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Transcript of Module6 WorkedOut Problems
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MODULE 6: Worked-out Problems
Problem 1:
For laminar free convection from a heated vertical surface, the local
convection coefficient may be expressed as hx=Cx-1/4, where hx is the
coefficient at a distance x from the leading edge of the surface and the
quantity C, which depends on the fluid properties, is independent of x.
Obtain an expression for the ratioxx hh /
, where
xh
is the average
coefficient between the leading edge (x=0) and the x location. Sketch the
variation of hx and xh
with x.
Schematic:
Analysis: It follows that average coefficient from 0 to x is given by
x
1/4
x
x
0
1/4
x
0
xx
h3
4Cx
3
4x3/4
x
C
3
4h
dxxxCdxh
x1h
===
==
Hence3
4
h
h
x
x =
Boundary layer,
hx=Cx-1/4 where C
is a constant
Ts
x
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The variation with distance of the local and average convection coefficient is
shown in the sketch.
Comments: note that
hhx / x =4/3, independent of x. hence the average
coefficients for an entire plate of length L isLh
=4/3L, where hL is the local
coefficient at x=L. note also that the average exceeds the local. Why?
Problem 2:
Experiments to determine the local convection heat transfer coefficient
for uniform flow normal to heated circular disk have yielded a radial
Nusselt number distribution of the form
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+==
n
o
Dr
ra1
k
h(r)DNu
Where n and a are positive. The Nusselt number at the stagnation
point is correlated in terms of the Reynolds number (ReD=VD/) and
Prandtl
0.361/2
Do Pr0.814Rek
0)Dh(rNu =
==
Obtain an expression for the average Nusselt number, kDhNuD
/
= ,
corresponding to heat transfer from an isothermal disk. Typically
boundary layer development from a stagnation point yields a decaying
convection coefficient with increasing distance from the stagnation
point. Provide a plausible for why the opposite trend is observed forthe disk.
Known: Radial distribution of local convection coefficient for flow
normal to a circular disk.
Find: Expression for average Nusselt number.
Schematic:
Assumptions: Constant properties.
Analysis: The average convection coefficient is
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o
o
s
r
0
n
o
2n2
3
o
n
o0
r
0
2
o
A
s
s
2)r(n
ar
2
r
r
kNuoh
]22r)a(r/r[1NuD
k
r
1h
hdAA
1h
+
+=
+=
=
+
Where Nuo is the Nusselt number at the stagnation point (r=0).hence,
( )
0.361/2
D
o
r
o
2n2
o
Pr2)]0.841Re2a/(n[1NuD
2)]2a/(n[1NuNuD
ro
r
2)(n
a
2
r/ro2Nu
k
DhNuD
o
++=
++=
++==
+
Comments: The increase in h(r) with r may be explained in terms of
the sharp turn, which the boundary layer flow must take around the
edge of the disk. The boundary layer accelerates and its thickness
decreases as it makes the turn, causing the local convection coefficient
to increase.
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Problem 3:
In a flow over a surface, velocity and temperature profiles are of the forms
u(y)=Ay+By2-Cy3 and T(y)=D+Ey+Fy2-Gy3
Where the coefficients A through G are constants. Obtain expressions
for friction coefficients Cf and the convection coefficient h in terms of
u, T and appropriate profile coefficients and fluid properties.
Known: form of the velocity and temperature profiles for flow over a
surface.
Find: expressions for the friction and convection coefficients.
Schematic:
Analysis: The shear stress at the wall is
.]32[ 02
0
ACyByAy
uy
y
s=+=
=
=
=
Hence, the friction coefficient has the form,
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2f
22
sf
u
2AC
u
2A
/2u
C
=
==
The convection coefficient is
=
=
=
+
=
=
TD
Ekh
TD
]3Gy2Fy[Ek
TT
y)T/(kh
f
0y
2
f
s
0yf
Comments: It is a simple matter to obtain the important surface
parameters from knowledge of the corresponding boundary layers
profile. However is rarely simple matter to determine the form of theprofile.
Problem 4:
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In a particular application involving airflow over a heated surface, the
boundary layer temperature distribution may be approximated as
=
v
yuPrexp1TT
TT
s
s
Where y is the distance normal to the surface and the Prandtl number,
Pr=cp/k=0.7, is a dimensionless fluid property. If T =400K, Ts=300K,
and u/v=5000m-1, what is the surface heat flux?
Known: Boundary layer temperature distribution
Find: Surface heat flux.
Schematic:
Properties:
Air ( kTS 300= ): k = 0.0263 W/m.k
Analysis:
Applying the Fouriers law at y=0, the heat flux is
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0y
"
sy
Tkq==
)PTk(Tq s"
s
=
1/m0.7x5000.K(100K)0.02063w/mq"
s =
Comments: (1) Negligible flux implies convection heat transited
surface (2) Note use of k at sT to evaluate from"
sq Fouriers law.
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Problem 5:
Consider a lightly loaded journal bearing using oil having the constant
properties =10-2 kg/s-m and k=0.15W/m. K. if the journal and the
bearing are each mentioned at a temperature of 400C, what is the
maximum temperature in the oil when the journal is rotating at 10m/s?
Known: Oil properties, journal and bearing temperature, and journal
speed for lightly loaded journal bearing.
Find: Maximum oil temperature.
Schematic:
Assumptions: (1) steady-state conditions, (2) Incompressible fluid
with constant properties, (3) Clearances is much less than journal
radius and flow is Couette.
Analysis: The temperature distribution corresponds to the result
obtained in the text example on Couette flow.
+=
2
2
oLy
LyU
2kTT(y)
The position of maximum temperature is obtained from
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==
2
2
L
2y
L
1U
2k
0
dy
dT
y=L/2.
Or,
The temperature is a maximum at this point since hence0.T/dyd 22 > air)( , kwater>>kair. Hence, Tmax,water Tmax,air .
Problem 7:
A flat plate that is 0.2m by 0.2 m on a side is orientated parallel to an
atmospheric air stream having a velocity of 40m/s. the air is at a
temperature of T=20C, while the plate is maintained at Ts=120C. The
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2.
NuL
NuL
and
k
2CL
k
(L)2CLNuL
Hence
2CLLL
2Cdxx
L
Cdxh
L
1h
where
k
LhNu
k
CL
k
)L(CL
k
LhNu
-
1/21/2-
1/21/2
L
0
1/2
L
0
xL
LL
1/21/2L
L
=
==
====
=
===
Comments: note the manner in which-
NuL is defined in terms of Lh
. Also
note that
L
0
xL dxNuL
1Nu
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Problem 8:
For flow over a flat plate of length L, the local heat transfer coefficient hx
is known to vary as x
-1/2
, where x is the distance from the leading edge ofthe plate. What is the ratio of the average Nusslet number for the entire
plate to the local Nusslet number at x=L (NuL)?
Known: Drag force and air flow conditions associated with a flat plate.
Find: Rate of heat transfer from the plate.
Schematic:
Assumptions: (1) Chilton-Colburn analogy is applicable.
Properties: Air(70C,1atm): =1.018kg/m3, cp=1009J/kg.K, pr=0.70,
=20.22*10-6m2/s.
Analysis: the rate of heat transfer from the plate is
)T(T(L)h2q s2
=
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Where
h may be obtained from the Chilton-Colburn analogy,
240Wq
C20)(120.K)(0.3m)2(30W/mq
israteheatThe
.K30W/mh
2/3.70)9J/kg.K)(0)40m/s(100(1.018kg/m105.76h
Prcu2
Ch
hence,
105.76/2(40m/s)1.018kg/m
(0.2m)(0.075N/2)
2
1
/2u
2
1
2
C
Prcu
htPrS
2
Cj
22
2-
34-
2/3
p
f-
4
23
2
2
sf
2/3
p
2/3f
H
==
=
=
=
===
===
Comments: Although the flow is laminar over the entire surface ()100.4/1022.20/2.0/40/Re 526 === smmsmLuL , the pressure gradient
is zero and the Chilton-Colburn analogy is applicable to average, as well as
local, surface conditions. Note that the only contribution to the drag force is
made by the surface shear stress.
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Problem 9:
Consider atmospheric air at 25C in parallel flow at 5m/s over both surfaces of 1-m-
long flat plate maintained at 75C. Determine the boundary layer thickness, the
surface shear stress, and the heat flux at the trailing edge. Determine the drag force onthe plate and the total heat transfer from the plate, each per unit width of the plate.
Known: Temperature, pressure, and velocity of atmospheric air in parallel flow over aPlate of prescribed length and temperature.
Find: (a) Boundary layer thickness, surface shear stress and heat flux at trailing edges,(b) drag force and total heat transfer flux per unit width of plate.
Schematic:
Assumptions: (1) Critical Reynolds number is 5*105, (2) flow over top and bottomsurfaces
Properties: (Tf=323K, 1atm) Air: =1.085kg/m3,=18.2*10-6m2/s,k=0.028W/m.K,pr=0.707
Analysis: (a) calculate the Reynolds number to know the nature of flow
5
26L102.75
/sm1018.2
m1m/s5
LuRe =
==
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Hence the flow is laminar, and at x=L
22
Ls,
1/252
3
1/2
L
2
Ls,
1/251/2
L
0.0172N/m.s0.0172kg/m
)1050.664/(2.7(5m/s)m
kg
2
1.085/2)0.664Re(
9.5mm)101m/(2.7555LRe
==
==
===
Using the correct correlation,
2
s
L
1/31/2
LL
L
.K(754.34W/m)ThL(Ts(L)q"
4.34W/m8W/m.K)/1m155.1(0.02h
hence,
0.332(2Pr0.332Rek
hLNu
==
==
===
(b) The drag force per unit area plate width isLs
LD
= 2' where the factor of two is
included to account for both sides of the plate. Hence with
/m2(1m)8.68W)T(T2Lhq
.K8.68W/m2hLhwithalso
0.0686N/m3N/m2(1m)0.034D
0.0343N/m
isdragThe
(1.085k/2)1.328Re(
s
-
L
"
2-
L
2'
2
Ls
1/2
L
2
Ls
==
==
==
=
==
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Problem 10:Engine oil at 100C and a velocity of 0.1m/s flows over both surfaces of a 1-m-long
flat plate maintained at 20C. Determine
a. The velocity and thermal boundary thickness at the trailing edge.
b. The local heat flux and surface shear stress at the trailing edge.c. The total drag force and heat transfer per unit area width of the plate.
Known: Temperature and velocity of engine oil Temperature and length of flat plate.
Find: (a) velocity and thermal boundary thickness at the trailing edge, (b)Heat flux and surface shear stress at the trailing edge, (c) total drag force and heattransfer per unit plate width.
Schematic:
Assumptions: engine oil (Tf=33K): =864kg/m3,=86.1*106m2/s, k=0.140W/m /K,Pr=1081.
Analysis: (a) calculate the Reynolds number to know the nature of flow
1161/sm10*86.1
1m0.1m/s
LuReL
26-=
==
Hence the flow is laminar at x=L, and
0.0143m)0.147(1081Pr
0.147m)5(1m)(11615LRe
1/31/3
t
1/21/2
L
===
===
(b) The local convection coefficient and heat flux at x=L are
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2
s
1/31/2
LL
C100).K(2016.25W/m)ThL(Tq"
0.3321m
0.140W/m.KPr0.3325LRe
L
kh
===
==
Also the local shear stress is
22
Ls
1/223
1/2
L
2
Ls
0.0842N/m.s0.0842kg/m
)0.664(1161(0.1m/s)2
864kg/m/2)0.664Re(
==
==
(c) With the drag force per unit width given byLs
LD
= 2' where the factor of 2 is
included to account for both sides of the plate, is follows that
5200W/mC100).K(20/m2(1m)32.5W)T(Th2Lq
thatfollowsalsoit.K,32.5W/m2hhwith
0.673N/m)1.328(1161(0.1m/s)/m2(1m)864kg/2)1.328Re(
2
sL
_
2
LL
_
1/2231/2
L
2
Ls
===
==
===
Comments: Note effect of Pr on (/t).
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Problem 11:
Consider water at 27C in parallel flow over an isothermal, 1-m-long, flat plate with a
velocity of 2m/s. Plot the variation of the local heat transfer coefficient with distance
along the plate. What is the value of the average coefficient?
Known: velocity and temperature of air in parallel flow over a flat plate of prescribedlength.
Find: (a) variation of local convection coefficient with distance along the plate, (b)Average convection coefficient.
Schematic:
Assumptions: (1) Critical Reynolds number is 5*105.
Properties: Water (300K): =997kg/m3,=855*10-6N.s/m2, =/=0.858*10-6m2/s,k=0.613W/m. K, Pr=5.83
Analysis: (a) With
5
26L 102.33/sm100.858
1m2m/s
Lu
Re =
==
Boundary layer conditions are mixed and
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40554240449148715514.K)(W/mh
1.00.80.60.40.0215x(m)
xPr
u0.0296kPr0.0296Re
x
khx0.215m,xfor
12061768hx(W/m2.K)
0.2150.1x(m)
xPr
u0.332kPr0.332Re
x
khx0.215m,xfor
0.215m)10/2.33101m(5)/ReL(Rex
x
0.21/3
4/5
1/34/5x
1/21/3
1/2
1/31/2
x
65Lcx,c
2
==>
==
===
The Spatial variation of the local convection coefficient is shown above
(b) The average coefficient is
.K4106W/mh
871(5.83))1033[0.0379(2.1m
0.613W/m.K871)Pr(0.037Re
L
kh
2L
_
1/34/561/34/5
LL
_
=
==
Problem 12:
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A circular cylinder of 25-mm diameter is initially at 150C and is quenched by
immersion in a 80C oil bath, which moves at a velocity of 2m.s in cross flow over the
cylinder. What is the initial rate of heat loss unit length of the cylinder?
Known: Diameter and initial temperature of a circular cylinder submerged in an oil bathf prescribed temperature and velocity.
Find: initial rate of heat loss unit per length.
Schematic:
Assumptions: (1) Steady-state conditions, (2) uniform surface temperature.
Properties: Engine oil (T=353K): =38.1*10-6m2/s, k=0.138W/m. K, Pr=501;(Ts=423K): Prs=98.
Analysis: The initial heat loss per unit length is
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( ) 8.8W/mC80)0.025m(150.K1600W/mq'
.K1600W.mh
98
501(501)0.26(1312)
0.025m
0.138W/m.K)1/4/Pr(PrPrCRe
D
kh
hence.7.4tablefrom0.6mand0.26CFind
1312/sm1038.1
m)2m/s(0.025VDRe
whenrelation.Zhukauskasthefromcomputedbemaywhere
)TD(Thq
2
2_
1/4
0.370.6
s
nm
D
_
26D
s
_
==
=
==
==
===
=
=
_
h
Comments: Evaluating properties at the film temperature, T f=388K(=14.0*10-6m2/s,k=0.135W/m. K, Pr=196), find ReD=3517.
Problem 13:
An uninsulated steam pipe is used to transport high-temperature steam from one
building to one another. The pipe is 0.5-m diameter, has a surface temperature of
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150C, and is exposed to ambient air at -10C.the air moves in cross flow over the pipe
with a velocity of 5m.s.What is the heat loss per unit length of pipe?
Known: Diameter and surface temperature of uninsulated steam pipe. Velocity and
temperature of air in cross flow.
Find: Heat loss per unit length.
Schematic:
Assumptions: (1) steady-state conditions, (2) uniform surface temperature
Properties: Air (T=263K, 1atm):=12.6*10-6m2/s, k=0.0233W/m. K, Pr=0.72;(Ts=423K,
1atm); Prs=0.649.
Analysis: the heat loss per unit length is
)TD(Thq s
_
=
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.K1600W.mh
98
501(501)0.26(1312)
0.025m
0.138W/m.K)1/4/Pr(PrPrCRe
D
kh
hence.7.4tablefrom0.6mand0.26CFind
1312/sm1038.1
m)2m/s(0.025
VDRe
whenrelation.Zhukauskasthefromcomputedbemaywhere
2_
1/4
0.370.6s
nmD
_
26D
=
==
==
=
==
=
_
h
Hence the heat rate is
( ) 4100W/mC10)(-0.5m(150.K16.3W/mq' 2 ==
Comments: Note that qDm, in which case the heat loss increases significantly with
increasing D.
Problem 14:
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Atmospheric air at 25C and velocity of 0.5m/s flows over a 50-W incandescent bulb
whose surface temperature is at 140C. The bulb may be approximated as a sphere of
50-mm diameter. What is the rate of heat loss by convection to the air?
Known: Conditions associated with airflow over a spherical light bulb of prescribed
diameter and surface temperature.
Find: Heat loss by convection.
Schematic:
Assumptions: (1) steady-state conditions, (2) uniform surface temperature.
Properties: Air (Tf=25C, 1atm): =15.71*10-6m2/s, k=0.0261W/m. /K.Pr=.
0.71,=183.6*10-7N.s/m2; Air (Ts=140C, 1atm): =235.5*10-7N.s/m2
Analysis:
)TD(Thq s
_
=
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10.3WC25)(140(0.05m).Km
W11.4q
israteheattheand
.K11.4W/mh
235.5
183.6](0.71)0.06(1591)[0.4(1591)2
0.05m
K0.0261W/m.h
hence
1591/sm1015.71
0.05m0.5m/s
VD
Re
Where
])(/)Pr0.06Re(0.4Re[2D
kh
whenrelation.Whitakerthefromcomputedbemaywhere
22
2_
1/4
0.42/31/2_
26D
1/4
s
0.42/3
D
1/2
D
_
==
=
++=
=
==
++=
_
h
Comments: (1) The low value of_
h suggests that heat transfer by free convection may
be significant and hence that the total loss by convection exceeds 10.3W
(2) The surface of the bulb also dissipates heat to the surroundings by radiation. Further,
in an actual light bulb, there is also heat loss by conduction through the socket.
(3) The Correlation has been used its range of application (/s )
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Problem 15:
Water at 27 C flows with a mean velocity of 1m/s through a 1-km-long
cast iron pipe at 0.25 m inside diameter.
(a) Determine the pressure drop over the pipe length and the
corresponding pump power requirement, if the pipe surface is
clean.
(b) If the pipe surface roughness is increased by 25% because of
contamination, what is the new pressure drop and pump power
requirement.
Known: Temperature and velocity of water in a cast iron pipe of prescribed
dimensions.
Find: pressure drops and power requirement for (a) a clean surface and (b) a
surface with a 25% larger roughness.
Schematic:
Assumptions: (1) Steady, fully developed flow.
Properties: Water (300K):=1000 kg/m3,=855*10-6N.s/m2.
Analysis: (a) from eq.8.22, the pressure drop is
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L2D
ufp
2
m=
e=2.6*10
-4
m for clean cast iron; hence e/D=1.04*10
-3
. With
5
326
mD 102.92
/1000kg/mN.s/m10855
0.25m1m/s
DuRe =
==
0.42barN/m104.2P
.mkg/s104.21000m2(0.25m)
(1m/s)(1000kg/m0.021P
,hence0.021.fthat8.3figfromfind
24
2423
==
==
)
The pump power requirement is
2.06kW1m/s/4)m0.25(N/m104.2P
/4)up(DVp.P
2224
m
2.
==
==
(b) Increasing by 25% it follows that =3.25*10-4m and e/D=0.0013. With
ReD unchanged, from fig 8.3, it follows that f 0.0225. Hence
2.21kW)P/f(fPbar0.45)/f(fP1122122 ==== 1p
Comments: (1) Note that L/D=4000>>(xfd,h/D) 10 for turbulent flow and
the assumption of fully developed conditions is justified.
(2) Surface fouling results in increased surface and increases operating costs
through increasing pump power requirements.
Problem 16:
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Hence, regardless of whether the hydrodynamic or thermal boundary layer is
fully developed, is follows that
Tm(x) is linear
T m,2 will be the same for all flow conditions.
The surface heat flux can be written as
(x)]Th[Tq ms"
s =
Under fully developed flow and thermal conditions, h=hfd is a constant.
When flow is developing h> hfd . Hence, the temperature distributions appearas below.
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Problem 17:
A thick-walled, stainless steel (AISI 316) pipe of inside and outsidediameter Di=20mm and Do=40m is heated electrically to provide a
uniform heat generation rate of 36.
/10 mWq = . The outer surface of the
pipe is insulated while water flows through the pipe at a rate of
skgm /1.0.
=
(a) If the water inlet temperature is Tm,1=20C and the desired outlet
temperature is Tm,o=40C, what is the required pipe length
(b) What are the location and value of the maximum pipe temperature?
Known: Inner and outer diameter of a steel pipe insulated on the outside and experiencing
uniform heat generation. Flow rate and inlet temperature of water flowing through thepipe.
Find: (a) pipe length required to achieve desired outlet temperature, (b)
location and value of maximum pipe temperature.
Schematic:
Assumptions: (1) steady-state conditions, (2) constant properties, (3)
negligible kinetic energy, potential energy and flow work changes, (4) one-
dimensional radial conduction in pipe wall, (5) outer surface is adiabatic.
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Properties: Stainless steel 316 (T 400K): k=15W/m.k; water ( 303KT.
m = );
cp=4178J/kg. K, k=0.617W/m. K,=803*10-6N.s/m2, Pr=5.45
Analysis: (a) performing an energy balance for a control volumeabout the inner tube, it follows that
8.87mL
](0.02)4m(/4)[(0.W/m10
C)20178(J/kg.K(0.1kg/s)4
)LD(/4)(q
)T(TmcL
)LD(/4)(qq)T(Tcm
22362
i
2
0
.
.
im,om,p
2
i
2
0
.
im,om,p
.
=
=
=
==
(b) The maximum wall temperature exists at the pipe exit (x=L) and the
insulated surface (r=ro). The radial temperature distribution in the wall is of
the form
Tnr2k
qr-r
4k
qCCnr
2k
qrr
4k
q-T)T(r:rr
2k
qrC
r
Cr
2k
q0
dr
dT;rr
;conditionsboundarythegconsiderin
CnrCr4k
qT(r)
si
.2
o2
i
.
22i
.2
o2
i
.
si1
.2
o
1o
1
o
.
rro
21
2
.
o
+=++===
=+==
=
++=
=
The temperature distribution and the maximum wall temperature (r=ro) are
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LD
Dq((/4)q
innertheTs,where
4
q-)T(rT
)r-(r4k
qT(r)
i
.2
o"
s
omaxw,
2
i
2
.
=
=
=
Where h is the local convection coefficient at the exit. With
7928N.s/m103(0.02m)80
0.1kg/s4
D
m4Re
26
i
.
D =
==
The flow is turbulent and, with (L/Di)=(8.87m/0.02m)=444>>(xfd/D)10,it is
also fully developed. Hence, from the Dittus-Boelter correlation,
.K1840W/m5.45)0.023(79280.02m
0.617W/m.K)pr(0.023Re
D
kh
20.44/50.44/5
D
i
===
Hence the inner surface temperature of the wall at the exit is
C52.4C48.20.01
0.02n
215W/m.K
(0.02)W/m10](0.01)[(0.02)
415W/m.K
W/m10T
and
C48.2C40.K(0.02m)180W/m4
](0.02m)[(0.04m)W/m10T
4D
)D(Dq
23622
36
maxw,
2
2236
om,
i
2
i
2
o
.
s
=++=
=+
=+
=
T
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Comments: The physical situation corresponds to a uniform surface heat
flux, and Tm increases linearly with x. in the fully developed region, Ts also
increases linearly with x.
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Problem 18:
The surface of a 50-mm diameter, thin walled tube is maintained thin
walled tube is maintained at 100C. In one case air is cross flow very the
tube with a temperature of 25C and a velocity of 30m/s. In another caseair is in fully developed flow through the tube with a temperature of 25C
and a mean velocity of 30m/s. compare the heat flux from the tube to the
air for the two cases.
Known: surface temperature and diameter of a tube. Velocity and
temperature of air in cross flow. Velocity and temperature of air in fully
developed internal flow.
Find: convection heat flux associated with the external and internal flows.
Schematic:
Assumptions: (1) steady-state conditions, (2) uniform cylinder surface
temperature, (3) fully developed internal flow
Properties: Air (298K): =15.71*10-6m2/s, k=0.0261W/m.K, Pr=0.71
Analysis: for the external and internal flow
4
261055.9
/1071.15
05./30Re =
===
sm
msmDuVD mD
From the Zhukauskas relation for the external flow, with C=0.26 and m=0.6
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223)1()71.0()1055.9(26.0)PrPr(Pr/Re 4/137.06.044/1_
=== sm
DD CuN
Hence, the convection coefficient and heat rate are
2"
__
./4.116)(
05.0
./0261.0
mWTThq
m
KmWNuD
D
kh
s ==
==
Using the Dittus-Boelter correlation, for the internal flow, which isTurbulent,
2
__
4.05/4_
100(./101)(
1905.0
./0261.0
9(023.0PrRe023.0
KmWTTh
KmWNu
D
kh
NU
ms
D
DD
==
==
==
"q
isfluxheattheand
Comments: Convection effects associated with the two flow conditions are
comparable.
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Problem 19:
Cooling water flows through he 25.4 mm diameter thin walled tubes of a
stream condenser at 1m/s, and a surface temperature of 350K is maintained
by the condensing steam. If the water inlet temperature is 290 K and the
tubes are 5 m long, what is the water outlet temperature? Water properties
may be evaluated at an assumed average temperature of 300K
Known: Diameter, length and surface temperature of condenser tubes. Water
velocity and inlet temperature.
Find: Water outlet temperature.
Schematic:
Assumptions: (1) Negligible tube wall conduction resistance, (2) Negligible
kinetic energy, potential energy and flow work changes.
Properties: Water (300K):=997kg/m3, cp=4179J/kg.K, =855*10-
6kg/s.m,k=0.613W/m.K, Pr=5.83
Analysis:
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29,618kg/s.m10855
54m(1m/s)0.02997kg/m
DuRe
]h)cm(DL)exp[T(TTT
6
3
mD
_
p
.
im,ssom,
=
==
=
The flow is turbulent. Since L/D=197, it is reasonable to assume fully
developed flow throughout the tube. Hence
323KK)(4179J/kg.(0.505kg/s
.K)5m(4248W/m(0.0254m)(60K)exp350KT
0.505kg/s4m)m/s)(0.025(1/m(/4)997k/4)(um
With
.K4248W/m4m)/m.K/0.025176(0.613WNuD(k/D)h
2
om,
232
m
.
2_
=
===
===
Comments: The accuracy of the calculations may be improved slightly by
reevaluating properties at 306.5KT_
m = .
Problem 20:
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The air passage for cooling a gas turbine vane can be approximated as a tube
of 3-mm diameter and 75-mm length. If the operating temperature of the
vane is 650C, calculate the outlet temperature of the air if it enters the tube
at 427C and 0.18kg/h.
Known: gas turbine vane approximation as a tube of prescribed diameter and
length maintained at an known surface temperature. Air inlet temperature
and flow rate.
Find: outlet temperature of the air coolant.
Schematic:
Assumptions: (1) Steady-state conditions, (2) negligible Kinetic and
potential energy changes.
Properties: Air (assume )1,780_
atmKTm = : cp=1094J/kg. K, k=0.0563 W/m.
K, =363.7*10-7 N.s/m2, Pr=0.706; Pr=0.706;(Ts=650C=923K, 1atm):
=404.2*10-7N.s/m2.
Analysis: For constant wall temperatures heating,
=
.
_
,
,exp
pims
oms
mc
hPL
TT
TT
Where P=D. for flow in circular passage,
N.s/m103.7(0.03m)36
/3600s/h)0.18kg/h(14
D
m4Re
7
.
D ==
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The flow is laminar, and since L/D=75mm/3mm=25, the Sieder-Tate
correlation including combined entry length fields.
.K87.5W/m10404.2
10363.7
25
0.7065841.86
0.003m
K0.0563W/m.h
L/DPrRe1.86
kDhNu
2
0.14
7
71/3_
0.14
s
1/3
D
__
D
=
=
==
Hence, the air outlet temperature is
=
1094J/kg.K)kg/s(0.18/3600
.K87.5W/m0.075m(0.003m)expC427)(650
T650 2om,
Tm,o=578C
Comments: (1) based upon the calculations for Tm,o=578C,_
mT =775K which
is in good agreement with our assumption to evaluate the thermo physical
properties.
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Problem 21
A household oven door of 0.5-m height and 0.7-m width reaches an average
surface temperature of 32C during operation. Estimate the heat loss to the
room with ambient air at 22C. If the door has an emissivity of 1.0 and thesurroundings are also at 22C, comment on the heat loss by free convection
relative to that by radiation.
Known: Oven door with average surface temperature of 32C in a room
with ambient temperature at 22C.
Find: Heat loss to the room. Also, find effect on heat loss if emissivity of
door is unity and the surroundings are at 22C.
Schematic:
Assumptions: (1) Ambient air in quiescent, (2) surface radiation effects
are negligible.
Properties: Air (Tf=300K, 1atm): =15.89*10-6 m2/s, k=0.0263W/m. K,
=22.5*10-6 m2/s, Pr=0.707, =1/Tf=3.33*10-3K-1
Analysis: the heat rate from the oven door surface by convection to the
ambient air is
)T(TAhq ss
_
=
Where_
h can be estimated from the free convection correlation for a
vertical plate,
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[ ]
2
8/279/16
1/6
L
__`
(0.492/Pr)1
0.387Ra0.825
k
LhLNu
++==
The Rayleigh number,
4
2626
322
s
L 101.142/sm1022.5/sm1015.89
m322)K0.52(1/300k))39.8m/s
)LT(TgRa =
=
=
Substituting numerical values into equation, find
[ ].K3.34W/m63.5
0.5m
K0.0263W/m.LNu
Lh
63.5
(0.492/Pr)1
)1020.387(1.140.825
k
LhLNu
2_`_
2
8/279/16
1/88_
_`
===
=
+
+==
kL
The heat rate using equation is
11.7W22)K(320.7)m.k(0.53.34W/mq 22 ==
Heat loss by radiation, assuming =1 is
21.4W]k22)(27332)[(273.KW/m5.67100.7)m1(0.5q
)T(TAq
4244282
rad
4
sur
4
ssrad
=++=
=
Note that heat loss by radiation is nearly double that by free convection.
Comments: (1) Note the characteristics length in the Rayleigh number is the
height of the vertical plate (door).
Problem 22
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An Aluminum alloy (2024) plate, heated to a uniform temperature of
227C, is allowed to cool while vertically suspended in a room where the
ambient air and surroundings are at 27C. The late is 0.3 m square with a
thickness of 15 mm and an emissivity of 0.25.
a. Develop an expression for the time rate of change of the platetemperature assuming the temperature to be uniform at any
time.
b. Determine the initial rate of cooling of the plate temperature is
227C.
c. Justify the uniform plate temperature assumption.
Known: Aluminum plate alloy (2024) at uniform temperature of 227C
suspended in a room where the ambient air and the surroundings are at 27C
Find: (1) expression for the time rate of change of the plate, (2) Initial
rate of cooling (K/s) when the plate temperature is 227C. (3) justify the
uniform plate temperature assumption.
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Schematic:
Properties: Aluminium alloy 2024 (T=500K):=2270 kg/m3,
k=186W/m.K, c=983 J/kg.K; Air (Tf=400K, 1atm):=26.41*10-
6m2/s,k=0.0338 W/m.K,=38.3*10-6 m2/s, Pr=0.690.
Analysis :( a) from an energy balance on the plate considering free
convection and radiation exchange..
stout EE = .
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)T(T)T(Th[c
2
dt
dTor
dt
dTA)T(T2A)T(T2Ah
4
sur
4
ssL
.
cs
4
sur
4
ssss
_
L +
==
Where Ts is the plate temperature assumed to be uniform at any time.
(b) To evaluate (dt/dx), estimate_
Lh . Find first the Rayleigh number
8
2626
322
s
L101.308
/sm10.33/sm1026.41
0.3m27)K27(1/400k)(29.8m/s
)LT(TgRa =
=
=
8
Substituting numerical values, find
[ ] [ ]5.55
)=
+
+=
++=
4/99/16
82
4/99/16
1/4
L_`
90)(0.492/0.61
1080.670(1.300.68
(0.492/Pr)1
0.670Ra0.68LNu
0.099K/s
.K)(500W/m100.25(5.6727)K.K(2276.25W/m983J/kg.K0.015m2770Kg/m
2
dt
dT
.K6.25W/mK/0.3m0.0338W/m.55.5k/LNuh
282
3
2_`
L
_
=
+
=
===
(c) The uniform temperature assumption is justified if the Biot number
criterion is satisfied. With == )/A.(A)(V/AL sssc and
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____
1.0/, =+= khBihhh totradconvtot . Using the linearized radiation coefficient
relation find,
.3.86W/m3002)K300)(5002)(500.K8W/m100.25(5.67)T)(TT(Th 23422sur2
ssursrad
_
=++=++=
Hence Bi=(6.25+3.86) W/m2.k (0.015m)/186W/m. K=8.15*10-4. Since
Bi
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Problem 23
The ABC Evening News Report in news segment on hypothermia research
studies at the University of Minnesota claimed that heat loss from the body
is 30 times faster in 1 0C water than in air at the same temperature. Isthat a realistic statement?
Known: Person, approximated as a cylinder, experiencing heat loss in
water or air at 10C.
Find: Whether heat loss from body in water is 30 times that in air.
Assumptions: (1) Person can be approximated as a vertical cylinder of
diameter D=0.3 m and length L=1.8m, at 25C, (2) Loss is only from the
lateral surface.
Properties: Air (_
T=(25+10) C/2=290K, 1atm); K=0.0293 W/m. K,
==19.91*10-6m2/s, =28.4*10-6 m2/s; Water (290K); k=0.598 W/m.K;
=vf=1.081*10-6m2/s, =k vf/cp =1.431*10-7m2/s, f=174*10-6K-1.
Analysis: in both water (wa) an air(a), the heat loss from the lateral
surface of the cylinder approximating the body is
)TDL(Thq s_
=
Where Ts and Tare the same for both situations. Hence,
_
a
wa
_
a
wa
h
h
q
q=
Vertical cylinder in air:
9
2626
323
L 105.228/sm28.410/sm10*19.91
10)k(1.8m)5(1/290K)(29.8m/s
gTRa =
==
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.K2.82W/mh173.4)100.1(5.228CRak
LhNu
1/3,nand0.1Cwith
2L
_1/39n
L
__
LL =====
==
Vertical cylinder in water.
11
2726
3162
L 109.643/sm101.431/sm101.081
10)K(1.8m)(25K101749.8m/sRa =
=
.K328W/mh978.9)100.1(9.643CRa
k
hLNu
1/3,nand0.1Cwith
2L
_1/311n
L
__
L =====
==
Hence, from this analysis we find
30ofclaimthewithpoorlycompareswhich117.K2.8W/m
.K328W/m
q
q2
2
a
wa ==
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Problem 24
In a study of heat losses from buildings, free convection heat transfer from
room air at 305 K to the inner surface of a 2.5-m-high wall at 295 K issimulated by performing laboratory experiments using water in a smaller test
cell. In the experiments the water and the inner surface of the test cell are
maintained at 300 and 290K, respectively. To achieve similarity between
conditions in the room and the test cell, what is the required test cell height?
If the average Nusselt number for the wall may be correlated exclusively in
terms of the Rayleigh number, what is the ratio f the average convection
coefficient for the room wall to the average coefficient for the test cell wall?
Known: Air temperature and wall temperature and height for a room. Water
temperature and wall temperature for a simulation experiment.
Find: Required test cell height for similarity. Ratio and height convection
coefficient for the two cases.
Schematic:
Assumptions: (1) Air and water are quiescent; (2) Flow conditions
correspond to free convection boundary layer development on an isothermal
vertical plate, (3) constant properties.
Properties: Air (Tf=300K, 1 atm); K=0.0293 W/m.K, ==15.9*10-6m2/s,
=22.5*10-6 m2/s; =1/Tf=3.33*10-3K-1, k=0.0263W/m.K; water
(Tf=295K):=998kg/m3, =959*10-6N.s/m2, cp=4181 J/kg.K, =227.5*10
-6K-
1, k=0.606 W/m.K; Hence=/ =9.61/s, =k / cp =1.45*10-7m2/s.
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Analysis: Similarity requires that RaL,a=RaL,w where
mmL
oHL
L
hence
w
a
airw
a
w
45.0)179.0(5.2
10*228.0
10*33.3
10*5.22*9.15
10*45.1*61.9
)(
)(
,
)L-T(TgRa
3
3
12
143/1
2
3
sL
==
=
=
=
3
w
a
a
w
w
_
_
a
_
wL,
_
aL,wL,aL,
10*7.810.606
0.0263
2.5
0.45
k
k
L
L
h
h
.henceNuNuthatfollowsit,RaRaif
===
==
Comments: Similitude allows us to obtain valuable information for one
system by performing experiments for a smaller system and a different
fluid.
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Problem 25
A square plate of pure aluminum, 0.5 m on a side and 16 mm hick, is
initially at 300C and is suspended in a large chamber. The walls of thechamber are maintained at 27C, as is the enclosed air. If the surface
emissivity of the plate temperature during the cooling process? Is it
reasonable to assume a uniform plate temperature during the cooling
process?
Known: Initial temperature and dimensions of an aluminum plate.
Conditions of the plate surroundings.
Find: (a) initial cooling rate, (2) validity of assuming negligible temperature
gradients in the plate during the cooling process.
Schematic:
Assumptions: (1) plate temperature is uniform; (2) chamber air is quiescent,
(3) Plate surface is diffuse-gray, (4) Chamber surface is much larger than
that of plate, (5) Negligible heat transfer from edges.
Properties: Aluminum (573k); k=232W/m.k, cp=1022J/kg.K, =2702 kg/m3:
Air (Tf=436K, 1 atm):=30.72*10-6m2/s,=44.7*10-6m2/s,k=0.0363W/m.K,
Pr=0.687, =0.00229K-1.
Analysis: (a) performing an energy balance on the plate,
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psurss
pstsurss
wcTTTThAdtdT
dtdTVcETTTThAq
/)]4()([2/
]/[)]4()([2
4_
.4
_
+=
==+=
8
2626
323s
L 105.58/sm017.44/sm10*30.72
)K(0.5m)721(300-0.00229K9.8m/s)L-T(TgRa =
==
KmWh
Ra
L
kh L
./8.5
])687.0/492.0(1[
)1058.5(670.068.0
5.0
0363.0
]Pr)/492.0(1[
670.068.0
2_
9/416/9
4/18
9/416/9
4/1_
=
+
+=
++=
Hence the initial cooling rate is
{ }
sKdt
dT
kkgJmmkg
KKKmWCKmW
dt
dT
/136.0
./1022)016.0(/2702
])300()573[(./1067.525.0)27300(./8.523
444282
=
+=
(b) To check the Validity of neglecting temperature gradients across the
plate thickness, calculate
"
(0.008m)(11W/m2.K)Bi
/)2/(
=
==
hence
qhwherekwhBi toeffeff
And the assumption is excellent.
Comments: (1) Longitudinal (x) temperature gradients are likely to me more
severe than those associated with the plate thickness due to the variation of h
with x. (2) Initially qconvqrad.
Problem 26
Beer in cans 150 mm long and 60 mm in diameter is initially at 27 C and is
to be cooled by placement in a refrigerator compartment at 4C. In the
interest of maximizing the cooling rate, should the cans be laid horizontally
or vertically in the compartment? As a first approximation, neglect heat
transfer from the ends.
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Known: Dimensions and temperature of beer can in refrigerator
compartment.
Find: orientation which maximum cooling rate.
Schematic:
Assumptions: (1) End effects are negligible, (2) Compartment air is
quiescent, (3) constant properties.
Properties: Air (Tf=288.5K, 1 atm): =14.87*10-6 m2/s, k=0.0254W/m.K,
=21.0*10-6m2/s, Pr=0.71, =1/Tf=3.47*10-3K-1.
Analysis: The ratio of cooling rates may be expressed as
h
v
s
s
h
v
qh
v
h
h
TT
TT
DL
DL
h
hq_
_
_
_
)(
)(=
==
For the vertical surface, find
639
L
393
26-26-
-1-323
s
L
1044.8)15.0(105.2Ra
L105.2L/s)m10/s)(21m10(14.87
C)(23K103.479.8m/s)L-T(TgRa
==
=
==
Using the correct correlation
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KmWm
KmW
L
kNuhh
NuL
LvL ./03.515.0
./0254.07.29hence,
7.29])71.0/492.0(1[
)1044.8(387.0825.0
2___
2
27/816/9
6/16
====
=
+
+=
Using the correct correlation,
97.018.5
03.5,
./18.506.0
./0254.024.12
24.12])71.0/559.0(1[
)104.5(387.060.0
2___
2
27/861/9
6/15_
==
====
=
+
+=
h
v
DhD
D
q
qhence
KmWm
KmW
D
kNuhh
Nu
Comments: in view of the uncertainties associated with equations and the
neglect of end effects, the above result is inconclusive. The cooling rates are
are approximately the same.