Solution manual for Heat Convection 2nd edition by Latif Jiji

All Chapters Covered
m m




SOLUTION MANUAL
m

, PROBLEM 1.1 m




Heat is removed from a rectangular surface by convection
m m m m m m m m
L
to an ambient fluid at T . The heat transfer coefficient is
m m m m m m
m
m m m m m




h. Surface temperature is given by
m m m m m m




A 0 x W
Ts = 1/ 2
x
m
m




where A is constant. Determine the steady state
m m m m m m




heat transfer rate from the plate.
m m m m m m




L
(1) Observations. (i) Heat is removed from the surface
dqs
m m m m m m




by convection. Therefore, Newton's law of cooling is
x
m m m m m m m m m m




applicable. (ii) Ambient temperature and heat transfer 0 m m m m m
W
coefficient are uniform. (iii) Surface temperature varies m m m m m m m




along the rectangle. m m
dx
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a
m m m m m m m m m m m m m m




plate with a variable surface area and heat transfer coefficient.
m m m m m m m m m m




(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection.
m m m m m m m m m m m m m




However, in this problem surface temperature is not uniform. This means that the rate of heat
m m m m m m m m m m m m m m m m




transfer varies along the surface. Thus, Newton’ s law should be applied to an infinitesimal area
m m m m m m m m m m m m m m m m




dAs and integrated over the entire surface to obtain the total heat transfer.
m m m m m m m m m m m m m




(4) Plan Execution. m




(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer m m m m m m m m m




m coefficient and (4) uniform ambient fluid temperature. m m m m m m




(ii) Analysis. Newton's law of cooling states that m m m m m m




qs = h As (Ts - T)
m
m m m m m (a)
where
As = surface area, m2
m m m m




h = heat transfer coefficient, W/m2-oC
m m m m m




qs = rate of surface heat transfer by convection, W
m
m m m m m m m m




Ts = surface temperature, oC
m m m m




T = ambient temperature, oC
m m m m




Applying (a) to an infinitesimal area dAs
m m m m m m m




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




A
Ts = 1/ 2 (c)
x
m
m

, PROBLEM 1.1 (continued) m m




The infinitesimal area dAs is given by
m m m m m m




dAs = W dx m m m (d)
where
x = axial distance, m
m m m m




W = width, m
m m m




Substituting (c) and into (b)
m m m m m




A
dq s = h( - T) Wdx
m m (e)
1/ 2
m




x m




Integration of (f) gives qs m m m m




 
L
q = dq
m m = hW ( Ax−1/2 −T
m m m m m m m
)dx (f)
s 
m

s
0
Evaluating the integral in (f) m m m m




qs = hW 2AL1/2 − LT
m
m 
m m m m m  m




Rewrite the above
 −T 
m m




mqs = hWL 2AL−1/2
m
m m m m m
 m
(g)
Note that at x = L surface temperature Ts (L) is given by (c) as
m m m m m m m m
m
m m m m m




Ts (L) = AL−1/2 m
m m m (h)
(h) into (g)
qs = hWL2Ts (L) −T 
m m




m
m m
m
m m
m
(i)

(iii) Checking. Dimensional check: According to (c) units of C areo C/m1/2 . Therefore units
m m m m m m m m m m m m m m m




qs in (g) are W.
m
m m m




Limiting checks: If h = 0 then qs = 0. Similarly, if W = 0 or L = 0 then qs = 0. Equation (i)
m m m m m m
m
m m m m m m m m m m m m
m
m m m




satisfies these limiting cases.
m m m m




(5) Comments. Integration is necessary because surface temperature is variable.. The m m m m m m m m m




msame procedure can be followed if the ambient temperature or heat transfer coefficient is non-
m m m m m m m m m m m m m m




uniform.

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