A 1 m long aluminum wire of diameter of 1 mm is submerged in an oil bath of temperature 28 °C. The wire has an electrical resistance of 0.02 Ω/m, which generates heat when current is applied to the wire. The convection coefficient between the wire and oil bath is 470 W/m2∙K. Assume no radiation effects. List and justify any assumptions you make.

Question

Physics

Patrick

3 October, 09:16

A 1 m long aluminum wire of diameter of 1 mm is submerged in an oil bath of temperature 28 °C. The wire has an electrical resistance of 0.02 Ω/m, which generates heat when current is applied to the wire. The convection coefficient between the wire and oil bath is 470 W/m2∙K. Assume no radiation effects. List and justify any assumptions you make.

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Answers (1)

Know the Answer?

Jayden Guzman · Answer 1

T = 163.45 C

t = 25.1 s

Explanation:

Given:

The complete question is given below:

A 1 mm long wire of diameter D=1 mm is submerged in an oil bath of temperature T∞=28∘C. The wire has an electrical resistance per unit length of R′e=0.02Ω/m. If a current of I=100 A flows through the wire and the convection coefficient is h=470W/m2⋅K,

The properties of the wire are ρ=8000kg/m3, c=500J/kg⋅K, and k=20W/m⋅K.

Find:

what is the steady-state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature that is within 1∘C of the steady-state value?

Solution:

- First we will check whether we can use Lumped capacitance method can be applied by calculating Biot Number Bi, as follows:

Bi = h*D / 4*k

Bi = 470*2.5*10^-4 / 4*20

Bi = 0.00146875 ... < 0.1

- From the condition validated above using Biot number we can apply Lumped capacitance method:

- Using Energy Balance we have:

Q_convec = Q_gen

A_wire*h * (T - T∞) = I^2*R′e

- Re - arrange for T:

T = T∞ + I^2*R′e / A_wire

- Where, A_wire is the cross sectional area of the wire as follows:

A_wire = pi*D

Hence,

T = T∞ + I^2*R′e / h*pi*D

Plug in values:

T = 28 + 100^2*0.02 / pi * (10^-3) * 470

T = 163.45 C

- Next using the derived results from Lumped capacitance for a rod we have:

dT/dt = 4*I^2*R′e / p*c_p*pi*D^2 - (4*h/p*c_p*D) * (T - T∞)

- The solution of the above differential equation is derived as:

Ln ((T - T∞ - (I^2*R′e / h*pi*D)) / (T_i - T∞ - (I^2*R′e / h*pi*D))) = - 4*h*t/p*c_p*D

- Plug in values and solve for t.

Ln ((163.45-28-135.4510154) / (28-28-135.4510154))) = - 4*470/8000*500*0.001

Ln (7.49643679*10^6) = - 0.47*t

t = 11.8 / 0.47

t = 25.1 s