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
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
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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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