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A-wire-that-is-highly-insulated-has-a-radius-of-2-1-mm-and-a-current-of-6-Agoes-through-it-The-material-used-in-insulation-has-thickness-of-2-1-mm-with-a-thermal-conductivity-of-0-2-W-Km-the-materia

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Question Number 99744 by Rio Michael last updated on 23/Jun/20
A wire that is highly insulated has a radius of 2.1 mm and a current of 6 Agoes through it. The material used in insulation has thickness of 2.1 mm with a thermal conductivity of 0.2 W/Km. the material used in constructing the wire has a resistivity of 4.2 × 10^(−7) Ωm. assume the the materials reach steady state. Find the difference in temperature between the outer suface and inner surface.
$$\mathrm{A}\:\mathrm{wire}\:\mathrm{that}\:\mathrm{is}\:\mathrm{highly}\:\mathrm{insulated}\:\mathrm{has}\:\mathrm{a}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{2}.\mathrm{1}\:\mathrm{mm}\:\mathrm{and}\:\mathrm{a}\:\mathrm{current} \\ $$$$\mathrm{of}\:\mathrm{6}\:\mathrm{Agoes}\:\mathrm{through}\:\mathrm{it}.\:\mathrm{The}\:\mathrm{material}\:\mathrm{used}\:\mathrm{in}\:\mathrm{insulation}\:\mathrm{has}\:\mathrm{thickness} \\ $$$$\mathrm{of}\:\mathrm{2}.\mathrm{1}\:\mathrm{mm}\:\mathrm{with}\:\mathrm{a}\:\mathrm{thermal}\:\mathrm{conductivity}\:\mathrm{of}\:\mathrm{0}.\mathrm{2}\:\mathrm{W}/\mathrm{Km}.\:\mathrm{the}\:\mathrm{material}\:\mathrm{used} \\ $$$$\mathrm{in}\:\mathrm{constructing}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{has}\:\mathrm{a}\:\mathrm{resistivity}\:\mathrm{of}\:\mathrm{4}.\mathrm{2}\:×\:\mathrm{10}^{−\mathrm{7}} \Omega\mathrm{m}.\:\mathrm{assume}\:\mathrm{the}\: \\ $$$$\mathrm{the}\:\mathrm{materials}\:\mathrm{reach}\:\mathrm{steady}\:\mathrm{state}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{in}\:\mathrm{temperature} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{outer}\:\mathrm{suface}\:\mathrm{and}\:\mathrm{inner}\:\mathrm{surface}. \\ $$
Answered by ajfour last updated on 23/Jun/20
P=I^2 R=((I^2 ρl)/(πa^2 )) R_T =∫(dr/(k(2πrl)))=(1/(2kπl))ln (b/a) ΔT=PR_T =((I^2 ρl)/(πa^2 ))((1/(2kπl)))ln (b/a) =(((36A^2 )(4.2×10^(−7) Ωm))/(2π^2 (2.1×10^(−3) m)^2 (0.2W/Km)))ln 2 ΔT =((36×4.2(ln 2))/(2π^2 ×4.41×2)) K = ((18(ln 2))/(2.1π^2 )) C^° ≈ 0.602 C °.
$${P}={I}^{\mathrm{2}} {R}=\frac{{I}^{\mathrm{2}} \rho{l}}{\pi{a}^{\mathrm{2}} }\: \\ $$$${R}_{{T}} =\int\frac{{dr}}{{k}\left(\mathrm{2}\pi{rl}\right)}=\frac{\mathrm{1}}{\mathrm{2}{k}\pi{l}}\mathrm{ln}\:\frac{{b}}{{a}} \\ $$$$\Delta{T}={PR}_{{T}} =\frac{{I}^{\mathrm{2}} \rho{l}}{\pi{a}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}{k}\pi{l}}\right)\mathrm{ln}\:\frac{{b}}{{a}} \\ $$$$\:\:\:\:=\frac{\left(\mathrm{36}{A}^{\mathrm{2}} \right)\left(\mathrm{4}.\mathrm{2}×\mathrm{10}^{−\mathrm{7}} \Omega{m}\right)}{\mathrm{2}\pi^{\mathrm{2}} \left(\mathrm{2}.\mathrm{1}×\mathrm{10}^{−\mathrm{3}} {m}\right)^{\mathrm{2}} \left(\mathrm{0}.\mathrm{2}{W}/{Km}\right)}\mathrm{ln}\:\mathrm{2} \\ $$$$\:\Delta{T}\:\:=\frac{\mathrm{36}×\mathrm{4}.\mathrm{2}\left(\mathrm{ln}\:\mathrm{2}\right)}{\mathrm{2}\pi^{\mathrm{2}} ×\mathrm{4}.\mathrm{41}×\mathrm{2}}\:{K} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\mathrm{18}\left(\mathrm{ln}\:\mathrm{2}\right)}{\mathrm{2}.\mathrm{1}\pi^{\mathrm{2}} }\:{C}^{°} \:\approx\:\mathrm{0}.\mathrm{602}\:{C}\:°. \\ $$
Commented by Rio Michael last updated on 23/Jun/20
thanks sir, please you got any other procedure than this??
$$\mathrm{thanks}\:\mathrm{sir},\:\mathrm{please}\:\mathrm{you}\:\mathrm{got}\:\mathrm{any}\:\mathrm{other}\: \\ $$$$\mathrm{procedure}\:\mathrm{than}\:\mathrm{this}??\: \\ $$

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