A 3 mm diameter and 5m long electrical wire is tightly wrapped with the 2-mm thick plastic cover whose thermal conductivity is 0.15 w/m-k. Electrical measurement indicate that current of 10 amps passes through the wire and there is a voltage drop of 8 volts along the wire. If the insulated wire exposed to a medium at 30c with a heat transfer coefficient of 12 w/m2-k. Determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature.
Answers
Answer:
According to the problem the diameter of the wire , r1
The outer diameter after wrapping , r2 = 7mm
The thickness of the plastic cover , t
Length of the wire = I
The Thermal conductivity of cover = = k
Current pass through the wire = I
Due to current passing voltage dropped, is V
The temperature = T
Heat transfer coefficient = h
The rate at which the heat is generated in the wire,
q= VI = 8 * 10 =80 Watt
Now for calculating the resistance for each medium .
x1 = 1/hA2 = 1/2 x 2π x 0.0035 x 5 = 0.76 K/ W
x2 = ln r2/r1/2πkl = ln 0.0035/0.0015 /2π 0.15 x 5 = 0.179 K/W
Therefore the total resistance ,
R = x1+x2 = 0.76 +0.179 = 0.939 K/W
Now to calculate the temperature at the interface of the wire,
q= R(T - t)
=> 80 = 0.939(T -30- 273)
T = 378.12 K
Critical radius is r(critical)
r(critical) = K/h = 0.15/12 = 0.0125 m or 12.5 mm
Answer:
According to the problem the diameter of the wire , r1
The outer diameter after wrapping
, r2 = 7mm
The thickness of the plastic cover , t
Length of the wire = I
The Thermal conductivity of cover
= = k
Current pass through the wire = I
Due to current passing voltage dropped, is V
The temperature = T
Heat transfer coefficient = h
The rate at which the heat is generated in the wire,
q= VI = 8 * 10 =80 Watt
Now for calculating the resistance for each medium .
x1 = 1/hA2 = 1/2 x 2π x 0.0035 x 5 = 0.76 K/ W
x2 = ln r2/r1/2πkl = ln 0.0035/0.0015 /2π 0.15 x 5 = 0.179 K/W
Therefore the total resistance ,
R = x1+x2 = 0.76 +0.179 = 0.939 K/W
Now to calculate the temperature at the interface of the wire,
q= R(T - t)
=> 80 = 0.939(T -30- 273)
T = 378.12 K
Critical radius is r(critical)
r(critical) = K/h = 0.15/12 = 0.0125 m or 12.5 mm