a 2,2 mm diameter and 14m long electric wire is tightly wrapped with a 1.0mm thi

Question

a 2,2 mm diameter and 14m long electric wire is tightly wrapped with a 1.0mm thick plastic insulation whose thermal conductivity is k=0,15W/m*K. Electrical measurements indicate that the current of 13A passes through the wire and there is a voltage drop of 8.0V along the wire.

The insulated wire is exposed to a surrounding medium at a temperature of 30C (celsius). The heat transfer coefficient is h=24W/m2*K. Determine the temperature at the interface of the wire and the plastic cover in steady operation.

Explanation / Answer

Heat generated in wire, Q = V*I = 8*13 = 104 W

Two types of resistance is offered for heat transfer between wire and surrounding.

Convection between plastic and surrounding:

Rconv = 1/(h*A)

outer radius, r2 = (1.1+1) mm = 0.0021 m

A = outer area = 2*pi*r2*L = 2*pi*(0.0021)*14= 0.18473 m2

Rcov = 1/(24*0.18473) = 0.22555 deg C/W

Conduction resistance,

Rcond = ln(r2/r1)/(2*pi*k*L) = ln(2.1/1.1)/(2*pi*0.15*14) = 0.04901 deg C/W

Since these are in series, Rtotal = Rcond + Rconv = 0.22555 + 0.04901 = 0.27456 deg C/W

Let T be temperature at interface.

then Q = (T-Tinfinity)/Rtotal

T = Tinfinity + Q*Rtotal = 30 + 104*0.27456 = 58.554 deg C

Temperature at plastic cover:

Let temperature at plastic cover be T1

Equation of heat transfer between plastic and surrounding gives:

Q = ( T1- Tinfinity)/Rconv

T1 = Tinfinity + Q*Rconv = 30 + 104*0.22555 = 53.4572 deg C

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