Heat TransferLecture
Jr.Safaa Hameed Faisal
Solved Problems
fffiIm
A load of fruit
at a temperature of 25"C is to be cooled down in a room at a constant
air temperature of toC. (a) How long the fruit wiII require to cool down to 2oC when the surface
heat transfer coefficient is 5.81 W /m2, K?
(b) What is the temperature of the fruit peas after a lapse of 10 min from the start of cooling?
(c) What air temperature must be used if the fruit were to be cooled down to 5oC in 30 min?
The fruit are supposed to have a sphere shape with average diameter of 8 rnm. Their density is
750 kglm3 and specific heat 3.35 kJ/kg K.
(T"o=\<' -('. ^- "r5 i ,f-:. t*
,
t--( -'-'^
Solution:
We have for a lumped heat-capacify system:
tts-
s-J,i(:-"u q'(t's'J-l' "
g
h,1r
T _T^,
- e
T, *T*
0i
pcr"
--f
- u-lli I;o
'Tt
Since the diameter of the peas is only 8 mrn?
\-a
. e*r
(- \.4 ,c
a(^
f\
-T*
i.zqot\--o,
we can neglect any temperature variation within the fruit.
_pxlr(ttiTJ3 = ptl _ 750x0.008-,
A 3x.1 fi(di}3 3x2
3x2
ht
S.ttlr
. 2'l
(a) In=---=---25-1 c' 3.35x10j
pN'
In 24
: -1.178 :
3.3-5
t
(b)
-
1832,4 s - 30,54
7--
25-t
?":
{c}
5.i$1
I
25 *
x t0j
min
Ans.
la)
i.'0{}
3350 = 0.353
,,'/,h{":*.
,4*s. {b}
a,/r_
5.lll x i0x{'0
I
T,*
I
x
t'r< t4'
-ra
-f,
- 0.044
135$
7'o,
0.956 7'xr
7_,
.li.rt
rlri {u,
-;'-i/( "^lt;""?
9.48='C
1-
1.
-a
- \ t ).'*
(lrrt'
5.81 r
-/,
= t).t)44
s-L
39
4.08"C lns. (c)
ar
(,1.-14 <'" li\
Iq
\/
"y'e1''"' tt!
d,-.-
i
/
€")
) r Jt .- c^,
i)\i)'s'"
\
Heat Transfer Lecture
Dr.Safaa Hameed Faisal
Mil!tr
A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is
quenched from 500 to 300 oC in a large reservoir of water at l0oC. Below 100.,C the heat transfer
coefficient is 1.5 kWm2 K. Above I00 "C, the heat transfer coefficient is 0.5 k|l//m2 K. The density
of steel is 7800 kg/m3 and the specific heat is 0.47 kJ/kg IC Neglecting internal thermal resistance of
the steel tube, determine the quenching time.
Solution:
d;
10 cm.
A:
4r:
12 srn. i = 20 cm.
n(tlr,- d,Il:7r{lt}-r
7rE
l': J (,i"_ dill=
I-:
l0'C
x 20 -.- l38l crn:
12)
. rt++_ l0tllx2{):(rglcrni
4\1
Cor:ling lrorrr 500"C to 100"(,
. 7' - T..
";r_ h,lr
_!n
ptt/
T, - T*
l0t) .- 10
--, _ .Sq0
_h r1n_=
'^'
500
- 10
90
r = 62.12
x l0-a x r
7800 x 470 x 691 x I0-6
500 x 1382
s
Cooling from 100'C to 30'C
" 30-10
-tlr-_=-
100*
l0
1500xi382x10-a xr
7800 x 170x691 x 10-6
t:
Tota} time I'or qucnching
18.38s
:62-12
*
ll{.38
: 80.5s .tx.r.
fmnl@Steel
ball bearings (k:50 \il/m.K. o:1.3* 10-5 m2ls) having a diameter of 40 mm are
treateA to a temperature of 650"C and then quenched"in a tank of oil at 55oC. If the heat transfer
coefficient between the baII bearings and oil is 308 W/m2 K. determine;
(a) the duration of time the bearings must remain in oil to reach a temperature of 200oC.
(b) the total amount of heat removed from eaeh bearing during this tirne.
(c) the instantaneous heat transfer rate from the bearings when they are first immersed in oil and
when they reach 200"C.
Solution: To determine whether the bearings have negligible resistance, we first check the magnitude of
the Biot number.
lrf
"1 -r0()x0.02
hl"
Bi :
=_=(i.04
-k & 3 v.5t.:
Since Bi < 0.1, internal resistancc may be neglectr-:d.
Fo=
(a)
I1:=(rio',-1'3xio-jl=0,2925r
3)r (0.0213)2
The time rcciuired ftrr the hall bearirlgs to reach 200oC is
-{ - n-""'u
0,t
200
*
650
-
ld{
-'' _
--a:..-t1.0.{ x t-1.i9i5 i _ '
59_5
5s
55
..-0.0i l7 r
6
;rSaf*a Hameed
t:
{b}
Heat Transfer Lecture
F'aisal
120.67 s, which corresponds to a Fourier number af
35.3
Aws-
Total amount of heat remor,Ed from eaeh bearing during I20.67 s
Q
= hA{F* L} =
J
hA(Ti*
7'*,
r1
: /'i,{#i
1
"-tsixro6,
hAt
F dr:hA4(l -r-Bi
f ."
Fuy
"k
rl
=
3S0
x 4:r{0.S?}? {650:
* {,{Hx:5-i} x
-5-{) {l
:
{c}
Instantane*us
5.79 x l$a W s $r-f : 57.f kJ Ans.
heat ffensfer rate at r: 0 {or Fo : S} is
l2fi.5
0.04 x 35.31
: hA{T;* Li : 30t} x 4wx (0.02}1 x {650*55) : 89? W
r : 1?fi * {Fo: 35.3i,
Q
and at
# = &J{4 * L}
:
:
u-Ri
Fer
3{i0 x 4rc x {0.fi?}? (650
2l fi S/
- 5CIi g-to'tt+} (35'}l
",4m,r.
average human body modeled as a 30_crn diameter, 170;p long cylinder. its
propertie,s may be taken as those of ry{gfit:fgom temperature: densityp : 10$0 kglm3, cp : 418CI
J/kg K and k:0.608 Wm K. A pEarson is found dead at 5 am in a roonr the temperature of which is
2{C. The temperature of the body is measured to be 25oC when found, and the heat transfer
coefficient is estimated to be I Wm2 K. Assuming the hody temperature of a living man is 37"C,
estimate the time of death ef the *bove person.
.'
V-=.
,/ ='
.l__ T",
Solution:
Characteristic length:
r r..l *I
,*,lt
"f, -\;"
<t/
-:
2n 4rL + 2nq.f
t\
e v(
.*'
a-c,
rrx{ti.l5}rxl.7
rv
<_{
P\
/
{],{i689 m
hL
k
B;
l*
A
!o'rt
?a{*,IsXt ?1+ 2n {fi.15J:
:
,,V
I x *"&{i8t}
i].6{i8
= 0.93
Sin*i: lJi > {i.i. the lurirpcd sysrem analysis is ru:t -tpplir:able. iiolu$r'sr. ire c&n siill usr it ta obtain
r*ugh eslimatc of ri:e tin:c of dcath.
tl,41
{}
$i
..
rt<'tr'
-
35-2U 5
l?-?s l7
* 44*5? s :
,17
:l1li
;:r:[.,
hAr
- 1.2238
1?"?4 /'t \z''?vl {t
)i c,D
";11--A body was found, and thus the time of death is 4.30 pm.
The person died about 12.5 h befbrAThe
r
a