Question

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.

Answer 1

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