【发布时间】:2021-09-07 22:11:38
【问题描述】:
我试图从 Simscape 中的热电等效电路模型中获取 10A 保险丝的温度-电流曲线。为此,我创建了一个自定义开关和可变电阻。
Thermal RC Cauer 模型像 FEM 一样分为不同的 RC 连接,然后从那里计算保险丝的温度。
- 与温度相关的电阻(在模型中:Variable_Resistance_Custom)
方程式:
- R = R0*(1+alpha*(T-T0))
- P_electric_loss = iRR
.ssc 脚本
component R_ele_variable
% Variable Resistor
% Resistor is an electrical component that reduces the electric current.
% The resistor's ability to reduce the current is called resistance.
%
% Resistance: Temperature Coefficient resistance could be expected to
% increase with temperature, since there will be more collisions.
% (R-R0)/R0 = alpha*(T-T0)
inputs
T = { 0.0, 'K' }; % T_RC:left
end
outputs
P_ele = {0, 'W'}; % P_ele:right
R_T = {0, 'Ohm'}; % R_T:right
end
nodes
p = foundation.electrical.electrical; % +:left
n = foundation.electrical.electrical; % -:right
end
parameters
R0 = {7,75e-3,'Ohm'}; % Nominal resistance
T0 = {296.15,'K'}; % Reference temperature
alpha = {3.527e-3,'1/K'}; % Temperature coefficient
end
variables
i = { 0, 'A' }; % Current
end
branches
i : p.i -> n.i;
end
equations
assert(R0>=0)
assert(T0>0)
assert(alpha>=0)
let
% Calculate R, protecting against negative values
Rdem = R0*(1+alpha*(T-T0));
R = if Rdem > 0, Rdem else {0,'Ohm'} end;
in
R*i == p.v-n.v; % Electrical equation
P_ele == R*i*i;
R_T == R;
end
end
end
-
控制保险丝的开关,每当温度达到熔化温度(在我的情况下约为 3.49 秒,它应该打开电路(在模型中:Switch_Custom),然后保险丝的温度将下降到环境温度(室温: 23°C)根据以下等式:
-
T = e^((-t)/(R∗C)),这里 R = R1 + R2 + R3 + R4 + R5 & C = C1+ C2 + C3 + C4 + C5(来自热模型)
.ssc 的 Switch 脚本
component switch_custom_tripping_1
% Switch_custom_tripping_1
% The block represents a switch controlled by an external physical
% signal. If the external physical signal PS is less than the threshold,
% then the switch is closed, otherwise the switch is open.
inputs
T = { 0.0, 'K' }; % T_RC:bottom
R = {0.0, 'Ohm'}; % R_T:top
end
nodes
p = foundation.electrical.electrical; % p:top
n = foundation.electrical.electrical; % n:bottom
end
parameters
T_melting = { 661.15, 'K' }; % Threshold
C_th = { 5035.9938, 'J/K' }; % Thermal Capacitance
R_th = { 215.45, 'K/W' }; % Thermal Resistance
end
variables
i = { 0, 'A' }; % Current
v = { 0, 'V' }; % Voltage
end
branches
i : p.i -> n.i;
end
equations
assert(T>0)
assert(T_melting>0)
assert(R>0)
v == p.v - n.v;
if T < T_melting % Switch is close
v == i*R;
else % Switch is open
T == T_melting*exp(-{1,'s'}/(R_th*C_th));
% R == {Inf, 'Ohm'};
end
end
end
在这个模型中给定默认电流:15A 熔断器熔化温度:388°C (661.15 Kelvin)
错误: 1) • 在时间 3.495038669135668 的瞬态初始化,解决一致的状态和模式,未能收敛。 • 非线性求解器:线性代数误差。使用迭代矩阵求解失败。
在达到熔化温度后,我怎样才能让这个温度恢复到环境室温?
我应该创建单独的函数来计算降额温度,还是可以将这个降额方程包含在 Variable_Resistor_Custom 中?
【问题讨论】:
标签: matlab fuse simscape automotive