Temperature evolution associated with phase transition from quasi static to dynamic loading

Abstract

Thermo-mechanical coupling is an intrinsic property of first order martensitic transformation. In this paper, we study the temperature evolution during phase transition at a wider strain rates from quasi static to impact loading to reveal the thermodynamic nature of the strain rate effect of phase transition materials. Based on the laws of thermodynamics and the principle of maximum dissipated energy, a thermal-mechanically coupled model was proposed. The model shows that, in the quasi static case, the temperature profile grades around the moving phase boundary, while for the dynamic case, thermal response of the specimen can be reached homogeneously due to random nucleation. The predicted results of the model are in good agreement with the experimental results, suggesting that the interaction between the self-heating effect and the temperature dependence of phase transition behavior plays a leading role in the process of the transformation deformation mechanism associated with the loading rate.

Appendices

Appendix 1: Thermodynamic basics

The first law of thermodynamics may be written in terms of the Gibbs free energy

$$\dot{G} + \dot{T}\eta + T\dot{\eta } = r - {\text{div}}q - {\dot{\upsigma}} :\upvarepsilon$$

(11)

where q is the heat influx vector and r is the density of internal heat generation. The entropy $\eta$ of the RVE is given by the following

$$\eta = - \frac{\partial G}{{\partial T}} =\upsigma :\alpha + C\ln (T/T_{0} ) - \eta_{0}$$

(12)

The second principle of thermodynamic or the Clausius–Duhem inequality gives:

$$D = T\dot{\eta } + {\text{div}}q - q\frac{\Delta T}{T} - r \ge 0$$

(13)

where D is the total dissipation.

Eliminating divq − r between Eqs. (11) and (13) gives

$$D = {\dot{\upsigma}} :\upvarepsilon - T\dot{\eta } - \dot{G} - q\frac{\Delta T}{T} \ge 0$$

(14)

Equation (14) emphasizes two dissipation contributions, $D_{1} = {\dot{\upsigma}} :\upvarepsilon - T\dot{\eta } - \dot{G}$ is the intrinsic dissipation and $D_{2} = - q\Delta T/T$ is the heat dissipation, which are both positive.

Given the Gibbs free energy G which is written as a function of control and internal variables, and substituting its time derivative into the expression of the intrinsic dissipation lead to:

$$D_{1} = {\dot{\upsigma}} :\upvarepsilon - T\dot{\eta } - \dot{G} = - \frac{\partial G}{{\partial \xi }}\dot{\xi }$$

(15)

From the expression of the intrinsic dissipation, one can write the following equation:

$$D_{1} = - \frac{\partial G}{{\partial \xi }}\dot{\xi } = {\text{div}}q - r + T\dot{\eta }$$

(16)

which leads, by taking into account the time derivative of the entropy, to:

$$D_{1} = - \frac{\partial G}{{\partial \xi }}\dot{\xi } = T\frac{{\partial^{2} }}{\partial T\partial \sigma }:\dot{\sigma } + T\frac{{\partial^{2} }}{\partial T\partial \xi }:\dot{\xi } + T\frac{{\partial^{2} }}{\partial T\partial T}:\dot{T} + {\text{div}}q - r$$

(17)

Heat conduction is taken into account through Fourier’s law,

$${\text{div}}( - k\nabla T) = - \rho C_{P} \dot{T} - \frac{\partial G}{{\partial \xi }}\dot{\xi } - \frac{{\partial^{2} G}}{\partial T\partial \xi }\dot{\xi }T + r$$

(18)

Or in a more condensed form:

$$\rho C_{P} \dot{T} = {\text{div}}(k\nabla T) + (\frac{\partial \eta }{{\partial \xi }}T - \frac{\partial G}{{\partial \xi }})\dot{\xi } + r$$

(19)

Assuming that the macroscopic transformation strain is expressed as the product of the internal variables. It leads to the following expression:

$${\dot{\varepsilon }}_{p} = {\Pi }{\text{sgn}} (\dot{\xi })\dot{\xi }$$

(20)

where sgn is defined as

$${\text{sgn}} (\dot{\xi }) = \left\{ {\begin{array}{*{20}c} 1 & {\dot{\xi } > 0} \\ { - 1} & {\dot{\xi } < 0} \\ \end{array} } \right.$$

(21)

Substituting Eq. (20) into Eq. (19) leads to the following explicit form of expression:

$$\rho C_{P} \dot{T} = {\text{div}}(k\nabla T) + (\frac{\partial \eta }{{\partial \xi }}T - \frac{\partial G}{{\partial \xi }})\frac{{{\dot{\varepsilon }}_{p} }}{{{\Pi }{\text{sgn}} (\dot{\xi })}} + r$$

(22)

The specific Gibbs free energy can be calculated by the rule of mixtures

$$G = \xi G_{M} + (1 - \xi )G_{A} ,\quad \eta = (1 - \xi )\eta_{A} + \xi \eta_{M}$$

(23)

Substituting Eq. (23) in Eq. (22), one can obtain:

$$\rho C_{P} \dot{T} = {\text{div}}(k\nabla T) + [T(\eta_{M} - \eta_{A} ) - (G_{M} - G_{A} )]\frac{{{\dot{\varepsilon }}_{p} }}{{{\Pi }{\text{sgn}} (\dot{\xi })}} + r$$

(24)

It can be seen that the effect of phase transformation deformation energy on the temperature evolution and distribution is not considered in Eq. (24). Therefore, by considering the dissipated energy W, the governing Eq. (24) can be modified as,

$$\rho C_{P} \dot{T} = {\text{div}}(k\nabla T) + [T(\eta_{M} - \eta_{A} ) - (G_{M} - G_{A} )]\frac{{{\dot{\varepsilon }}_{p} }}{{{\Pi }{\text{sgn}} (\dot{\xi })}} + W + r$$

(25)

Appendix 2: Solution of temperature evolution and distribution

The dynamic system of the above equations consists of an unknown temperature field T(x, t) and an unknown interface position S(t), which is mathematically called a free boundary problem. Now, replacing the variables

$$u(x) = T(x + \dot{S}t,t) - T_{0}$$

(26)

then, Eq. (1) is transformed into

$$u_{xx} + \frac{{\rho C_{p} \dot{S}}}{k}u_{x} - \frac{2h}{{Rk}}u = 0$$

(27)

Its solution is

$$u(x) = \left\{ \begin{array}{ll} c_{1} e^{{\lambda_{1} x}} &\quad x > 0 \\ c_{2} e^{{\lambda_{2} x}}&\quad x \le 0 \\ \end{array} \right.$$

(28)

where

$$\lambda_{1,2} = - \frac{{\rho C_{p} v}}{2k} \mp \frac{1}{2}\sqrt {(\frac{{\rho C_{p} v}}{k})^{2} + \frac{8h}{{Rk}}}$$

(29)

In Eq. (26), let x = 0, then

$$u(0) = T(S,t) - T_{0} = T_{{\text{int}}}$$

(30)

T_int reflects the relative temperature at the moving phase interface S(t).

In Eq. (28), let x = 0, then

$$c_{1} = c_{2} = T_{{\text{int}}}$$

(31)

and

$$u(x) = \left\{ {\begin{array}{*{20}l} {T_{{\text{int} }} e^{{\lambda _{1} x}} } \hfill &\quad {x > 0} \hfill \\ {T_{{\text{int} }} e^{{\lambda _{2} x}} } \hfill &\quad {x \le 0} \hfill \\ \end{array} } \right.$$

(32)

Furthermore,

$$T(x,t) = \left\{ {\begin{array}{*{20}l} {T_{{\text{int} }} e^{{\lambda _{1} (x - \dot{S}t)}} + T_{0} } \hfill &\quad {x > \dot{S}t} \hfill \\ {T_{{\text{int} }} e^{{\lambda _{2} (x - \dot{S}t)}} + T_{0} } \hfill &\quad {x \le \dot{S}t} \hfill \\ \end{array} } \right.$$

(33)

By substituting Eq. (33) into phase boundary condition Eq. (5), we can get

$$l^{*} \dot{S} = T_{{\text{int}}} \sqrt {(\rho C_{p} v)^{2} + 8hk/R}$$

(34)

and, the temperature gradient at the interface,

$$T_{{\text{int}}} = \frac{{l^{*} }}{{\rho C_{p} \sqrt {1 + 8\frac{h}{{R\rho C_{p} }} \cdot \frac{k}{{\rho C_{p} }}\frac{1}{{\dot{S}^{2} }}} }}$$

(35)

Substituting Eq. (35) into Eq. (33), we can get

$$T(x,t) = \left\{ {\begin{array}{*{20}l} {T_{0} + \frac{{l^{*} }}{{\sqrt {(\rho C_{p} )^{2} + 8hk/\dot{S}^{2} R} }}e^{{\lambda_{1} (x - \dot{S}t)}} } \hfill & {x > \dot{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\cdot}$}}{S} }t} \hfill \\ {T_{0} + \frac{{l^{*} }}{{\sqrt {(\rho C_{p} )^{2} + 8hk/\dot{S}^{2} R} }}e^{{\lambda_{2} (x - \dot{S}t)}} } \hfill & {x \le \dot{S}t} \hfill \\ \end{array} } \right.$$

(36)

Equation (36) describes the temperature distribution of the both sides of the moving phase boundary, where $x > \dot{S}t$ represents the region in front of the phase boundary, and $x < \dot{s}t$ represents the region behind the phase boundary.

According to the mass conservation condition, the relationship between the propagation velocity and the strain rate of the phase boundary is as follows

$$\dot{S} = \frac{{L_{0} \dot{\varepsilon }}}{\prod }$$

(37)

where Π is the completion strain of the phase transition.

Substituting Eq. (37) into Eqs. (35) and (36), then the temperature distribution across the phase front obeys such rule

$$T(x,t) = \left\{ {\begin{array}{*{20}l} {T_{0} (1 + He^{{\lambda_{1} (x - \dot{S}t)}} )} \hfill & {x > \dot{S}t} \hfill \\ {T_{0} (1 + He^{{\lambda_{2} (x - \dot{S}t)}} )} \hfill & {x \le \dot{S}t} \hfill \\ \end{array} } \right.$$

(38)

where

$$H = \frac{{l^{*} }}{{\rho C_{p} T_{0} \sqrt {1 + 8\frac{h}{{R\rho C_{p} }} \cdot \frac{k}{{\rho C_{p} L_{0}^{2} }} \cdot \left( {\frac{\prod }{{\dot{\varepsilon }}}} \right)^{2} } }}$$

(39)

$$\lambda_{1,2} = - \frac{1}{2}\frac{1}{{L_{0} }}\frac{{\dot{\varepsilon }}}{\prod }\frac{{\rho C_{p} L_{0}^{2} }}{k} \mp \frac{1}{2}\frac{1}{{L_{0} }}\sqrt {\left( {\frac{{\dot{\varepsilon }}}{\prod }\frac{{\rho C_{p} L_{0}^{2} }}{k}} \right)^{2} + \frac{{8hL_{0}^{2} }}{Rk}}$$

(40)