﻿ 一种密闭腔体的加热设计与分析
 舰船科学技术  2017, Vol. 39 Issue (9): 169-173 PDF

Heating design and analysis of a closed cavity
WANG Bo, DONG Guang-kun
Huazhong Institute of Electro-Optics, Wuhan National Laboratory for Optoelectronics, Wuhan 430223, China
Abstract: Some optical systems make use of large aperture plastic lenses. To ensure their optical parameters remain stable under low temperature, it’s necessary to heat these lenses’ working environment. This article designs a closed chamber based on theory of heat transfer. This closed cavity contains heating component and temperature control component. According the heating demand, structure design of this cavity is completed. At the same time analysis and calculation of the cavity’s heating is finished. Heating experiment is carried out based on the designed closed cavity. The experimental results are analyzed.It shows,the mathematical model can reflect the heating process of the cavity. And analyzing the experimental result proves that the design of this closed cavity can meet the working environment requirement of this kind of lenses.
Key words: closed cavity     plastic lenses     heating design     heating analysis and calculation
0 引 言

1 密闭腔体结构及加热设计 1.1 热传导方式

1.2 腔体设计

 图 1 密闭腔体外形及内部元件示意图 Fig. 1 Sketch of closed cavity’s shell and interior component

2 加热分析计算

2.1 腔体加热的数学模型

2.1.1 加热腔体所需的能量。

 ${Q_{\rm{1}}} = {Q_a} + {Q_b} + {Q_c} + {Q_d} + {Q_e} + {C_1} \cdot ({T_2} - {T_1})\text{，}$ (1)

1）物体传热计算

 $Q = c \cdot m \cdot ({T_2} - {T_1})\text{，}$ (2)

 ${Q_a} = {c_1} \cdot {m_1} \cdot \Delta T\text{，}$ (3)
 ${Q_b} = {c_2} \cdot {m_2} \cdot \Delta T\text{，}$ (4)
 ${Q_c} = {c_3} \cdot {m_3} \cdot \Delta T\text{，}$ (5)

2）气体传热计算

 $Q = \frac{m}{M} \cdot c \cdot ({T_2} - {T_1})\text{，}$ (6)

 $pV = \frac{m}{M} \cdot R \cdot {T_0}\text{，}$ (7)

 ${Q_d} = \frac{{{p_N}{V_N}}}{{R{T_0}}} \cdot {c_N} \cdot \Delta T\text{。}$ (8)

3）耗散能量计算

 $\varPhi = hA\Delta {T_d}\text{，}$ (9)

 $\varPhi = {h_1}{A_1}\Delta {T_d} + {h_2}{A_2}\Delta {T_d} + {h_3}{A_3}\Delta {T_d}\text{，}$ (10)

 $\begin{split}{Q_e} = \int_0^{{t_{\text{总}}}} {\Phi {\rm d}t} =& \int_0^{{t_{\text{总}}}} {({h_1}{A_1}\Delta {T_d} + {h_2}{A_2}\Delta {T_d} +}\\ &{ {h_3}{A_3}\Delta {T_d}){\rm d}t}\text{，} \end{split}$ (11)

 ${Q_e} = \frac{1}{2}({h_1}{A_1} + {h_2}{A_2} + {h_3}{A_3})\Delta T \cdot t\text{。}$ (12)

 $\begin{split}{Q_{\rm{1}}} = & {c_1} \cdot {m_1} \cdot \Delta T + {c_2} \cdot {m_2} \cdot \Delta T + {c_3} \cdot {m_3} \cdot \Delta T + \\ &\frac{{{p_N}{V_N}}}{{R{T_0}}} \cdot {c_N} \cdot \Delta T + \frac{1}{2}({h_1}{A_1} + {h_2}{A_2} + {h_3}{A_3})\Delta T \cdot t + \\ &{C_1}\! \cdot \!\Delta T \!=\! ({c_1} \!\cdot \! {m_1} \!+\! {c_2} \!\cdot \!{m_2} \!+\! {c_3} \cdot {m_3} \!+ \!\frac{{{p_N}{V_N}}}{{R{T_0}}} \cdot {c_N}\! + \\ &\frac{1}{2}({h_1}{A_1} + {h_2}{A_2} + {h_3}{A_3}) \cdot t + {C_1})\Delta T\text{。}\end{split}$ (13)
2.1.2 电阻器提供的能量

 ${Q_2} = {c_t} \cdot P \cdot {t_{\text{总}}}\text{。}$ (14)

2.1.3 加热时间与加热温度之间的关系式

 $\begin{split}{c_t} \cdot P \cdot t = & ({c_1} \cdot {m_1} + {c_2} \cdot {m_2} + {c_3} \cdot {m_3} + \frac{{{p_N}{V_N}}}{{R{T_0}}} \cdot {c_N} + \\ &\frac{1}{2}({h_1}{A_1} + {h_2}{A_2} + {h_3}{A_3}) \cdot t + {C_1})\Delta {T_1}\text{，}\end{split}$ (15)

 $t \!=\! \frac{{({c_1}\! \cdot \!{m_1} \!+\! {c_2} \!\cdot \!{m_2}\! +\! {c_3} \!\cdot \!{m_3}\! +\! \displaystyle\frac{{{p_N}{V_N}}}{{R{T_0}}} \!\cdot\! {c_N}\! +\! {C_1})\Delta T}}{{{c_t} \cdot P - \displaystyle\frac{1}{2}({h_1}{A_1} + {h_2}{A_2} + {h_3}{A_3})\Delta T}}\text{。}\!\!\!\!$ (16)

2.2 特定腔体加热所需时间计算

 \begin{aligned}& {h_1} = 1.77{\left( {\Delta T} \right)^{1/4}};{A_1} = 0.2159\;{{\rm{m}}^2};\\ & {h_2} = 2.49{\left( {\Delta T} \right)^{1/4}};{A_2} = 0.0213\;{{\rm{m}}^2};\\ & {h_3} = 1.31{\left( {\Delta T} \right)^{1/4}};{A_3} = 0.0213\;{{\rm{m}}^2}\text{。}\end{aligned}

 $t = \frac{{(1876.842 + 0.03 \times {c_3} + {C_1})\Delta T}}{{200{c_t} - 0.2315\Delta {T^{5/4}}}}\text{。}$ (17)

3 腔体加热试验 3.1 加热试验目的

3.2 加热试验装置

 图 2 密闭腔体加热试验装置 Fig. 2 Heating experiment device of designed closed cavity

3.3 加热试验过程

3.4 加热试验结果及分析

 $t = \frac{{1849.3 \times \Delta T}}{{110 - 0.2315\Delta {T^{5/4}}}}\text{。}$ (18)

 图 3 加热时间理论值与实际值随加热温度的变化曲线图 Fig. 3 Curve of heating time theory value change with actual value along with heating temperature

$\mu = \left| {\displaystyle\frac{{{t_{\text{实际}}} - {t_{\text{理论}}}}}{{{t_{\text{理论}}}}}} \right|$ 表示实际加热时间与理论加热时间的相对误差，从上述试验数据可以计算得到，μ的最大值不超过5%。这说明所建立的数学模型与试验结果吻合较好，能较好的反映实际腔体的热传递过程。另外，从–28 ℃加热到36 ℃的时间约为30 min，说明本文设计的密闭腔体能够较好较快的实现所需的加热功能，从而满足塑料透镜的工作环境要求。

4 结 语

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