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准确推导并求解激波层内部的辐射传输方程是研究高温激波层热辐射和传输特性的基础。由于气动热效应,红外窗口会被加热产生强红外辐射,因此,在求解激波层辐射传输方程时应将窗口辐射也考虑进来。在进行激波层辐射传输方程推导前需要做如下假设:
(1) 在目标辐射传输过程中,认为激波层对目标辐射的散射作用很小,仅考虑其对目标辐射的吸收作用;
(2) 激波层不仅是削弱辐射的介质,而且它本身也发射辐射,因此,在考虑激波层吸收作用的同时,也须考虑其自身辐射;
(3) 认为辐射场在每一时刻都可以看成是与发射源和吸收源的瞬时分布,即与激波层的温度和密度的瞬时分布相适应的准定常态,因而可以在辐射传输方程中忽略辐亮度对时间的导数。
因此,激波层的辐射传输方程可以表示为:
$$ \frac{{{\rm{d}}{L_\lambda }}}{{{\rm{d}}s}} = - k_\lambda ^{'}\left[ {{L_\lambda } - {B_\lambda }\left( T \right)} \right] $$ (1) 式中:λ为波长;Lλ为光谱辐射亮度,W∙cm−2∙sr−1∙μm−1;s为传输路径长度;kλ’为考虑诱导辐射后的吸收系数,cm−1[18-19],且
$$ k_\lambda ^{'}= {k_\lambda }\left[ {1 - {{{\rm{e}}} ^{ - \frac{{{c_2}}}{{\lambda T}}}}} \right] $$ (2) 式中:kλ为原始吸收系数,cm−1,可通过HITEMP数据库获得相关参数并根据实际流场信息进行修正和计算得到;Bλ(T)为普朗克函数, W∙cm−2∙sr−1∙μm−1:
$$ {B_\lambda }\left( T \right) = \frac{{{c_1}}}{{\pi {\lambda ^5}}}{\left( {{{\rm{e}}^{\frac{{{c_2}}}{{\lambda T}}}} - 1} \right)^{ - 1}} $$ (3) 式中:c1为第一辐射常数;c2为第二辐射常数。
在驻点区域附近,垂直于飞行器壁面方向的流场参数变化梯度较大,而切线方向的参数变化梯度较小[19-20]。因此,可以沿垂直飞行器壁面方向将激波层等效为每一层内的流场性质都均匀的无限大的分层结构,忽略切线方向的流场变化。从理论上来说,激波层分层数足够多,就可以对非均匀的激波层进行足够的近似。激波层内辐射传输过程如图1所示。到达激波层前,目标源辐射亮度为Lλ,t,之后其经过激波层和光学窗口的吸收作用,最终抵达探测器。
定义目标信号入射边界的坐标为x=0,飞行器窗口靠近激波层一侧的坐标为x=xw;在激波层内,垂直窗口且指向探测器的方向为正向,反方向为负向;θ为辐射传输方向与正向之间的夹角。
$\delta $ 为光学厚度:$$ \delta \left( x \right) = \int_0^x {k_\lambda ^{'}{\rm{d}}x} $$ (4) 正向初始辐射亮度为目标辐射Lλ,t,负向初始辐射亮度为窗口辐射Lλ,w,因此辐射传输方程(1)的解为[18-19]:
$$ \begin{split} L_\lambda ^{\text{ + }}\left( {x,\theta } \right) = & {L_{\lambda ,t}}{{\rm{e}}^{ - \delta \left( x \right)/\cos \theta }} +\\[-5pt] & \int_0^x {k_\lambda ^{'}\left( {{x^{'}}} \right){B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right]{{\rm{e}}^{ - \left( {\delta \left( x \right) - \delta \left( {{x^{'}}} \right)} \right)/\cos \theta }}\frac{{{\rm{d}}{x^{'}}}}{{\cos \theta }}} \\ \end{split} $$ (5) $$ \begin{split} L_\lambda ^{\text{ - }}\left( {x,\theta } \right) = &{L_{\lambda ,w}}{{\rm{e}}^{ - \left( {\delta \left( x \right){\text{ - }}\delta \left( {{x_w}} \right)} \right)/\cos \theta }} +\\[-5pt] & \int_{{x_w}}^x {k_\lambda ^{'}\left( {{x^{'}}} \right){B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right]{{\rm{e}}^{ - \left( {\delta \left( x \right) - \delta \left( {{x^{'}}} \right)} \right)/\cos \theta }}\frac{{{\rm{d}}{x^{'}}}}{{\cos \theta }}} \\ \end{split} $$ (6) 方程(5)为正向(
${\text{0}} \leqslant \theta \leqslant \pi /2$ )辐射传输方程的解,等号右边第一项代表目标辐射被激波层吸收后,在x处剩余的辐射亮度;等号右边第二项为在x处激波层的辐射亮度。方程(6)为辐射传输方程负向($\pi /2 \leqslant \theta \leqslant \pi $ )的解,根据方程(5)~(6)便可得到激波层内部任意位置的辐射场分布。 -
积分方程(4)~(6)需要用离散求和代替积分后进行求解。到达第n层流场时,光学厚度可用如下的分段近似表示:
$$ \begin{split} \delta \left( {{x_n}} \right) = &\int_0^{{x_n}} {k_\lambda ^{'}} \left( {{x^{'}}} \right){\rm{d}}{x^{'}} \approx \sum\limits_{k = 1}^{n - 1} {\int_{{x_k}}^{{x_{k + 1}}} {k_\lambda ^{'}\left( {{x^{'}}} \right)} } {\rm{d}}{x^{'}}\approx \\[-5pt] &\frac{1}{2}\sum\limits_{k = 1}^{n - 1} {\left[ {k_\lambda ^{'}\left( {{x_k}} \right) + k_\lambda ^{'}\left( {{x_{k + 1}}} \right)} \right]\left( {{x_{k + 1}} - {x_k}} \right)} \\ \end{split}$$ (7) 如果用每层中间位置xk,mid的流场性质代表该层流场状态,上式可简化为:
$$ \delta \left( {{x_n}} \right) \approx \sum\limits_{k = 1}^{n - 1} {k_\lambda ^{'}\left( {{x_{k,mid}}} \right)} \left( {{x_{k + 1}} - {x_k}} \right) $$ (8) 对于辐射亮度积分方程(5)~(6),由于有指数项,为了提高积分精度,需要假设每层流场内光学厚度近似线性[19],即
$$ \delta \left( x \right) = {a_k}x + {b_k} $$ (9) 因此,方程(5)离散后可以写为:
$$ \begin{split} L_\lambda ^ + \left( {{x_n},\theta } \right) =& {L_{\lambda ,t}}{{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }} + \int_0^{{x_n}} {k_\lambda ^{'}\left( {{x^{'}}} \right){B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right]} {{\rm{e}}^{ - \left( {\delta \left( {{x_n}} \right) - \delta \left( {{x^{'}}} \right)} \right)/\cos \theta }}{\rm{d}}\frac{{{x^{'}}}}{{\cos \theta }} =\\ & {L_{\lambda ,t}}{{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }} + \sum\limits_{k = 1}^{n - 1} {k_\lambda ^{'}\left( {{x_{k,mid}}} \right){B_\lambda }\left[ {T\left( {{x_{k,mid}}} \right)} \right]{{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }}\int_{{x_k}}^{{x_{k + 1}}} {{{\rm{e}}^{\delta \left( {{x^{'}}} \right)/\cos \theta }}{\rm{d}}\frac{{{x^{'}}}}{{\cos \theta }}} } =\\ & {L_{\lambda ,t}}{{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }} + \sum\limits_{k = 1}^{n - 1} {k_\lambda ^{'}\left( {{x_{k,mid}}} \right){B_\lambda }\left[ {T\left( {{x_{k,mid}}} \right)} \right]\frac{{{{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }}}}{{{a_k}}}\int_{{x_k}}^{{x_{k + 1}}} {{\rm{d}}{{\rm{e}}^{\delta \left( {{x^{'}}} \right)/\cos \theta }}} } =\\ & {L_{\lambda ,t}}{{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }} + \sum\limits_{k = 1}^{n - 1} {k_\lambda ^{'}\left( {{x_{k,mid}}} \right){B_\lambda }\left[ {T\left( {{x_{k,mid}}} \right)} \right]} \frac{{{{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }}}}{{{a_k}}}\left( {{{\rm{e}}^{\frac{{\delta \left( {{x_{k + 1}}} \right)}}{{\cos \theta }}}} - {{\rm{e}}^{\frac{{\delta \left( {{x_k}} \right)}}{{\cos \theta }}}}} \right) \\[-10pt] \end{split}$$ (10) ak可由每层流场两端的光学厚度推出:
$$\begin{gathered} \left\{ \begin{gathered} \delta \left( {{x_{k + 1}}} \right) = {a_k}{x_{k + 1}} + {b_k} \hfill \\ \delta \left( {{x_k}} \right) = {a_k}{x_k} + {b_k} \hfill \\ \end{gathered} \right. \Rightarrow {a_k} = \frac{{\delta \left( {{x_{k + 1}}} \right) - \delta \left( {{x_k}} \right)}}{{{x_{k + 1}} - {x_k}}} = k_\lambda ^{'}\left( {{x_{k,mid}}} \right) \end{gathered}$$ (11) 将公式(11)代入公式(10),离散后正方向辐射亮度进一步表达为:
$$\begin{split} L_\lambda ^ + \left( {{x_n},\theta } \right) =& {L_{\lambda ,t}}{{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }}+\\ & \sum\limits_{k = 1}^{n - 1} {{B_\lambda }\left[ {T\left( {{x_{k,mid}}} \right)} \right]} {{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }}\left( {{{\rm{e}}^{\frac{{\delta \left( {{x_{k + 1}}} \right)}}{{\cos \theta }}}} - {{\rm{e}}^{\frac{{\delta \left( {{x_k}} \right)}}{{\cos \theta }}}}} \right) \end{split}$$ (12) 同理,反方向离散后辐射亮度表示为:
$$\begin{split} L_\lambda ^ - \left( {{x_n},\theta } \right) = &{L_{\lambda ,w}}{{\rm{e}}^{ - \left( {\delta \left( {{x_n}} \right) - \delta \left( {{x_w}} \right)} \right)/\cos \theta }} - \\ &\sum\limits_{k = n}^{{k_{\max }} - 1}{{B_\lambda }\left[ {T\left( {{x_{k,mid}}} \right)} \right]} {{\rm{e}}^{ - \delta \left( {{x_n}} \right)/\cos \theta }}\left( {{{\rm{e}}^{\frac{{\delta \left( {{x_{k + 1}}} \right)}}{{\cos \theta }}}} - {{\rm{e}}^{\frac{{\delta \left( {{x_k}} \right)}}{{\cos \theta }}}}} \right) \end{split}$$ (13) -
令方程(5)中x=xw(或令方程(12)中
$n = {k_{\max }}$ ),可以得到激波层对θ方向上单根目标光束的透过率以及该方向上激波层的辐射亮度,如下式所示:$$ \begin{split} L_\lambda ^{\text{ + }}\left( {{x_w},\theta } \right) =& {L_{\lambda ,t}}{{\rm{e}}^{ - \delta \left( {{x_w}} \right)/\cos \theta }} + \\ & \int_0^{{x_w}} {k_\lambda ^{'}\left( {{x^{'}}} \right){B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right]{{\rm{e}}^{ - \left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)/\cos \theta }}\frac{{{\rm{d}}{x^{'}}}}{{\cos \theta }}} \\ \end{split} $$ (14) 式中:
${{\rm{e}}^{ - {\delta _0}/\cos \theta }}$ 即为θ (${\text{0}} \leqslant \theta \leqslant \pi /2$ )方向激波层的透过率。等号右边第二项为激波层内气体在θ方向上产生的辐射亮度Lλ,s:$$ {L_{\lambda ,s}}\left( {{x_w},\theta } \right) = \int_0^{{x_w}} {k_\lambda ^{'}{B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right]{{\rm{e}}^{ - \left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)/\cos \theta }}\frac{{{\rm{d}}{x^{'}}}}{{\cos \theta }}} $$ (15) 离散后激波层辐射亮度为:
$$ \begin{split} {L_{\lambda ,s}}\left( {{x_{{k_{\max }}}},\theta } \right) =&\sum\limits_{k = 1}^{{k_{\max }} - 1} {\{ {B_\lambda }\left[ {T\left( {{x_{k,mid}}} \right)} \right]} \times \\ &{{\rm{e}}^{ - \delta \left( {{x_{{k_{\max }}}}} \right)/\cos \theta }}({{\rm{e}}^{\frac{{\delta \left( {{x_{k + 1}}} \right)}}{{\cos \theta }}}} - {{\rm{e}}^{\frac{{\delta \left( {{x_k}} \right)}}{{\cos \theta }}}})\} \\ \end{split} $$ (16) 当目标辐射为平行辐射时,激波层透过率为:
$$ {\tau _{\lambda ,s}} = {{\rm{e}}^{ - \delta \left( {{x_w}} \right)/\cos \theta }} $$ (17) 若目标辐射不可视为平行辐射,需要将公式(14)中第一项对立体角进行积分,通过求解目标源穿过激波层后剩余的辐射通量密度来获得激波层的透过率。
$$ {E_n}(X) = \int_1^\infty {{{\rm{e}}^{ - \xi X}}\frac{{{\rm{d}}\xi }}{{{\xi ^n}}}} $$ (18) 显然,
$\dfrac{{{\rm{d}}{E_n}\left( X \right)}}{{{\rm{d}}X}} = - {E_{n - 1}}\left( X \right)$ 。根据辐射通量计算公式:$$ M = \int {L\cos \theta {\rm{d}}\Omega } $$ (19) 则目标信号穿过激波层后辐射通量密度为:
$$ \begin{split} {M_{\lambda ,t}} =& \int {{L_{\lambda ,t}}{{\rm{e}}^{ - \delta \left( {{x_w}} \right)/\cos \theta }}\cos \theta {\rm{d}}\Omega } \hfill =\\ & \int_0^{2\pi } {\int_0^{\frac{\pi }{2}} {{L_{\lambda ,t}}{{\rm{e}}^{ - \delta \left( {{x_w}} \right)/\cos \theta }}\cos \theta \sin \theta {\rm{d}}\theta {\rm{d}}\varphi } } \hfill =\\ & 2\pi \cdot {L_{\lambda ,t}} \cdot \int_0^1 {{{\rm{e}}^{ - \delta \left( {{x_w}} \right)/\cos \theta }}\cos \theta {\rm{d}}\cos \theta } \hfill \\ \end{split} $$ (20) 令
$\cos \theta = \dfrac{1}{\xi }$ ,则$$ \begin{split} \int_0^1 {{{\rm{e}}^{ - \delta \left( {{x_w}} \right)/\cos \theta }}\cos \theta {\rm{d}}\cos \theta } \hfill =&\int_1^\infty {{{\rm{e}}^{ - \delta \left( {{x_w}} \right)\xi }}} \frac{{{\rm{d}}\xi }}{{{\xi ^3}}} \\ =&{E_3}\left( {\delta \left( {{x_w}} \right)} \right) \hfill \\ \end{split} $$ (21) 将上式代入公式(20)中:
$$ \begin{gathered} {M_{\lambda ,t}} = 2\pi \cdot {L_{\lambda ,t}} \cdot {E_3}\left( {\delta \left( {{x_w}} \right)} \right)= 2{E_3}\left( {\delta \left( {{x_w}} \right)} \right) \cdot \pi {L_{\lambda ,t}} \\ \end{gathered} $$ (22) 式中:
$2{E_3}\left( {\delta \left( {{x_w}} \right)} \right)$ 即为激波层的透过率。综上,激波层透过率为:
$$ {\tau }_{\lambda ,s}=\left\{\begin{array}{l}{{\rm{e}}}^{-\delta \left({x}_{w}\right)/\mathrm{cos}\theta }\text{=}{{\rm{e}}}^{-\delta \left({x}_{{k}_{\mathrm{max}}}\right)/\mathrm{cos}\theta }\;\;\;\;\;\;\;\;\;\;\;\;\;\;平行辐射\\ 2{E}_{3}\left(\delta \left({x}_{w}\right)\right)=2{E}_{3}\left(\delta \left({x}_{{k}_{\mathrm{max}}}\right)\right)\;\;\;\;漫射辐射\end{array}\right. $$ (23) -
公式(15)对立体角积分,可以得到高温激波层射向探测器方向的辐射出射度Mλ,s:
$$ \begin{split} & {{M_{\lambda ,s}} = \displaystyle\int {\left\{ {\displaystyle\int_0^{{x_w}} {k_\lambda ^{'}\left( {{x^{'}}} \right){B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right]{{\rm{e}}^{ - \left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)/\cos \theta }}\dfrac{{{\rm{d}}{x^{'}}}}{{\cos \theta }}} } \right\}\cos \theta {\rm{d}}\Omega} =}\\ &{\displaystyle\int_0^{2\pi } {\displaystyle\int_0^{\tfrac{\pi }{2}} {\left\{ {\displaystyle\int_0^{{x_w}} {k_\lambda ^{'}\left( {{x^{'}}} \right){B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right]{{\rm{e}}^{ - \left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)}\right)\cos \theta }}\dfrac{{{\rm{d}}{x^{'}}}}{{\cos \theta }}} } \right\}\cos \theta \sin \theta {\rm{d}}\theta } {\rm{d}}\varphi } }=\\ &{ 2\pi \cdot \displaystyle\int_0^{{x_w}} {\left\{ {k_\lambda ^{'}\left( {{x^{'}}} \right){B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right] \cdot \displaystyle\int_0^1 {{{\rm{e}}^{ - \left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)/\cos \theta }}{\rm{d}}\cos \theta } } \right\}} {\rm{d}}{x^{'}}} \end{split} $$ (24) 令
$\cos \theta = \dfrac{1}{\xi }$ ,则$$ \begin{split} \int_0^1 {{{\rm{e}}^{ - \left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)/\cos \theta }}{\rm{d}}\cos \theta }=& \int_1^\infty {{{\rm{e}}^{ - \left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)\xi }}} \frac{{{\rm{d}}\xi }}{{{\xi ^2}}} =\\ &{E_2}\left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right) \end{split}$$ (25) 根据公式(9)、(11)、(18)可得:
$$\begin{split} {E_2}\left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right){\rm{d}}{x^{'}} =& \frac{{{E_2}\left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right){\rm{d}}\left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)}}{{ - {a_k}}} =\\ & \frac{{ - {\rm{d}}{E_3}\left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)}}{{ - k_\lambda ^{'}\left( {{x^{'}}} \right)}} =\\ & \frac{{{\rm{d}}{E_3}\left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)}}{{k_\lambda ^{'}\left( {{x^{'}}} \right)}} \\[-6pt] \end{split}$$ (26) 将公式(25)、(26) 代入公式(24) 中,并进行离散化,可得激波层内高温气体产生的辐射出射度为:
$$ \begin{split} {M_{\lambda ,s}}=& 2\pi \cdot \int_0^{{x_w}} {k_\lambda ^{'}\left( {{x^{'}}} \right){B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right] \cdot {E_2}\left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right)} {\rm{d}}{x^{'}} =\\ &2\pi \cdot \int_0^{{x_w}} {{B_\lambda }\left[ {T\left( {{x^{'}}} \right)} \right]} {\rm{d}}{E_3}\left( {\delta \left( {{x_w}} \right) - \delta \left( {{x^{'}}} \right)} \right) =\\ &2\pi \cdot \sum\limits_{k = 1}^{{k_{\max }} - 1} {{B_\lambda }\left[ {T\left( {{x_{k,mid}}} \right)} \right]\int_{{x_k}}^{{x_{k + 1}}} {{\rm{d}}{E_3}\left( {\delta \left( {{x_{{k_{\max }}}}} \right) - \delta \left( {{x^{'}}} \right)} \right)} } =\\ &2\pi \cdot \sum\limits_{k = 1}^{{k_{\max }} - 1} {B_\lambda }\left[ {T\left( {{x_{k,mid}}} \right)} \right]\left[ {E_3}\left( {\delta \left( {{x_{{k_{\max }}}}} \right) - \delta \left( {{x_{k + 1}}} \right)} \right) -\right. \\ & {{{E_3}\left( {\delta \left( {{x_{{k_{\max }}}}} \right) - \delta \left( {{x_k}} \right)} \right)}\left. \right]} \\[-5pt] \end{split} $$ (27) 如果整个激波层近似认为温度恒定,即激波层仅分一层时,上式可以进一步简化为:
$$ \begin{split} {M_{\lambda ,s}} =& { 2\pi \cdot {B_\lambda }\left( T \right)\displaystyle\sum\limits_{k = 1}^{{k_{\max }} - 1} {\left[ {{E_3}\left( {\delta \left( {{x_{{k_{\max }}}}} \right) - \delta \left( {{x_{k + 1}}} \right)} \right) - {E_3}\left( {\delta \left( {{x_{{k_{\max }}}}} \right) - \delta \left( {{x_k}} \right)} \right)} \right]} =}\\ & {2\pi \cdot{B_\lambda }\left( T \right) \left[ {{E_3}\left( {\delta \left( {{x_{{k_{\max }}}}} \right) - \delta \left( {{x_{{k_{\max }}}}} \right)} \right) - {E_3}\left( {\delta \left( {{x_{{k_{\max }}}}} \right) - \delta \left( {{x_1}} \right)} \right)} \right] =}\\ & 2\pi \cdot {B_\lambda }\left( T \right)\left[ {{E_3}\left( 0 \right) - {E_3}\left( {\delta \left( {{x_{{k_{\max }}}}} \right)} \right)} \right] =\\ & \pi \cdot {B_\lambda }\left( T \right)\left[ {1 - 2{E_3}\left( {\delta \left( {{x_{{k_{\max }}}}} \right)} \right)} \right] \\[-7pt] \end{split} $$ (28) -
此节给出不同辐射源(目标源、激波层和窗口)在探测器光敏面产生的辐射通量及其占比的计算模型,以此来分析激波层和窗口热辐射噪声在不同波段的占比变化,为寻求探测器最优工作波段提供数值分析依据。
文中仅考虑图2所示的面目标源情况,且忽略大气衰减影响。其中,At为发射面积即目标源面积,Ao为光学入瞳的面积, Ad为光敏面的面积,ω1为入瞳面对目标中心所张立体角,ω为系统瞬时视场角,ω2为光敏面的接收立体角,R为目标源与入瞳面之间的距离,f为光学系统焦距。
此时,探测器入瞳面只能接收到目标源部分表面发出的辐射。因此,目标源在入瞳处产生的辐照度为:
$$ {E_{\lambda ,t}} = {\tau _{\lambda ,w}} \cdot {\tau _{\lambda ,s}} \cdot {L_{\lambda ,t}} \cdot \omega $$ (29) 式中:
${\tau _{\lambda ,w}}$ 为窗口透过率;${\tau _{\lambda ,s}}$ 为激波层透过率。目标源在探测器光敏面产生的辐射通量为:
$$ \begin{split} {\varPhi _{\lambda ,t}} &= {E_{\lambda ,t}} \cdot {A_o} \cdot {\tau _o}= {\tau _{\lambda ,w}} \cdot {\tau _{\lambda ,s}} \cdot {\tau _o} \cdot {L_{\lambda ,t}} \cdot \omega \cdot {A_o} \\ \end{split} $$ (30) 式中:
${\tau _o}$ 为光学系统效率。近似认为在系统瞬时视场角范围内,激波层的辐射亮度为常数,等于
$ \theta = 0 $ 时的辐射亮度,即:$$ {L_{\lambda ,s}}\left( {{x_{{k_{\max }}}},0} \right) = \sum\limits_{k = 1}^{{k_{\max }} - 1} {{B_\lambda }\left[ {T\left( {{x_{k,mid}}} \right)} \right]{{\rm{e}}^{ - \delta \left( {{x_{{k_{\max }}}}} \right)}}} \left( {{{\rm{e}}^{\delta \left( {{x_{k + 1}}} \right)}} - {{\rm{e}}^{\delta \left( {{x_k}} \right)}}} \right) $$ (31) 激波层和窗口辐射均可以视为近场辐射噪声[22],参照目标辐射通量计算公式(30),可以得到高温激波层在探测器光敏面产生的辐射通量为:
$$ {\varPhi _{\lambda ,s}} = {\tau _{\lambda ,w}} \cdot {\tau _o} \cdot {L_{\lambda ,s}} \cdot \omega \cdot {A_o} $$ (32) 高温窗口在探测器光敏面产生的辐射通量为:
$$ {\varPhi _{\lambda ,w}} = {\tau _o} \cdot {L_{\lambda ,w}} \cdot \omega \cdot {A_o} $$ (33) 因此,在不考虑背景辐射时,目标源产生的辐射通量占光敏面接收到的总辐射通量的比例为:
$$ \begin{split} {\eta _{\lambda ,t}} &= \frac{{{\varPhi _{\lambda ,t}}}}{{{\varPhi _{\lambda ,t}} + {\varPhi _{\lambda ,s}} + {\varPhi _{\lambda ,w}}}} = \\ &\frac{{{\tau _{\lambda ,w}} \cdot {\tau _{\lambda ,s}} \cdot {L_{\lambda ,t}}}}{{{\tau _{\lambda ,w}} \cdot {\tau _{\lambda ,s}} \cdot {L_{\lambda ,t}} + {\tau _{\lambda ,w}} \cdot {L_{\lambda ,s}} + {L_{\lambda ,w}}}} \\ \end{split} $$ (34) 激波层产生的辐射通量占光敏面接收到的总辐射通量的比例:
$$ \begin{split} {\eta _{\lambda ,s}} =& \frac{{{\varPhi _{\lambda ,s}}}}{{{\varPhi _{\lambda ,t}} + {\varPhi _{\lambda ,s}} + {\varPhi _{\lambda ,w}}}} =\\ & \frac{{{\tau _{\lambda ,w}} \cdot {L_{\lambda ,s}}}}{{{\tau _{\lambda ,w}} \cdot {\tau _{\lambda ,s}} \cdot {L_{\lambda ,t}} + {\tau _{\lambda ,w}} \cdot {L_{\lambda ,s}} + {L_{\lambda ,w}}}} \\ \end{split} $$ (35) 高温窗口产生的辐射通量占光敏面接收到的总辐射通量的比例:
$$ \begin{split} {\eta _{\lambda ,w}} =& \frac{{{\varPhi _{\lambda ,w}}}}{{{\varPhi _{\lambda ,t}} + {\varPhi _{\lambda ,s}} + {\varPhi _{\lambda ,w}}}} =\\ &\frac{{{L_{\lambda ,w}}}}{{{\tau _{\lambda ,w}} \cdot {\tau _{\lambda ,s}} \cdot {L_{\lambda ,t}} + {\tau _{\lambda ,w}} \cdot {L_{\lambda ,s}} + {L_{\lambda ,w}}}} \\ \end{split} $$ (36) 从公式(26)~(28)可以看出,计算激波层和窗口热辐射占比,无需计算其在光敏面上产生的辐射通量,仅需计算各辐射源的辐射亮度以及传输介质的透过率即可。
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文中以某型超声速红外制导导弹为研究对象,重点研究飞行高度为h=1 km、飞行速度为Ma=3,4,5时激波层的红外辐射与传输特性。高温激波层内的H2O、CO2等气体是红外探测的主要辐射噪声源。文中为说明切线平板计算方法,仅以CO2气体为例,计算其红外辐射和传输特性,并假设空气中的CO2含量为0.033%,忽略气体的电离、离解。
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图3~5所示为h=1 km、Ma=3,4,5时某导弹的流场仿真结果。根据仿真结果将激波层驻点区域进行分层,并假设每层流场状态恒定。文中为说明方法同时简化计算,仅将流场分为两层。
图 3 Ma=3、h=1 km时激波层流场仿真结果。(a) 压力;(b) 温度;(c) 流场近似
Figure 3. Flow field simulation results in the shock layers at Ma=3, h=1 km. (a) Pressure; (b) Temperature; (c) Flow field approximation
图 4 Ma=4、h=1 km时激波层流场仿真结果。(a) 压力;(b) 温度;(c) 流场近似
Figure 4. Flow field simulation results in the shock layers at Ma=4, h=1 km. (a) Pressure; (b) Temperature; (c) Flow field approximation
图 5 Ma=5、h=1 km时激波层流场仿真结果。(a) 压力;(b) 温度;(c) 流场近似
Figure 5. Flow field simulation results in the shock layers at Ma=5, h=1 km. (a) Pressure; (b) Temperature; (c) Flow field approximation
图3(a)、(b)分别为Ma=3、h=1 km时激波层的压力和温度分布图,图3(c)为该飞行条件下的激波层近似流场。第一层流场坐标区间为x1=−19 mm~x2=−9 mm,该层温度为T=516 K,压力为p=4.5 atm,厚度为10 mm;第二层坐标区间为x2=−9 mm~x3=0 mm,该层温度为T=777 K,压力为p=9.7 atm,厚度为9 mm。
图4(a)、(b)分别为Ma=4、h=1 km时激波层的压力和温度分布图,图4(c)为该飞行条件下激波层近似流场,第一层坐标区间为x1=−14 mm~x2=−7 mm,该层温度为T=688 K,压力为p=10.58 atm,厚度为7 mm;第二层坐标区间为x2=−7 mm~x3=0 mm),该层温度为T=1161 K,压力为p=17.3 atm,厚度为7 mm。
图5(a)、(b)分别为Ma=5,h=1 km时激波层的压力和温度分布图,图5(c)为该飞行条件下激波层近似流场,第一层坐标区间为x1=−11 mm~x2=−6 mm,该层温度为T=956 K,压力为p=14.55 atm,厚度为5 mm;第二层坐标区间为x2=−6 mm~x3=0 mm,该层温度为T=1667 K,压力为p=26.4 atm,厚度为6 mm。
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根据理论分析和流场仿真结果,计算得到h=1 km、 Ma=3,4,5时激波层CO2的透过率、光谱辐射亮度和辐射出射度曲线,如图6所示。其中图6(a)、(b)分别为2.5~3 μm和4.1~4.8 μm激波层的透过率曲线。从图6(a)中可以看出,短波波段激波层的透过率几乎为1,可以忽略CO2的吸收作用。对比图6(a)、(b)可以看出,与短波波段相比,中波波段激波层的透过率有所下降。Ma=3、h=1 km时,透过率最小值为0.81;Ma=4、h=1 km时,透过率最小值为0.88;Ma=5、h=1 km时,透过率最小值为0.94。总体来说,激波层对目标辐射的吸收作用不是很强。
图 6 Ma=3,4,5、h=1 km时激波层内CO2 (a) 2.5~3 μm透过率;(b) 4.1~4.8 μm透过率;(c) 2.5~3 μm光谱辐射亮度;(d) 4.1~4.8 μm光谱辐射亮度;(e) 2.5~3 μm辐射出射度;(f) 4.1~4.8 μm辐射出射度
Figure 6. (a) 2.5-3 μm transmissibity; (b) 4.1-4.8 μm transmissibity; (c) 2.5-3 μm spectral radiance; (d) 4.1-4.8 μm spectral radiance; (e) 2.5-3 μm spectral radiant emittance; (f) 4.1-4.8 μm spectral radiant emittance of carbon dioxide in shock layer at h=1 km, Ma=3, 4, 5 respectively
图6(c)、(d)为θ =0时,2.5~3 μm和4.1~4.8 μm波段激波层的光谱辐射亮度。从图6(d)中可以看出,在中波波段,Ma=3、h=1 km时辐射亮度峰值为0.005 W∙cm−2∙sr−1∙μm−1,辐射亮度大于零区间为4.18~4.45 μm;Ma=4、h=1 km时辐射亮度峰值为0.013 W∙cm−2∙sr−1∙μm−1,辐射亮度大于零区间为4.18~4.6 μm;Ma=5,h=1 km时辐射亮度峰值为0.023 W∙cm−2∙sr−1∙μm−1,辐射亮度大于零的区间为4.18~4.8 μm。飞行高度相同时,飞行速度越快激波层的辐射亮度越强,且辐射亮度大于零的区间明显向右拓宽。对比图6(c)、(d)可以看出,相同飞行条件下,短波波段的辐射亮度小于中波波段。根据公式(27)可以得到激波层的光谱辐射出射度,如图6(e)、(f)所示。激波层的光谱辐射出射度与光谱辐射亮度具有相似的变化规律。
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此节分别计算h=1 km、 Ma=3,4,5时目标、激波层和窗口辐射占比。假设导引头窗口材料为蓝宝石,窗口厚度为5 mm,目标源为温度300 K、发射率为0.9的灰体。
根据公式(34)~(36)分别计算上述三种飞行条件下各辐射源在光敏面的辐射占比变化,如图7所示。其中图7(a)~(c)中飞行速度分别为Ma=3, Ma=4, Ma=5。通过对比不同飞行条件下的各辐射源占比曲线可知,随着飞行速度的增加,激波层内的CO2在长波波段的辐射占比逐渐增加。
表1给出了4.2~4.4 μm和4.4~4.6 μm波段各辐射分量占比的均值。从图7和 表1中可以看出,当Ma=3、h=1 km时,4.4 μm以上波段激波层内CO2的辐射占比小于目标辐射,截止波长为4.4 μm的滤波器可以滤除激波热辐射噪声干扰。而Ma=4、h=1 km和Ma=5、h=1 km时,4.4 μm以上波段CO2的辐射占比远大于目标辐射,说明激波层辐射噪声会淹没目标信号,截止波长为4.4 μm的滤波器无法适用于高速、低空飞行。
表 1 中波波段目标、激波和窗口辐射占比
Table 1. Proportion of target, shock layers and window radiation in mid-wave band
Conditions Proportion Wavelength/μm 4.2-4.4 4.4-4.6 Ma=3
h=1 km`ηt 0.12 0.72 `ηs 0.86 0.17 `ηw 0.02 0.11 Ma=4
h=1 km`ηt 0.03 0.30 `ηs 0.96 0.65 `ηw 0.01 0.05 Ma=5
h=1 km`ηt
`ηs
`ηw0.02
0.98
0.000.06
0.92
0.02
A numerical study of carbon dioxide radiation and transmission property in high temperature shock layer
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摘要: CO2是中、短波红外探测的主要辐射噪声源,目前针对高温CO2辐射和传输特性的研究尚不明晰。重点研究了激波层中CO2的吸收作用,并对CO2热辐射及光敏面各辐射分量占比进行了计算与分析。采用改进的切线平板法求解辐射传输方程,并考虑了目标与窗口辐射对激波层内辐射场分布的影响,据此推导了激波层红外辐射和传输特性计算模型,最后给出了探测器光敏面各辐射分量及占比计算模型。对某型超声速红外制导导弹飞行高度h=1 km、飞行速度Ma=3~5时激波层特性进行仿真计算,结果表明:中波波段激波层透过率低于短波波段,但总体来说可忽略激波层吸收作用;飞行速度增加导致CO2辐射噪声区间向长波方向拓宽,Ma≥4、h=1 km时,4.4 μm以上波段有严重的CO2辐射噪声淹没目标信号,截止波长为4.4 μm的滤波器无法适用。Abstract: CO2 is the main source of radiation noise in mid-infrared and short-infrared detection. The research on radiation and transmission property of CO2 at high temperature is still not clear at present. The absorption effect of CO2 in the shock layer was mainly studied, and the thermal radiation of CO2 and radiation component proportion on the photosensitive surface were analyzed and calculated. The improved tangent-slab approximation method was used to solve the radiative transport equation, and the influences of target and window radiation on the radiation field distribution in the shock layer were considered. Based on this, the radiation and transmission property of shock layer were deduced. Finally, the calculation model of each radiation component and proportion of the photosensitive surface was given. The property of the shock layer of a supersonic missile at an altitude of h=1 km and flight speed of Ma=3-5 was simulated and calculated. The results showed that the transmittance of shock layer in mid-wave bands is lower than that in short-wave band, but shock layer absorption can be ignored in general. The increase of flight speed leads to a broadening of the CO2 radiated noise region to the long-wave direction, and the target signal is submerged seriously by CO2 radiation noise in the band above 4.4 μm under condition of Ma≥4 and h=1 km, the filter with a cutoff wavelength of 4.4 μm is not suitable.
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Key words:
- supersonic /
- high temperature shock layer /
- radiative transport /
- radiation noise
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图 6 Ma=3,4,5、h=1 km时激波层内CO2 (a) 2.5~3 μm透过率;(b) 4.1~4.8 μm透过率;(c) 2.5~3 μm光谱辐射亮度;(d) 4.1~4.8 μm光谱辐射亮度;(e) 2.5~3 μm辐射出射度;(f) 4.1~4.8 μm辐射出射度
Figure 6. (a) 2.5-3 μm transmissibity; (b) 4.1-4.8 μm transmissibity; (c) 2.5-3 μm spectral radiance; (d) 4.1-4.8 μm spectral radiance; (e) 2.5-3 μm spectral radiant emittance; (f) 4.1-4.8 μm spectral radiant emittance of carbon dioxide in shock layer at h=1 km, Ma=3, 4, 5 respectively
表 1 中波波段目标、激波和窗口辐射占比
Table 1. Proportion of target, shock layers and window radiation in mid-wave band
Conditions Proportion Wavelength/μm 4.2-4.4 4.4-4.6 Ma=3
h=1 km`ηt 0.12 0.72 `ηs 0.86 0.17 `ηw 0.02 0.11 Ma=4
h=1 km`ηt 0.03 0.30 `ηs 0.96 0.65 `ηw 0.01 0.05 Ma=5
h=1 km`ηt
`ηs
`ηw0.02
0.98
0.000.06
0.92
0.02 -
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