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Three rods of materialX andthree rodsof materail Y are connected as shown in figure. All therodsare of idental lengthand cross- sectionareas. If theend Ais maintained at 60^(@) C and the junction E at 10^(@)C , calculate the temperature of thejunction B, C and D . The thermalconductivityof X is 9.2 xx 10^(-2)kilocal/m""^(@)C s andthat of Y is 4.6 xx 10^(-2) kilocal/m""^(@)C s. |
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Answer» SOLUTION :Let `theta_(B), theta_(C)`and `theta_(D)`be thetemperature of junctionB,C and Drespectively at staady STATE . Let `K_(x)` and `k_(y)` be thermalconductivites of rodsof materialX and Y respectively . The concept in solving the problem is that in steady state heat enteringper SECONDAT anyjunctionmust be EQUALTO thatleaving the junction.  As `K_(x) = 9.2 xx 10^(-2)` ans `K_(y) = 4.6 xx 10^(-2) kilocal m^(-1) (""^(@)C)^(-1) (sec)^(-1)` i.e..,`(K_(x))/(K_(y)) = (9.2 xx 10^(-2))/(4.6 xx 10^(-2))2 ` Let `K_(x) = 2K, K_(y) = K` (or) `(K_(y)A(60-theta_(B)))/(l) = (K_(x)A(theta_(B) - theta_(C)))/(l) + (K_(y) A(theta_(B) - theta_(C)))/(l)` (or)`K (60 - theta_(B)) = 2K(theta_(B) - theta_(C))+ K(theta_(b) - theta_(D)) therefore (60-theta_(B)) = 2(theta_(B) - theta_(C)) + theta_(B) - theta_(D)` `therefore4 theta_(B) - 2 theta_(C) - 2 theta_(D) = 60...........(1)` Forjunction `C_(1),H_(1) = H_(3) + H_(5) therefore(K_(y)A(theta_(B) - theta_(C)))/(l) = (K_(x)A(theta_(c) - 10))/(l) + (k_(y)A(theta_(D) - 10))/(l)` Thisgives `therefore theta_(B) - 3 theta_(C)+ theta_(D) = - 10..........(2)` For junction D, `H_(2) + H_(3)= H_(5)therefore(k_(y) A(theta_(B) - theta_(D)))/(l) + (k_(y)A(theta_(C) - theta_(D)))/(l) = (k_(y) A(theta_(D) - 10))/(l)` (or) ` K(theta_(B) - theta_(D)) + 2K(theta_(C) - theta_(D)) = k(theta_(D) - 10)` Thisgives `theta_(B) +2theta_(C) - 4 theta_(D)= - 10..............(3)`from (2) and (3) `THETA _(B)-3theta_(C) - theta_(D) = theta_(B) + 2 theta_(C) - 4 theta_(D)` i.e., `5theta_(C) = 5 theta_(D) "" or "" theta_(C) = theta_(D) ............(4)`Substituting this in (1) and (2), we get . ` 4 theta_(B) - 3 theta_(C) = 60.............(5) , theta_(B) - 2theta_(C) = - 10............(6)` Solving (5) and (6), we get `theta_(B) = 30^(@)C, theta_(C) = 20^(@)C` As `theta_(c) = theta_(D)"" therefore "" theta_(C) = theta_(D) = 20^(@)C` `therefore` Thus the temperatureof junction , B, Cand D are `30^(@),20^(@)C and 20^(@)C`respectively. |
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