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In the following situations, the length and area of cross-section of each rod is same. Find temperature theta at junction of rods. (a) (b) (c ) (d) The thermal conductivity of each rod is K. |
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Answer» Solution :(a) Thermal resistance ofrod 1 `R_(1) = (L)/(2KA) = R_(0)` Thermal resistance of rod 2 `R_(2) = (L)/(KA) = 2 R_(0)` Heat CURRENT in rods is same `i= (Delta theta)/(R )` `i= (100 - theta)/(R _(1)) = (theta - 2 0)/(R_(2))` `(100 - theta)/(R_(0)) = (theta- 20)/(2 R_(0))` `200 - 2 theta = theta - 20` `3 theta = 220` `theta = (220)/(3).^(@)C` (B) `R_(1) (L)/(KA) = R_(0)` `R_(2) = (L)/(2KLa) = (R_(0))/(2)` `R_(3) = (L)/(4 KA) = (R_(0))/(4)` `i = i_(1) + i_(2)` `(100 - theta)/(R_(1)) = (theta - 40)/(R_(2)) + (theta - 10)/(R_(3))` `(100 - theta)/(R_(0)) = (theta - 40)/(R_(0)//2) + (theta - 10)/(R_(0)//4)` `100 - theta = 0 (theta - 40) + 4 (theta - 10)` `= 2 theta- 80 + 40 - 40` `7 theta = 220` `theta = (220)/(7).^(@)C` (c ) `R_(1) = (L//2)/(KA) = (L)/(2KA) = R_(0)` `R_(2) = (L)/(KA) = 2 R_(0)` `R_(3) = (L//2)/(2KA) = (L)/(4KA) = (R_(0))/(2)` `i = i_(1) + i_(2)` `(100 - theta)/(R_(1)) = (theta - 20)/(R_(2)) + (theta - 40)/(R_(3))` `(100 - theta)/(R_(0)) = (theta - 20)/(2 R_(0)) + (theta - 40)/(R_(0) // 2)` `100 - theta = (theta - 20)/(2) + 2 (theta - 40)` `200 - 2 theta = theta - 20 + 4 theta - 160` `7 theta = 380` `theta = (380)/(7).^(@)C` Let `(L)/(KA) = R` `i = i_(1) + i_(2)` `(T_(1) - theta)/(R ) = (theta - T_(2))/(R + (R)/(2)) + (theta - T_(3))/(R + (R)/(2))` `T_(1) - theta = (2)/(3) (2 theta - T_(2) - T_(3))` `3 T_(1) - 3 theta = 4 theta - 2 T_(2) - 2 T_(3)` `7 theta = 3 T_(1) + 2 T_(2) +2 T_(3)` `theta = (3 T_(1) + 2 T_(2) + 2 T_(3))/(7)`    
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