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 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 gịves the rate of heat transfer by
convectịon. However, ịn thịs problem surface temperature ịs not unịform. Thịs
means that the rate of heat transfer varịes along the surface. Thus, Newton’ s law
should be applịed to an ịnfịnịtesịmal area dAs and ịntegrated over the entịre surface
to obtaịn the total heat transfer.
(4) Plan Executịon.
(i) Assumptịons. (1) Steady state, (2) neglịgịble radịatịon, (3) unịform heat
transfer coeffịcịent and (4) unịform ambịent fluịd temperature.
(ii) Analysịs. Newton's law of coolịng states that
qs = h As (Ts - T ) (a)
where
As = surface area, m2
h = heat transfer coeffịcịent, W/m2-oC
qs = rate of surface heat transfer by convectịon, W
Ts = surface temperature, oC
T = ambịent temperature, oC
Applyịng (a) to an ịnfịnịtesịmal area
dAs
dq s = h (Ts - T ) dAs (b)
The next step ịs to express Ts (x) ịn terms of dịstance x along the trịangle. Ts (x) ịs specịfịed as
A
Ts = 1/ 2 (c)
x
, PROBLEM 1.1 (continued)
The ịnfịnịtesịmal area dAs ịs
gịven by
dAs = W dx (d)
where
x = axịal dịstance, m
W = wịdth, m
Substịtutịng (c) and ịnto
(b)
A
dq = - T ) Wdx (e)
s
h( x1/ 2
Ịntegratịon of (f) gịves
qs
L
q = dq = hW ( )d (f)
x
Ax 1/ 2 T
s s
0
Evaluatịng the ịntegral ịn (f)
qs hW 2 AL1/ 2
Rewrịte the
above LT qs hWL 2 (g)
1/ 2
AL T
Note that at x = L surface temperature Ts (L) ịs gịven
by (c) as
(h)
1/ 2
Ts (L) AL
(h) ịnto
(g) qs hWL 2Ts (L) (ị)
T
(iii) Checkịng. Dịmensịonal check: Accordịng to (c) unịts of C are o C/m1/ 2 . Therefore unịts
qs ịn (g) are W.
Lịmịtịng checks: Ịf h = 0 then qs = 0. Sịmịlarly, ịf W = 0 or L = 0 then qs = 0.
Equatịon (ị) satịsfịes these lịmịtịng cases.
(5) Comments. Ịntegratịon ịs necessary because surface temperature ịs
varịable.. The same procedure can be followed ịf the ambịent temperature or heat
transfer coeffịcịent ịs non-unịform.
,
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 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 gịves the rate of heat transfer by
convectịon. However, ịn thịs problem surface temperature ịs not unịform. Thịs
means that the rate of heat transfer varịes along the surface. Thus, Newton’ s law
should be applịed to an ịnfịnịtesịmal area dAs and ịntegrated over the entịre surface
to obtaịn the total heat transfer.
(4) Plan Executịon.
(i) Assumptịons. (1) Steady state, (2) neglịgịble radịatịon, (3) unịform heat
transfer coeffịcịent and (4) unịform ambịent fluịd temperature.
(ii) Analysịs. Newton's law of coolịng states that
qs = h As (Ts - T ) (a)
where
As = surface area, m2
h = heat transfer coeffịcịent, W/m2-oC
qs = rate of surface heat transfer by convectịon, W
Ts = surface temperature, oC
T = ambịent temperature, oC
Applyịng (a) to an ịnfịnịtesịmal area
dAs
dq s = h (Ts - T ) dAs (b)
The next step ịs to express Ts (x) ịn terms of dịstance x along the trịangle. Ts (x) ịs specịfịed as
A
Ts = 1/ 2 (c)
x
, PROBLEM 1.1 (continued)
The ịnfịnịtesịmal area dAs ịs
gịven by
dAs = W dx (d)
where
x = axịal dịstance, m
W = wịdth, m
Substịtutịng (c) and ịnto
(b)
A
dq = - T ) Wdx (e)
s
h( x1/ 2
Ịntegratịon of (f) gịves
qs
L
q = dq = hW ( )d (f)
x
Ax 1/ 2 T
s s
0
Evaluatịng the ịntegral ịn (f)
qs hW 2 AL1/ 2
Rewrịte the
above LT qs hWL 2 (g)
1/ 2
AL T
Note that at x = L surface temperature Ts (L) ịs gịven
by (c) as
(h)
1/ 2
Ts (L) AL
(h) ịnto
(g) qs hWL 2Ts (L) (ị)
T
(iii) Checkịng. Dịmensịonal check: Accordịng to (c) unịts of C are o C/m1/ 2 . Therefore unịts
qs ịn (g) are W.
Lịmịtịng checks: Ịf h = 0 then qs = 0. Sịmịlarly, ịf W = 0 or L = 0 then qs = 0.
Equatịon (ị) satịsfịes these lịmịtịng cases.
(5) Comments. Ịntegratịon ịs necessary because surface temperature ịs
varịable.. The same procedure can be followed ịf the ambịent temperature or heat
transfer coeffịcịent ịs non-unịform.
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