Explanation Given: The radius of the container is r o . The volumetric heat generation rate is q ˙ = q ˙ o [ 1 − ( r r o ) 2 ] . The thermal conductivity of the radioactive waste is k . The temperature of the liquid is T ∞ . The convection coefficient of the liquid is h . Formula used: The heat equation for uniform heat generation in the container is given as, 1 r 2 d d r ( r 2 d T d r ) + q ˙ k = 0 Here, r is the radius inside the container and T is the temperature at the required r . The Boundary conditions are given as, d T d r | r = 0 = 0 and [ Q cond = Q conv ] r = r o Here, Q cond is the conduction heat transfer, Q conv is the convection heat transfer and r o is the radius of the container. The conduction heat transfer is given as, Q cond = − k d T d r The convection heat transfer is given as, Q conv = h [ T ( r ) − T ∞ ] Here, T ( r ) is the temperature at a radius r in the container. Calculation: The temperate distribution inside the container can be calculates as, 1 r 2 d d r ( r 2 d T d r ) + q ˙ k = 0 d d r ( r 2 d T d r ) = q ˙ o [ 1 − ( r r o ) 2 ] r 2 k d d r ( r 2 d T d r ) = q ˙ o k [ r 2 − ( r 4 r o 2 ) ] Integrate the above equation with respect to r , ∫ d d r ( r 2 d T d r ) = − ∫ q ˙ o k [ r 2 − ( r 4 r o 2 ) ] ( r 2 d T d r ) = − q ˙ o k [ r 3 3 − r 5 5 r o 2 ] + C 1 d T d r = − q ˙ o k [ r 3 − r 3 5 r o 2 ] + C 1 r 2 Here, C 1 is the integration constant. Integrate the above equation again with respect to r , ∫ d T d r = ∫ [ − q ˙ o k [ r 3 − r 3 5 r o 2 ] + C 1 r 2 ] T ( r ) = − q ˙ o k [ r 2 6 − r 4 20 r o 2 ] − C 1 r + C 2

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ISBN: 9780470501962

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3, Problem 3.103P

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Solution Summary: The author describes the temperature distribution in the container, the volumetric heat generation rate, and the thermal conductivity of the radioactive waste.

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