Solution Manual for Heat Convection (2nd Edition) by Latif M. Jiji – Complete Problem Solutions and Explanations

All Chapters Covered
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SOLUTION MANUAL
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, PROBLEM 1.1 ty




Heat is removed from a rectangular surface by
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convection to an ambient fluid at T . The heat
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ty
t y ty


transfer coefficient is h. Surface temperature is given
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by
ty x W
A 0
Ts = 1/ 2 t y

x ty




where A is constant. Determine the steady state
t y t y t y t y t y t y


heat transfer rate from the plate.
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L
(1) Observations. (i) Heat is removed from the surface ty t y ty ty ty ty

by convection. Therefore, Newton's law of cooling is
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dqs
applicable. (ii) Ambient temperature and heat transfer t y t y t y t y t y
x 0 W
coefficient are uniform. (iii) Surface temperature variest y ty t y t y ty t y t y


along the rectangle. ty ty
dx
(2) Problem Definition. Find the total heat transfer rate by convection from the surface
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of a plate with a variable surface area and heat transfer coefficient.
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(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection.
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However, in this problem surface temperature is not uniform. This means that the rate of
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heat transfer varies along the surface. Thus, Newton’ s law should be applied to an
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infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer.
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(4) Plan Execution. ty




(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat t y t y t y t y t y t y t y t y


t y transfer coefficient and (4) uniform ambient fluid temperature.
t y ty ty ty ty ty ty




(ii) Analysis. t y Newton's law of cooling states that ty ty ty ty ty




qs = h As (Ts
ty
ty ty ty t y - T)ty (a)
where
As = surface area, m2
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h = heat transfer coefficient, W/m2-oC
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qs = rate of surface heat transfer by convection, W
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Ts = surface temperature, oC
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T = ambient temperature, oC
ty ty ty ty




ty Applying (a) to an infinitesimal area ty ty ty ty t y



t y dAs
dq s t y t y
= h (Ts ty ty t y - T) dAs
ty ty (b)
The next step is to express Ts (x) in terms of distance x along the triangle. Ts (x) is specified as
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ty
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ty
t y ty ty


A
Ts = 1/ 2 (c) t y
x ty

, PROBLEM 1.1 (continued) ty ty




The infinitesimal area dAs is given
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by
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dAs = W dx ty ty ty (d)

where
x = axial distance, m
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W = width, m ty ty ty




ty Substituting (c) and into (b) ty ty ty ty



A
dqs = - T) Wdx ty ty (e)
1/ 2
h( x ty
ty




Integration of (f) gives ty ty ty


qs
t y





q = dq = hW ( Ax1/ 2

L
t y ty t y ty ty ty ty )dx (f)
T
ty
ty ty

s s 
0
Evaluating the integral in (f) ty ty ty ty



qs  hW 2AL1/2  LT
ty
ty ty ty ty ty ty 
ty 
Rewrite the above 

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ty qs  hWL 2AL1/2  T
ty
ty ty ty ty ty (g)


ty

Note that at x = L surface temperature Ts (L) is given by (c)
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ty
ty ty ty ty


as
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(h)
1/2
Ts (L)  AL
ty
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(h) into (g)
ty ty 
qs  hWL 2Ts (L)  T
t y
ty ty
ty
ty ty (i)

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(iii) Checking. Dimensional check: According to (c) units of C are o C/m1/2 . Therefore units
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qs in (g) are W.
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Limiting checks: If h = 0 then qs = 0. Similarly,
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ty
ty t y t y if t y W = 0 or L = 0 then qs = 0.
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ty
ty



t Equation (i) satisfies these limiting cases.
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(5) Comments. Integration is necessary because surface temperature is variable.. t y t y t y t y t y t y t y t y


The same procedure can be followed if the ambient temperature or heat transfer coefficient
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is non-uniform.
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,