Answer :

Correct answer is : 0.11

Wire radius (r) = 1 mm, Heat generation (Q / L) = 5 W/m, Wire surface temperature (Ts1) = 75°C

Surrounding temperature (T_∞) = 25°C

After applying insulation :

P_VC insulation thickness, t = 1 mm

New wire surface temperature, T_s2 = 55°C

Now, using concept 1) & Equation i)

$\left(\frac{Q}{L}\right)=\left(2\pi r_1\right)h\left(T_{s1}-T_{\mathrm{\infty }}\right)$

S = h(2π × 10^-3) (75 - 25)

h = 15.9 W/m²

Now,

Using concept 2) & equation ii)

$\left(\frac{Q}{L}\right)=\frac{T_{s2}-T_{\mathrm{\infty }}}{\frac{\ln \left(r_2/r_1\right)}{2\pi K}+\frac{1}{h\left(2\pi r_2\right)}}$

Now,

After applying insulation,

r₂ = r₁ + t = 2 mm

$S=\frac{55-25}{\frac{\ln \left(\frac{2}{1}\right)}{2\pi K}+\frac{1}{15.9\times \left(2\pi \times 2\times {10}^{-3}\right)}}$

$\frac{\ln 2}{2\pi K}+\frac{250}{15.9\pi }=6$

$k=\frac{\ln 2}{2\pi }=0.11W/mK$