A 2 m long wire of resistance 4 ohm and diameter 0.64 mm is coated with plastic insulation of thickness 0.06 mm . When a current of 5 ampere flows through the wire, find the temperature difference across the insulation in steady-state if K = 0.16 × 10- 2 cal cm- 1 ° C- 1 s- 1 .

Question

A 2 m long wire of resistance 4 ohm and diameter 0.64 mm is coated with plastic insulation of thickness 0.06 mm . When a current of 5 ampere flows through the wire, find the temperature difference across the insulation in steady-state if K = 0.16 × 10- 2 cal cm- 1 ° C- 1 s- 1 . (A) 1 ° C (B) 2 ° C (C) 3° C (D) 4° C

Tardigrade Admin · Accepted Answer

Consider a coaxial cylindrical shell of radius r and thickness dr as shown in diagram. The radial rate of flow of heat through this shell in steady state will be
Solution

H = \frac{d Q}{d t} = - K A \frac{d θ}{d r}

[Negative sign is used as with increase in

r, θ

decreases.] Now as for cylindrical shell

A = 2 π r L

,

H = - 2 π r L K \frac{d θ}{d r}

or

a \int b \frac{d r}{r} = - \frac{2 π L K}{H} θ_{1} \int θ_{2} d θ

which on integration and simplification gives:

H = \frac{d Q}{d t} = \frac{2 π K L θ _{1} - θ _{2}}{l o g _{e} b / a} \dots ...

(i)

Here

H = \frac{1 ^{2} R}{4.2} = \frac{5 ^{2} \times 4}{4.2} = 24 c a l / s, L = 2 m = 200 c m

r_{1} = (0.64/2) mm = 0.032 c m

and

r_{2} = r_{1} + d = 0.032 + 0.006 = 0.038 c m

So

θ_{1} - θ_{2} = \frac{24 \times l o g _{e} 38/32}{2 \times 3.14 \times 200 \times 0.16 \times 1 0 ^{- 2}}

= \frac{24 \times 2.3026 [ l o g _{1 0^{38}} - l o g _{1 0^{32}} ]}{3.14 \times 0.64}

or

θ_{1} - θ_{2} = \frac{55 \times [ 1.57 - 1.50 ]}{2} ≃ 2^{\circ} C

$1^{\circ} C$

$2^{\circ} C$

$3^{\circ} C$

$4^{\circ} C$

Q. A 2m long wire of resistance 4ohm and diameter 0.64mm is coated with plastic insulation of thickness 0.06mm . When a current of 5 ampere flows through the wire, find the temperature difference across the insulation in steady-state if K=0.16×10−2calcm−1∘C−1s−1 .

1∘C

2∘C

3∘C

4∘C

$1^{\circ} C$

$2^{\circ} C$

$3^{\circ} C$

$4^{\circ} C$