Mechanical Engineering - Advanced Manufacturing Processes

Full exam

ADVANCED MANUFACTURING PROCESSES A Prof. Annoni SURNAME NAME PERSONAL ID Exam – 10 th July 202 4 Question n. 1 According to the lumped heat cap acity model applied to the laser cutting process, which of the following statements is TRUE? o The kerf width is always equal to the beam waist diameter of the laser beam o If the sheet thickness is increased the velocity must be increased proportionally o In the case material evaporation occurs the cutting speed must be reduced o The model allows to estimate the heat affected zone Question n. 2 Which of the following statements is TRUE regarding laser welding process es ? o Fiber laser sources are always the preferred option o When performing butt joints it is prefer able to have high gap between the sheets o A plume cutter allows to reduce plasma formation thus increasing the process stability o A higher beam quality factor generates lower beam waist diameter Question n. 3 Taking into account laser texturing processes, indicate which of the following statements is WRONG ? o With pulsed wave lasers, higher emission power increase s the ablation diameter o Shorter pulse durations allow to increase the peak emission power o The use of optics with longer focal length reduces the beam waist diameter o Both polishing and roughing operations can be performed Question n. 4 Considering the Laser Powder Bed F usion process (also known as Selective Laser Melting) indicate which of the following is TRUE: o Diode laser sources cannot be employed due to their elevated efficiency o Lack of fusion porosity is obtained at high levels of energy density o A scanner head with galvanometric mirrors is employed to deflect the beam along the intended trajectory o A vacuum chamber is typically employed to prevent oxidation of the material during processing Question n. 5 Which is the solution for reducing the waterjet cutting kerf width? o Reducing the focusing tube diameter o Increasing the water pressure o Increasing the abrasive mass flow rate o Decreasing the feed rate Question n. 6 When using the Bernoulli’s equation for calculating the waterjet velocity in waterjet applications, why do we neglect the gravity potential energy? o The orifice dimensions are negligible respect to the water column that generates the pressure o We make the assumption according to which the cutting head is fixed in space o The specific weight of water is negligible respect to the specific weight of abrasives o The water weight is supported by the pipes Question n. 7 Which of the following sentences are TRUE for face milling? o The mill axis is parallel to the machined surface o ap corresponds to the stock allowance thickness on top of the machined surface o The mill axis position is not important to distinguish face milling from peripheral milling o ap corresponds to the stock allowance width on top of the machined surface Question n. 8 What is the meaning of “minimum chip thickness” effect in mechanical microcutting? o It means that the chip thickness must be small to reduce the cutting forces o It means that the chip thickness must be smaller than a certain critical value to form the chip o It is a valid concept only for turning operations o It means that if the chip thickness is too small, ploughing takes place at the workpiece surface Question n. 9 Which of the following sentences is TRUE for EDM? o The mechanical characteristics of the target material are very important for the MRR o Forces on the wire in WEDM are null o It mainly removes material due to the Joule effect in the workpiece o It cannot work if the target material is not conductive Question n. 10 In Plasma Arc Cutting, what is the role of the nozzle diameter? o If the nozzle diameter reduces, resistance increases and power increases o If the nozzle diameter increases, more power is discharged on the target material o Nozzle diameter does not play a role on power o HyDefinition Plasma counts on a larger orifice diameter Question n. 11 Which is the role of Poisson’s ratio in flexible tools for flexible forming? o Water and elastomers never have the same value of the Poisson’s ratio o Flexible media with a Poisson’s ratio equal to 0.5 makes it easier to foresee the final shape of the part o It does not play a relevant role o Plastic materials cannot count on a Poisson’s ratio equal to 0 .33 since this is the value for metals Question n. 12 Which sentence is TRUE for ultrasonic welding? o It can weld thick metallic parts o It can weld dissimilar materials together o It gets the best results on steel o It welds because of local melting at the interface Question n. 13 Laser welding exercise The company you work for nee ds to design a welding system for the joining of zinc -coated steel sheets in lap joint configuration (as shown in the figure) . In order to successfully perform this type of joint, you are required to leave a 1 mm gap in between the two sheets being joined (to allow vaporized zinc to flow through this area). The thickness of each sheet is t = 3 mm. After preliminary trials, you identify that the best solution in terms of focal position is to have the beam waist diameter positioned on the top of the first sheet in the lap joint configuration. The laser system at your disposal is a 6 kW CO 2 laser source with a beam parameter product of 8 mm*mrad , a collimated beam diameter of 15 mm and a remote scanner head equipped with a n F -theta focusing lens with a 250 mm focal length . A. Calculate the irradiance on the lower surface of the sheet in order to understand if you are operating in keyhole mode considering that the process absorptivity was estimated in 82% : o I = 0.33*10^6 W/cm2 o I = 0.62*10^6 W/cm2 o I = 1.02*10^6 W/cm2 o I = 1.36 *10^6 W/cm2 o I = 1.77 *10^6 W/cm2 o I = 2.31*10^6 W/cm2 o None of these values B. The average width of the weld bead may be related to the maximum beam diameter with the following empirical formula (where K = 1.3 and units are in mm) : ����� ,���� = ���1.22 Consider the lumped heat capacity model under the hypothesis that no material evaporation occurs and that t he material properties can be considered as follows: - Density, ρ=7800 kg/m 3 - Specific heat capacity, cp= 430 J/kgK - Melting temperature, T m=1550 °C - Ambient temperature, T amb =25°C - Latent heat of fusion, L m=280 000 J/kg - Latent heat of vaporization , L v=6400000 J/kg Calculate the maximum weld speed of the process : o v = 13 mm/s o v = 42 mm/s o v = 87 mm/s o v = 139 mm/s o v = 248 mm/s o v = 342 mm/s o v = 532 mm/s o v = 1234 mm/s o None of these values Solution of Question n. 1 3 A. To solve the first point of the exercise , we must at first determine the beam quality factor M 2 that can be calculated as follows: ��� = �� 2 � � 2= ��� ∙� � = 2.37 We can then calculate the beam waist diameter as: �0= 4�� 2 � ���� ���� = 533 �� We need to determine the beam diameter on the lower sheet, which can be calculated as: ��= √�02+ Δ������2∙(���� ���� ) 2 = 679 �� Where Δz = 7 mm since the beam waist diameter is positioned on the upper sheet and we must consider the thickness of the two sheets plus the presence of the gap of 1 mm. Δ������= 2∙�+ � = 7 �� Therefore , it is possible to calculate the beam irradiance as: �= � ∙� ���2 4 = 1.36 ∙10 6 � �� 2 B. The second point can be determined by first calculating the width of the weld bead as: ����� ,���� = ���1.22 = 0.810 �� where the value of ds must be inserted in mm as indicated in the text. Using the lumped heat capacity model , we can estimate the weld velocity. We must first determine the specific energy required to melt the material. � = �{��(�� − ��)+ �� + � ′��} Given the hypothes es, the latent heat of vaporization must not be considered (m’=0). � = 7.3∙10 9 � � 3 The weld speed can be determined as: � = � ∙� ��� = 139 �� /� where the thickness in this case corresponds to the sum of the two sheets ( t = 6 mm). In this case, we do not consider the gap as it does not contribute as a material to the energy balance assumed by the model). Question n. 14 EDM exercise A plunge EDM system is employed to drill holes on gas turbine blades . The operating frequency of the employed circuit is f = 1 kHz and it is possible to consider a power fraction at the cathode Fc = 0.2 . The hole diameter is 3 mm and 50 -mm -long electrodes are employed to perform such an operation . The material properties of the steel being machined are as follows: - Density : ρ = 7545 kg/m 3 - Non -dimensional erosion energy : W s = 1.3 - Erosion enthalpy : hef = 4788000 J/kg The system has the following characteristic curves for the roughing operation. A. Indicate the pulse duration (ton) which maximises the workpiece/electrode wear ratio : o ton = 10 µs o ton = 20 µs o ton = 30 µs o ton = 40 µs o None of these values B. Calculate the power consumption of the EDM process when the highest productivity is required : o P = 1431 W o P = 2643 W o P = 3042 W o P = 3705 W o P = 4372 W o None of these values C. You are asked to evaluate the monthly operative costs of the machine when operating at maximum productivity. In such case, the electrical power required at the grid corresponds to 45 00 W (taking into account al l of the power requirements including the EDM power calculated at the previous point ). Consider that the machine is operated on two shifts of 8 hours each for 20 days per month . Take into account the following cost parameters : - Electricity cost : 0.5 €/kWh - Electrode cost : 2 €/part - Operator cost : 15 €/h o € 2432 o € 4 832 o € 5242 o € 6134 o € 7 476 o € 8432 o € 10231 o € 18 203 o None of these values Solution of Question n. 1 4 A. The condition which maximises the workpiece/electrode wear ratio is ton = 40 us , which can be calculated from the provided graph. B. The operative condition which maximises the process produ ctivity is the one with the highest MRR at the cathode , which corresponds to 20 us. To calculate the power consumption , we must first calculate the duty cycle. Since we know the frequency at which the circuit operates and the pulse duration , we can calculate: ������ = �∙��� = 0.02 The material removal rate is defined as: ��� = ��∙�� ∙������∙�� �ℎ�� The equation can be rearranged to calculate the required power , which is the product between current and tension: �� = ��� ∙�ℎ�� ���������� = 3705 � C. To calculate the monthly cost of operations , we must first determine the number of working hours. �ℎ���� ,���� ℎ= �ℎ���� /�ℎ�������� ∙��ℎ�������� /�������� ∙���������� /���� ℎ= 320 ℎ���� /���� ℎ There are three cost items to consider: • Electricity • Consumables • Operator We must consider the power consumption provided in the text : ��� = 4500 � We have the electricity cost per kWh. Hence , we must calculate the consumption in kWh: ����������������� = �ℎ���� ,���� ℎ∙��� = 1440 �� ℎ The monthly electricity cost is: ��������������������������� ,���� ℎ= ����������������� ∙��������������������������� ,������� ℎ= € 720 We must now calculate the number of electrodes consumed in a month to perform the machining operations . The volume of a single electrode is: ���������� = �����������2 4 ∙ℎ��������� = 353 .4 � � 3 The operative condition requires ton = 20 µs (highes t MRR at the cathode from the graph) . Hence , the MRR at the electrode is 18 mm 3/min from the graph . We can thus calculate the duration of an electrode: ���������� = ���������� �� ���������� = 0.33 h Given the electrode duration , it is possible to calculate the number of electrodes consumed in 1 month (round ed up value ): ���������� = �ℎ���� ,���� ℎ ���������� = 978 Hence, it is possible to calculate the monthly cost of the consumables : ���������� ,���� ℎ= ���������� ∙���������� = € 1956 We can also determine the operator cost , which is: ��������� ,���� ℎ= �ℎ���� ,���� ℎ∙��������� ,ℎ��� = €4800 Finally , it is possible to determine the overall cost as: ���� ,���� ℎ= ��������� ,���� ℎ+ ���������� ,���� ℎ+ ��������������������������� ,���� ℎ= € 7476 Question n. 1 5 Waterjet exercise Your company needs to evaluate the cutting costs for the nest ed parts shown in the figure below from a 25 -mm -thick titanium sheet. In total , there are 130 parts per sheet and each geometry has a cut length of 1446 mm . The nest ed parts of the picture correspond to a batch of the product your company wants to sell. The machine operates with the following parameters: - Pressure : p = 320 MPa - Abrasive mass flow rate : ma = 400 g/min - Orifice diameter : dn = 0.3 mm - Discharge c oefficient : cd = 0.6 - Velocity c oefficient : cv = 0.9 6 - Compressibility coefficient : ψ = 0.97 - Intensifier efficiency : η = 0.8 A. Calculate the electric power consumption of the system o P = 6.2 kW o P = 8.3 k W o P = 10.9 kW o P = 11.7 k W o P = 13.6 kW o None of these values B. The following cost parameters must be considered to estimate the cost per unit length: - Abrasive: ca = 0.5 €/kg - Electric energy: ce = 0.2 €/kWh - Focuser cost: cm = 285 €/part - Focuser duration: tm = 150 h - Orifice cost: cn = 30 €/part - Orifice duration: tn = 150 h - Water cost: cw = 2 €/m 3 Given that the following curve is provided to determine the feed rate as a function of the part thickness, calculate the cost per unit of length of operating the AWJ cutting system o 0.21 €/m o 2.10 €/m o 3.69 €/m o 5.14 €/m o 6.72 €/m o 7.14 €/m o 8.82 €/m o None of these values C. Determine the minimum selling price of the batch to cover the direct costs of production o 50 € o 83 € o 231 € o 421 € o 544 € o 694 € o None of these values Solution of Question n. 1 5 A. We can start by calculating the theoretical velocity of the waterjet : ��ℎ= √2� �1 = 800 � /� We can then determine the water flow rate �� = ���0��ℎ= ��(��2 4 ∙�)��ℎ= 3.4� 3 � = 2.04 �/min = 3.39 ∗10 −5� 3 � Given the water flow rate we can determine the p ower required by the pump ����� = �∙�� = 10 .9 �� Finally , we can calculate the electricity consumption by considering the intensifier efficiency: ���������������������������� = ����� �������������������������������� = 13 .6 �� B. To solve this point , we should use the cost model for AWJ . First , we must determine the feed rate from the provided graph which corresponds to vf = 77 mm/min = 0.00128 m/s It is possible to calculate the cost of the abrasive per unit of length as: �� = ��∙� �̇ �� = 2.597 €/� The cost of water corresponds to: �� = �� ∙�� �� = 0.053 €/� The cost of electricity corresponds to: �� = ��∙� �� ∙ 1 3600 = 0.588 €/� The cost of the consumables can be determined as: �� = (�� �� + �� �� )∙1 �� ∙ 1 3600 = 0.455 €/� The overall cost corresponds to: ���� = �� + �� + �� + �� = 3.69 €/� C. The cost per part can be calculated as: ����� = ����� ∙���� = € 5.34 The minimum selling price of the batch to cover the production costs is: ����� ℎ= ����� ∙����� = € 694