Question 1 (Answers: a) 67.9°C; b) 61.7°C) A 12-m-long electric wire is insulate

Question

Question 1 (Answers: a) 67.9°C; b) 61.7°C) A 12-m-long electric wire is insulated with a plastic layer, which has a thermal conductivity of 0.13 W/(m oc). The electrical conductor has a diameter of 2.3 mm whereas the outside diameter of the plastic insulation is 4.3 mm. For the application, a current of 13.4 A passes through the wire with a voltage dro measured at 28°C and the convective heat transfer on the outside surface of the wire insulation is p of 8 V along the length of the wire. The air surrounding the wire is estimated at 20 W/(m* °C). Assuming steady-state conditions and one-dimensional radial heat transfer Calculate the temperature at the interface between the conductor and the plastic insulation. What will be the temperature at the same location if the thickness of the plastic insulation were doubled? Does your answer make sense? Discuss briefly. a) b)

Explanation / Answer

given radius of wire = r = 2.3/2 = 1.15 mm
radius of wire + insulator, R = 4.3/2 = 2.15 mm

consider ar radius x
r < x < R
consider thicness dx
then heat conducted through this portion
Q = k*2*pi*x*l*dT/dx [ where L is length of the wire]
Qdx/x = 2*k*pi*ldT
integrating from x = r to x = R
Q*ln(R/r) = 2*k*pi*l*(To - Ti)
Q = 2*k*pi*l(To - ti)/ln(R/r)

now given L = 12 m
k = 0.13 W/(m C)
CURRent I = 13.4 A
VOltage drop V = 8 V
To = 28 C
convective heat transfer rate = Qc = 20 W/m^2*C

at steady state

a) let temperature of the interface between conductor and plastic be T
   convective heat loss rate for the wire Q = 20*2*pi*R*L*To = 40*pi*2.15*10^-3*12*28 = 90.7334

   so at steady state this is the heat being transferred through the wire insulation
   90.7334 = 2*0.13*pi*12(To - T)/ln(2.15/1.15)
   T = 33.794 C

b) if R' = 2R
   then Q = 40*pi*2.15*10^-3*12*28*2 = 181.4668 W
   181.4668 = 2*0.13*pi*12(28 - T)/ln(2.15*2/1.15)
   T = 52.429 C

   this is because the increased surface area of the wire increases the heat dissipation rate but this increased heat dissipation means that the thicker weire needs to have a higher temperature gradiuent across it for this to be acheived as evident from the naswer

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