Given Information:
diameter of wire = d = 1 mm = 0.001 m
Ambient Temperature = T∞= 33° C
Resistance of wire = R = 0.01 Ω/m
Current = I = 97 A
density = ρ = 8000 kg/m3
specific heat = c = 500 J/kg·K
Thermal conductivity = k= 20 W/m·K.
convection coefficient = h= 500 W/m²·K
Required Information:
a) Steady-state temperature = T = ?
b) Time to reach within 1° C of steady-state temperature = t = ?
Answer:
a) Steady-state temperature = 92.9° C
b) Time to reach within 1° C of steady-state temperature = 23.61 seconds
Step-by-step explanation:
a) What is the steady-state temperature of the wire?
The steady-state temperature can be found using
T = T∞ + I²R/πdh
Where T∞ is the ambient temperature of the wire, d is the diameter, and h is the convection coefficient.
T = 33° + (97)²0.01/π(0.001)500
T = 33° + 59.9°
T = 92.9° C
b) How long does it takes to reach a temperature that is within 1°C of the steady-state value?
Taking derivative of the above equation yield,
T - T∞ - (I²R/πdh)/Ti - T∞ - (I²R/πdh) = e^(-4h/ρcd)t
Where T∞ =Ti= 33° C and T = 92.9° C
substituting the values
92.9 - 33 - (97²*0.01/π*0.001*500)/33 - 33 - (97²*0.01/π*0.001*500)
= e^(-4*500/8000*500*0.001)t
-0.000445/-59.89 = e^(-0.5t)
-0.00000743 =e^(-0.5t)
Taking ln on both sides
ln(0.00000743) = -0.5t
-11.80 = -0.5t
t = 23.61 seconds
