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Question
The three rods shown in figure have identical geometrical dimensions. Heat flows from the hot end at a rate of 40 W in the arrangement (a). Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c). Thermal condcutivities of aluminium and copper are 200 W m−1°C−1 and 400 W m−1°C−1 respectively.
Solution
For arrangement (a),
Temperature of the hot end ,T1 = 100°C
Temperature of the cold end ,T2 = 0°C
`R_s = R_1 + R_2 + R_3`
`(l)/(K_{AI}A) + l/(K_{cu}A) + {l}/(K_{Al}A)`
`(l)/(A) ( 1/200 +1/400 + 1/200)`
`1/A (5/400)`
`1/Axx 1/80`
`(dQ)/d = q = Rate of flow of heat = (T_1 - T_2)/R_s `
`= (100 - 0)/((1/Axx1/(80))`
Given :
q = 40W
`40 = (100)/(1/Axx 1/80)`
`⇒ l/A = 200`
`⇒ l/A = 1/200 `
For arrangement (b),
`"R_net" = R_[AI} + (R_cuxxR_{AI})/(R_cu = R_{AI)}`
`= l / (K_AI A) + (l/(K_cuA)xxl/K_(AI))/(l / (K_{cu}A9 )+ (l)/ (K_{AL}A)`
`= (l)/(A.K_Al) +( l )/ (A(K_AL + K_cu)) `
`= l/A (1/200 + 1/"200 + 400")`
`=l/A ( 1/200 + 1/600)`
= `4/600 1/A`
Rate of flow of heat is given by
`q = (T_1 - T_2)/ R_"net"`
`= ((100 - 0))/ (4 l) 600 A`
`= (100xx600)/4 xx1/200`
= 75 W
For arrangement (c),
`1 / (R"net") = (1)/(R_{AI)) + 1/R_(cu)+ 1/(R_AI) `
`= K_{K_{AI}}/ (l)+ K_(cuA)/l + (K_{Al}A)/l`
`1/R_"net" = (K_Al + K_( cu) + K_(Al))A/l`
`l /R_"(net)" = (200 + 400 +200)`
`1/ R_"net" = 200/800`
`=(100)/(1/4)`
rate of heat flow = `(DeltaT)/R_"net"`
`= 100/1/4`400 W
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