Chapter 2 Thermal Analysis of Friction Welding

Transcription

1 Chater 2 Thermal Analysis of Friction Welding Abstract Thermal energy is generated during the friction welding rocess. In this case, the solid surfaces rub against each other and heat is generated as a result of friction. Since the rocess is trasient and involves with axis-symmetric heating situation, formulation of the thermal analysis becomes essential. In this chater, thermal analysis based on Fourier heat conduction is introduced and the solution of conduction equation is obtained for aroriate boundary conditions. Keywords Friction welding Heating analysis Temerature 2.1 Introduction Welding of solid materials is achieved by roviding thermal energy in the form of heat for melting or softening the interface between the two materials and bringing or ressing them together. Friction is one of the methods of generating the required thermal energy for welding rocess. As the solid surfaces rub against each other heat is generated as a result of friction. The heat generated due to friction subsequently diffuses through the bulk of the contacting solid materials. As the heat is necessary for obtaining sound welds, it also affects the mechanical as well as the microstructural roerties of the welded materials in the vicinity of the welding interface. Thermal analysis of friction welding is carried out to determine the resulting temerature distribution around the welding interface and thus allows determination of the high temerature effects on the micro-structure of the materials as well as the quality of the weld. The thermal analysis related to the friction welding is carried out in line with the revious studies [1 12]. As an elementary examle, consider two solid objects with flat surfaces ressed together with a force F and sliding against each other with a relative velocity of V. The ower consumed against the frictional force F f = lf is converted to thermal energy generation at the interface. The rate of thermal energy generation is given by _Q ¼ F f V ¼ lfv ð2:1þ B. S. Yilbas and A. Z. Sahin, Friction Welding, SringerBriefs in Manufacturing and Surface Engineering, DOI: / _2, Ó The Author(s)

2 6 2 Thermal Analysis of Friction Welding where l is the coefficient of friction. If the contact area is A then the rate of thermal energy generation er unit area becomes q ¼ _ Q A ¼ lfv A : ð2:2þ Examle 1 A solid block of 100 kg mass slides along a horizontal concrete avement with a seed of 2 m/s. The coefficient of friction between the block and the avement is estimated to be 0.4. Determine the rate of thermal energy generation in W. Solution: The force F acting on the avement due to the weight of the block is F ¼ mg ¼ 100 9:81 ¼ 981 N Therefore, the thermal energy generation as a result of friction between the solid block and the avement is obtained by using Eq. (2.1) _Q ¼ lfv ¼ 0: ¼ 784:8 W The thermal energy generated at the interface diffuses through the solid objects. The amount of thermal energy diffusion in each of the solids deends on the thermal conductivity of the solids. The temerature distribution in solids can be determined by solving the heat conduction equation. If the thermo-hysical roerties are assumed constant and no hase change (i.e. no melting of solids) is considered the heat conduction equation is given by ot ot ¼ ar2 T ð2:3þ where a ¼ k=qc P is the thermal diffusivity, k is the thermal conductivity, q is the density and C P is the secific heat. Solution of Eq. (2.3) with aroriate initial and boundary conditions yields the transient temerature distribution inside the solids. 2.2 Infinite Medium Instantaneous Release of Heat One of the fundamental solutions of the heat conduction equation in relation to the welding rocess is for the instantaneous oint source in an infinite medium (Fig. 2.1). In this case the heat liberated at a oint diffuses in all directions in the medium. The solution for the transient temerature distribution in this case is given by

3 2.2 Infinite Medium 7 Fig. 2.1 Coordinate system in an infinite medium z 0 x y q r2 Tðr; tþ ¼T i þ ex ð2:4þ 3=2 qc P ð4atþ 4at where q (J) is the amount of heat suddenly resealed at time t = 0 at the origin (r = 0), the radial coordinate is r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 þ z 2 and the initial temerature is T i. This solution is useful to study the thermal exlosion that occurs in a homogeneous medium and the resulting temerature distribution in the medium as function of time and distance. Figure 2.2 shows the temerature distribution in a large steel samle as a result of instantaneous heat release of 1000 J of thermal energy at origin at different times after the initial release of the energy. The initial temerature of the steel material is 300 K. The density, the secific heat and the thermal diffusivity of the steel samle are 7800 kg/m 3, 473 J/kg K and m 2 /s, resectively. Each of the curves reresent the temerature variation in the material at times 0.25, 0.5, 0.75, 1.0, and 1.25 s after the thermal energy release. As a result of heat diffusion in the material the temerature rofiles are flattened in an exonential manner as given in Eq. (2.4). Examle 2 1 kj of oint energy is released in a large block of steel at time = 0 that is initially at 300 K. Determine the temerature at a location 5 mm away from the location of the heat release after 2 s. Solution: Using Eq. (2.4) and substituting q = 1000 J, r = m and t = 2s 1000 Tð0:005; 2Þ ¼300 þ ð4 1: Þ ex 0: =2 4 1: ¼ 341 K If the instantaneous heat release occurs along the z-axis rather than a oint, then the solution of the heat conduction equation yields q 0 r2 Tðr; tþ ¼T i þ ex ð2:5þ qc P ð4atþ 4at where q 0 (J/m) is the amount of heat released instantaneously along the z-axis er unit length at time t = 0 and r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 is the radial distance from the z-axis.

4 8 2 Thermal Analysis of Friction Welding Fig. 2.2 The temerature distribution in a large size steel material after an instantaneous thermal energy release of 1 kj at the origin. (t = 0.25, 0.5, 0.75, 1.0 and 1.25 s) Fig. 2.3 The temerature distribution in a large size steel material after a thermal energy release of 100 kj/m along the z-axis at time = 0. (t = 0.25, 0.5, 0.75, 1.0 and 1.25 s) Figure 2.3 shows the temerature rofiles in a large size steel material after an instantaneous release of 100 kj/m of heat along the z-axis at time t = 0. Each of the curves in Fig. 2.3 reresents the temerature rofile at different times after the release of the heat, namely, at t = 0.25, 0.5, 0.75, 1.0, and 1.25 s. The temerature on the z axis at t = 0.25 s is more than 1000 K and it decreases sharly as a result of heat diffusion. Similarly, if the instantaneous heat release occurs on the entire y-z lane, then the heat diffuses along the x direction. Thus the solution of the heat conduction equation yields a transient one-dimensional temerature distribution in the form q 00 x2 Tðx; tþ ¼T i þ ex ð2:6þ 1=2 qc P ð4atþ 4at where q 00 (J/m 2 ) is the amount of heat released instantaneously over the y-z lane er unit area at time t = 0. Figure 2.4 shows the temerature rofiles in this case

5 2.2 Infinite Medium 9 Fig. 2.4 The temerature distribution in a large size steel material after a thermal energy release of 10 MJ/m 2 on the y-z lane at time = 0. (t = 0.25, 0.5, 0.75, 1.0 and 1.25 s) for the instantaneous heat release of 10 MJ/m 2 on the y-z lane at time t = 0. The material considered in this case is also steel with the same thermohysical characteristics mentioned above. The temerature rofiles show similar behavior to the cases mentioned above, however, the eak values of temerature are different. This is because the instantaneous heat release takes lace in a larger area and the diffusion of heat occurs in a larger volume. In this case the change (or decrease) of eak temerature during the time eriod t = 0.25 to 1.25 s is around 120 K. The fundamental solutions given by the above three equations rovide information on the diffusion of heat in the solid medium with no boundaries. However, they do not adequately describe the thermal energy diffusing during the welding rocess. This is because the thermal energy generation in a tyical welding rocess is not sontaneous and the domain is normally finite Continuous Release of Heat Now, consider a steady (continuous) case of thermal energy generation in an infinite medium. In this case there is no transient term in the governing heat conduction equation and therefore Eq. (2.3) simlifies to r 2 T ¼ 0 ð2:7þ If the thermal energy generation occurs at a rate _q (W) at the origin (r = 0) the temerature distribution in the medium is obtained to be _q Tðr; tþ ¼T i þ qc P ð4arþ erfc r 2 ffiffiffiffi ð2:8þ at where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 þ z 2. It should be noted that as t!1the term erfc (0) = 1 and therefore the steady-state temerature distribution in the medium is obtained to be

6 10 2 Thermal Analysis of Friction Welding Fig. 2.5 The radial temerature distribution at different times in a steel material for a continuous release of 100 W rate of heat at the origin (t = 10-4,10-3, 10-2,10-1 and 1 s) _q TðrÞ ¼T i þ qc P ð4arþ : ð2:9þ Figure 2.5 shows the temerature distribution in a steel material for the case of a continuous thermal energy release of 100 W at the origin. The steady temerature distribution in the material varies inversely with the radial distance as given in Eq. (2.9). Examle 3 Consider the oint energy released in the large steel material as given in Examle 1 to be continuous, i.e. 1 kw. Determine the temerature at a location 5 mm away from the location of the heat release after 2 s. Solution: In this case the solution is given by Eq. (2.8) _q Tðr; tþ ¼T i þ qc P ð4arþ erfc r 2 ffiffiffiffi at 1000 Tð0:005; 2Þ ¼300 þ ð4 1: Þ erfc 0:005 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1: ¼ 317 K The temerature in this case is lower when comared with the solution given in Examle 2. The reason for this is because the heat release in the current case is not instantaneous and therefore it is slower that the case in Examle 2. Therefore, the temerature of the material around the sot where the heat is released continues to increase and aroaches a steady value. This value for t? infinity can be shown to be K. For the case of instantaneous heat release, however, the temerature at the given location increases initially and then decreases as the wave of heat asses through that oint.

7 2.2 Infinite Medium 11 Fig. 2.6 The radial temerature distribution at different times in a steel material for a continuous release of 10 kw/m rate of heat along the z-axis (t = 0.01, 0.1, 1, 10 and 100 s) In case of continuous heat release along the z-axis at a rate of _q 0 (W/m) the temerature distribution in the infinite medium becomes _q 0 Tðr; tþ ¼T i þ 4qC P a Ei r 2 ð2:10þ 4at where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2. Since for t!1 the term Eið0Þ!1 there is no steady temerature distribution in this case. Figure 2.6 shows the temerature rofiles at various times (t = 0.01, 0.1, 1, 10 and 100 s) in a steel material when a continuous release of 10 kw/m rate of heat takes lace along the z-axis. The heat diffuses through the bulk of the material and therefore the temerature in the material continues to increase. Examle 4 A high resistance electric wire is located in a large steel block releasing 10 kw/m heat energy. The initial temerature of the block is 25 C. Estimate the temerature 5 mm away from the wire after 1 min of heating. Solution: According to Eq. (2.10) the temerature is obtained after substituting the given information: Tð0:005; 60Þ ¼298 þ : Ei 0: : ¼ 374:5 K For the case of thermal energy generation over the y-z lane at a rate of _q 00 (W/ m 2 ) the temerature distribution in the infinite medium is given by " rffiffiffiffiffiffi # Tðx; tþ ¼T i þ _q00 4at ex x2 x xerfc 2k 4at 2 ffiffiffiffi ð2:11þ at

8 12 2 Thermal Analysis of Friction Welding Fig. 2.7 The temerature distribution at different times in a steel material for a continuous release of 10 MW/m 2 rate of over the y- z lane (t = 1, 2, 3, 4 and 5s) It is clear that there is no steady-state solution for this case either since the first ffiffi term in the equation is roortional to t. The temerature anywhere in the medium including that on the y-z lane increases continuously with time roortional to t. For x = 0 the temerature on the y-z lane is obtained to be ffiffi rffiffiffiffi Tð0; tþ ¼T i þ _q00 at ð2:12þ k Figure 2.7 shows the temerature distribution in the direction of x-axis for the case of continuous rate of heat release of 10 MW/m 2 on the y-z lane. Each of the curves in Fig. 2.7 reresents the temerature distribution at time t = 1, 2, 3, 4, and 5 s, resectively. The temerature in the bulk of the material increases in a continuous manner as a result of diffusion of heat away from the y-z lane. The ffiffi temerature increase at the y-z lane is roortional to t as shown in Fig Moving Sources of Heat In most ractical situations the heat source moves along the medium. Therefore, for accurate determination of the temerature distribution in the medium, moving heat sources must be considered. When the heat source or the medium moves along the x-direction the governing equation for conduction heat transfer becomes V ot ox ¼ ar2 T ð2:13þ

9 2.2 Infinite Medium 13 Fig. 2.8 The temerature variation with time on the y-z lane in the steel material for a continuous release of 100 kw/m 2 rate of over the y-z lane Fig. 2.9 Temerature distribution in a steel material along the x-axis subjected to the moving oint source of 100 W along the x-axis with a seed of 0.01 m/s Moving Point-Source Now, consider a steadily moving oint heat source in the x-direction with a seed of V in an infinite medium. The temerature distribution on the coordinates moving with the oint source is named the quasi-steady state solution. This solution can be exressed as _q Vðr þ xþ Tðx; y; zþ ¼T i þ ex ð2:14þ qc P ð4arþ 2a where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 þ z 2. It should be noted that when the velocity goes to zero the solution aroaches the steady-state solution for the case of continuous oint source given in Eq. (2.9). Figure 2.9 shows the temerature distribution along the x-axis in a steel material in which a moving oint release of heat at a rate of 100 W occurs along the x-axis. The seed of moving oint heat source along the x- direction is 0.01 m/s.

10 14 2 Thermal Analysis of Friction Welding Examle 5 A oint source of 100 W moves in a large steel material with a seed of 10 mm/s. The temerature of the material far away from the source of heat release is 300 K. Determine the temerature 2 mm behind and 2 mm in front of the moving front. Solution: Equation (2.14) can be written in the one-dimensional form as: _q Vðjjþ x xþ TðxÞ ¼T i þ ex qc P ð4ajj x Þ 2a 2 mm in front: 100 0:01 Tð0:002Þ ¼300 þ ð4 1: ex ð 0:002 þ 0:002 Þ 0:002Þ 2 1: ¼ 316:8 K 2 mm behind: 100 0:01 Tð 0:002Þ ¼300 þ ð4 1: ex ð 0:002 0:002 Þ 0:002Þ 2 1: ¼ 392:5 K Moving Line-Source If the moving source is a line heat source along the z direction and moving along the x-axis the quasi-steady state solution for the temerature on the moving coordinate system is obtained to be Tðx; yþ ¼T i þ _q0 Vx ex 2k 2a K 0 Vr 2a ð2:15þ Figure 2.10 shows the temerature distribution in the steel material along the x- axis when a line source of heat of 100 kw/m (along the z-axis) moves in the direction of x-axis. The seed of the line heat source is taken as 0.01 m/s Moving Plane-Source Finally when the heat source is uniformly distributed on the y-z lane that is moving along the x-axis with a velocity V the quasi steady state solution of the temerature on the moving coordinate axis becomes _q00 TðxÞ ¼T i þ qc P V Vx ex a 1 þ sgnðxþ 2 ð2:16þ

11 2.2 Infinite Medium 15 Fig Temerature distribution in a steel material along the x-axis subjected to the moving line source of 100 kw/m on the z-axis with a seed of 0.01 m/s along the x-axis Fig Temerature variation along the x- direction in a steel material subjected to a lane moving source in the direction of x- axis where sgn(x) is the Signum function which is defined as 8 sgnðxþ ¼ x < 1 if x\0 jj x ¼ 0 if x ¼ 0 : þ1 if x [ 0 ð2:17þ Figure 2.11 shows the variation of temerature in the direction of x-axis when a lane heat source of 10 MW/m 2 (on the y-z lane) moves along the x direction with a velocity of 0.01 m/s. On the other hand the solution on the stationary coordinates yields the transient temerature on the stationary lane at x = 0as _q00 Tð0; tþ ¼T i þ 2qC P V " r ffiffiffiffiffiffiffi! # 1 ex V2 t V erfc 2 t a a ð2:18þ

12 16 2 Thermal Analysis of Friction Welding In this case the lane heat source moves away from the origin in the direction of x and the temerature at the stationary y-z lane decreases with time. The solutions for the infinite medium resented above are useful in analyzing local temerature variations far away from the boundaries of the hysical medium. They also describe the early stages of the transient behavior in the finite domain before the time the heat diffusion reaches the boundaries of the hysical medium. These solutions rovide an insight of the exected temerature history during the actual welding rocess. However, in the actual welding rocess the intensity of the heat generation is high and there may be sufficient time for the diffusion of heat to reach the hysical boundaries of the medium. Therefore, the finite size medium with its boundaries must be considered in order to make more accurate redictions for the temerature distribution in most of the welding rocesses. The first ste to consider the hysical boundaries is to study the semi-infinite medium that is considered in the following section. 2.3 Semi-Infinite Medium When the surface of the semi-infinite medium is insulated, in other words, when the heat transfer from the surface is neglected the temerature distribution in the domain and its mirror image across the surface of the semi-infinite medium (Fig. 2.12) matches the solution of the temerature distribution in the infinite medium. Therefore, the solutions for the semi-infinite medium with insulated boundary condition can easily be obtained by using the solutions for the infinite medium. All the heat released on the insulated surface of the semi-infinite medium will have to diffuse towards the deth of the semi-infinite medium as oosed to the infinite medium where the heat released diffuses in all directions Instantaneous Point Heat Release q (J) on the Surface The solution for the temerature distribution in the case of instantaneous oint heat release q (J) on the surface of the semi-infinite medium follows Eq. (2.4), i.e. 2q r2 Tðr; tþ ¼T i þ ex ð2:19þ 3=2 qc P ð4atþ 4at where q (J) is the amount of heat suddenly resealed at time t = 0 at the origin (r = 0), the radial coordinate is r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 þ z 2, the initial temerature is T i, and the factor 2 in front of the second term on the right hand side is due to the fact that all the heat released has to diffuse only towards the deth of the semi-infinite medium (z [ 0) and no heat loss is assumed from the surface. Figure 2.13 shows

13 2.3 Semi-Infinite Medium 17 Fig Coordinate system in semi-infinite medium y 0 x z Fig Temerature variation with resect to time at a distance 5 mm away from the origin where an instantaneous heat release of 100 J takes lace the temoral variation of temerature in a steel material 5 mm away from the origin where an instantaneous heat release of 100 J occurs at time t = 0. As a result of heat diffusion the temerature at r = 5 mm initially rises and reaches at a eak value at around t = 0.5 s and then decreases afterwards. Examle 6 1 kj of oint energy is released on the surface of a large semi infinite block of steel at time = 0 that is initially at 300 K. Determine the temerature inside the steel 5 mm below the location of the heat release after 2 s. Solution: Equation (2.19) can be written in the z direction as: 2q z2 Tðz; tþ ¼T i þ ex 3=2 qc P ð4atþ 4at Therefore, for z = 5 mm the temerature is Tð0:005; 2Þ ¼300 þ ð4 1: Þ ex 0: =2 4 1: ¼ 382 K

14 18 2 Thermal Analysis of Friction Welding Fig Temerature variation wit time at the origin after 100 J instantaneous heat release at the origin It can also be shown that the temerature at the sot where the heat is released is K after the time the heat is released. The temerature at the origin (r = 0) decreases with time (Fig. 2.14) as 2q Tð0; tþ ¼T i þ : ð2:20þ 3=2 qc P ð4atþ Instantaneous Line Heat Release q 0 (J/m) on the Surface For the case of the instantaneous line heat release along the y-axis direction the transient temerature distribution is obtained from Eq. (2.5) as 2q 0 r2 Tðr; tþ ¼T i þ ex ð2:21þ qc P ð4atþ 4at where q 0 (J/m) is the amount of heat released instantaneously along the y-axis er unit length at time t = 0, r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ z 2 is the radial distance from the y-axis and z [ 0. Figure 2.15 shows the temerature variation with time at a distance of 5 mm away from the y-axis after 100 kj/m heat release along the y-axis. The temerature initially rises sharly and after reaching a eak value it starts decreasing. The temerature along the y-axis (r = 0) decreases with time as Tð0; tþ ¼T i þ qc P ð2atþ : ð2:22þ In this case the temerature along the y-axis decreases inversely roortional with time after the sudden release of heat along the y-axis. q 0

15 2.3 Semi-Infinite Medium 19 Fig Temerature variation with time at a location 5 mm away from the y-axis after an instantaneous heat release of 100 kj/m along the y-axis Fig Temoral variation of temerature 5 mm below the surface of the semi-infinite steel material after an instantaneous and uniform heat release of 1 MJ/m 2 at the surface of the material Instantaneous Plane Heat Release q 00 (J/m 2 ) on the Surface If the instantaneous heat release occurs uniformly on the surface of the semiinfinite medium (z = 0 lane) the solution for the temerature distribution can be written from Eq. (2.6) as 2q 00 z2 Tðz; tþ ¼T i þ ex ð2:23þ 1=2 qc P ð4atþ 4at where q 00 (J/m 2 ) is the amount of heat released instantaneously over the x-y lane er unit area at time t = 0. The temerature variation with time at a deth of 5 mm from the surface is shown in Fig after an instantaneous heat release of 1 MJ/m 2 on

16 20 2 Thermal Analysis of Friction Welding the surface. As the heat is diffused through the deth of the steel material the temerature initially increases and then decreases after reaching a eak value. The temerature on the insulated surface of the semi-infinite medium (z = 0) decreases with time as Tð0; tþ ¼T i þ : ð2:24þ 1=2 qc P ðatþ In this case the temerature variation with time is inversely roortional with the square root of time. Examle 7 The surface of a semi infinite steel material is exosed to an instantaneous heat release of 1 MJ/m 2. Determine the maximum temerature at a location 5 mm below the surface and the time at which this occurs. Solution: The time at which a maximum temerature is reached at a given z location in the material can be obtained by differentiating Eq. (2.23) with resect to t and setting the result equal to zero, i.e. otðz; tþ ot ¼ ot ot q 00 " 2q 00 # z2 T i þ ex 1=2 qc P ð4atþ 4at Carrying out the differentiating the time for the maximum temerature is obtained as ¼ 0 Substituting the values we get t ¼ z2 2a : 0:005 2 t ¼ ¼ 1:067 s 2 1: Therefore the maximum temerature becomes Tð0:005; 1:067Þ ¼300 þ ð4 1: :067Þ ex 0: =2 4 1: :067 ¼ 326:2 K In most welding alications the heat release is continuous for a certain eriod of time rather than sontaneous. Therefore the case of continuous release of heat must be analyzed to describe the thermal behavior in such welding rocesses.

17 2.3 Semi-Infinite Medium 21 Fig Temerature variation in a semi infinite steel material at a location 5 cm away from the origin where a continuous heat release of 10 kw occurs Continuous Point Heat Release _q (W) on the Surface The most elementary examle in this case is the continuous oint heat release _q (W) on the insulated surface of the semi-infinite medium at the origin (r = 0). The temerature distribution in this case is obtained to be 2_q Tðr; tþ ¼T i þ qc P ð4arþ erfc r 2 ffiffiffiffi ð2:25þ at where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 þ z 2 and z [ 0. The temerature variation with time in a semi-infinite steel material at a location 5 cm away from the origin is shown in Fig In this case a continuous heat release of 10 kw occurs at the origin. The temerature increases in a continuous manner as the heat diffusion occurs. As t!1the steady-state temerature distribution in the medium becomes _q TðrÞ ¼T i þ qc P ð2arþ : ð2:26þ Figure 2.18 shows the steady temerature variation with r in a semi-infinite steel substance which is subjected to 1 kw continuous oint heat release at the origin. The temerature varies inversely roortional with the distance from the origin. In this case all the heat released at the origin diffuses through the deth of the material and a steady temerature variation is obtained. In other words, the temerature at any oint in the material starts increasing after the enetration of heat reaches at the oint as shown in Fig and then the temerature aroaches a constant steady value asymtotically. Examle 8 1 kw of continuous oint heat release starts taking lace on the surface of a large steel samle whose temerature before the heat release was 300 K.

18 22 2 Thermal Analysis of Friction Welding Fig Steady temerature variation in a semi-infinite steel substance subjected to 1 kw of continuous oint heat release at the origin on the surface Determine the maximum temerature at a oint 20 mm below the sot at which the heat release occurs. Solution: The maximum temerature is reached when t? infinity, i.e. the steady state situation. Therefore, using Eq. (2.26), we have _q TðzÞ ¼T i þ qc P ð2azþ or 1000 Tð0:02Þ ¼300 þ ð2 1: :02Þ ¼ 484 K Continuous Line Heat Release _q 0 (W/m) on the Surface In the case of continuous heat release _q 0 (W/m) along y-axis the temerature distribution in the semi-infinite medium is found to be Tðr; tþ ¼T i þ 2_q0 4qC P a Ei r 2 ð2:27þ 4at where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ z 2 and z [ 0. The temerature in the medium increases with time t but decreases with distance r. Figure 2.19 shows the temerature variation with time in a semi-infinite steel material at a distance of 5 cm from the y-axis

19 2.3 Semi-Infinite Medium 23 Fig Temerature variation with time in a semiinfinite steel medium at a distance of 5 cm from the y- axis where a continuous line heat release of 10 kw/m occurs where a continuous line heat release of 10 kw/m takes lace. The temerature starts increasing after the enetration time of about 10 s. The increase of temerature continues and no steady temerature is reached Uniform Heat Flux _q 00 (W/m 2 ) on the Surface For the case of uniform heat release _q 00 (W/m 2 ) on the insulated surface of the semi-infinite medium the temerature distribution along the deth z [ 0 of the medium is found to be Tðz; tþ ¼T i þ _q00 k " rffiffiffiffiffiffi # 4at ex z2 z z erfc 4at 2 ffiffiffiffi at ð2:28þ Figure 2.20 shows the temerature variation with time at a deth of 5 cm of a semi-infinite steel medium subjected to 1 MW/m 2 uniform surface heat flux. The temerature at this deth starts increasing after the enetration time and the increase of temerature takes lace continuously. Steady-state solution does not ffiffi exist since the first term in the bracket is roortional to t. The temerature on ffiffi the surface of the semi-infinite medium z = 0 also varies with t as Tð0; tþ ¼T i þ 2_q00 k rffiffiffiffi at : ð2:29þ Examle 9 The surface of a large steel block which is at 300 K is suddenly exosed to 1 MW/m 2 heat flux. Determine the time needed for the surface temerature to reach 1500 K.

20 24 2 Thermal Analysis of Friction Welding Fig Temerature rise inside a semi infinite steel medium at a deth of 5 cm that is subjected to a continuous heat flux of 1 MW/m 2 at the surface Solution: Solving t from Eq. (2.29) t ¼ a k T T 2 i 2_q 00 ¼ Tð0; tþ ¼T i þ 2_q00 k 1: rffiffiffiffi at ¼ 178:4 s Continuous Stri (Along y-axis) Heat Release on the Surface of Semi-Infinite Medium When the heat release _q 00 (W/m 2 ) occurs continuously over the long stri of 2b width on the surface of the semi-infinite medium the initial and surface boundary condition can be exressed as T ¼ T i at t ¼ 0 k ot oz ¼ q_ 00 z ¼ 0; b\x\ þ b ð2:30þ 0 z ¼ 0; x [ b; x\ b In this case the transient temerature variation over the surface of the semi-infinite medium at z = 0 is given by

21 2.3 Semi-Infinite Medium 25 Fig Temerature variation with time on the surface at x = 5cm 8 1 þ x q Tð0; x; tþ ¼T i þ _ ffiffiffiffiffiffiffi erf 00 b Fo 2 ffiffiffiffiffiffiffi 1 x þ erf Fo b 2 ffiffiffiffiffiffiffi 1 þ x " 2 Fo b 2 ffiffiffiffiffiffiffi Fo Ei 1 þ x # 2 9 >< b 2 ffiffiffiffiffiffiffi Fo b >= b ffiffiffi k 1 x " 2 2 ffiffiffiffiffiffiffi Fo Ei 1 x # 2 >: b 2 ffiffiffiffiffiffiffi Fo >; b ð2:31þ where Fo b ¼ at b 2. Figure 2.21 shows the temerature variation on the surface at x = 5 cm when a continuous stri heat release of 1 MW/m 2 takes lace on the stri of 10 cm width Continuous Circular Disk Area (of Radius R) Heat Release on the Surface of Semi-Infinite Medium If the continuous heat release _q 00 (W/m 2 ) occurs uniformly over the circular area of radius R over the surface of the semi-infinite medium the initial and boundary conditions are given by T ¼ T i at t ¼ 0 k ot oz ¼ q _ 00 z ¼ 0; 0\r\R ð2:32þ 0 z ¼ 0; r [ R In this case the transient temerature variation along the z-axis going through the center (r = 0) of the circular disk area is found to be 8 0rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 19 2 q Tð0; z; tþ ¼T i þ 2 _ 00 R ffiffiffiffiffiffiffiffi >< 1 þ 1 R Fo R ierfc k 2 ffiffiffiffiffiffiffiffi z ierfcb R 2 ffiffiffiffiffiffiffiffi >= C Fo A ð2:33þ >: R >;

22 26 2 Thermal Analysis of Friction Welding Fig Temerature variation with time in a semiinfinite steel substance at a deth of 10 cm below the origin for the case of continuous circular disk heat release of 10 MW/m 2 on the surface with the radius of disk 10 cm where Fo R ¼ at R and r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2. Figure 2.22 shows the temerature variation 2 with time inside the semi-infinite steel material 10 cm below the origin that is located on the surface. The continuous disk area heat release is considered to be 10 MW/m 2 on the disk of radius 10 cm. As can be seen from the figure, the temerature starts increasing after a enetration time of about 45 s. The increase of temerature is continuous and no steady temerature is exected since the temerature variation is roortional to the square root of time as can be seen from Eq. (2.33). Examle 10 The flat surface of a large steel samle initially at 300 K is subjected to a circular disk heat release of 10 MW/m 2 with the radius of the disk being 10 cm. Determine the temerature 10 cm below the center of the disk after 2 min. Solution: The Fourier number is Fo R ¼ at R 2 ¼ 1: :1 2 ¼ 0:14 Therefore the temerature is obtained using Eq. (2.33) as 8 0qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 19 Tð0; 0:1; 120Þ ¼300 þ :1 ffiffiffiffiffiffiffiffiffi< 1 0:14 ierfc 43 2 ffiffiffiffiffiffiffiffiffi 1 þ 0:1 2 = 0:1 ierfc@ 0:14 2 ffiffiffiffiffiffiffiffiffi A : 0:14 ; ¼ 554 K

23 2.3 Semi-Infinite Medium 27 Fig Temerature distribution along x direction on the surface of steel substance with a moving oint heat source of 100 W along the x-axis Moving Point-Source on the Surface If the heat source is not stationary but is moving along the x-axis with a velocity of V then the quasi-steady state solution on the moving coordinate system is obtained as 2_q Vðr þ xþ Tðx; y; zþ ¼T i þ ex ð2:34þ qc P ð4arþ 2a where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 þ z 2 and z [ 0. Figure 2.23 shows the temerature distribution along the x-axis on the surface of a semi-infinite steel material with a moving oint heat source of 100 W along the x direction. The seed of the moving heat source is taken as 10 mm/s. As the velocity tends to zero (V = 0) the temerature distribution becomes _q Tðx; y; zþ ¼T i þ ð2:35þ qc P ð2arþ as shown in Fig The temerature distribution becomes symmetrical with resect to the origin and decreases with r in an inversely roortional manner. Examle 11 A continuous oint heat source of 100 W moves along the surface of a large steel samle with a velocity of 10 mm/s. The temerature of the samle far away from the heat source is 300 K. Determine the temerature on the surface 5 mm behind and 5 mm in front of the heat source location. Solution: Equation (2.34) can be written in one dimensional form as 2_q Vðjjþ x xþ TðxÞ ¼T i þ ex qc P ð4ajj x Þ 2a

24 28 2 Thermal Analysis of Friction Welding Fig Temerature distribution along the radial direction when the velocity of the moving heat source is zero Therefore, the temerature 5 mm behind the heat source is :01 ð0:005 0:005Þ Tð 0:005Þ ¼300 þ ð4 1: ex 0:005Þ 2 1: ¼ 373:6 K and the temerature 5 mm in front of the heat source is :01 ð0:005 þ 0:005Þ Tð0:005Þ ¼300 þ ð4 1: ex 0:005Þ 2 1: ¼ 301 K Note that the temerature in front of the heat source is very low since the enetration of heat has not yet occurred at that location. It can be shown using Eq. (2.35) that when the velocity of the heat source is set to zero, then the temeratures at both locations become K Moving Line-Source on the Surface In the case of moving y-axis line source _q 0 (W/m) in the direction of x-axis with a velocity of V the quasi-steady state temerature variation on the moving coordinate system in the semi-infinite medium is obtained as Tðx; zþ ¼T i þ _q0 Vx ex k 2a Vr 2a K 0 ð2:36þ where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ z 2 and z [ 0. Figure 2.25 shows the temerature distribution on the surface of the semi-infinite steel material along the x-axis. The y-axis line heat

25 2.3 Semi-Infinite Medium 29 Fig Temerature distribution on the surface of a semi-infinite steel material along the x-axis subjected to a moving line heat source of 10 kw/m in the x direction source is 10 kw/m and moves with a seed of 10 mm/s along the x-axis. Temerature gradient is higher on the right side because of the travel of the heat source in that direction Moving Infinite y-stri Source (in x Direction) on the Surface of a Semi-Infinite Solid If the infinite y-axis line source _q 00 (W/m 2 ) acts uniformly on a stri of finite width 2b and moves in the direction of x-axis with a velocity of V, then the initial and boundary conditions can be written as T ¼ T i at t ¼ 0 k ot oz ¼ q_ 00 z ¼ 0; b\x\ þ b ð2:37þ 0 z ¼ 0; x [ b; x\ b The quasi-steady state temerature distribution in the semi-infinite medium on the moving coordinate system is given by Z Tðx; zþ ¼T i þ _q00 a 2 XþB ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi exðkþk 0 Z kv 2 þ k 2 dk ð2:38þ X B where X ¼ Vx 2a, Z ¼ Vz Vb 2a, and B ¼ 2a. In all case considered above the convection heat transfer from the surface of the medium is neglected. The case of convective boundary condition is considered in the following.

26 30 2 Thermal Analysis of Friction Welding Moving Infinite y-stri Source (in x Direction) on the Surface of a Semi-Infinite Solid with Convection Boundary If the convection heat transfer is taken into consideration in the above case, the initial and boundary conditions can be written as T ¼ T 1 at t ¼ 0 k ot oz ¼ _q 00 z ¼ 0; b\x\ þ b hðt 1 TÞ z ¼ 0; x [ b; x\ b ð2:39þ T ¼ T 1 at x!1; z!1 The quasi-steady state solution for the temerature distribution on the moving coordinate axis in this case is obtained to be Tðx; zþ ¼T 1 þ 2a_q00 kv 8 Z XþB ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 exð kþk 0 Z 2 þ k 2 dk X B >< 2 3 X þ B >= Z 1 erf þ s H exðhzþ s exðh 2 s 2 Z 2s Þerfc 0 2s þ Hs X B 5 ds >: erf þ s >; 2s ð2:40þ where X ¼ Vx 2a Z ¼ Vz 2a B ¼ Vb 2a H ¼ 2ah kv When the convection heat transfer coefficient is small, i.e. h? 0 the solution becomes as exected. Tðx; zþ ¼T i þ _q00 a kv 2 Z XþB X B ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi exðkþk 0 Z 2 þ k 2 dk ð2:41þ 2.4 Slab (Plate) of Thickness L When the medium has a finite thickness L rather than being semi-infinite the boundary conditions on both sides of the domain affect the solution (Fig. 2.26). For the case of continuous uniform heat flux _q 00 (W/m 2 ) on one surface for t [ 0 and constant temerature on the other surface the initial and boundary conditions are given as T ¼ T i at t ¼ 0

27 2.4 Slab (Plate) of Thickness L 31 Fig Schematic view of late geometry y 0 x z k ot oz ¼ q00 z ¼ 0 ð2:42þ T ¼ T i z ¼ L The solution for the transient temerature distribution in the late satisfying the above conditions can be obtained as " Tðz; tþ ¼T i þ _q00 kl 1 z L þ 8 X 1 ð 1Þ n!# ð2n þ 1ÞðL zþ sin ex ð2n þ 1Þ2 2 at 2 ð2n þ 1Þ 2L 4L 2 n¼0 n¼0 ð2:43þ As time goes to infinity the steady state solution is reached. The above solution yields the steady state solution as h TðzÞ ¼T i þ _q00 kl 1 z i ð2:44þ L On the other hand the temerature variation on the surface z = 0 which is subjected to uniform and continuous heat flux is obtained to be "!# Tð0; tþ ¼T i þ _q00 kl 1 þ 8 X 1 1 ð2n þ 1Þ 2 ex ð2n þ 1Þ2 2 at 4L 2 ð2:45þ Moving Point Source _q (W) on the Surface of an Insulated Infinite Plate Now consider an infinite late of thickness L that is insulated on both sides is subjected to a moving oint source _q (W) in direction of x-axis with a velocity of V on one side of the medium. The quasi-steady state solution for the temerature on the moving coordinate axis is given by

28 32 2 Thermal Analysis of Friction Welding 8 Tðx; y; zþ ¼T i þ _q < 1 kl 2 K Vr 0 þ 1 : 2a where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2. X 1 K 0 n¼1 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3 Vr 1 þ 2an 2 4 2a VL 5 cos nz L 9 ex Vx = 2a ; ð2:46þ Moving Line Source on the Surface of an Insulated Infinite Plate In the case of the infinite y-axis line source moving along the x-axis with a velocity of V the quasi-steady state solution for the temerature distribution on the moving coordinate system is 8 Tðx; zþ ¼T i þ _q0 a >< VkL 1 þ 2 X1 >: n¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 15 cos nz 1 þ 2an 2 L VL ex Vx 2a 9 >= >; ð2:47þ The heat loss through convection from the surfaces of the late is neglected in the above solution. 2.5 Thin Slab (Sheet) When the thickness of the late is negligible the variation of the temerature across the thickness of the late is neglected. In this case thin slab assumtion is used Sot Welding Sot welding can be characterized as the instantaneous line heat release q across the thin sheets of total thickness d to be welded between the two electrodes. The transient temerature distribution in the sheets is given as Tðr; tþ ¼T i þ ðq=dþ r2 ex ð2:48þ 4kt 4at where r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 :

29 2.5 Thin Slab (Sheet) 33 Fig Temerature rofiles in the 2 mm steel sheet after a sot welding at various times Figure 2.27 shows the temerature distribution in a 2 mm thick steel sheet after a sot welding oeration in which 100 J heat release across the thickness of the sheet at the origin. The eak temerature at the weld sot decreases sharly as a result of heat diffusion that takes lace in the radial direction in the sheet. Examle 12 In the sot welding oeration on a 2 mm steel sheet an instantaneous heat release of 1 kj occurs. Determine the maximum temerature at a location of 5 mm away from the sot and at what time this maximum temerature is reached. Solution: The time at which a maximum temerature is reached at a given r location in the sheet material can be obtained by differentiating Eq. (2.48) with resect to t and setting the result equal to zero, i.e. Tðr; tþ ¼T i þ ðq=dþ r2 ex 4kt 4at otðr; tþ ot ¼ ot ot T i þ q=d r2 ex ¼ 0 4at 4at Carrying out the differentiating the time for the maximum temerature is obtained as Substituting the values we get t ¼ r2 4a : t ¼ 0:005 2 ¼ 0:53 s 4 1:

30 34 2 Thermal Analysis of Friction Welding Fig Temerature versus time at the sot of a steel sheet of 2 mm thickness after a heat release of 100 W at the sot Therefore the maximum temerature 5 mm away from the sot becomes Tð0:005; 0:53Þ ¼300 þ 1000=0: :53 ex 0: : :53 ¼ 938 K Temerature at the sot varies with time as Tð0; tþ ¼T i þ ðq=dþ 4kt : ð2:49þ The temerature decreases inversely roortional with time as shown in Fig Moving Line Heat Release q Across the Thin Sheets of Total Thickness d to Be Welded Between the Two Electrodes Along x-direction In the case of the line heat release across the thin sheets is continuous and moves along x-axis with a velocity V the quasi-steady state solution on moving coordinates is given as Tðx; yþ ¼T i þ ðq=dþ Vx ex 2k 2a K 0 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 V 2 4 þ h 1 þ h 2 5 ð2:50þ 2a kd where h 1 and h 2 are the convection heat transfer coefficients on both each of the sides of the sheets. Figure 2.29 shows the temerature distribution in a 2 mm steel sheet. The line heat release is considered to be 100 W/m and the velocity of the

31 2.5 Thin Slab (Sheet) 35 Fig Temerature rofile in the 2 mm steel sheet subjected to a line heat release of 100 W/m and a seed of 10 mm/s in the direction of the x-axis line heat release is 10 mm/s in the direction of the x-axis. The convection heat transfer coefficients h1 and h2 are both assumed to be 100 W/m 2 K. Examle 13 In a seam welding oeration of steel sheets that can be characterized by a moving line heat source of 1 kw with a seed of 10 mm/s in the direction of x-axis the total thickness of the sheets to be welded is 4 mm. Determine the temerature 2 mm behind and 2 mm in front of the heat source location during this welding rocess. Assume that the convection heat transfer coefficients on both the to and bottom surfaces are 100 W/m 2 K. Solution: Equation (2.50) can be written along the direction of x-axis (i.e. for y = 0) as TðxÞ ¼T i þ ðq=dþ Vx ex 2k 2a K 0 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3 V 2 4x þ h 1 þ h 2 5 2a kd So, the temerature at the location 2 mm behind the heat source is Tð 0:002Þ ¼300 þ ð1000=0:004þ 0:01 ð 0:002Þ ex : sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3 0:01 2 K þ 100 ð 0:002Þ 2 1: þ :004 ¼ 1428:5 K

32 36 2 Thermal Analysis of Friction Welding Similarly, the temerature 2 mm in front of the heat source becomes Tð0:002Þ ¼300 þ ð1000=0:004þ 0:01 ð0:002þ ex : sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3 0:01 2 K þ 100 ð0:002þ 2 1: þ :004 ¼ 504:8 K 2.6 Solid Rod Friction Welding of Long Rods For the case of moving lane source _q(w) (over the cross sectional area of the rod) in the axial direction (x) of an infinite rod the quasi-steady state solution for the temerature distribution on the moving coordinate system is given as TðxÞ ¼T 1 þ _q a ka U ex Vx jju x r where U ¼ 2a ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi V 2 þ 4a 2 h ka _q ðwþ ð2:51þ T 1 is the surrounding temerature. Figure 2.30 shows the temerature variation in the axial direction or a rod subjected to the moving lane source of 1 kw with a seed of 1 mm/s in the direction of the x-axis. The temerature in the back side of the moving from decreases slightly in the negative direction of the x-axis as a result of the convection cooling. The temerature on the right side of the moving front shows a stee temerature gradient since the heat diffusion is yet to reach this art as the moving front aroaches. For a secial case when the velocity is zero (V = 0) the temerature variation is obtained as rffiffiffiffiffi! _q TðxÞ ¼T 1 þ 2 ffiffiffiffiffiffiffiffiffiffi h ex jj x ð2:52þ hka ka This solution is useful when radius of the rod is small and the radial temerature variation in the rod is negligible. Figure 2.31 shows the temerature variation in a rod of 2 cm radius with a lane source of 1 kw at the origin erendicular to the axial direction. The temerature variation is symmetric around the origin. It decreases exonentially in both directions as a result of diffusion and convection. The convection heat transfer coefficient is considered to be 1000 W/m 2 K.

33 2.6 Solid Rod 37 Fig Temerature distribution in the steel rod of 2 mm radius subjected to 1 kw moving lane source with a velocity of 1 mm/s Fig The temerature variation in a rod of radius 2 cm and subjected to stationary lane source of 1kW Examle 14 In a friction welding rocess of cylindrical steel rods of diameter 4 cm a temerature of 1000 K is needed at the interface. The convection heat transfer coefficient is estimated to be 10 W/m 2 K. Determine the rate of heat needed for this rocess at the interface of the rods. If the equvalent relative radial seeds of rods at the interface is 10 m/s and the friction coefficion is assumed to be 0.2, determine the comression force needed at the interface to achieve the friction welding. Solution: At the interface (x = 0) Eq. (2.50) becomes _q Tð0Þ ¼T 1 þ 2 ffiffiffiffiffiffiffiffiffiffi hka

34 38 2 Thermal Analysis of Friction Welding Therefore the rate of heat needed is ffiffiffiffiffiffiffiffiffiffi _q ¼ 2 hka½tð0þ T 1 Š ¼ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 100 0: :02 2 ½ Š ¼3648 W: From Eq. (2.1) the comression force needed at the interface is F ¼ _q lv ¼ 3648 ¼ 1216 N: 0: Time Variable Heat Source in Rod In the actual friction welding rocess the source of heat is time-varying in the exonential form. When stationary coordinate system is used the formulation of the roblem in this case can be written as 1 ot a ot ¼ o2 T h ox2 ka ðt T 1Þþ _q000 o k exðatþdðxþ ð2:53þ where d(x) is Dirac delta function and _q 000 o (W/m3 ). The initial and boundary conditions are ot Tðx; 0Þ ¼T 1 ð1; tþ ¼0 ox ð2:54þ The last term on the right hand side considers the variation of the strength of the heat source in the exonential form. The solution for the transient temerature variation in the rod is given by 8 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi Tðx; tþ ¼T 1 þ _q ffiffi ex m 2 þ a x x erfc o a exðatþ 2kA 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 ffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 >< ðm 2 a þ aþt at >= m 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ a ex m 2 þ a x x erfc a 2 ffiffiffiffi þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi >: ðm 2 a þ aþt >; at ð2:55þ qffiffiffiffi h where m ¼ ka and _q o (W). For the steady heat release at the interface for t [ 0 the exonent a = 0 and the solution for the transient temerature distribution becomes Tðx; tþ ¼T 1 þ _q o 4 ffiffiffiffiffiffiffiffiffiffi hka ex½ mxšerfc x 2 ffiffiffiffi m ffiffiffiffi at at ex½mxšerfc x 2 ffiffiffiffi þ m ffiffiffiffi at at ð2:56þ The temerature rofiles in a transient heating of 10 cm diameter steel rod subjected to a steady lane heat source of 10 kw is shown in Fig The rofiles corresond to times 10, 20, 30, 40, and 50 s.

35 2.6 Solid Rod 39 Fig Temerature rofiles during the transient friction welding oeration of steel rods of 10 cm diameter subjected to 10 kw of lane heat source at the origin at different times from 10 to 50 s Fig The temerature rise at the origin in a 10 cm diameter rod during the transient heating due to steady lane heat source at the interface x = 0 The temerature at the interface x = 0 is obtained to be _q o Tð0; tþ ¼T 1 þ 2 ffiffiffiffiffiffiffiffiffiffi erf m ffiffiffiffi ffiffi at T1 þ _q000 o t ffiffiffiffiffiffiffiffiffiffiffiffiffi exð m 2 atþ ð2:57þ hka A qc k in which the aroximation erf ðxþ 2 ffiffi x exð x 2 Þ is used. The rise of temerature at the interface of the rod for the above case is shown in Fig Examle 15 Determine the time needed for the interface temerature to reach 1000 K in the friction welding rocess given in examle 14.

36 40 2 Thermal Analysis of Friction Welding Fig The steady temerature variation in a steel rod of 10 cm diameter subjected to a steady lane heat source of 10 kw at the interface x = 0 with e heat transfer coefficient 1 kw/m 2 K Solution: Equation (2.57) is _q o Tð0; tþ ¼T 1 þ 2 ffiffiffiffiffiffiffiffiffiffi erf m ffiffiffiffi _q o at ¼ T1 þ hka 2R ffiffiffiffiffiffiffiffiffiffi erf 2hkR ¼ 300 þ 2 0:02 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi erf :02 s ffiffiffiffiffiffiffiffiffiffiffiffiffi! 2h qc R t r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! :02 t Solving for t using trial and error method the time needed for the friction welding is obtained to be t = 29 s. On the other hand the steady state temerature distribution is obtained as time goes to infinity as _q o TðxÞ ¼T 1 þ 2 ffiffiffiffiffiffiffiffiffiffi exð mxþ: ð2:58þ hka Figure 2.34 shows the steady temerature variation in the steel rod of 10 cm diameter subjected to 10 kw lane heat source and heat transfer coefficient of 1 kw/m 2 K. Because of the convection heat transfer the temerature reaches a steady rofile. The eak temerature occurs at the interface and it decreases sharly in both sides reaching almost the ambient temerature 10 cm away from the interface.