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67 Verify the temperature distribution for case 2 in Section 2-9, i.e., that T – T. То — Т cosh m(L-x) + (h/mk) sinh m(L – x) cosh mL+ (h/mk) sinh m L Subsequently show that the heat transfer is sinh mL+ (h/mk) cosh mL cosh m L+ (h/mk) sinh mL q=/h PkA (To – T∞) that of the surrounding fluid. CASE 2 The fin is of finite length and loses heat by convection from its end.

67 Verify the temperature distribution for case 2 in Section 2-9, i.e., that T – T. То — Т cosh m(L-x) + (h/mk) sinh m(L – x) cosh mL+ (h/mk) sinh m L Subsequently show that the heat transfer is sinh mL+ (h/mk) cosh mL cosh m L+ (h/mk) sinh mL q=/h PkA (To – T∞) that of the surrounding fluid. CASE 2 The fin is of finite length and loses heat by convection from its end.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
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-67 Verify the temperature distribution for case 2 in Section 2-9, i.e., that T – T cosh m(L – x) +(h/mk) sinh m(L – x) То — То cosh mL+ (h/mk) sinh m L Subsequently show that the heat transfer is sinh mL+ (h/mk) cosh m L cosh mL+ (h/mk) sinh mL q=/h PkA (To – T∞) that of the surrounding fluid. CASE 2 The fin is of finite length and loses heat by convection from its end. If we let m-h P/kA, the general solution for Equation (2-306) may be written 6=Cje +C 12-31] For case I the boundary conditions are e = at x=0 at x00 CHAPTER 2 Steady-Sute Conduction One Dimension and the solution becomes e T-Tx 12-32] To -T For case 3 the boundary conditions are e = at x0 de -0 at x=L dx Thus C +C2 0mm(-Cjemt +Czey Solving for the constants C, and C2, we obtain + 12-33a] cosh [m(L-x)1 12-336| cosh ml. The hyperbolic functions are defined as e +e 2 sinh x cosh x sinh x - - cosh x tanh x= e te The solution for case 2 is more involved alecbraically and the result is T-T _cosh m (L -x) + (h/mk) sinh m (L -x) L 12-34 cosh mL. + (h/mk) sinh m All of the heat lost by the fin must be conducted into the base at x=0. Using the equations for the temperature distribution, we can compute the heat loss from =-kA An altemative method of integrating the convection heat loss could be used: hPT - Tx) dx= [*. In most cases, however, the first equation is easier to apply. For case 1, q=-kA (-mee(0)= APRAG 12-35] For case 3, 12-36| - VA PRAA tanh ml. 2-10 Fins The heat flow for case 2 is sinh mL+ (h/mk) cosh m L qVhPRA (To-T) cosh mL+ (h/mk) sinh mL 12-37]
Transcribed Image Text:-67 Verify the temperature distribution for case 2 in Section 2-9, i.e., that T – T cosh m(L – x) +(h/mk) sinh m(L – x) То — То cosh mL+ (h/mk) sinh m L Subsequently show that the heat transfer is sinh mL+ (h/mk) cosh m L cosh mL+ (h/mk) sinh mL q=/h PkA (To – T∞) that of the surrounding fluid. CASE 2 The fin is of finite length and loses heat by convection from its end. If we let m-h P/kA, the general solution for Equation (2-306) may be written 6=Cje +C 12-31] For case I the boundary conditions are e = at x=0 at x00 CHAPTER 2 Steady-Sute Conduction One Dimension and the solution becomes e T-Tx 12-32] To -T For case 3 the boundary conditions are e = at x0 de -0 at x=L dx Thus C +C2 0mm(-Cjemt +Czey Solving for the constants C, and C2, we obtain + 12-33a] cosh [m(L-x)1 12-336| cosh ml. The hyperbolic functions are defined as e +e 2 sinh x cosh x sinh x - - cosh x tanh x= e te The solution for case 2 is more involved alecbraically and the result is T-T _cosh m (L -x) + (h/mk) sinh m (L -x) L 12-34 cosh mL. + (h/mk) sinh m All of the heat lost by the fin must be conducted into the base at x=0. Using the equations for the temperature distribution, we can compute the heat loss from =-kA An altemative method of integrating the convection heat loss could be used: hPT - Tx) dx= [*. In most cases, however, the first equation is easier to apply. For case 1, q=-kA (-mee(0)= APRAG 12-35] For case 3, 12-36| - VA PRAA tanh ml. 2-10 Fins The heat flow for case 2 is sinh mL+ (h/mk) cosh m L qVhPRA (To-T) cosh mL+ (h/mk) sinh mL 12-37]
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