Answer:
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
