# Calculation of Some Integrals Arising in the Samara-Valencia Solution for Dry Flat Grinding

Authors: J. L. González-Santander
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

In the framework of the Samara-Valencia solution for heat transfer in grinding, two nontabulated integrals involving Macdonald’s function of zeroth order are calculated.


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## Body

## 1. Introduction

Grinding is a machining process that consists in the material removal from a workpiece by an abrasive wheel that rotates at a high speed over its surface [1] (see Figure 1). This kind of machining is very common in industry, so any improvement represents a significant reduction in production costs. In grinding, almost all the energy is converted into heat which is concentrated in the contact zone between the wheel and the workpiece [2]. Thermal damage occurs when temperature surpasses the phase transformation of the surface material. For over 50 years, we can find an extensive literature about thermal damage in grinding [3–9]. Usually, in order to prevent thermal damage, liquid coolant is injected onto the workpiece surface, so power generation by friction is reduced and cooling by convection occurs [10].Figure 1
Two-dimensional modelling of flat grinding.Traditionally, most of the studies of dry grinding temperature are based on Jaeger’s model [11, 12], where a heat source of constant strength sliding over the surface of a half-plane is considered. DesRuisseaux and Zerkle [13] extended Jaeger’s model to include the effect of surface cooling. Andrews et al. [14] investigate the well posedness of mathematical models for heat transfer in dry grinding. More recently, the Samara-Valencia solution [15] has been proposed. As in [16], this solution is assumed to be two-dimensional; see Figure 1. The differential equation that governs the heat transfer is the convective heat equation. Also, the heat flux profile entering the workpiece and the action of the coolant are considered in the boundary condition. In [15], this two-dimensional boundary-value problem is transformed into an integral equation that is useful for the numerical evaluation of the heat transfer in intermittent wet grinding [17]. However, in the case of dry grinding, this integral equation can be reduced to a two-dimensional integral [18]. The latter is called T0 theorem.This paper is written from the point of view in which an engineering problem provides a new mathematical result. In fact, this point of view is very classical. For instance, Euler was the first one to give a proof (although not rigorous) of the equality of mixed partial derivatives, when he was working in hydrodynamics [19]. Another example is the engineering problem of measuring the coastline, which initiates the study of fractals in mathematics [20]. In the framework of grinding, comparing Jaeger’s solution with the Samara-Valencia one, a new representation of the Dirac delta [21] and two new nontabulated integrals [22] have been obtained recently. The scope of this paper is just to provide another two nontabulated integrals by using T0 theorem.This paper is organized as follows. Section2 describes the Samara-Valencia solution and the T0 theorem. Section 3 derives from T0 theorem two integrals involving Macdonald’s function of zeroth order that seem not to be found in the usual literature [23–25].


## 2. Samara-Valencia Solution

Figure1 shows the setup assumed in the Samara-Valencia grinding model. The differential equation that governs the heat transfer is (1)∂Tt,x,y∂t=k∂2Tt,x,y∂x2+∂2Tt,x,y∂y2-vd∂Tt,x,y∂x,where T(t,x,y) is the rise temperature of the workpiece with respect to the room temperature, k the thermal diffusivity, and vd the workpiece velocity with respect to the grinding wheel. According to Figure 1, the Cartesian coordinates x and y are referred to as the grinding wheel. The initial condition is(2)T0,x,y=0,and the boundary condition is(3)k0∂Tt,x,0∂y=bt,xTt,x,0+dt,x,where k0 is the thermal conductivity of the workpiece, b(t,x) is the heat transfer coefficient over the surface of the workpiece, and d(t,x) is the heat flux entering the workpiece. Also -∞<x<∞ and t,y≥0. Physically speaking, b(t,x) takes into account the convection that occurs over the workpiece surface due to the wheel and the lubricant fluid (in wet grinding); meanwhile d(t,x) takes into account the heat that the wheel generates by friction over the workpiece surface. However, experiments suggest that the convective action of the wheel can be neglected, because, in dry grinding, Jaeger’s model (where b(t,x) is zero) is used to fit the temperature field inside the workpiece, considering a triangular heat flux profile for d(t,x) [26]. In any case, the heat transfer in grinding is a complex process, so the set of (1)–(3) provides just a model, that is, an approximation to reality.According to [15], the solution of the boundary-value problem (1)–(3) can be expressed as(4)Tt,x,y=T0t,x,y+T1t,x,y,where we have defined(5)T0t,x,y=-14πk0∫0texp⁡⁡-y24ksdss×∫-∞∞dt-s,x′exp⁡-x′-x-vds24ksdx′,(6)T1t,x,y=14π∫0texp⁡⁡-y24ksdss∫-∞∞y2ks-bt-s,x′k0×Tt-s,x′,0exp⁡-x′-x-vds24ksdx′.As aforementioned, in dry grinding, we may assume that the heat transfer coefficient is zero:(7)bt,x=0,so (6) is reduced to(8)T1t,x,y=y8πk∫0texp⁡⁡-y24ksdss2×∫-∞∞Tt-s,x′,0exp⁡-x′-x-vds24ksdx′.Moreover, according to [18], we have (9)Tt,x,y=2T0t,x,y;thus, taking into account (5), we obtain the temperature field in terms of a two-dimensional integral:(10)Tt,x,y=-12πk0∫0t‍exp⁡-y24ksdss×∫-∞∞‍dt-s,x′exp⁡-x′-x-vds24ksdx′.The result given in (10) is called T0 theorem.


## 3. Calculation of New Integrals fromT0 Theorem

According to (4) and (9), we have(11)T0t,x,y=T1t,x,y.Defining the function(12)mt,x,y=1texp⁡-x+vdt2+y24ktand taking into account expressions (5) and (8) for T0 and T1, we may rewrite (11) as(13)T0t,x,y=-14πk0∫0tds∫-∞∞dt-s,x′ms,x-x′,ydx′=-14π∫0tds∫-∞∞Tt-s,x′∂∂yms,x-x′,ydx′.Using (9) and exchanging the integration order, we have(14)T0t,x,y=-14πk0∫-∞∞dx′∫0tdt-s,x′ms,x-x′,yds=-12π∫-∞∞dx′∫0tT(0)t-s,x′,0∂∂yms,x-x′,yds.Applying the Laplace transform in (14),(15)f~τ=Lft=∫0∞‍e-τtftdt,and taking into account the convolution theorem for this integral transform [24, Eq. 17.12.5](16)LftLgt=L∫0t‍ft-sgsds,(14) becomes (17)T~0t,x,y=-14πk0∫-∞∞‍d~τ,x′m~τ,x-x′,ydx′=-12π∫-∞∞‍T~0τ,x′,0∂∂ym~τ,x-x′,ydx′.Performing the Fourier transform in (17)(18)f^w=Ff(x)=12π∫-∞∞‍eiwxfxdxand taking into account the convolution theorem for this integral transform [27, Eq. 33.9] (19)FfxFgx=F12π∫-∞∞‍fx′gx-x′dx′,we may rewrite (17) as(20)T~^0τ,w,y=-12k0d~^τ,wm~^τ,w,y(21)=-T~^0τ,w,y∂∂ym~^τ,w,y.Takingy=0 in (20) and substituting the result in (21), we arrive at (22)m~^(τ,w,y)=-m~^(τ,w,0)∂∂ym~^(τ,w,y).Solving the previous ODE (22), we get (23)m~^τ,w,y=m~^τ,w,0exp⁡-ym~^τ,w,0,y≥0,where, according to Figure 1  y≥0. Later on, we will refer to (23) as the “master equation.”


### 3.1. First Integral

First, let us perform the Laplace transform of them(t,x,y) function, defined in (12),(24)m~τ,x,y=∫0∞‍e-τtmt,x,ydt=∫0∞‍e-τttexp⁡-x+vdt2+y24ktdt=exp⁡-xvd2k·∫0∞‍exp⁡-x2+y24kt-vd24k+τtdtt.Performing the change of variables4ks=vd2+4kτt results in(25)m~τ,x,y=exp⁡-xvd2k·∫0∞‍exp⁡-x2+y2vd2+4kτ16k2s-sdss.Knowing that an integral representation of Macdonald’s function of zeroth order is [28, Eq. 5.10.25] (26)K0(z)=12∫0∞‍exp⁡-z24σ-σdσσ,argz<π4,we express (25) as(27)m~(τ,x,y)=2exp⁡-xvd2kK0x2+y2vd2+4kτ2k.Thus(28)m~(τ,x,0)=2exp⁡-xvd2kK0xvd2+4kτ2k.On the one hand, applying the Fourier transform to (28), we have(29)m~^τ,w,0=12π∫-∞∞‍m~τ,x,0eiwxdx=1π∫-∞∞‍exp⁡-αxK0βxdx,where we have set(30)α=vd2k-iw,(31)β=vd2+4kτ2k.We may split (29) into two summands(32)m~^τ,w,0=1π∫-∞0‍exp⁡-αxK0-βxdx+1π∫0∞‍exp⁡-αxK0βxdx.Changingx by -x in (32), we rewrite (29) as follows:(33)m~^τ,w,0=2π∫0∞‍cosh⁡αxK0βxdx=1β2-α2,β>α,where we have applied the result given in Appendix (A.17).On the other hand, applying the Fourier transform to (27) and taking into account the definitions of α and β  (30)-(31), we have(34)m~^τ,w,y=12π∫-∞∞‍m~τ,x,yeiwxdx=1π∫-∞∞‍exp⁡-αxK0βx2+y2dx=2π∫0∞‍cosh⁡αxK0βx2+y2dx.Inserting now (33) and (34) into the “master equation” (23), we arrive at(35)∫0∞‍cosh⁡αxK0βx2+y2dx=πexp⁡-yβ2-α22β2-α2β>α,y∈R,where we have extended the values of y, from y≥0 to y∈R, by taking y in the right hand side of (35). As far as the author’s knowledge, (35) is not given in the literature. The equation most similar to (35) we have found in the the literature is [24, Eq. 6.677.5](36)∫0∞‍cos⁡αxK0βx2+y2dx=πexp⁡-yβ2+α22β2+α2Reβ>0,Rey>0,α>0.See that, takingα=0 and β=1 in (35), we have(37)∫0∞‍K0x2+y2dx=π2e-y,y∈R.Surprisingly, (37) is not reported in the usual literature.


### 3.2. Second Integral

Performing in the inverse Fourier transform in (22) by using the convolution theorem for Fourier transforms (19), we obtain(38)m~τ,w,y=-12π∫-∞∞‍m~τ,x′,0∂∂ym~τ,x-x′,ydx′.According to the definition ofβ  (31), we rewrite (27) as(39)m~τ,w,y=2exp⁡-xvd2kK0βx2+y2;thus,(40)m~τ,w,0=2K0βx,(41)∂∂ym~(τ,x,y)=-2yβexp⁡-xvd2kK1βx2+y2x2+y2.Substituting (39)-(40) into (38), we have(42)K0βx2+y2=yβπ∫-∞∞‍K0βx′K1βx-x′2+y2x-x′2+y2dx′.Takingξ=βx, υ=βy and performing the change of variables ξ′=βx′, we arrive at(43)K0ξ2+υ2=υπ∫-∞∞‍K0ξ′K1ξ-ξ′2+υ2ξ-ξ′2+υ2dξ′.The integral given in (43) has been already calculated by using a complex integration contour [22]. However, we can get another new integral, integrating in (43) with respect to the υ variable between 0 and ∞, obtaining(44)∫0∞‍K0ξ2+υ2dυ=1π∫-∞∞‍K0ξ′dξ′×∫0∞‍υK1ξ-ξ′2+υ2ξ-ξ′2+υ2dυ.Taking into account (37) and (41), it turns out that(45)π2e-ξ=1π∫-∞∞‍K0ξ′K0ξ-ξ′2+υ20∞dξ′.Since [28, Eq. 5.16.5](46)limx→∞K0x=limx→∞π2xe-x=0,we obtain the following integral:(47)∫-∞∞‍K0ξ′K0ξ-ξ′dξ′=π22e-ξ.As in (35), (47) seems not to be found in the usual literature. Notice that taking ξ=0 in (47) we recover the following integral given in the literature [24, Eq. 6.511.13]:(48)∫0∞‍K02ξ′dξ′=π24.


## 3.1. First Integral

First, let us perform the Laplace transform of them(t,x,y) function, defined in (12),(24)m~τ,x,y=∫0∞‍e-τtmt,x,ydt=∫0∞‍e-τttexp⁡-x+vdt2+y24ktdt=exp⁡-xvd2k·∫0∞‍exp⁡-x2+y24kt-vd24k+τtdtt.Performing the change of variables4ks=vd2+4kτt results in(25)m~τ,x,y=exp⁡-xvd2k·∫0∞‍exp⁡-x2+y2vd2+4kτ16k2s-sdss.Knowing that an integral representation of Macdonald’s function of zeroth order is [28, Eq. 5.10.25] (26)K0(z)=12∫0∞‍exp⁡-z24σ-σdσσ,argz<π4,we express (25) as(27)m~(τ,x,y)=2exp⁡-xvd2kK0x2+y2vd2+4kτ2k.Thus(28)m~(τ,x,0)=2exp⁡-xvd2kK0xvd2+4kτ2k.On the one hand, applying the Fourier transform to (28), we have(29)m~^τ,w,0=12π∫-∞∞‍m~τ,x,0eiwxdx=1π∫-∞∞‍exp⁡-αxK0βxdx,where we have set(30)α=vd2k-iw,(31)β=vd2+4kτ2k.We may split (29) into two summands(32)m~^τ,w,0=1π∫-∞0‍exp⁡-αxK0-βxdx+1π∫0∞‍exp⁡-αxK0βxdx.Changingx by -x in (32), we rewrite (29) as follows:(33)m~^τ,w,0=2π∫0∞‍cosh⁡αxK0βxdx=1β2-α2,β>α,where we have applied the result given in Appendix (A.17).On the other hand, applying the Fourier transform to (27) and taking into account the definitions of α and β  (30)-(31), we have(34)m~^τ,w,y=12π∫-∞∞‍m~τ,x,yeiwxdx=1π∫-∞∞‍exp⁡-αxK0βx2+y2dx=2π∫0∞‍cosh⁡αxK0βx2+y2dx.Inserting now (33) and (34) into the “master equation” (23), we arrive at(35)∫0∞‍cosh⁡αxK0βx2+y2dx=πexp⁡-yβ2-α22β2-α2β>α,y∈R,where we have extended the values of y, from y≥0 to y∈R, by taking y in the right hand side of (35). As far as the author’s knowledge, (35) is not given in the literature. The equation most similar to (35) we have found in the the literature is [24, Eq. 6.677.5](36)∫0∞‍cos⁡αxK0βx2+y2dx=πexp⁡-yβ2+α22β2+α2Reβ>0,Rey>0,α>0.See that, takingα=0 and β=1 in (35), we have(37)∫0∞‍K0x2+y2dx=π2e-y,y∈R.Surprisingly, (37) is not reported in the usual literature.


## 3.2. Second Integral

Performing in the inverse Fourier transform in (22) by using the convolution theorem for Fourier transforms (19), we obtain(38)m~τ,w,y=-12π∫-∞∞‍m~τ,x′,0∂∂ym~τ,x-x′,ydx′.According to the definition ofβ  (31), we rewrite (27) as(39)m~τ,w,y=2exp⁡-xvd2kK0βx2+y2;thus,(40)m~τ,w,0=2K0βx,(41)∂∂ym~(τ,x,y)=-2yβexp⁡-xvd2kK1βx2+y2x2+y2.Substituting (39)-(40) into (38), we have(42)K0βx2+y2=yβπ∫-∞∞‍K0βx′K1βx-x′2+y2x-x′2+y2dx′.Takingξ=βx, υ=βy and performing the change of variables ξ′=βx′, we arrive at(43)K0ξ2+υ2=υπ∫-∞∞‍K0ξ′K1ξ-ξ′2+υ2ξ-ξ′2+υ2dξ′.The integral given in (43) has been already calculated by using a complex integration contour [22]. However, we can get another new integral, integrating in (43) with respect to the υ variable between 0 and ∞, obtaining(44)∫0∞‍K0ξ2+υ2dυ=1π∫-∞∞‍K0ξ′dξ′×∫0∞‍υK1ξ-ξ′2+υ2ξ-ξ′2+υ2dυ.Taking into account (37) and (41), it turns out that(45)π2e-ξ=1π∫-∞∞‍K0ξ′K0ξ-ξ′2+υ20∞dξ′.Since [28, Eq. 5.16.5](46)limx→∞K0x=limx→∞π2xe-x=0,we obtain the following integral:(47)∫-∞∞‍K0ξ′K0ξ-ξ′dξ′=π22e-ξ.As in (35), (47) seems not to be found in the usual literature. Notice that taking ξ=0 in (47) we recover the following integral given in the literature [24, Eq. 6.511.13]:(48)∫0∞‍K02ξ′dξ′=π24.



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