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.

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

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.