Tugas 03_ Andriyansa.docx

September 19, 2017 | Author: Andri | Category: Welding, Heat Transfer, Heat, Materials Science, Crystalline Solids
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Tugas – 03 Welding Metallurgy Nama: Andriyansa NIM: 1506775071 2.1 In one welding experiment, 50-mm-thick steel plates were joined using electro-slag welding. The current and voltage were 480A and 34V, respectively. The heat losses to the water-cooled copper shoes and by radiation from the surface of the slag pool were 1275 and 375 cal/s, respectively. Calculate the heat source efficiency. Diketahui: Welding method Electro slag Base metal = 50 mm-thick steel plate Current = 480 A Voltage = 34 V Heat losses = 1275 cal/s to the water cooled copper shoes 375 cal/s radiation from the surface of the slag pool Heat source efficiency


Q base metal. t weld Qnominal .t weld ¿

( 480 .34 )−( ( 1275+375 ) . 4.184) 480 .34

= 0.577=57.7%

2.2 It has been reported that the heat source efficiency in electro-slag welding increases with increasing thickness of the work-piece. Explain why. Ketika volume weld metal meningkat hal itu akan membuat laju pendinginan menurun. Laju pendinginan menurun dapat membuat distribusi panas dibatasi pada area dan membuat heat input lebih fokus pada satu spot area. Meningkatkan ketebalan potongan pekerjaan akan meningkatkan ukuran weld pool, sehingga akan meningkatkan volume logam las. 2.3 (a) Consider the welding of 25.4-mm-thick steel plates. Do you prefer to apply Rosenthal’s two- or three-dimensional heat flow equation for fullpenetration electron beam welds? What about beadon-plate gas–tungsten arc welds? (b) Suppose you are interested in studying the solidification structure of the weld metal and you wish to calculate the temperature distribution in the weld pool. Do you expect Rosenthal’s equations to provide reliable thermal information in the pool? Why or why not? (c) In multi-pass welding do you expect a higher or lower cooling rate in the first pass than in the subsequent passes? Why? 2.3 (a) Untuk penetrasi penuh pada electron beam weld, lebih memilih menggunakan aplikasi perhitungan heat flow Rosenthal 3 dimensi. Karena

dapat memproduksi pengelasan dengan penetrasi penuh untuk ketebalan plat 25.4 mm. Sedangkan untuk GTAW, lebih memilih menggunakan perhitungan heat flow 2 dimensi, karena GTAW akan memproduksi pengelasan dengan penetrasi tidak penuh dan membuat ukuran HAZ yang besar. 2.3 (b) Perhitungan Rosenthal dapat memberikan informasi thermal untuk analisis struktur solidification. Karena struktur solidification membutuhkan informasi tentang temperatur awal dan laju pendingan pada bagian tersebut. Perhitungan Rosenthal akan memberikan informasi tentang temperatur pada satu tempat selama proses pengelasan. Informasi tersebut berarti temperatur awal dimana efektif pada proses pengelasan. Dengan variabel X dan R, dapat mendeskripsikan besaran T pada tempat R. Besarnya T memberikan informasi tentang kestabilan dari spot selama pengelasan pada melt atau solid. 2.3 (c) Laju pendinginan pada laju awal lebih besar dari laju berikutnya, karena temperatur awal dari bidang kerja masih mendapatkan pengaruh dari temperatur sekitar, sehingga terjadi perbedaan yang besar dari temperatur weld metal dan bidang kerja lebih kecil dari perbedaan semua laju awal, jadi laju pendinginan akan menurun. 2.4 Large aluminum sheets 1.6 mm thick are butt welded using GTAW with alternating current. The current, voltage, and welding speed are 100A, 10V, and 2mm/s, respectively. Calculate the peak temperatures at distance of 1.0 and 2.0 mm from the fusion boundary. Assume 50% arc efficiency. Diketahui: Welding method GTAW dengan AC Current = 100 A Voltage = 10 V Kec. Las = 2 mm/s Ketebalan = 1.6 mm Jarak = 1.0 mm dan 2.0 mm Effisiensi = 50% Peak Temperature?

1 4.13V Yg ρC 1 1 = + → Tp= +¿ Tp−T 0 Qxh Tm−T 0 4.13VYgρC 1 + Qx h Tm−¿


Tp ,1 mm=




4.13 x 0.002 x 0.001 x 2.7 . 106 1 + 100 x 10 x 50 933−298


−298=905.47 K

Tp ,2 mm=




4.13 x 0.002 x 0.002 x 2.7 . 10 1 + 100 x 10 x 50 933−298


−298=880.23 K

2.5 Bead-on-plate welding of a thick-section carbon steel is carried out using 200A, 20V, and 2mm/s. The preheat temperature and arc efficiency are 100°C and 60%, respectively. Calculate the cross sectional area of the weld bead. Diketahui Current = 200 A Voltage = 20 V Welding speed = 2 mm/s Preheat temperature = 1000C Arc effisiensi = 60% Rosenthal equation for two dimension:

2 π (T −¿) 2 x 3.14 x ( 1800−373 ) x 41.0 [ 0 ] Vx Vr =exp Ko → =e Q 2a 2a 200 x 20 x 60

[ ] [ ]

r = (x2+y2)1/2 x=0, y=0, r=0

 153.0933 g = 1  g = 0.006532 m = 0.257 in

2.6 (a) Do you expect to have difficulty in achieving steady-state heat flow during girth (or circumferential) welding of tubes by keeping constant heat input and welding speed? Explain why. What is the consequence of the difficulty? (b) Suggest two methods that help achieve steady-state heat flow during girth welding. 2.6(a) The fusion zone increased significantly in size as welding processed. This is because heat continued to build up during welding and the areas yet to be welded were preheated, even though both the heat input and the rotation speed were kept constant throughout the entire welding process. At the beganing of welding, there was no preheating and the weld bead was too small to have full penetration. 2.6(b) In the case of automatic girth welding, the welding current and hence the heat input can be preprogrammed through the use of progammable power source. Girth welding especially steady-state heat flow. The heat input per unit length of weld should be high at the beganing and reduced continuously as welding processed. 2.7 A cold-rolled AISI 1010 low-carbon steel sheet 0.6 mm thick was tested for surface reflectivity in CO2 laser beam welding under the following

different surface conditions: (a) as received; (b) oxidized in air furnace at 1000°C for 20 s; (c) oxidized in air furnace at 1000°C for 40 s; (d) covered with steel powder. In which order does the reflectivity rank in these surface conditions and why?

Source dari journal J.xie Reflexifity yang tertinggi dari as received dengan range 65-80%, adapun perbedaan reflectivity ini karena tingginya elektrik DC current resistivity pada permukaan lapisan oksida. Sedangkan reflectivity terendah range 20-35% karena reflection laser beam yang irregular oleh steel powder. 2.8 It was observed in YAG laser beam welding of AISI 409 stainless steel that under the same power the beam size affected the depth–width ratio of the resultant welds significantly. Describe and explain the effect. Efek ini terjadi karena efek melting ratio yang bertambah sehingga thermal conduction pada material dari weld zone bertambah pada nilai overlapping yang tertinggi. Dan akan mengurangi keefektifan sumber panas. Adanya input energy yang bertambah tidak digunakan untuk melelehkan material karena energi ini hilang melalui konduksi ke material tersebut. 2.9 Calculate the thermal cycle at the top surface of a very thick carbon steel plate at 5 mm away from the centerline of the weld surface. Power of the arc is 2kW, the arc efficiency 0.7, the travel speed 2mm/s, and the preheat temperature 100°C. Diketahui : Thickness = 5mm Power = 2000 W Arc Efficiency = 0.7 Speed = 2 mm/s Preheat = 373 K

Rosenthal’s equation for three dimension


2 π (T −¿) kR −V ( R−x ) =exp Q 2a






−V ( R−x ) Q exp +¿ 2 π kR 2a

x = V.t y= 5mm, z= 0, R = (25 x 10-6 + (Vt)2)1/2

Dengan grafiknya sebagai berikut,

Thermal cycle for carbon steel welding at 5 mm from bead

Temperature (K)

Time (second)

2.10 Is the transverse cross section of the weld pool at a fixed value of x perfectly round according to Rosenthal’s three-dimensional heat flow equation? Explain why or why not based on the equation. What does your answer tell you about the shape of the transverse cross section of a weld based on Rosenthal’s three-dimensional equation? Tidak, karena sesuai dengan persamaan; pada arah potongan melintang pada arah transversal pada kondisi dimana proses pengelasan terjadi secara isothermal, yang meliputi proses fusion boundary dan boundary luar dari daerah HAZ yang mengakibatkan bentuknya semicircular.

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