Odpowiedź:
A=5m²
T1=180°C
T4=30°C
λ1=0,12 V/m*K
d1=8mm=0,008m
λ2=46,2 V/m*k
d2=6mm=0,006m
λ3=0,15 V/m*K
d3=20mm=0,02m
ΔT(1-4)=150°C
ΔT(1-4)=ΔT1+ΔT2+ΔT3
ΔT=d*q/A*λ
ΔT1=q*0,008m/5m²*0,12V/m*K
ΔT2=q*0,006m/5m²*46,2V/m*K
ΔT3=q*0,02m/5m²*0,15V/m*K
150=q*0,008m/5m²*0,12V/m*K + q*0,006m/5m²*46,2V/m*K + q*0,02m/5m²*0,15V/m*K
150=q (0,008m/5m²*0,12V/m*K + 0,006m/5m²*46,2V/m*K + 0,02m/5m²*0,15V/m*K)
q=150/(0,008m/5m²*0,12V/m*K + 0,006m/5m²*46,2V/m*K + 0,02m/5m²*0,15V/m*K)
q=3747,5V
ΔT1=3747,5V*0,008m/5m²*0,12V/m*K
ΔT1=49,96°C
ΔT3=3747,5V*0,02m/5m²*0,15V/m*K
ΔT3=99,9°C
T2=130,03°
T3=129,93°C
λ3=0,15 V/m*K
d3=20mm
λ4=0,05 V/m*K
d3/λ3=d4/λ4
d4=d3*λ4/λ3
d4=20mm*0,05/0,15
d4=6,6mm