Depth of thermal penetration in straight grinding

Abstract

Unlike the usual numerical FEM approach to determine the thermally affected layer during the grinding process, we propose a simple analytical approach to estimate the depth of thermal penetration. For this purpose, the one-dimensional definition of depth of thermal penetration is applied to the two-dimensional heat transfer models of straight grinding. A method for computing the depth of thermal penetration in these two-dimensional models is derived and compared to the one-dimensional approximation. For dry grinding, it turns out that the one-dimensional approximation is quite accurate when we consider a moderate percentage in the temperature fall beneath the surface, regardless the type of heat flux profile entering into the workpiece (i.e., constant, linear, triangular, or parabolic). In wet grinding, the latter is true if we consider a constant heat flux profile and a high Peclet number, i.e., Pe > 5. Finally, the one- and two-dimensional approaches calculating analytically the depth of thermal penetration have been compared to the temperature field numerically evaluated by a three-dimensional FEM simulation given in the literature, obtaining a quite good agreement.

Appendix. One-dimensional solution for wet grinding

We have to solve the following boundary value problem:

$$\begin{array}{@{}rcl@{}} \frac{\partial T\left( t,y\right) }{\partial t} &=&k\frac{\partial^{2}T\left( t,y\right) }{\partial y^{2}}, \end{array} $$

(87)

$$\begin{array}{@{}rcl@{}} T\left( 0,y\right) &=&T_{\infty },\qquad y>0, \end{array} $$

(88)

$$\begin{array}{@{}rcl@{}} \frac{\partial T}{\partial x}\left( t,0\right) &=&-q_{0}+h_{0}\left[ T\left( t,0\right) -T_{\infty }\right] ,\quad t>0. \end{array} $$

(89)

Performing the change of variables,

$$ \vartheta \left( t,y\right) =T\left( t,y\right) -T_{\infty }+q_{0}y, $$

(90)

the problems stated in Eqs. 87–89 become

$$\begin{array}{@{}rcl@{}} \frac{\partial \vartheta }{\partial t} &=&k\frac{\partial^{2}\vartheta }{ \partial y^{2}}, \end{array} $$

(91)

$$\begin{array}{@{}rcl@{}} \frac{\partial \vartheta }{\partial y}\left( t,0\right) &=&h_{0}\vartheta \left( t,0\right) , \end{array} $$

(92)

$$\begin{array}{@{}rcl@{}} \vartheta \left( 0,y\right) &=&q_{0}y. \end{array} $$

(93)

In [9, Eqn. 14.2(7)], the solution to the problem

$$\begin{array}{@{}rcl@{}} \frac{\partial v}{\partial t} &=&k\frac{\partial^{2}v}{\partial x^{2}},\\ \frac{\partial v}{\partial y}\left( t,0\right) -h\,v\left( t,0\right) &=&h\,\phi \left( t\right) ,\\ v\left( 0,x\right) &=&f\left( x\right) , \end{array} $$

is given by

$$\begin{array}{@{}rcl@{}} &&v\left( t,x\right)\\ &\,=\,&{\int}_{0}^{\infty }f\left( \xi \right) \left\{ \frac{e^{-\left( x-\xi \right)^{2}/4kt}+e^{-\left( x+\xi \right)^{2}/\left( 4kt\right) }}{2\sqrt{ \pi kt}}\right. \\ &&\,-\,\left. h\,e^{kth^{2}+h\left( x+\xi \right) }\text{erfc}\left( \frac{ x+\xi }{2\sqrt{kt}}+h\sqrt{kt}\right) \right\} d\xi \\ &&\!+kh{{\int}_{0}^{t}}\phi \left( \tau \right) \left\{ \frac{e^{-x^{2}/4k\left( t-\tau \right) }}{\sqrt{\pi k\left( t-\tau \right) }}\right. \\ &&\quad \,-\,\left. h\,e^{kh^{2}\left( t-\tau \right) +hx}\text{erfc}\left( \frac{x}{2\sqrt{k\left( t-\tau \right) }}+h\sqrt{k\left( t-\tau \right) } \right) \right\} d\tau . \end{array} $$

(94)

Therefore, applying the solution given in Eq. 94 to the boundary value problem stated in Eqs. 91–93, taking ϕ(t) = 0 and f(y) = q₀y, we have

$$\begin{array}{@{}rcl@{}} &&\vartheta \left( t,y\right) \end{array} $$

(95)

$$\begin{array}{@{}rcl@{}} &=&q_{0}\left\{ \frac{1}{2\sqrt{\pi kt}}{\int}_{0}^{\infty }\left( e^{-\left( y-\xi \right)^{2}/4kt}+e^{-\left( y+\xi \right)^{2}/4kt}\right) \xi d\xi \right. \end{array} $$

(96)

$$\begin{array}{@{}rcl@{}} &&-\left. h_{0}e^{kt{h_{0}^{2}}}{\int}_{0}^{\infty }e^{h_{0}\left( y+\xi \right) }\,\text{erfc}\left( \frac{y+\xi }{2\sqrt{kt}}+h_{0}\sqrt{kt}\right) \xi d\xi \right\} . \end{array} $$

(97)

The integral given in Eq. 96, denoted by I₁ hereafter can be calculated by using the following result (see [31, Eqn. 3.462.5]):

$$\begin{array}{@{}rcl@{}} &&{\int}_{0}^{\infty }x\,e^{-\mu x^{2}-2\nu x}dx=\frac{1}{2\mu }-\frac{\nu }{ 2\mu }e^{\nu^{2}/\mu }\,\text{erfc}\left( \frac{\nu }{\sqrt{\mu }}\right) , \\ &&\left\vert \arg\,\nu \right\vert <\frac{\pi }{2},\,\mathrm{Re\,} \mu >0, \end{array} $$

thereby,

$$\begin{array}{@{}rcl@{}} I_{1} &=&\frac{1}{2\sqrt{\pi kt}}{\int}_{0}^{\infty }\left( e^{-\left( y-\xi \right)^{2}/4kt}+e^{-\left( y+\xi \right)^{2}/4kt}\right) \xi d\xi\\ &=&2\sqrt{\frac{kt}{\pi }}e^{-y^{2}/4kt}+y\,\text{erfc}\left( \frac{y}{2 \sqrt{kt}}\right) . \end{array} $$

(98)

The infinite integral given in Eq. 97, denoted by I₂, can be expressed in terms of the indefinite integral I₂(ξ) as

$$ I_{2}=\lim_{\xi \rightarrow \infty }I_{2}\left( \xi \right) -I_{2}\left( 0\right) $$

(99)

In order to calculate I₂(ξ), perform the substitution $u=\frac {y+\xi }{2\sqrt {kt}}+h_{0}\sqrt {kt}$ and set $\alpha = 2h_{0}\sqrt {kt}$ , to arrive at

$$\begin{array}{@{}rcl@{}} I_{2}\left( \xi \right) &=&-h_{0}e^{kt{h_{0}^{2}}}\int e^{h_{0}\left( y+\xi \right) }\,\text{erfc}\left( \frac{y+\xi }{2\sqrt{kt}}+h_{0}\sqrt{kt} \right) \xi d\xi\\ &=&-\alpha \left\{ \frac{\alpha }{h_{0}}e^{-\alpha^{2}/4}\int u\,e^{\alpha u}\,\text{erfc}\left( u\right) du\right.\\ &&\quad -\left. \left( \alpha \sqrt{kt}+y\right) \int e^{\alpha u}\,\mathrm{ erfc}\left( u\right) du\right\} . \end{array} $$

(100)

Both integrals given in Eq. 100 can be calculated with the following results (see [32, Eqns. 1.5.2(3)-(4)]):

$$\int \!e^{by}\,\mathrm{\!erfc}\left( ay\right) dy\,=\,\frac{e^{by}}{b}\text{erfc} \left( ay\right) +\frac{e^{b^{2}/4a^{2}}}{b}\,\text{erfc}\left( \!ay\!-\frac{b }{2a}\right) , $$

and

$$\begin{array}{@{}rcl@{}} &&\int y\,e^{by}\,\text{erfc}\left( ay\right) dy \\ &=&\frac{e^{by}}{b}\left( y-\frac{1}{b}\right) \text{erfc}\left( ay\right) \\ &&+\frac{e^{b^{2}/4a^{2}}}{b}\left[ \left( \frac{1}{2a^{2}}-\frac{1}{b} \right) \text{erfc}\left( ay-\frac{b}{2a}\right) -\frac{e^{-\left( 2ay-b/a\right)^{2}/4}}{a\sqrt{\pi }}\right] . \end{array} $$

Therefore, Eq. 100 is calculated as

$$\begin{array}{@{}rcl@{}} I_{2}\left( \xi \right) &=&2\sqrt{\frac{kt}{\pi }}e^{-\left( y+\xi \right)^{2}/4kt}+\frac{1+h_{0}y}{h_{0}}\,\text{erf}\left( \frac{y+\xi }{2\sqrt{kt}}\right)\\ &&+\frac{1-h_{0}y}{h_{0}}e^{{h_{0}^{2}}kt+h_{0}\left( y+\xi \right) }\,\mathrm{ erfc}\left( \frac{y+\xi }{2\sqrt{kt}}+h_{0}\sqrt{kt}\right) . \end{array} $$

(101)

Notice that since y, t, k, h₀ > 0, we have

$$\begin{array}{@{}rcl@{}} \lim\limits_{\xi \rightarrow \infty }I_{2}\left( \xi \right) &=&\frac{1+h_{0}y}{ h_{0}}-\lim\limits_{\xi \rightarrow \infty }\xi \,e^{h_{0}\xi }\,\text{erfc} \left( \frac{\xi }{2\sqrt{kt}}\right) \\ &=&\frac{1+h_{0}y}{h_{0}}, \end{array} $$

(102)

where we have used the asymptotic expansion (see [27, Eqn. 7.12.1])

$$\text{erfc}\left( z\right) \sim \frac{e^{-z^{2}}}{\sqrt{\pi }z} \sum\limits_{m = 0}^{\infty }\left( -1\right)^{m}\frac{\left( 2m-1\right) !!}{\left( 2z^{2}\right)^{m}},\quad z\rightarrow \infty . $$

Therefore, taking into account (101) and (102) in Eq. 99, we have

$$\begin{array}{@{}rcl@{}} I_{2} &=&\frac{1+h_{0}y}{h_{0}}\,\text{erfc}\left( \frac{y}{2\sqrt{kt}} \right) -2\sqrt{\frac{kt}{\pi }}e^{-y^{2}/4kt}\\ &&-\frac{e^{{h_{0}^{2}}kt+h_{0}y}}{h_{0}}\,\text{erfc}\left( \frac{y}{2\sqrt{ kt}}+h_{0}\sqrt{kt}\right) . \end{array} $$

(103)

Substituting back in Eqs. 95 and 90 the results obtained in Eqs. 98 and 103, we finally arrive at

$$\begin{array}{@{}rcl@{}} &&T\left( t,y\right) -T_{\infty } \\ &=&\frac{q_{0}}{h_{0}}\left\{ \,\text{erfc}\left( \frac{y}{2\sqrt{kt}} \right) -e^{{h_{0}^{2}}kt+h_{0}y}\,\text{erfc}\left( \frac{y}{2\sqrt{kt}} +h_{0}\sqrt{kt}\right) \right\} . \end{array} $$