q1 =- K1•A•(ΔT1/Δx1),
q2 =- K2•A•(ΔT2/Δx2),
q1=q2, =>
K1•A•(ΔT1/Δx1)= K2•A•(ΔT2/Δx2),
ΔT1= 60 –T, ΔT2 = T -10,
(60 –T)/(T-10) = (K2/K1) •(Δx1/Δx2) =
=(0.02/400) •(0.02/0.04) = 1•10^-4.
T =60.001/(1+1•10^-4) = 59.995 degr.
ΔT1/Δx1 = (60 –T)/Δx1 = 0.005/0.004 =1.25 degr/m,
ΔT2/Δx2 = (T-10)/Δx2 = 49.995/0.002 =24997.5 degr/m.
a metal cylinder containing water at 60ged C, has a thickness of 4mm.And the thermal conductivity 400W per m per K. it is lagged by felt of thickness 2mm and thermal conductivity 0.002W per m per K. the room temperature is 10deg C.Find the temperature gradient g for for the metal and the felt
2 answers
thanks :)