Aufgabe 2

Gegeben:

clear all;
rho = 1.7e-2*(1e-3)^2 %% 1.7e-2 Ohm mm^2 / m
rho = 1.7000e-08
d = 1.8e-3 % 1,8mm
d = 0.0018
l = 5 %m
l = 5
T = 50 + 273 %K  , 50 °C
T = 323
alpha = 3.9e-3 %1/K
alpha = 0.0039
I = 2 %A
I = 2
T_Z = 293 %K, 20°C Zimmertemperatur
T_Z = 293

Kabelwiderstand

A = (d/2)^2 * pi
A = 2.5447e-06
Rinnen = rho * l / A
Rinnen = 0.0334

Temperaturerhöhung

deltaT = T - T_Z
deltaT = 30
rho_T = rho * (1 + alpha*deltaT)
rho_T = 1.8989e-08
Rinnen_T = rho_T * l / A
Rinnen_T = 0.0373

Leistung

U = Rinnen * I^2
U = 0.1336