Researching | Simulation and analysis of the time evolution of laser power and temperature in static pulsed XPALs

Journals >High Power Laser Science and Engineering >Volume 7 >Issue 3 >Page 03000e44 > Article

High Power Laser Science and Engineering
Vol. 7, Issue 3, 03000e44 (2019)

Simulation and analysis of the time evolution of laser power and temperature in static pulsed XPALs

Chenyi Su, Binglin Shen, Xingqi Xu, Chunsheng Xia, and Bailiang Pan^†

Author Affiliations

Department of Physics, Zhejiang University, Hangzhou 310027, China

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DOI: 10.1017/hpl.2019.28 Cite this Article Set citation alerts

Chenyi Su, Binglin Shen, Xingqi Xu, Chunsheng Xia, Bailiang Pan. Simulation and analysis of the time evolution of laser power and temperature in static pulsed XPALs[J]. High Power Laser Science and Engineering, 2019, 7(3): 03000e44 Copy Citation Text

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Abstract

A theoretical model is established to describe the thermal dynamics and laser kinetics in a static pulsed exciplex pumped Cs–Ar laser (XPAL). The temporal behaviors of both the laser output power and temperature rise in XPALs with a long-time pulse and multi-pulse operation modes are calculated and analyzed. In the case of long-time pulse pumping, the results show that the initial laser power increases with a rise in the initial operating temperature, but the laser power decreases quickly due to heat accumulation. In the case of multi-pulse operation, simulation results show that the optimal laser output power can be obtained by appropriately increasing the initial temperature and reducing the thermal relaxation time.

Keywords

excimer lasers simulation theoretical model

$$\begin{eqnarray}\displaystyle \displaystyle \frac{\text{d}n_{0}}{\text{d}t} & = & \displaystyle k_{10}n_{1}-k_{01}n_{0}[\text{Ar}]+\frac{n_{3}}{\unicode[STIX]{x1D70F}_{D1}}\nonumber\\ \displaystyle & & \displaystyle +\,(n_{3}-2n_{0})\unicode[STIX]{x1D70E}_{D1}\frac{f_{p}(P_{l}^{+}+P_{l}^{-})}{h\unicode[STIX]{x1D708}_{l}},\end{eqnarray}$$

(1)

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$$\begin{eqnarray}\displaystyle \displaystyle \frac{\text{d}n_{1}}{\text{d}t} & = & \displaystyle -k_{10}n_{1}+k_{01}n_{0}[\text{Ar}]\nonumber\\ \displaystyle & & \displaystyle -\,(n_{1}-n_{2})\int _{0}^{\infty }\frac{\unicode[STIX]{x1D70E}_{D2}(\unicode[STIX]{x1D708})P_{p}(\unicode[STIX]{x1D708})}{h\unicode[STIX]{x1D708}_{p}}\,\text{d}\unicode[STIX]{x1D708},\end{eqnarray}$$

(2)

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$$\begin{eqnarray}\displaystyle \displaystyle \frac{\text{d}n_{2}}{\text{d}t} & = & \displaystyle -k_{23}n_{2}+k_{32}n_{3}[\text{Ar}]\nonumber\\ \displaystyle & & \displaystyle +\,(n_{1}-n_{2})\int _{0}^{\infty }\frac{\unicode[STIX]{x1D70E}_{D2}(\unicode[STIX]{x1D708})P_{p}(\unicode[STIX]{x1D708})}{h\unicode[STIX]{x1D708}_{p}}\,\text{d}\unicode[STIX]{x1D708},\end{eqnarray}$$

(3)

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$$\begin{eqnarray}\displaystyle \displaystyle \frac{\text{d}n_{3}}{\text{d}t} & = & \displaystyle k_{23}n_{2}-k_{32}n_{3}[\text{Ar}]-\frac{n_{3}}{\unicode[STIX]{x1D70F}_{D1}}\nonumber\\ \displaystyle & & \displaystyle -\,(n_{3}-2n_{0})\unicode[STIX]{x1D70E}_{D1}\frac{f_{p}(P_{l}^{+}+P_{l}^{-})}{h\unicode[STIX]{x1D708}_{l}},\end{eqnarray}$$

(4)

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$$\begin{eqnarray}\displaystyle \displaystyle n_{\text{Cs}}\frac{T_{0}}{T} & = & \displaystyle n_{0}+n_{1}+n_{2}+n_{3},\end{eqnarray}$$

(5)

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$$\begin{eqnarray}\unicode[STIX]{x1D70E}_{D2}(\unicode[STIX]{x1D708})=\unicode[STIX]{x1D70E}_{D2,0}\frac{(\unicode[STIX]{x0394}\unicode[STIX]{x1D708}_{\text{abs}}/2)^{2}}{(\unicode[STIX]{x1D708}-\unicode[STIX]{x1D708}_{\text{abs}})^{2}+(\unicode[STIX]{x0394}\unicode[STIX]{x1D708}_{\text{abs}}/2)^{2}},\end{eqnarray}$$

(6)

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$$\begin{eqnarray}\unicode[STIX]{x1D70E}_{D2,0}=k_{\text{abs}}\frac{[\text{Ar}]}{f_{10}}=\frac{g_{0}}{g_{1}}\frac{k_{\text{abs}}}{4\unicode[STIX]{x1D70B}R_{0}^{2}\unicode[STIX]{x0394}R\exp (-\unicode[STIX]{x0394}E_{10}/k_{b}T)},\end{eqnarray}$$

(7)

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$$\begin{eqnarray}f_{p,l}(x,y,z)=\frac{c_{2}}{\unicode[STIX]{x1D70B}w_{p,l}(z)^{2}}\exp \left\{-c_{2}\left[\frac{x^{2}+y^{2}}{w_{p,l}(z)^{2}}\right]\right\},\end{eqnarray}$$

(8)

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$$\begin{eqnarray}w_{p,l}(z)=w_{0,p,l}\sqrt{\left[\frac{(z-z_{0})c_{p,l}l_{p,l}}{pw_{0,p,l}}\right]^{2}+1},\end{eqnarray}$$

(9)

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$$\begin{eqnarray}P_{p}=P_{p,0}\frac{2}{\unicode[STIX]{x1D708}_{p}}\sqrt{\frac{\ln 2}{\unicode[STIX]{x1D70B}}}\exp \left[-4\ln 2\frac{(\unicode[STIX]{x1D708}-\unicode[STIX]{x1D708}_{p})^{2}}{\unicode[STIX]{x0394}\unicode[STIX]{x1D708}_{p}^{2}}\right],\end{eqnarray}$$

(10)

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$$\begin{eqnarray}\displaystyle \displaystyle P_{p}(z+\unicode[STIX]{x0394}z) & = & \displaystyle P_{p}(z)\mathop{\sum }_{-R\leqslant x,y\leqslant R}f_{p}(x,y,z)\nonumber\\ \displaystyle & & \displaystyle \times \exp [-(n_{1}-n_{2})\unicode[STIX]{x1D70E}_{D2}(\unicode[STIX]{x1D708})\unicode[STIX]{x0394}z]\unicode[STIX]{x0394}x\unicode[STIX]{x0394}y,\qquad\end{eqnarray}$$

(11)

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$$\begin{eqnarray}\displaystyle \displaystyle P_{l}^{\pm }(z+\unicode[STIX]{x0394}z) & = & \displaystyle P_{l}^{\pm }(z)\mathop{\sum }_{-R\leqslant x,y\leqslant R}f_{l}(x,y,z)\nonumber\\ \displaystyle & & \displaystyle \times \exp [-(n_{3}-2n_{0})\unicode[STIX]{x1D70E}_{D1}\unicode[STIX]{x0394}z]\unicode[STIX]{x0394}x\unicode[STIX]{x0394}y.\end{eqnarray}$$

(12)

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$$\begin{eqnarray}\displaystyle & \displaystyle P_{l}^{-}(0)=P_{l}/[T_{l}(1-R_{oc})], & \displaystyle\end{eqnarray}$$

(13)

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$$\begin{eqnarray}\displaystyle & \displaystyle P_{l}^{+}(0)=P_{l}T_{l}R_{oc}/(1-R_{oc}). & \displaystyle\end{eqnarray}$$

(14)

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$$\begin{eqnarray}c_{p}\unicode[STIX]{x1D70C}\frac{\unicode[STIX]{x2202}T}{\unicode[STIX]{x2202}t}=K\unicode[STIX]{x0394}T+Q,\end{eqnarray}$$

(15)

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$$\begin{eqnarray}c_{p}\unicode[STIX]{x1D70C}\frac{\unicode[STIX]{x2202}T}{\unicode[STIX]{x2202}t}=K\left(\frac{\unicode[STIX]{x2202}^{2}T}{\unicode[STIX]{x2202}x^{2}}+\frac{\unicode[STIX]{x2202}^{2}T}{\unicode[STIX]{x2202}y^{2}}+\frac{\unicode[STIX]{x2202}^{2}T}{\unicode[STIX]{x2202}z^{2}}\right)+Q(x,y,z,t).\end{eqnarray}$$

(16)

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$$\begin{eqnarray}\left\{\begin{array}{@{}l@{}}\displaystyle \frac{\unicode[STIX]{x2202}T}{\unicode[STIX]{x2202}t}=\frac{T_{i,j,k}^{t+1}-T_{i,j,k}^{t}}{\unicode[STIX]{x0394}t},\\ \displaystyle \frac{\unicode[STIX]{x2202}^{2}T}{\unicode[STIX]{x2202}x^{2}}=\frac{T_{i+1,j,k}^{t}-2T_{i,j,k}^{t}+T_{i-1,j,k}^{t}}{\unicode[STIX]{x0394}x^{2}},\\ \displaystyle \frac{\unicode[STIX]{x2202}^{2}T}{\unicode[STIX]{x2202}y^{2}}=\frac{T_{i,j+1,k}^{t}-2T_{i,j,k}^{t}+T_{i,j-1,k}^{t}}{\unicode[STIX]{x0394}y^{2}},\\ \displaystyle \frac{\unicode[STIX]{x2202}^{2}T}{\unicode[STIX]{x2202}z^{2}}=\frac{T_{i,j,k+1}^{t}-2T_{i,j,k}^{t}+T_{i,j,k-1}^{t}}{\unicode[STIX]{x0394}z^{2}}.\end{array}\right.\end{eqnarray}$$

(17)

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$$\begin{eqnarray}\displaystyle T_{i,j,k}^{t+1} & = & \displaystyle (1-4\unicode[STIX]{x1D6FC}-2\unicode[STIX]{x1D6FD})T_{i,j,k}^{t}+\frac{\unicode[STIX]{x0394}t}{c_{p}\unicode[STIX]{x1D70C}}Q_{i,j,k}^{t}\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D6FD}(T_{i,j,k+1}^{t}+T_{i,j,k-1}^{t})\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D6FC}(T_{i+1,j,k}^{t}+T_{i-1,j,k}^{t}+T_{i,j+1,k}^{t}+T_{i,j-1,k}^{t}).\qquad\end{eqnarray}$$

(18)

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$$\begin{eqnarray}\displaystyle Q_{i,j,k}^{t} & = & \displaystyle (k_{01}n_{0_{i},j,k}^{t}[\text{Ar}]-k_{10}n_{1_{i},j,k}^{t})\unicode[STIX]{x0394}E_{10}\nonumber\\ \displaystyle & & \displaystyle +\,(-k_{32}n_{3_{i},j,k}^{t}[\text{Ar}]+k_{23}n_{2_{i},j,k}^{t})\unicode[STIX]{x0394}E_{23}.\qquad\end{eqnarray}$$

(19)

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$$\begin{eqnarray}c_{p}\unicode[STIX]{x1D70C}=\frac{c_{p-\text{Ar}}n_{\text{Ar}}}{N_{A}},\end{eqnarray}$$

(20)

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References (18)

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Chenyi Su, Binglin Shen, Xingqi Xu, Chunsheng Xia, Bailiang Pan. Simulation and analysis of the time evolution of laser power and temperature in static pulsed XPALs[J]. High Power Laser Science and Engineering, 2019, 7(3): 03000e44

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