Graphs Junctions will transmit common effort and sum flow (0) or transmit common flow and sum effort (1)   V Wheel Motor Gear box Battery   I   Friction 𝜏 = 10 + 5 = 15𝑁𝑚 SE V  15  TF GY 1 I 1:100  15:1 100 100 10 R 0.1 5 100 1 Numerical Solution of Bond Graphs The 15:1 transformer will multiply the flow by 15 and divide the effort by 15  V Motor Gear box Battery  I   Friction 1Nm =15/(15/1) SE 1 V  15 TF GY I 1:100  15:1 100 1500 1500 rad/s = 100*(15/1)   1 100 10 R 0.1 5 100 1 Wheel Numerical Solution of Bond Graphs The motor (a gyrator) has its own equation to calculate effort input from output flow 𝑒1 𝑓2 = 𝑘 Battery V I Motor   1 𝑒1 = 𝑘 ∗ 𝑓2 = ∗ 1500 = 15𝑉 100 15V V SE GY 1 TF Gear box   Friction 15 I 1:100 1500 15:1 100 1 100 10 R 0.1 5 100 1   Wheel Numerical Solution of Bond Graphs This provides a steady-state solution for the system I is assumed to be unlimited by an “ideal” source  V Motor Gear box Battery  I   15𝑉 SE Friction V 15 1 TF GY I 1:100 1500 15:1 100 1 100 10 R 0.1 5 100 1   Wheel Tutorial – self study Produce a bond graph representation of this gear box word graph. The gear ratio is 15:1 and the rotational friction is 0.08Nms 0.08Nms Friction Engine Gear box Wheel 15:1 Given that the output gear shaft should produce 20 Nm torque at an angular velocity of 100 rad/s, work out the required  and  What is the efficiency (output power/input power) of this gear system?