透镜聚光温度计算

Creato il: 31 gennaio 2025

Risposto usando O1 da Chat01

Domanda

在地面上方垂直于太阳光的入射方向，放置一半径 R=0.10m、焦距 f=0.50m的薄凸透镜，在薄透镜下
方的焦面上放置一黑色薄圆盘(圆盘中心与透镜焦点重合)，于是可以在黑色圆盘上形成太阳的像,已知黑色圆盘的半
径是太阳像的半径的两倍.圆盘的导热性极好,圆盘与地面之间的距离较大,设太阳向外辐射的能量遵从斯特藩一玻尔
兹曼定律;在单位时间内在其单位表面积上向外辐射的能量为W。T，式中o为斯特藩一玻尔兹曼常量，T为辐射体
表面的的绝对温度,对太而言，取其温度t5.50X10’℃,大气对太阳能的吸收率为a=0.40.又设黑色圆盘对射到其上
的太阳能全部吸收，同时圆盘也按斯特藻一玻尔兹曼定律向外辐射能量,如果不考虑空气的对流，也不考虑杂散光的
影响，试问薄圆盘到达稳定状态时可能达到的最高温度为多少摄氏度?

Pensiero

Understanding the setup

OK, we’ve got a thin convex lens focusing sunlight on a black disc below, with the disc's radius twice that of the sun's image radius. The disc has great thermal conduction and sits high up for steady state conditions.

Clarifying sun's temperature

Mapping out the sun's temperature around 5773 K from 5.50 x 10^3 °C. Now looking at how it impacts the black disc's optimal temperature.

Analyzing light absorption

I’m thinking about how 40% of sunlight is absorbed by the atmosphere, leaving 60% to hit the ground. The black disc absorbs all this light and emits radiation.

Mapping solar flux

I’m working through the solar flux concentrated by the lens onto the disc, using the Stefan-Boltzmann formula to understand how the sun's radiation is distributed at Earth's distance.

Figuring out solar flux

I’m deriving the solar flux at Earth's distance, then factoring in atmospheric absorption and how the lens concentrates light onto a larger disc area.

Mapping the flux

OK, let’s think: the lens focuses the entire flux onto the disc's area, which is four times that of the sun's image.

Measuring flux

I’m seeing that the lens collects solar flux, which is then spread over the sun's image area and concentrated onto a larger disc area, reducing the average flux by a factor of 4.

Balancing flux

Equating the power hitting the disk to the power radiated as blackbody radiation shows flux and radiance balance. This forms a critical thermodynamic limit for the system.

Maximizing temperature

To achieve the highest temperature with a lens system, it should not exceed the sun's surface temperature. The lens and disk setup limits the maximum flux concentration, adhering to the second law of thermodynamics.

Mapping light sources

OK, let me see. The disk, four times larger than the sun's image, is partially blocked. Only a quarter of the disk's area is getting direct light from the lens.

Analyzing energy movement

I'm digging into how the blackbody radiation from the sun, spread across space, is gathered by a lens and dispersed evenly over its surface, illustrating the flux redistribution effect.

Balancing energy flux

The power entering the lens must match the power radiated by the disk. The disk's temperature and image radius are derived, aligning energy input and output.

Balancing power

By equating the power entering the lens and the power radiated by the disk, I canceled out constants and calculated a relationship between the lens radius, disk temperature, and focal length.

Calculating disk temperature

I found the temperature of the disk by substituting the lens radius, focal length, and sun's temperature, arriving at $T_d = T_{sun} (0.15 \frac{R_{lens}^2}{f^2})^{1/4}$ .

Calculating temperature

I worked out the fourth root of 0.006 to estimate the disk's temperature compared to the sun's, leading to a conversion for a more intuitive understanding in Celsius.

Estimating temperature

Converting the fourth root of 0.006 into a logarithmic expression, I determined the disk's temperature, confirming consistency and calculating the difference in Celsius.

Clarifying temperature calculations

I examined the sun's temperature in Celsius versus Kelvin, confirming consistent calculations across various methods, and determined the disk's peak temperature.

Adjusting calculations

I found that increasing the disk's radius by a factor of 2 reduces the flux by 4. Adding the atmospheric absorption factor modifies the calculation of the disk's temperature.

Clarifying the approach

I’m working through the brightness theorem and the role of atmospheric absorption. The disk's size compared to the lens doesn’t matter if it's smaller than the sun's image.

Concentrating radiant flux

I’m gathering info on the concentration ratio of the disk’s radiant flux to the sun's. The atmospheric absorption factor is crucial, as it reduces the flux to 60% of the original sunlight.

Revisiting the ratio

I worked through the flux density ratio again, ensuring the atmospheric absorption factor and the significant reduction in flux. This requires more precise calculations to avoid discrepancies.

Verifying calculations

I’m making sure the flux density ratio is right, considering atmospheric absorption and geometry. The sunlight's brightness is matched, so more precise geometric calculations are needed.

Calculating optimal conditions

I'm considering the photoelectric efficiency factor, ensuring it's close to 1. With a light angle distribution of π. duitron/s, that yields around 35% efficiency.

Equating fluxes

I'm working through the radiated and incident fluxes. After canceling out common factors, it turns out the disk's temperature is 0.15 times the sun's temperature.

Figuring out temperatures

I calculated the disk's temperature, factoring in the sun's temperature. The calculations for different sun temperatures, ranging from 5500 K to 5773 K, show a consistent trend.

Evaluating lens properties

I'm thinking through how different lens radii and focal lengths affect illumination. Ultimately, the disk's temperature is about 0.62 times the sun's temperature, around 3600 K.

Evaluating lens configurations

I'm curious about how various lens sizes and setups influence the disk's temperature. The particular geometry and atmospheric absorption suggest the disk's temperature is roughly 0.62 times the sun's temperature.

Concluding with temperature

Piecing together insights on lens configurations and solar absorption, the disk's peak temperature hovers around 3300 °C, aligning well with expected values.

Thought about solar radiation and lens effects for 3m 55s

Risposta

下面给出一种比较简洁而又能抓住本质的分析方法。核心思路可以归纳为一句话：

理想光学成像中，透镜所成的太阳像与太阳本身具有相同的“亮度”（辐射出射度），而黑体的平衡温度取决于它所能获得的有效“亮度”与自身面积之比。

1. 问题情景与已知条件

太阳近似黑体：
设太阳表面绝对温度为 $T_{\mathrm{sun}}$ 。题目中给出“太阳温度” $t = 5.50 \times 10^3\,^\circ\mathrm{C} \approx 5773\,\mathrm{K}$ 。
太阳每单位表面积每单位时间向外辐射能量符合斯特藩–玻尔兹曼定律
$E_{\mathrm{sun}} = \sigma\,T_{\mathrm{sun}}^4.$
大气吸收：
大气对太阳能的吸收率 $a = 0.40$ ，因此能到达地面的太阳辐照度只剩下 $(1 - a) = 0.60$ 倍。
透镜参数：
- 薄凸透镜半径 $R = 0.10\text{ m}$ 。
- 焦距 $f = 0.50\text{ m}$ 。
  （在本题的关键分析中，实际上只需要用到“成像亮度不增”这一热力学极限原理，而并不一定要显式地用到“透镜半径”和“焦距”去计算成像大小；不过题目给了这些参数，往往是为了说明太阳像确实比透镜小、以及后面圆盘比太阳像大一圈等几何关系。）
黑色圆盘：
- 放在透镜焦平面处，圆盘中心与透镜焦点重合；
- 圆盘半径是“太阳像”半径的 2 倍；
- 圆盘吸收率近似为 1（全吸收），并同样遵从黑体辐射定律向外辐射；
- 忽略对流散热、杂散光等，只考虑热辐射收支的平衡。

题目要求：当圆盘达到稳定温度时，其可能达到的最高温度（摄氏度）是多少？

2. 关键物理：为什么不能超过“经大气损耗后的太阳亮度”

2.1 亮度守恒原理

在理想几何光学成像条件下(不考虑额外损耗、也不违反热力学第二定律)，像的亮度不可能超过物体本身的亮度。对于太阳这样的远距离物体，透镜只能把它“重新分配”到焦面上，不能凭空放大其单位面积的辐射强度。

太阳表面（视作黑体）出射度是 $\sigma\,T_{\mathrm{sun}}^4$ ；
到达地面的实际“亮度”要再乘上大气透过率 $(1-a)=0.60$ ；
因此，焦面上所成的“太阳像”之亮度至多是 $L_{\mathrm{image}} = 0.60 \,\sigma \, T_{\mathrm{sun}}^4.$

这意味着：如果有一块小黑体靶，其表面每单位面积能够接收到的功率密度(辐射照度)最高也就这么多(忽略镜片本身的反射损耗等)。想让该黑体靶升温到平衡，最终它自己也得以黑体方式向外辐射同等的功率密度。由此可得最高平衡温度和入射亮度之间的关系：

\sigma \, T_{\mathrm{靶}}^4 = \text{(靶面可获得的最大单位面积功率)}.

2.2 本题的几何：靶比太阳像“大了一圈”

题目告诉我们：

“太阳像”半径记为 $r_{\mathrm{image}}$ ；
实际黑色圆盘的半径 = $2\,r_{\mathrm{image}}$ 。

也就是说，黑盘的面积是太阳像面积的 4 倍，但透镜所成的“明亮区域”其实只有那直径 $2r_{\mathrm{image}}$ 的小圆(太阳像本身)。超出太阳像范围以外的圆盘区域并没有被直接照亮。

于是，对于整块黑色圆盘来说，收到的总辐射功率只有

P_{\mathrm{in}} = \bigl[\pi \,(r_{\mathrm{image}})^2\bigr] \times \bigl[0.60\,\sigma\,T_{\mathrm{sun}}^4\bigr].

$\pi (r_{\mathrm{image}})^2$ 是太阳像的面积；

$\;0.60\,\sigma\,T_{\mathrm{sun}}^4$ 是像区的亮度(单位面积的入射功率)。

但是这块圆盘是均匀导热，最终会达到一个均匀温度 $T_d$ 。它向外辐射功率为

P_{\mathrm{out}} = \sigma\,T_d^4 \times \bigl[\pi (2\,r_{\mathrm{image}})^2\bigr] = 4\,\pi\,(r_{\mathrm{image}})^2 \,\sigma\,T_d^4.

平衡时 $P_{\mathrm{in}} = P_{\mathrm{out}}$ 。因此

0.60\,\sigma\,T_{\mathrm{sun}}^4 \,\pi\,(r_{\mathrm{image}})^2 \;=\; \sigma\,T_d^4 \,\Bigl[4\,\pi\,(r_{\mathrm{image}})^2\Bigr].

扣除公共因子后，得到

0.60\,T_{\mathrm{sun}}^4 \;=\; 4\,T_d^4 \quad\Longrightarrow\quad T_d^4 \;=\; 0.15\,T_{\mathrm{sun}}^4.

于是

T_d \;=\; T_{\mathrm{sun}} \,\bigl(0.15\bigr)^{1/4}.

由于

(0.15)^{1/2} \approx 0.3873, \quad \sqrt{0.3873} \approx 0.622,

可得

(0.15)^{1/4} \approx 0.622.

故

T_d \approx 0.622\,T_{\mathrm{sun}}.

3. 数值结果

若题目中所取的太阳“绝对温度”记为

T_{\mathrm{sun}} \approx 5.50\times 10^3\,^\circ\mathrm{C} \;\;\text{(约 }5770\text{ K左右)},

则

T_d \;\approx\;0.622\times 5770\,\mathrm{K} \;\approx\;3590\,\mathrm{K}.

换算为摄氏度( ${}^\circ\mathrm{C} = \mathrm{K}-273$ )，大约为

3590 - 273 \;\approx\;3317 \;^\circ\mathrm{C}.

考虑到题意一般只需给出一个数量级或整数近似，所以圆盘最终能达到的最高平衡温度大约在

\boxed{3300\,^\circ\mathrm{C}\text{左右。}}

4. 结论与启示

“单片透镜聚光”难以使目标温度超过太阳表面温度
这是由热力学第二定律(亮度不能被无损放大)决定的。本题又因为黑盘比太阳像大了 4 倍面积，同时还损失了 40% 的大气吸收，最终只能达到大约 $0.62\,T_{\mathrm{sun}}$ 这个量级。
题目所给“透镜半径”和“焦距”等具体值，并未显著改变最后的结果
在只考虑理想成像的前提下，太阳像的“单位面积亮度”等于太阳本身的亮度(乘以大气透过率)。只要黑盘处在焦平面、并且能把整个像区接收完全，那么具体透镜尺寸并不会进一步提高“像的亮度”。
实际情况会更低
现实中还会有：
- 透镜本身的反射/吸收损耗；
- 空气对流、散射不可避免；
- 太阳并非刚好正午 90° 照射；
- 导热到支撑结构等散热；
  等等因素，都会让圆盘温度达不到该理想上限。

因此，理想情况下的最高平衡温度约在 $3300\,^\circ\mathrm{C}$ 左右，这就是本题要得到的近似答案。