﻿ 基于智能化仿真试验平台的货运列车再充气特性
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 同济大学学报(自然科学版)  2019, Vol. 47 Issue (10): 1508-1513.  DOI: 10.11908/j.issn.0253-374x.2019.10.017 0

### 引用本文

YING Zhiding, CHEN Jiamin. Re-inflating Characteristics of Wagon Based on Intelligent Simulation Test Platform[J]. Journal of Tongji University (Natural Science), 2019, 47(10): 1508-1513. DOI: 10.11908/j.issn.0253-374x.2019.10.017

### 文章历史

Re-inflating Characteristics of Wagon Based on Intelligent Simulation Test Platform
YING Zhiding , CHEN Jiamin
Institute of Railway Transit, Tongji University, Shanghai 201804, China
Abstract: Based on the gas flow theory, according to the gas state equation and mass equation and the flow characteristics, the mathematical model of the re-inflation process of wagon brake system was established. Then, the model was verified on an intelligent simulation test platform. The experimental curve is basically in agreement with the simulation curve. The re-inflation process of the first and tail vehicles of 150 trains was investigated, and the variation characteristics of the vehicle re-inflation mitigation process at different positions of the train were obtained, and the time difference between train tube and brake cylinder at different positions was tested.
Key words: wagon brake    gas flow    brake    re-inflation characteristics

1 再充气缓解工况仿真模型的建立 1.1 再充气缓解工况

“再充气缓解”是加速缓解风缸对列车管的局部增压.列车再充气缓解时，首先让制动缸里的空气打开加速缓解风缸向列车管逆流的充气通路，然后排入大气.初充气工况下，加速缓解风缸与列车管的压力相同；制动时，加速缓解风缸由于止回阀的作用空气不能逆流回副风缸而保持压力恒定.列车管在制动过程中排气使得列车管压力降低，导致列车管压力低于加速缓解风缸压力.当制动系统再充气缓解时，两者的压力差使得加速缓解风缸内空气可以通过加速缓解阀逆流到列车管，形成对列车管的局部增压作用.加速缓解作用提高了后部车辆制动系统列车管的增压速度，从而缩短了再充气缓解的时间.再充气缓解简化模型如图 1所示.

 图 1 再充气缓解模型 Fig.1 Re-inflation mitigation model
1.2 制动缸缓解数学模型

 $p V=m R T$ (1)

 $\mathrm{d} m_{2}=-q_{m_{2}} \mathrm{d} t$ (2)

dt时间内，从容器内放出的气体质量

 $\mathrm{d} m_{2}=-q_{m_{2}} \mathrm{d} t$ (3)

 $q_{m}=q_{m}^{*} \sqrt{1-\left(\frac{p_{2} / p_{1}-b}{1-b}\right)^{2}}, 1 \geqslant \frac{p_{2}}{p_{1}}>b$ (4)

 $q_{m}=0.04 S \frac{p_{1}}{\sqrt{T}} \sqrt{1-\left(\frac{p_{2} / p_{1}-b}{1-b}\right)^{2}}, 1 \geqslant \frac{p_{2}}{p_{1}}>b$ (5)

 图 2 制动缸放气简化模型 Fig.2 Simplified model of brake cylinder venting

 $q_{m}=-\frac{\mathrm{d} m}{\mathrm{d} t}=-\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{p_{{\rm z}} V_{{\rm z}}}{R T}\right)=-\frac{V_{{\rm z}}}{R T} \frac{\mathrm{d} p_{{\rm z}}}{\mathrm{d} t}$ (6)

 $t=\frac{1.4603 V_{z}(1-b)}{S \sqrt{1-2 b} \sqrt{R T}}\left(\ln \left(\frac{\sqrt{\left(1-\frac{p_{\mathrm{a}}}{p_{\mathrm{z} 0}}\right)\left(1-2 b+\frac{p_{\mathrm{a}}}{p_{\mathrm{z0}}}\right)}+\sqrt{1-2 b}}{p_{\mathrm{a}} / p}+\frac{b}{\sqrt{1-2 b}}\right)-\right.\\ ~~~~~\left. {\ln \left( {\frac{{\sqrt {\left( {1 - \frac{{{p_{\rm{a}}}}}{{{p_{\rm{z}}}}}} \right)\left( {1 - 2b + \frac{{{p_{\rm{a}}}}}{{{p_{\rm{z}}}}}} \right)} + \sqrt {1 - 2b} }}{{{p_{\rm{a}}}/{p_{\rm{z}}}}} + \frac{b}{{\sqrt {1 - 2b} }}} \right)} \right)$ (7)

 ${p_{\rm{z}}} = \frac{{2{p_{\rm{a}}}}}{{\sqrt {\frac{{A_1^2}}{{{{\rm{e}}^{2{A_2}\left( {t - {t_1}} \right)}}}} - 1.3164\frac{{{A_1}}}{{{{\rm{e}}^{{A_2}\left( {t - {t_1}} \right)}}}} + 5.93162} - \frac{{{A_1}}}{{{{\rm{e}}^{{A_2}\left( {t - {t_1}} \right)}}}} + 0.658}}$ (8)

 \begin{aligned} A_{1}=& \frac{\sqrt{\left(1-\frac{p_{{\rm a}}}{p_{{\rm z} 0}}\right)\left(0.6+\frac{p_{{\rm a}}}{p_{{\rm z} 0}}\right)}+0.7746}{p_{\mathrm{a}} / p_{{\rm z0}}}+\\ & 0.2582 \\ A_{2}=& \frac{S \sqrt{T}}{8.9026 \times 10^{-2} V_{{\rm z}}} \end{aligned}
1.3 列车管再充气及局部增压数学模型

 图 3 列车管再充气简化模型 Fig.3 Simplified model of train tube refilling

 $q_{m}=\frac{\mathrm{d} m}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{p_{2} V}{R T}\right)=\frac{V}{R T} \frac{\mathrm{d} p_{2}}{\mathrm{d} t}$ (9)

 $t=\frac{1.4603 V(1-b)}{S \sqrt{R T}}\left(\sin ^{-1}\left(\frac{p_{2} / p_{1}-b}{1-b}\right)-\sin ^{-1}\left(\frac{p_{{\rm z0}} / p_{1}-b}{1-b}\right)\right)$ (10)

 $p_{2}=p_{1}\left(\begin{array}{c}{\left.(1-b) \arcsin \frac{1.4603 V(1-b)}{S \sqrt{R T }t-1.4603 V(1-b) \sin ^{-1}\left(\frac{p_{{\rm 00}} / p_{1}-b}{1-b}\right)}+b\right)}\end{array}\right.$ (11)

 $V = \frac{{{\rm{ \mathsf{ π} }}D_1^2}}{4}N{L_1} + \frac{{{\rm{ \mathsf{ π} }}D_2^2}}{4}N{L_2}$ (12)

 ${p_1} = {p_{\rm{s}}}\left( {0.8\arcsin \frac{{{\rm{ \mathsf{ π} }}{D^2}NL}}{{58.0055S\sqrt T t - {\rm{ \mathsf{ π} }}{D^2}NL{{\sin }^{ - 1}}\left( {\frac{{{p_{{\rm l}0}}}}{{0.8{p_{\rm{s}}}}} - 0.25} \right)}} + 0.2} \right)$ (13)

 $\begin{array}{l} {p_1} = \left( {{p_{\rm{s}}} + {p_{\rm{j}}}} \right)\\ \left( {\begin{array}{*{20}{c}} {0.8\arcsin }&{\frac{{{\rm{ \mathsf{ π} }}{D^2}NL}}{{58.0055S\sqrt T \left( {t - {t_1}} \right) - {\rm{ \mathsf{ π} }}{D^2}NL{{\sin }^{ - 1}}\left( {\frac{{{p_{{\rm{l}}1}}}}{{0.8\left( {{p_{\rm{s}}} + {p_{\rm{j}}}} \right)}} - 0.25} \right)}} + 0.2} \end{array}} \right) \end{array}$ (14)

 $\begin{array}{l} {p_1} = {p_{\rm{s}}}\\ \left( {0.8\arcsin \frac{{{\rm{ \mathsf{ π} }}{D^2}NL}}{{58.0055S\sqrt T \left( {t - \left( {{t_1} + {t_2}} \right)} \right) - {\rm{ \mathsf{ π} }}{D^2}NL{{\sin }^{ - 1}}\left( {\frac{{{p_{{\rm{l}}2}}}}{{0.8{p_{\rm{s}}}}} - 0.25} \right)}} + 0.2} \right) \end{array}$ (15)
1.4 缓解波和缓解波速

 $W_{\mathrm{a}}=\frac{L_{\mathrm{HB}}}{t_{\mathrm{HB}}}=\frac{L_{\mathrm{HB}}}{\left(t_{\mathrm{e}}-t_{\mathrm{s}}\right)}$ (16)

2 单车模型仿真与试验

 图 4 智能化仿真试验平台 Fig.4 Intelligent simulation test platform

 图 5 再充气缓解曲线 Fig.5 Re-inflation relief curve

3 150辆编组列车模型仿真分析

 图 6 不同位置处车辆列车管与制动缸压力试验曲线 Fig.6 Pressure curve of vehicle train pipe and brake cylinder at different positions

(1) 第1、50、100、150辆车列车管开始增压的时间分别为1.0、3.2、5.5、7.6 s，基本呈均匀分布，说明缓解工况下空气波在列车管内具有均匀传播的特性.

(2) 各个位置处车辆制动缸开始减压时间比相同位置处列车管开始增压时间略晚.四个位置处车辆制动缸与列车管作用时间差约为1.2、1.7、2.0、2.2 s，这是由于120阀主活塞两侧需要一定的压力差才能发生规定的动作.

(3) 第1、50、100、150辆车列车管达到系统平衡压力的时间分别约为48、69、82、91 s.造成这种现象的根本原因是制动系统列车管内壁、折角塞门及连接软管的阻尼使得空气波沿着列车车长方向能量越来越弱，导致由前至后列车管增压速度越来越慢.

 图 7 不同位置处车辆列车管与制动缸作用时间差 Fig.7 Time difference of vehicle train pipe and brake cylinder at different positions

4 结语

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