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A 2m long wire of resistance 4 ohm and diameter 0.64 mm is coated wilth plastic insulation of thickness 0.06 mm. When a current of 5A flows through the wrie, find the temperature difference across the insultion insteady state. If K=0.16xx10^(-2) cal/cm ""^(@)C sec. |
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Answer» Solution :Consider a concenric CYLINDRICAL shell of radius r and thickness dr as shown in figure. The radial rate of flow of heat through this shell in STEADY STATE will be `H=(dQ)/(dt)=-KA(d theta)/(dr)` Negative sign isused as with Increase in r `theta` decreases Now as for cylindrical shell `A=2 pi rL` `H=-2piLK(d theta)/(dr)` or `int_(a)^(b)(dr)/r=-(-2piK)/Hint_(theta_(1))^(theta_(2))d theta` Which on integration and simplification gives `H=(dQ)/(dt)=-(-2piLK(theta_(1)-theta_(2)))/("In"(b//a))`..........(1)  Here `H=(I^(2)R)/4.2=((5)^(2)xx4)/4.2=24("CAL")/("sec"),L=2m=200cm` `r_(1)=(0.64//2)mm=0.032cm` and `r_(2)=r_(1)+d=0.032+0.006=0.038cm` `(theta_(1)-theta_(2))=(24xx"In"(38//32))/(2xx3.14xx200xx0.16xx10^(-2))` `implies(theta_(1)-theta_(2))=(55xx[1.57-1.50])/2=2^(@)C` |
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