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All Chapters Covered
SOLUTION MANUAL
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PROBLEM 1.1
Heat is removed from a rectangular surface by L
convection to an ambient fluid at T . The heat transfer
coefficient is h. Surface temperature is given by
A 0 x W
Ts = 1/ 2
x
where A is constant. Determine the steady state heat
transfer rate from the plate.
L
(1) Observations. (i) Heat is removed from the surface
by convection. Therefore, Newton's law of cooling is dqs
applicable. (ii) Ambient temperature and heat transfer 0 x W
coefficient are uniform. (iii) Surface temperature varies
along the rectangle. dx
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a
plate with a variable surface area and heat transfer coefficient.
(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection.
However, in this problem surface temperature is not uniform. This means that the rate of heat
transfer varies along the surface. Thus, Newton’ s law should be applied to an infinitesimal area
dAs and integrated over the entire surface to obtain the total heat transfer.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer
coefficient and (4) uniform ambient fluid temperature.
(ii) Analysis. Newton's law of cooling states that
qs = h As (Ts - T ) (a)
where
As = surface area, m2
h = heat transfer coefficient, W/m2-oC
qs = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
T = ambient temperature, oC
Applying (a) to an infinitesimal area dAs
dq s = h (Ts - T ) dAs (b)
The next step is to express Ts (x) in terms of distance x along the triangle. Ts (x) is specified as
A
Ts = 1/ 2 (c)
x
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The infinitesimal area dAs is given by
dAs = W dx (d)
where
x = axial distance, m
W = width, m
Substituting (c) and into (b)
A
dq s = h( - T ) Wdx (e)
1/ 2
x
Integration of (f) gives qs
L
q = dq = hW ( )dx (f)
Ax 1/ 2 T
s s
0
Evaluating the integral in (f)
qs hW 2 AL1/ 2
Rewrite the above
LT qs hWL 2 (g)
1/ 2
AL T
Note that at x = L surface temperature Ts (L) is given by (c) as
1/ 2
Ts (L) AL (h)
(h) into (g)
qs hWL 2Ts (L) (i)
T
(iii) Checking. Dimensional check: According to (c) units of C are o C/m1/ 2 . Therefore units
qs in (g) are W.
Limiting checks: If h = 0 then qs = 0. Similarly, if W = 0 or L = 0 then qs = 0. Equation (i)
satisfies these limiting cases.
(5) Comments. Integration is necessary because surface temperature is variable.. The same
procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.
,
All Chapters Covered
SOLUTION MANUAL
, @SOLUTIONSSTYDY
PROBLEM 1.1
Heat is removed from a rectangular surface by L
convection to an ambient fluid at T . The heat transfer
coefficient is h. Surface temperature is given by
A 0 x W
Ts = 1/ 2
x
where A is constant. Determine the steady state heat
transfer rate from the plate.
L
(1) Observations. (i) Heat is removed from the surface
by convection. Therefore, Newton's law of cooling is dqs
applicable. (ii) Ambient temperature and heat transfer 0 x W
coefficient are uniform. (iii) Surface temperature varies
along the rectangle. dx
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a
plate with a variable surface area and heat transfer coefficient.
(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection.
However, in this problem surface temperature is not uniform. This means that the rate of heat
transfer varies along the surface. Thus, Newton’ s law should be applied to an infinitesimal area
dAs and integrated over the entire surface to obtain the total heat transfer.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer
coefficient and (4) uniform ambient fluid temperature.
(ii) Analysis. Newton's law of cooling states that
qs = h As (Ts - T ) (a)
where
As = surface area, m2
h = heat transfer coefficient, W/m2-oC
qs = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
T = ambient temperature, oC
Applying (a) to an infinitesimal area dAs
dq s = h (Ts - T ) dAs (b)
The next step is to express Ts (x) in terms of distance x along the triangle. Ts (x) is specified as
A
Ts = 1/ 2 (c)
x
, @SOLUTIONSSTYDY
The infinitesimal area dAs is given by
dAs = W dx (d)
where
x = axial distance, m
W = width, m
Substituting (c) and into (b)
A
dq s = h( - T ) Wdx (e)
1/ 2
x
Integration of (f) gives qs
L
q = dq = hW ( )dx (f)
Ax 1/ 2 T
s s
0
Evaluating the integral in (f)
qs hW 2 AL1/ 2
Rewrite the above
LT qs hWL 2 (g)
1/ 2
AL T
Note that at x = L surface temperature Ts (L) is given by (c) as
1/ 2
Ts (L) AL (h)
(h) into (g)
qs hWL 2Ts (L) (i)
T
(iii) Checking. Dimensional check: According to (c) units of C are o C/m1/ 2 . Therefore units
qs in (g) are W.
Limiting checks: If h = 0 then qs = 0. Similarly, if W = 0 or L = 0 then qs = 0. Equation (i)
satisfies these limiting cases.
(5) Comments. Integration is necessary because surface temperature is variable.. The same
procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.
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