Präsentation der Baugruppe Kurbeltrieb in Englisch Mechatroniker Mittelstufe Max-Eyth-Schule in Kassel.

Slides:



Advertisements
Similar presentations
Unit C: Agricultural Power Systems
Advertisements

AND ELEMENTS OF MACHINES
Training Presentation
USER MAINTENANCE.
Engine Cooling And Lubrication
Engine Fundamentals TRF 210.
Purpose Body Making module is designed for bending the laminated foil film in the shape of a tube, for welding the side seam and for cutting bodies with.
ENGINE OPERATION CHAPTER 3.
HYDRAULICS & PNEUMATICS
CHAPTER 3 INTERNAL COMBUSTION ENGINES
Small Engines.
Engine Systems and Components
Build Basic Knowledge of Internal Combustion Engines (I.C.E)
Section 5: Commercial Refrigeration
Diesel Engine 4 Stroke Cycle model
CHAPTER (1) General Introduction Dr. Ahmed Abou El-Wafa.
Actuators.
NAZARIN B. NORDIN What you will learn: Strength, elasticity, ductility, malleability, brittleness, toughness, hardness Ferrous/ non-ferrous.
Engineer Training Mechanical Overview. Engineer Training Confidential 2 XL1500 is both Roll-to-Roll and Roll-to-Sheet Printer MachineXL1500-2mXL1500-3mXL1500-5m.
Drawing Process.
New Mexico FFA Agricultural Mechanics Career Development Event Engine Power.
CHAPTER 4 CPB 20004: Plant Utility n Maintenance
STRUCTURES Outcome 3 Gary Plimer 2008 MUSSELBURGH GRAMMAR SCHOOL.
1 Agricultural Power Systems Identifying Engine Systems and Their Components.
Aaron’s Engine Anatomy
A Presentation On Construction Of Turbo Generator
A Presentation on construction of Turbo Generator.
Chapter 41 Lubrication System Fundamentals.
Submitted by Pradeep pataskar Roll no.11031M02034.
ENM208 INTRODUCTION to MACHINING ANADOLU UNİVERSITY Industrial Engineering Department.
Prepared By Rohit G. Sorte M.Tech
Air Cylinders Convert the energy contained in the compressed air
Cooling System Get the engine up to optimum operating Temperature as quickly as possible and maintains it at that temperature. Controls the heat produced.
Manoj kumar 2009JE0406 Mining machinery engineering.
EXTRUSION BASIC CONCEPTS
LECTURE 1.
IDENTIFY THE MAJOR SYSTEMS OF AN ENGINE!. NEXT GENERATION SCIENCE/COMMON CORE STANDARDS ADDRESSED! CCSS.ELA Literacy.RST.9‐ 10.3 Follow precisely a complex.
MCQ. 1. the process of converting of row material in to finished product using machine is called as __________ a. Finishing process b. Metal cutting process.
Pistons and piston rings
Types of material handling equipment
Fixed Orifice Tube Cycling Clutch System (FOTCC)
Unit III Class II Pneumatic actuators.
Engineering II – August 2016
Metal forming processes
CGS Ground School Technical The piston engine
Engine Lubrication Systems
FERROUS AND NON FERROUS ALLOYS
Automotive Heating And Air Conditioning
4/27/2018 2:57 AM TOOL STEELS © 2007 Microsoft Corporation. All rights reserved. Microsoft, Windows, Windows Vista and other product names are or may.
GOVERNMENT ENGINEERING COLLEGE DAHOD
EMF Induced in a Moving Conductor (“Motional EMF”)
BEARING MATERIALS.
Engine Lubrication Systems
FERROUS AND NON FERROUS ALLOYS
Air Distribution Control
Friction Welding Welding Technology/ 3.5 Friction Welding.
ME390 Automotive Mechanics
BTE 1013 ENGINEERING SCIENCEs
Induction Motor. Induction motor Main parts are Stator Rotor.
by Norman E. Dowling. ISBN
Industrial Electronics
Chapter 5 Power Estimation in Extrusion and Wire-rod Drawing
Diesel Engine 4 Stroke Cycle model
Steel.
Applied Technology High School (ATHS)
Automotive Technology Principles, Diagnosis, and Service
Presentation transcript:

Präsentation der Baugruppe Kurbeltrieb in Englisch Mechatroniker Mittelstufe Max-Eyth-Schule in Kassel

Ziel: Ziel: Durchführung der Präsentation der Baugruppe Kurbeltrieb des Lernträgers Kolbenkompressor in Verbindung mit dem Lernfeld 2 – Herstellen mechanischer Teilsysteme, dem Lernfeld 5 – Kommunizieren mit Datenverarbeitungssystemen, dem Lernfeld 13 – Übergabe von mechatronischen Systemen an den Kunden, bzw. Vorbereitung für das Fachgespräch in der Abschlussprüfung und dem Fach Englisch. Durchführung der Präsentation der Baugruppe Kurbeltrieb des Lernträgers Kolbenkompressor in Verbindung mit dem Lernfeld 2 – Herstellen mechanischer Teilsysteme, dem Lernfeld 5 – Kommunizieren mit Datenverarbeitungssystemen, dem Lernfeld 13 – Übergabe von mechatronischen Systemen an den Kunden, bzw. Vorbereitung für das Fachgespräch in der Abschlussprüfung und dem Fach Englisch. Gruppeneinteilung: Gruppeneinteilung: Für diese Präsentation, deren Vorbereitung in Power Point mit der Übersetzung in Englisch (gesprochene Texte und Folientexte) ca. 30 Schulstunden aufgewendet wurden,sind folgende 5 Gruppen entstanden: Für diese Präsentation, deren Vorbereitung in Power Point mit der Übersetzung in Englisch (gesprochene Texte und Folientexte) ca. 30 Schulstunden aufgewendet wurden,sind folgende 5 Gruppen entstanden: Moderation Moderation Funktionsanalyse Funktionsanalyse Weg des Öls Weg des Öls Werkstoffkunde Werkstoffkunde Lagerarten Lagerarten Bewertung der Präsentation: Bewertung der Präsentation: Die Vortragenden sind nach Absprache und Zustimmung der gesamten Klasse mit Hilfe der IHK-Kriterien „Bewertungskriterien Präsentation“ von allen Zuhören (Schülern, Lehrern und Ausbildern) einzeln für das Lernfeld 13 beurteilt worden. Die Vortragenden sind nach Absprache und Zustimmung der gesamten Klasse mit Hilfe der IHK-Kriterien „Bewertungskriterien Präsentation“ von allen Zuhören (Schülern, Lehrern und Ausbildern) einzeln für das Lernfeld 13 beurteilt worden. Weg zum Ziel: Weg zum Ziel: Das Lernfeld 2, 5 und das Fach Englisch wurde durch die beteiligten Kollegen/in Herrn Erythropel und Frau Scheidemann benotet ( Max-Eyth-Schule Kassel, Juni 2003). Das Lernfeld 2, 5 und das Fach Englisch wurde durch die beteiligten Kollegen/in Herrn Erythropel und Frau Scheidemann benotet ( Max-Eyth-Schule Kassel, Juni 2003). Präsentation der Baugruppe Kurbeltrieb in Englisch Mechatroniker Mittelstufe Max-Eyth-Schule in Kassel

Presentation of the crank mechanism on the piston bulk module Mechatronik Middle School II Daimler Chrysler AG, Deutsche Bahn AG, Kasseler Verkehrs- und Versorgungsgesellschaft

Introduction

Content Introduction Presentation of the piston bulk module Crank mechanism and oil route Material technology Type and strength of bearing Discharge

Presentation of the piston bulk module Text

Electronics Locking device Line safety switch Safety contact socket Engine safety switch Emergency switch On / Off switch Ammeter Pressure Demarcation Pressure Control Stator / Rotor Text

Mechanics Rotor Rotor belt pulley Fan belt Fan belt pulley Crank drive Connecting rod Cylinder head Inlet- / Outlet valve Text

Pneumatics Cylinder head attracts the air Air filter Air pressuring Pressure control Pressure demarcation Airboiler with condensation recycling and safety valve ( pressure acccumulator) Performance valve Consumer Text

Crank mechanism Presented by Michael Machuletz & Christian Erdmann MTM 2 / KVG & StW

Main Function 1.Initiation of revolutions by flywheel on the crankshaft. 2.Transformation into excentric revolutions 3.Energy transmission at connection causing a straight movement.

Crank mechanism Partial function Deep groove ball bearing: Taking up and positioning the shaft on the end of deep groove ball bearing. 1.Crankshaft: power transmission from flywheel on to the connection. Taking up connecting and upper part of the connecting.

The production of a crankshaft

The components of a crankshaft The crankshaft consists of 16MnCr5: 16MnCr5 is an alloyed case of hardened steel. It consists of …. …. 0.16% C (carbon) …. 1.25% Mn (manganese) and a small chrome plated part

Alloys Alloys are combinations of several metals or metallic and non metallic material. The alloys are produced by mixing molten metals. What are alloys?

The production The crankshaft is die-forged. The work piece is then removed by pressureof air. It changes its structure. But why does it change its structure? Well, the materials must be red hot. The material smelts and is stretched. This is the reason why it changes its structure. The blank of the crankshaft consists of an upper and a lower mould. That shape is equiped with hollow spaces which corresponds to the finished work piece. There is sufficient space in these hollows for the surplus material, the burr. The last step is deburring the workpiece.

Line of oil In the crank case by Christoph Bieniasch and Stefan Breunung MTM 2

Problem: Friction on moving parts Friction is caused when two moving parts come into contact. Text

Components for lubrication Grooved ball bearing Oil ring tappet Oil catch ringCrank shaft Oil ring Text

Single parts which safeguard lubrication Oil ring form of a ring is controlled by oil ring tappet transports oil to the crank shaft Text

Single parts which safeguard lubrication Oil ring tappet form of a diabolo mounted on the crank shaft transports oil outward Text

Single parts which safeguard lubrication Oil transporting ring retains oil from oil ring tappet form of a dish with no base transports oil to inside the crank shaft Text

The way of oil to the bearing oil bath Text

Lubrication takes place by rotation in oil bath. Lubrication circuit available Lubrication takes place in a depressurised atmosphere Lubricant circulation from oil bath to the bearings Lubricant is in the bearing case Text

Lubrications Problems with lubricants: Friction attenuation Curb from brunts Rebuff from worn parts Anticorrosive Heat dissipation Text

Assortment of lubricants Important points to observe regarding types of lubricants Operating condition of the bearing (engine speed) Behaviour of the bearings at differing temperatures Life expectation of the engine Text

Material technology by Florian Funke, Andreas Kulick, Sascha Schmekel and Dirk Strämke

Contents 1) Material standardization I. Unalloyed steels II. Unalloyed steels < 5% III. Alloyed steels > 5% 2) Material science I. Crankshaft II. Oil ring tappet III. Oil ring 3) Stress - strain - diagram I. Reason II. Tensile test III. Examples

Piston compressor Crank mechanism Oil ring tappet Crankshaft Text

1)Material standardization There are 3 different steel categories:

I. Unalloyed steels E 295 Yield point Re = 295 N / mm² Steel for mechanical engineering C 15 C 30 } Quenched and tempered steels Carbon A content of C of 30 : 100 = 0,3% C Quenched and tempered means hardening and tempering. Text

II. Unalloyed steels < 5% 16 Mn Cr 5 A content of Mn: 5:4 = 1,25% Mn A small content of chromium. Manganite A content of C: 16:100 = 0,16% C ElementDivisor Cr, Co, Mn, Ni, Si, W4 Al, Be, Cu, Mo, Nb, Pb, Ta, Ti, V, Zr10 Ce, N, P, S, C100 B1000 Text

III. Alloyed steels > 5% X 5 Cr Ni A content of Ni: 10% Ni A content of Cr: 18% Cr Nickel Chromium A content of C: 5:100 = 0,05% C A content of alloys >5% Text

2) Material science Assembly crank mechanism:

I. Crankshaft 16 MnCr 5 Case-hardened steel with a content of 0,16% C and 1,25% Mn as well as a small content of chromium. Outer zones hardened Tenacious Core

II. Oil ring tappet 9 SMn 28 K Free-cutting steel with a content of 0,09% C, 0,28% S as well as a small content of Mn; cold drawn The higher content of sulphur and phosphorus guarantees easier processing because of small chips.

I. Why S – S diagram? The S – S diagram is the graph of a tensile test, which shows the mechanical characteristics of a material. Text

II. Tensile test Text

III. Examples of diagrams Text

Crank mechanism

Name (term) of the grooved ball bearing from the crank drive

Definition of DIN 625 – 6302 (6303)  DIN 625 – 6302 (6303) Drilling mark 02 (03) Bearing type 6 Diameter series 3 Grooved ball bearing German industrial norm

Data block DBr max h min Basic number Bearing range , d 15 17

! Function of the Lever law ! - The Lever law is used to calculate forces which combine at a certain lever length.

To calculate from length we use the point of rotation, length of arm and force. Point of rotation Length of arm Force

Lever law Formula ∑ Ml = ∑ Mr Sum of anti-clockwise elements = Sum of clockwise elements M = F * L Turning point = Force* Lever arm F1 * L1 = F2 * L2 Force 1 * Lever arm 1 = Force 2 * Lever arm 2

Analysis of the assembly shows that the parallel grooved ball bearings are rather large. From the drawing, we see that the connecting rod is not placed exactly in the middle but between the bearings of the crank shaft. What is the reason for this?

Force FA?Force FB? [Geg: 8bar (80N/cm²) ; d = 50mm] 41mm31mm 72mm Calculation of the Lever law

Calculation: F1 41mm 72mm 1.Presumption:FB is point of rotation.. FB FA Geg: F1 = 1570N Ges: FA 2.F1 clockwise, is positive. 3.FA anti-clockwise, is negative. F1 * L1 = FA * L2 1570N * 31mm = FA*72mm FA = 676N 31mm FB * L1 = F1 * L2 4.Point of rotation:FA 5. F1 clockwise, is positive. 6. FB anti-clockwise, is negative. FB * 72mm = 1570N * 41mm FB = 1570N * 41mm / 72mmFA = 1570N * 31mm / 72mm FB = 894N

Thank you for your interest MTM2 Mechatronics Auswertung-Leerbogen

Finish Please press ESC to close the Presentation

The spoken Part of the Presentation

Electronics Electric energy is sent from a source via the emergency switch and the main switch to an ammeter, which, in turn, opens the pressure control valve and the demarcation valve leading to the stator. The electronic magnetic field between the stator and rotor set the rotor in motion. zurück

Piston bulk module The piston bulk module is a machine to produce pressurised air. The controlled release is variously used for impulse compressor motors and tools - for example pneumatic hammer and spray tins. There are many different types of air compressor: frequently used varieties are piston and pegtop bulk modules. zurück

Mechanics The rotor moves the rotor belt pulley, which sets the fan belt pulley in motion via a fan belt. The crank drive converts the rotation into a straight movement. When the connecting rod moves the piston is raised and lowered and air is sucked in. zurück

Pneumatics This air is then cleaned by a filter and sent via an inlet valve into the cylinder head. When the piston moves up, the cylinder head encloses the air which is thus pressurised. There is a safety valve above the cylinder head. This valve creates only one certain pressure. The pressurised air is expelled through the outlet valve to the air boiler. The pressure control is in the outlet valve. This valve controls the pressure of air flow to the air boiler. This means only the required pressure is sent to the air storage module. The pressure demarcation switches off, if too much air is located in the boiler. The air storage module retains the air at a constant pressure until required. In the air there are water particles, which form during the pressurising process. This water collects at the base of the air boiler.The pressurised filtered air in the storage module can now be forwarded to the control valve, where it is regulated to the required working pressure and subsequently utilised. zurück

Sliding friction On the crank shaft we have a problem, that the designer also faced. How will the bearing on the crank shaft be lubricated? Here it is presented in red. Our aim is to avoid corrosion and possible erosion. zurück

The solution are 3 different components: First of all we have the crank shaft. So that the oil reaches the right place, we have followings components: first the oil catch ring, next the oil ring tappet and oil ring Another two components are placed on the crank shaft: the ball bearings. zurück

Here are the separate parts. First we have the oil ring, which is circular. It is controlled by an oil ring tappet and transports oil to the crank shaft. zurück

Next we have the oil ring tappet. It has the form of a diabolo and is secured on the crank shaft. Through the centrifugal force the oil is transported outward. zurück

Finally we have the oil retaining ring. It is in the shape of a dish with no base. It holds oil from the oil ring tappet and transports oil to the inside of the crank shaft. zurück

Let me explain the path of the oil from the oil bath to the floating bearing. The oil ring transports oil from the base of the crank case and over the crank shaft. From there it flows by virtue of centrifugal force to the oil retaining ring. From there it flows to the oil ring tappet and then through a hole to a balancing weight to the floating bearing, which it lubricates. Afterwards the used oil flows back to the oil bath and the circular flow starts again. zurück

Now we come to the lubrication by circulating oil. The lubrication circulation is assured by the oil ring. The complete lubrication takes place without pressure and stops when the crank shaft stops. The oil ring transports the oil from the oil bath to the final bearing. The oil for lubrication is available in the box. zurück

I would like to explain specific problems of lubricants. Lubricants are used to minimize friction. Next they lessen shocks and protect parts from corrosion and guarantee the rebuff of heat. Also they transport wear particles to ensure no dirt and that the machine is working smoothly. zurück

Now we come to the assortment of lubricants. Some points must be observed - for example the operating conditions of the bearings or which rotation speed the lubricants are suitable for. 2. The second point is the temperature. You must check if the lubricants can withstand the temperature of the bearings. 3. You must check if your lubricants can sustain long running times and high stress. zurück

Piston compressor Now you can see the complete crank mechanism with the oil ring tappet. My workmate will show you which materials the components consist of, and why these materials are chosen. zurück

First, I would like to tell you something about unalloyed steels. There are two definitions. The first for example is E 295. The capital letter E means steel for mechanical engineering and the number means the Yield point Re, in this case Re is 295 N/mm². (square mm) The explanation of the yield point will be shown by my colleague later. The second for example are C15 and C30. These are quenched and tempered steels. Quenched and tempered means hardening which is followed by tempering. The letter C stands for carbon and the figure means a content of Carbon, in this case 15:100=0,15% (O point 15 divided by) and 30:100=0,3% of C. zurück

The second category is unalloyed steels under 5%. This 5% means the content of alloys. One example is 16MnCr5. The first figure means a content of carbon, in this case 16:100=0,16% of C. Mn stands for manganite and Cr means a small content of chromium. The last figure refer to the first letter, in this case the 5 refer to the Mn and means a content of Mn 5:4=1,25% of Mn. The divisor 5 or 100 results from this table. This table stands for all unalloyed steels which can also be found on the sheet, too. zurück

And the third category is alloyed steels over 5%. An example for this category can be this: X5CrNi The X means a content of alloyed steels over 5%. The first figure means a content of carbon, in this case 5:100=0,05% of C. Cr stands for chromium. Ni stands for nickel. The first figure after the letters means a content of the first element and the second figure after the letters means a content of the second element. In this case, there are 18% of Cr and 10% of Ni. zurück

The stress – strain diagram is the graph of a tensile – test, which shows the mechanical characteristics of a material. That means the stress – strain – diagram is the graph which accompanies the results of the tensile – test. This graph shows how elastic, flexible and useable a material is. With its help, engineers can see if the material meets the requirements. zurück

At the top of this page you can see how a specimen for the tensile test changes. First you can see the specimen in its original shape. The specimen has changed in length, but up to this point, with no plastic deformation. When, however, the sample is plastically deformed certain tolerances are exceeded in the centre part of the work piece and it will snap. zurück

Here are some examples of different materials. I will attempt to familiarise you with the diagram by use of the examples of copper and S.235 JR ( or in old standardisation ST – 37 – 2 ). The verticle axis shows the tensile stress in Newton per mm^2, the horizontal axis the extention in percent. Copper is a very soft material. That means it has a very small plastic area.Up to the relevant point it is deformed, but it does not break as early. Furthermore, in comparison to ST – 37 – 2, copper has nearly no Re (the yield point). Here you can see that ST – 37 – 2 has a large yield point. This is the tolerance point at which the material breaks. zurück