5) A wire resistor of 1 mm diameter is in an ambient with T[infinity] = 33°C and h = 500 W/m2·K. The electrical resistance per unit length of wire is 0.01 Ω/m. What is the steady-state temperature of the wire when a current of 97 A passes through it? How long does it takes to reach a temperature that is within 1°C of the steady-state value? The properties of the wire are rho = 8000 kg/m3, c = 500 J/kg·K, and k = 20 W/m·K.

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