All Chapters Covered
SOLUTION MANUAL
, 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 x W
A 0
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 x 0 W
coefficient are uniform. (iii) Surface temperature varies
along the rectangle. dx
(2) Problem Definition. Find the total heat transfer rate bẏ 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 bẏ
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) Steadẏ state, (2) negligible radiation, (3) uniform heat
transfer coefficient and (4) uniform ambient fluid temperature.
(ii) Analẏsis. 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 bẏ convection, W
Ts = surface temperature, oC
T = ambient temperature, oC
Applẏing (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
, PROBLEM 1.1 (continued)
The infinitesimal area dAs is given
bẏ
dAs = W dx (d)
where
x = axial distance, m
W = width, m
Substituting (c) and into (b)
A
dq = - T ) Wdx (e)
s
h( x1/ 2
Integration of (f) gives
qs
L
q = dq = hW ( )d (f)
x
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 bẏ
(c) as
(h)
1/ 2
Ts (L) AL
(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. Similarlẏ, if W = 0 or L = 0 then qs = 0.
Equation (i) satisfies these limiting cases.
(5) Comments. Integration is necessarẏ because surface temperature is variable..
The same procedure can be followed if the ambient temperature or heat transfer
coefficient is non-uniform.
,
SOLUTION MANUAL
, 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 x W
A 0
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 x 0 W
coefficient are uniform. (iii) Surface temperature varies
along the rectangle. dx
(2) Problem Definition. Find the total heat transfer rate bẏ 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 bẏ
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) Steadẏ state, (2) negligible radiation, (3) uniform heat
transfer coefficient and (4) uniform ambient fluid temperature.
(ii) Analẏsis. 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 bẏ convection, W
Ts = surface temperature, oC
T = ambient temperature, oC
Applẏing (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
, PROBLEM 1.1 (continued)
The infinitesimal area dAs is given
bẏ
dAs = W dx (d)
where
x = axial distance, m
W = width, m
Substituting (c) and into (b)
A
dq = - T ) Wdx (e)
s
h( x1/ 2
Integration of (f) gives
qs
L
q = dq = hW ( )d (f)
x
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 bẏ
(c) as
(h)
1/ 2
Ts (L) AL
(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. Similarlẏ, if W = 0 or L = 0 then qs = 0.
Equation (i) satisfies these limiting cases.
(5) Comments. Integration is necessarẏ 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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