CA10B解放牌汽車調(diào)整臂外殼工藝規(guī)程及專用夾具設(shè)計(jì)【車φ60孔及端面】【鉆φ13.8和φ16沉孔】【說(shuō)明書+CAD+3D】

CA10B解放牌汽車調(diào)整臂外殼工藝規(guī)程及專用夾具設(shè)計(jì)【車φ60孔及端面】【鉆φ13.8和φ16沉孔】【說(shuō)明書+CAD+3D】,車φ60孔及端面,鉆φ13.8和φ16沉孔,說(shuō)明書+CAD+3D,CA10B解放牌汽車調(diào)整臂外殼工藝規(guī)程及專用夾具設(shè)計(jì)【車φ60孔及端面】【鉆φ13.8和φ16沉孔】【說(shuō)明書+CAD+3D】,CA10B American Journal of Applied Sciences 6 (2): 251-262, 2009 ISSN 1546-9239 2009 Science Publications Corresponding Author: Brian Boswell, Department of Mechanical Engineering, Curtin University of Technology, GPO Box U1987, Perth Western Australia 6845 Tel: (08) 9266 3803 Fax (08) 9266 2681 251 Air-Cooling Used For Metal Cutting Brian Boswell and Tilak T Chandratilleke Department of Mechanical Engineering, Curtin University of Technology, GPO Box U1987, Perth Western Australia 6845 Abstract: Air-cooling and dry machining are both being trialled as possible solutions to the metal cutting industrys long running problems of extending tool life, reducing tool failure and minimising the heat generation at the tool tip. To date, large amounts of expensive coolant which cause both environmental damage and health hazards have had to be used. The introduction of dry machining is the goal of todays metal cutting industry that tirelessly endeavours to reduce machining costs and impact from chemicals in the environment. Modern tool tips are already capable of maintaining their cutting edge at higher temperatures, but even with these improvements in tool materials, the cutting edge will eventually break down. Applying cold air to the tool interface of these modern tool tips will also help prolong their tool life reducing the cost of metal cutting. Dry machining incorporating air being directed on to the tool interface is considered in this paper as a possible alternative for harmful liquid-based cooling. However, low convective heat removal rates associated with conventional air-cooling methods are generally inadequate for dissipating intense heat generation in the cutting processes and suitable improved cooling methodologies have yet to be established. Key words: Vortex tube, tool life, flank wear, cold fraction, coefficient of performance, air-cooled, environmentally friendly INTRODUCTION In this research examines the operational effectiveness of a Ranque-Hilsch vortex tube being used to cool tool tip during machining. The Ranque-Hilsch vortex effect was discovered in the early 1930s when it caused considerable excitement, as it demonstrated that it was possible to produce hot and cold air by supplying compressed air to a tube. At first it is hard to believe that such a device can produce hot and cold air and at a useful flow rate. The vortex tube is a simple device with no moving parts, which simultaneously produces cold and hot air streams. However, to date, there is little research in determining the efficiency of using a vortex tube in cooling tool tips. Therefore, to establish the effectiveness of the heat transfer process on the tool tip a series of experimental investigations has been carried out. These tests will determined the most suitable parameters to use, like mass flow rate of cold and hot air, cold and hot tube diameter with respect to tube length, to achievable minimum cold air temperatures. Air-cooling has never been taken seriously by the manufacturing industry due to the fact that for many years traditional cutting fluid has been shown to be effective in cooling tool tips during the machining processes. The outcome of this research will prove that air-cooling can replace traditional cutting fluid for many machining applications, without any reduction in tool life or reduction in quality of work piece surface finish. The introduction of using a Ranque-Hilsch Vortex Tube to provide cold air to the tool interface is shown to significantly improve the performance of air-cooling. Recorded tool tip interface temperatures clearly indicate that there is a highly significant reduction in tool tip temperature. This reduction in temperature slows the wear mechanisms as shown by the reduced flank wear when examined under a microscope. Therefore, monitoring the growth of the flank wear indicates the increased tool life when being air-cooled. The Ranque-Hilsch vortex tube1 is a remarkable device that is able to separate airflow into two different streams simultaneously, one hotter than the inlet air and the other cooler, without any moving parts being involved. The mechanism producing the temperature separation of cold air and hot air when passing through the vortex tube is not yet fully understood. This device has been described as Maxwells demon, a fanciful means of separating heat from cold without work. The Am. J. Applied Sci., 6 (2): 251-262, 2009 252 vortex tube basically consists of three pipes and a supply of compressed air to achieve a moderately low temperature at the cold outlet. Ranque2 attempted to exploit the commercial potential for this strange device that produced hot and cold air with no moving parts. Unfortunately, this venture failed and the vortex tube slipped into obscurity. The mechanism underlying the energy transfer from the cold to the hot flow remains elusive. However, there is debate even as to the basic physics of the phenomenon, while the majority of researchers suggest the mechanism is based on the interactions of turbulence, compressibility and shear work as shown by the analysis of Deissler and Perlmutter3. Recent research has been divided into two categories. The first category termed as external studies were concerned with the performance of the tubes. It was found by Gulyaev4 that the minimum ratio of the length of the tube to that of its diameter was thirteen. Other research suggested a ratio of forty to fifty for optimum operation. As for the diaphragm, the optimum dimension is a ratio of 2:3 for the diaphragm diameter to tube diameter. The vortex tube consists of three important parts the mid-section where the air enters into the vortex generator (which increases the speed of the air), the cold tube and the hot tube as shown in Fig. 1. Normally the hot tube is about 350 mm long and at the end there is a conical valve which controls the amount of hot air escaping. On the right side of the vortex generator is the cold tube exit. Between the vortex generator and the cold tube there is a diaphragm, with a central hole that can be easily changed. Diaphragms with large or small holes can also increase or decrease the temperature obtained at the cold exit. Considering the above vortex tube, the compressed air is supplied circumferentially into the tube at sonic speed and creates a cyclone (vortex) spinning at a million revolutions per minute. The air is forced to spin inward to the centre where it then escapes up along the hot tube as this path presents the least resistance to the airflow. The air continues to spin as it travels along the tube until it meets the conical valve where it turns part of the spinning air column (vortex) inside itself. The slower moving air inside column of the spinning air gives up its heat to the faster spinning outside column of air. The cold air travelling down the spinning air is now directed out the cold end of the vortex generator and the hot air is exhausted out of the other end of the vortex tube. Adjusting the conical valve built into the hot air exhaust can change the temperature of these two air streams to as low as 55C as shown by Fig. 2. Fig. 1: Diagram of the Hilsch Vortex Tube5 -60-50-40-30-20-10010050100150Time (s)Nozzle exit temperature (C) Fig. 2: Temperature recoded at cold nozzle exit having an inlet pressure of 1Mpa VORTEX THEORY Currently no one can definitively explain why the vortex tube operates as it does: the process itself is straightforward as outlined by Lewins and Bejan6. The inlet nozzle is tangential to the vortex generator and therefore can provide a high speed rotating airflow inside the vortex generator. Subsequently, there is a radial temperature gradient increasing from the inner core of the tube to the outside wall of the tube. This is primarily because of the potential energy of compressed air converting to kinetic energy due to the forced vortex caused by the external torque near the tangential air inlet. Therefore the high-speed swirling flow inside the tube and away from the walls is created. The existing air inside the vortex hot tube is normally at the atmospheric temperature and so, when the rotating flow enters the vortex tube it expands and its temperature drops to a temperature lower than the ambient temperature. The difference between these two temperatures will lead to a temperature gradient along the tube producing colder peripheral air than the core air. As a result, the central air molecules will lose heat to those in the outer region as shown in Fig. 3. It is notable that this system is a dynamic system due to the nature of the airflow in the tube and so will not reach equilibrium. Hence the peripheral air has a higher kinetic energy (hotter) than the inner air (colder). Am. J. Applied Sci., 6 (2): 251-262, 2009 253 Fig. 3: Radial heat convection in vortex tube due to the expansion of the compressed air The existence of a major pressure gradient due to the forced vortex in the radial direction will provide a centripetal force for circular swirling and therefore it will lead to a high pressure at the tube wall and low pressure at the centre. When the air enters to peripheral region (A), as it expands, the outer air will be cooled due to its expansion. Consequently, the inner core air (B) will get warm because it is compressed by the expansion of the peripheral air. Heat is then transferred from the inner core (B) to the outer core (A). As the inner air is being compressed, it naturally tries to push against the periphery by expanding. Work is therefore done on the outer core air, which then gets heated and the difference in pressures results in the expansion and contraction of the air, which causes work to be done on the peripheral air. Therefore, heat is transferred radially outward as shown in Fig. 4. When the air continues to swirl along the tube the more energy separation will occur by axial convection while it moves towards the hot end. During this progression, the heat will be transferred from the core air to the outer air. As the airflow reaches the hot end a fraction of the air will exhaust through the conical valve, which is located at the hot end and the remaining air flow will spin back towards the cold end due to the adverse pressure gradient near the centre as shown in Fig. 5. The remaining portion of the warm air preserves its direction of motion in the vertical flow that is either in a clockwise or anticlockwise manner around the circumference of the tube. Furthermore, this air stream resides at the inner core of the tube where the air pressure there is lower. If the angular velocities of both the air streams are preserved, it means that any two particles taken from Fig. 4: Schematic positions of the peripheral and inner core air Fig. 5: A diagram of the airflow pattern in vortex tube both the air streams will take the same time to complete a revolution around the circumference of the tube. From the principle of conservation of angular momentum, it seems that the angular velocity of the inner core molecules would increase, by the Eq: 2a aaam rconstant = (1) The equation implies that in the inner core, where the value of ra (radial distance measured from the centre of the tube to the particular molecule in concern) is small, there should be a corresponding increase in the molecules angular velocity, wa, to allow for the conservation of the total angular momentum in the system. This is assuming that there is negligible mass difference, ma, between any two-air molecules in the tube. However, the angular velocity of a particular molecule in the inner core remains unchanged. This means that angular momentum has actually been lost from the inner core of the vortex tube. Angular momentum of the inner core is not preserved or more specifically decreases, due to heat transferred to the outer core. This results in the transfer of energy from the inner core to the outer core. The loss in heat energy Am. J. Applied Sci., 6 (2): 251-262, 2009 254 Fig. 6: Schematic vortex tube diagram showing tangential air inlet from the inner core goes into heating up of the air molecules in the outer core. Hence, the outer core becomes hotter and the inner core becomes cooler. Upon reaching the hot end, the hotter peripheral air escapes through the small openings between the conical valve and the tube wall (hot outlet). However, the central air that is cooler is deflected by the tapered valve spindle and continues its travel from the hot end towards the cold tube. Only the innermost air molecules pass through the diaphragm and exit through the cold end where it is collected. As a result, the air molecules are separated into a hot stream and cold stream through the hot and cold ends of the vortex tube respectively. The Fig. 6 shows a good view of the vortex tube. It is important to note that separation takes place specifically at the hot end tube. The purpose of the tapered spindle (conical valve) is to direct the cold air to the axial region of the tube in a counter flow. The diaphragm (orifice) on the other hand is used to block the peripheral air, so that the central flow will escape through the cold end. The absence of moving parts in the vortex tube may create this wrong supposition that this phenomenon is violating thermodynamics law. The fact that without doing any work at room temperature, a stream of air can be divided into two different steams, one cooler and one hotter, seems to contradict the second law of thermodynamics. However, it is important to mention that despite this misleading belief the physics remains intact. Although, the physics of the vortex tube is complicated, the study of the basic principles of thermodynamics can help to gain a better understanding of what is happening inside a vortex tube. The first law of thermodynamics is about the conservation of energy. According to this law, during a reaction between a system and its ambient, the energy that can be received from the ambient to the system is exactly equal to the energy that is lost from the system to the ambient. This energy can be seen in two different states: Heat and Work. Hence, for every thermodynamics system with a specific control volume: Fig. 7: Schematic control volume of a vortex tube 2iC.Viii2eeeec.vvQm (hgz )2vm (hgz )W2+=+? (2) where cvQ?is the rate of heat flow, which transfers through the control volume boundary and cvW?is the work that can be done by the system on its ambient, m ? is the mass flow rate, h is the enthalpy of the air stream, v is the air stream velocity, z is the distance between the air stream and a source point and the subscripts i and e refer to inlet and outlet streams. Assuming that the vortex tube is well insulated, then the heat transfer between the system and the ambient can be taken as cvQ?equal to zero, with vi and ve also equal to zero, as can the work W. Considering the control volume as shown in Fig. 7 for the vortex tube then Eq. (2) can be simplified as: iieem hm h=? (3) The expanding air can be treated as ideal gas and hence with no Joule-Thomson heating or cooling effect. Also, assuming air obeys the ideal gas laws and having constant specific heat capacity Cp, you can write: ipihC T= (4) ipicpchphm C Tm C Tm C T=+? (5) Am. J. Applied Sci., 6 (2): 251-262, 2009 255 by defining cf as the cold fraction: ccfimm=? (6) from the continuityEq: chimmm+=? (7) Combining Eq. 5, 6 and 7 gives the relationship between inlet stream temperature and cold stream temperature, (hot stream temperature and cold fraction). Hence Eq: ()icfchTT1T= + (8) by assuming, hcTTT= (9) cicTTT= (10) hhiTTT= (11) where, ?T is the difference between the hot and the cold air stream temperatures, ?Tc is the temperature difference between the inlet and cold air streams and finally ?Th is the temperature difference between the inlet and hot air streams. The equation can be written as: hcTTT= (12) Combining E q. 8 and 12 allows you to determine ?Tc and ?Th theoretically by measuring the cold fraction, cf and total temperature difference, ?T. Therefore: ()ccfT1T= (13) hcfTT= (14) Equation 13 and 14 can be used to show the consistency of the first law of thermodynamics for test carried out on the vortex tube. VORTEX TUBE EFFICIENCY In practical cases of refrigeration it is so important to determine the coefficient of performance of the cooling device. Hence, it seems only logical to determine the coefficient of performance of the vortex tube and compare it with the conventional refrigeration coefficient of performance to determine its efficiency in use. The vortex tube can be used as a refrigeration device when the cold pipe wall is used to reduce the temperature or as a heating device when the hot pipe wall is used to increase the temperature of an enclosure. It should be noted that opposite to what is normally viewed in thermodynamics, the vortex tube in this case is an open control volume device. If the system was assumed to be steady state, then from the first law of thermodynamics: HQ=? (15) where, H? is the system enthalpy change and Q? is the heat exchanged between the system and its surroundings. Lets assume that Q? is approximately zero even though the cold tube may have frost on it and the hot tube is very warm. If this is the case then: cHHHH0= + = (16) where, ?Hc is the enthalpy change of cold stream and ?HH is the enthalpy change of hot stream. Assuming the air as an ideal gas, the total enthalpy change can be written as: ()()cpcihphiHm C(TTm CTT0=+= (17) where, mc is mass flow rate at cold tube, mh is mass flow rate at hot tube, Tc is cold air temperature, Ti is inlet air temperature, Th is hot air temperature and Cp is specific heat of air at constant pressure and assumes the process as reversible and adiabatic. By applying the second law of thermodynamics to the above: 1Sdq0T =? (18) where, ?S is total entropy change, q is heat transfer and T is absolute temperature. The actual entropy change of the control volume at steady state is: chSSS = + (19) where, ?Sc and ?Sh are the entropy change from entrance to exit of the portion of entering air which leaves the cold tube and the portion of entering are Am. J. Applied Sci., 6 (2): 251-262, 2009 256 which leaves the hot tube, respectively. For an ideal gas (air) with constant specific heat, the entropy change can be written as: ccihhippiiciihmTPmTPSC lnRlnC lnRlnmTPmTP? =+?(20) where the subscripts i, c and h are respectively inlet stream, cold stream and hot stream and R is the ideal gas (air) constant. Since the appearance of a cold (or hot) effect upon the pipe wall without moving parts would attempt to consider this device as competition for a refrigerator (or heat pump), it is useful to estimate its coefficient of performance (COP). Focusing on the cooling effect that can be achieved by placing the cold pipe within an enclosure, the coefficient of performance can be calculated by: cHCOPW=? (21) Where cH? is obtained from: ()ccicHmTT=? (22) cH? is equal to the heat that is transferred to the cold stream through the cold pipe wall (like a heat exchanger) from some source (like the cold box in a refrigerator) and W? in the present case is the work done to compress the air from atmospheric pressure and temperature to the inlet conditions of the tube. Assuming reversible compression (isentropic, minimum work), W?is then obtained from: ()21mR TT nWn1=? (23) where, T2 is the compressor exit temperature and T1 is the compressor inlet temperature (reversible, polytropic process; air: n = 1.4). If we consider a complete system, P1 and T1 are the atmospheric pressure and temperature, P2 and T2 are the compressor exit conditions, 2211n1PTnPT= (24) After the air is compressed, it is kept in the high-pressure tank where then it cools down to the atmosphere temperature, T1 so the inlet temperature of the sonic nozzle Ti, is equal to T1 noting that: pRnCn1= (25) Equation (23) can be simplified to: ()ip21Wm CTT=?(26) with T2 calculated from Eq. (24). This is an ideal work value so it