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Example 1 Defrosting ice from an automobile windshield To defrost ice accumulated on the outer surface of an automobile windshield, warm air is blown over the inner surface of the windshield. Consider an automobile windshield (kw = 1.4 W/m.K). With an overall height of 0.5m and thickness of 5mm. The outside air (1 atm) ambient temperature is -20°C and the average air flow velocity over the outer windshield surface is 80 km/h, while the ambient temperature inside the automobile is 25°C. Determine the value of the convection heat transfer coefficient, for the warm air blowing over the inner surface of the windshield, necessary to cause the accumulated ice to begin melting. Assume the windshield surface can be treated as aa flat plate surface. Assumptions: the outside air pressure is 1 atm and the critical Reynolds number is Recr = 5 x105
.KW/m75.510.5m
.K0.02288W/m1131
L
k1131h
11310.738787110875.8037.0Pr871Re037.0k
LhNu
is windshield theof surface outer the on tcoefficien
transfer heataverage the number,Nusselt the for relation proper theUsing
flow. turbulent and laminar combined a is flow the 10Re105 Since
10875.8s/m10252.1
0.5m80/3.60m/sVL Re
table) (from 0.7387 Pr and /sm101.252v .K,0.02288W/m k are
C102
C0C20Tof etemperatur film the at airof properties The
2
0
3
18.053
18.0
L0
o
7
L
5
5
25L
25-
f
e.temperatur and rate
flow air warm theadjusting by done is This well. as varied be should ice the melt to
necessaryair warm theof tcoefficien transfer heatconvection the Therefore, speed.
automobile the and conditions weather with vary automobile the outside tcoefficien
transfer heatconvective the and etemperatur ambient the situation practical In
.KW/m6.481.4W/m.K
m005.0
C020
C250
.KW/m75.51
1
k
t
TT
TT
h
1h
be; must shield wind theof surface inner the overblowing air warm the
for tcoefficien transfer heatconvection the Then .C0 least at be should )(T
windshield theof etemperatur surface outer the melting, begin to ice the For
h1kt
TT
h1
TT
as; written be can
thickness windshield the through transfer heatthe balance energy the From
2
1
2
1
wo,so,
i,o,s
i
os,
iw
i,o,s
o
o,so,
Example 2
Cooling of Plastic Sheets by Forced Air The forming section of a plastics plant puts out a continuous sheet of plastic that is 1.2 m wide and 0.1 cm thick at a velocity of 9 m/min. The temperature of the plastic sheet is 95°C when it is exposed to the surrounding air, and a 0.6-m-long section of the plastic sheet is subjected to air flow at 25°C at a velocity of 3 m/s on both sides along its surfaces normal to the direction of motion of the sheet. Determine (a) the rate of heat transfer from the plastic sheet to air
by forced convection and radiation (b) the temperature of the plastic sheet at the end of the
cooling section. Take the density, specific heat, and emissivity of the plastic sheet to be ρ = 1200 kg/m3, Cp = 1.7 kJ/kg · °C, and ε = 0.9.
is sheet plastic the
across flow air theof end the at numberReynolds the m, 1.2 L thatNoting sheet. plastic
theof drop etemperatur the for account to ryif necessa nscalculatio the repeat will We
started. get to C95 at isothermal be to sheet plastic the assume we Therefore, drop.
thatof magnitude the know notdo we point this at but section,cooling long -m-0.6 the
through flows it as somewhat drop to sheet plastic theof etemperatur the expect We
/sm 101.896
0.7202 Pr
C · W/m 0.02808 k
table) in (given are pressure atm 1 and
C60 25)/2 (95 )/2T (T Tof etemperatur film the at airof properties The
air. room theof etemperatur the at are surfacesg surroundin the and atm 1
is pressure catmospheri local The .10 5 Re is numberReynolds Critical
25-
sf
5
cr
1380WW687612WQQQ
is radiation
and convection combined by sheet plastic theof cooling of rate the Therefore,
W687K298368m44.1K.W/m10.6759.0TTAQ
612WC2595m44.1C.W/m07.6TTAhQ
m44.1sides2m2.1m6.0A
C.W/m07.63.2591.2m
C.0.02808W/mNu
L
kh
Then,
3.2590.720210899.1664.0PrRe664.0k
hLNu
be to plate flat a for relations flow
laminar the from determined is numberNusselt the and sheet, entire the over
flow laminar havewe Thus, number.Reynolds critical the than less is which
10899.1s/m10896.1
1.2mm/s 3VL Re
radconvtotal
442428-4
surr
4
ssrad
22
ssconv
2
s
2
3
15.053
15.0
L
5
25L
Example 3 Heat Loss from a Steam Pipe in Windy Air A long 10-cm-diameter steam pipe whose external surface temperature is 110°C passes through some open area that is not protected against the winds. Determine the rate of heat loss from the pipe per unit of its length when the air is at 1 atm pressure and 10°C and the wind is blowing across the pipe at a velocity of 8 m/s.
4
25L
25-
f
10219.4s/m10896.1
0.1m8m/sVL Re
0.7202 Pr
/sm101.896v
C.0.02808W/m k
table) (from are pressure atm 1 and
C602
C10C110Tof etemperatur film the at airof properties The
m. in pipe theof length the by above value
theg multiplyin by obtained be can pipe entire the from loss of heat rate The
W0931C10110.314m0C.4.8W/m3TTAhQ
m314.0m1m1.0DLpLA
becomes
length itsof unitper pipe the from transfer of heat rate the Then
C.W/m8.341240.1m
C.0.02808W/mNu
D
kh
Then,
124282000
10219.41
7202.04.01
7202.010219.462.03.0
282000
Re1
Pr4.01
PrRe62.03.0
k
hDNu
from determined is numberNusselt The
22
ssconv
2
s
2
5
4
8
54
4
1
3
2
3
12
14
5
4
8
5
4
1
3
2
3
1
2
1
Example 4 Cooling of a Steel Ball by Forced Air A 25-cm-diameter stainless steel ball ( 8055 kg/m3, Cp 480 J/kg·°C) is removed from the oven at a uniform temperature of 300°C. The ball is then subjected to the flow of air at 1 atm pressure and 25°C with a velocity of 3 m/s. The surface temperature of the ball eventually drops to 200°C. Determine the average convection heat transfer coefficient during this cooling process and estimate how long the process will take.
Assumptions • Radiation effects are negligible. • The outer surface temperature of the ball is uniform at all times. • The surface temperature of the ball during cooling is changing. Therefore, the
convection heat transfer coefficient between the ball and the air will also change. To avoid this complexity, we take the surface temperature of the ball to be constant at the average temperature of (300 + 200)/2 = 250°C in the evaluation of the heat transfer coefficient and use the value obtained for the entire cooling process.
1351076.2
10849.10.729610802.406.010802.44.02
PrRe06.0Re4.02k
hDNu
is numberNusselt The
10802.4s/m10562.1
0.25m3m/sVL Re
0.7296 Pr
/sm101.562
skg/m101.849v
C.0.02551W/m k
are atm 1 and C25of etemperatur
stream-free the at airof properties The kg/m·s. 10 2.76
is etemperatur surface average the at airof viscosity dynamic The
41
5
54.0324214
41
s
4.03221
4
25L
25-
5-
5
C250 @s
J3,163,000 C200) - C)(300 J/kg·kg)(480 (65.9 )T -(TmC Q
kg 65.9 m) (0.25 )kg/m (8055 D6
1 V m
:C200 to C300 from cools it as ball theof energy the in change
the simply is which ball, the from dtransferre heattotal the determine we Next
W 610 C25) - )(250m C)(0.1963 · W/m (13.8 )T -(T hAQ
m 0.1963 m) (0.25 D A
is, That e.temperatur surface average the
usingbycooling of law sNewton’ from transfer of heat rate average the
determine we C,200 to C300 from ball theof cooling of time the estimate to order In
C.W/m8.131350.25m
C.0.02551W/mNu
D
kh
becomes tcoefficien transfer heatconvection average the Then
12ptotal
333
22
ave s,save
22
2s
2
min 26 h1 s 5185 J/s610
J3,163,000
Q
Qt
ave
The time of cooling could also be determined more accurately using the transient temperature charts or relations introduced previously. But the simplifying assumptions we made above can be justified if all we need is a ballpark value. It will be naive to expect the time of cooling to be exactly 1 h 26 min, but, using our engineering judgment, it is realistic to expect the time of cooling to be somewhere between one and two hours.
In this calculation, we assumed that the entire ball is at 200°C, which is not necessarily true. The inner region of the ball will probably be at a higher temperature than its surface. With this assumption, the time of cooling is determined to be
Example 5 Heating of water in a tube by steam Water enters a 2.5 cm internal diameter thin cooper tube of a heat exchanger at 15 °C at a rate of 0.3kg/s and is heated by steam condensing outside at 120 °C. If the average heat transfer coefficient is 800W/m2. °C, determine the length of the tube required in order to heat the water to 115 °C.
The conduction resistance of a cooper tube is negligible so that the inner surface temperature of the tube is equal to the condensation temperature of steam. The specific heat of water at the bulk mean temperature of (15+115)/2=65 °C is 4187J/kg. °C. The heat of condensation of steam at 120 °C is 2203kJ/kg
m61
m025.0
4.78m
D
ALDLA
becomes length tube required Then
m78.4C85.32C0.8kW/m
kW 125.6
Th
QAT hAQ
is area surface transfer heatThe
C85.321055ln
1055
TTln
TTT
C510C15-C120TTT
C5C115-C120TTT
etemperatur mean clogarithmi The
kW 125.6 C15) - C)(115kJ/kg· )(4.187skg/ (0.3 )T -(Tcm Q
2
ss
2
2
ln
aveslnsave
ie
ieln
isi
ese
iep
The bulk mean temperature of water during this heating process is 65°C and thus the arithmetic mean temperature difference is ΔTam=120-65=55°C. Using ΔTam instead of ΔTln would give L=36m which is grossly in error. This shows the importance of using the logarithmic mean temperature in calculation.
Example 6 Flow of Oil in a Pipeline through a Lake Consider the flow of oil at 20°C in a 30-cm-diameter pipeline at an average velocity of 2 m/s. A 200-m-long section of the pipeline passes through icy waters of a lake at 0°C. Measurements indicate that the surface temperature of the pipe is very nearly 0°C. Disregarding the thermal resistance of the pipe material, determine a) the temperature of the oil when the pipe b) leaves the lake c) the rate of heat transfer from the oil d) the pumping power required to overcome the
pressure losses and to maintain the flow of the oil in the pipe
Assumptions The thermal resistance of the pipe is negligible. The inner surfaces of the pipeline are smooth. The flow is hydrodynamically developed when the pipeline reaches the lake.
We do not know the exit temperature of the oil, and thus we cannot determine the bulk mean temperature, which is the temperature at which the properties of oil are to be evaluated. The mean temperature of the oil at the inlet is 20°C, and we expect this temperature to drop somewhat as a result of heat loss to the icy waters of the lake. We evaluate the properties of the oil at the inlet temperature, but we will repeat the calculations, if necessary, using properties at the evaluated bulk mean temperature. At 20°C we read
666
/sm10019
0.3m2m/sDV Re
C880J/kg1 c
10400 Pr
/sm10019
C0.145W/m. k
888kg/m
26-
hmL
p
26-
3
Note that this Nusselt number is considerably higher than the fully developed value of 3.66. Then,
C71.19C) · J/kg kg/s)(1880 (125.5
)m C)(188.5 · W/m (18.0exp C]20) - [(0 - C0
cmhA-)expT -(TTT
from oilof etemperatur exit the determine we Next
kg/s 125.5 s/m2m) (0.34
1 )kg/m (888 VA m
m 188.5 m) (200 m3.0DL Lp A
,Also
C.W/m0.183.370.3m
C0.145W/m.Nu
D
kh
22
psisse
23
mc
2
s
2
Thus, the mean temperature of oil drops by a mere 0.29°C as it crosses the lake. This makes the bulk mean oil temperature 19.86°C, which is practically identical to the inlet temperature of 20°C. Therefore, we do not need to reevaluate the properties.
422
lnsave
is
es
eiln
1074.6C85.19188.5mC18.0kW/m T hAQ
C85.19
020
71.910ln
71.9102
TT
TTln
TTT
The logarithmic mean temperature difference and the rate of heat loss from the oil are;
Therefore, the oil will lose heat at a rate of 67.4 kW as it flows through the pipe in the icy waters of the lake. Note that ΔTln is identical to the arithmetic mean temperature in this case, since Δ Ti ≈Δ Te. The laminar flow of oil is hydrodynamically developed. Therefore, the friction factor can be determined from
0961.0666
64
Re
64f
kW 16.1kg/m 888
)N/m 10 kg/s)(1.14 (125.5PmW
m/N1014.12
m/s) )(2kg/m (888
m 0.3
m 2000.0961
2
V
D
LfP
3
25
pump
25
232
m
Then the pressure drop in the pipe and the required pumping power become
We will need a 16.1-kW pump just to overcome the friction in the pipe as the oil flows in the 200-m-long pipe through the lake. If the pipe is long, a more optimal design to reduce pump work would be to increase the pipe diameter D, since pump work is inversely proportional to D5. Optimal pipe sizing requires an economic life-cycle cost analysis
Example 7 Heating of Water by Resistance Heaters in a Tube Water is to be heated from 15°C to 65°C as it flows through a 3-cm-internaldiameter 5-m-long tube. The tube is equipped with an electric resistance heater that provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of 10 L/min, determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit.
The properties of water at the bulk mean temperature of Tb=(Ti + Te)/2 = (15+ 65)/2 = 40°C.
kg/s 0.1654 kg/min 9.921 /min)m )(0.01kg/m (992.1 Vm
becomes rate flow mass the
Then /min.m 0.01 L/min 10 V as given is waterof rate flow volume The
m 0.471 m) m)(5 (0.03 DL pL A
m 10 7.069m03.04
1D
4
1A
are areas surface transfer heatand sectional cross The
C 4179J/kgc
4.32 Pr
/sm 10 0.658
C0.631W/m. k
992.1kg/m
33
3
2
s
2422
c
p
26
3
2
2
c
s
s
msmss
iep
kW/m 73.46m 0.471
kW 34.6
A
from determined be can value
its and case, this in constant is flux heatsurface The location. that at fluid
theof etemperatur mean the is Tm and tcoefficien transfer heatthe is hwhere
h
qT T )T - h(Tq
from
determined be can location any at tube theof Ts etemperatur surface The
kW. 34.6 be must heatertheof rating
power the Therefore, heater.resistance the from come must energy thisof All
kW 34.6 kJ/s 34.6 C15) - C)(65kJ/kg 9kg/s)(4.17 (0.1654 )T - (TCmQ
of rate a at water the to
supplied be must heatC,65 to C15 from rate flow mass this at water the heatTo
69.5 (4.34) 60)0.023(10,7 Pr Re 0.023k
hD Nu
from number
Nusselt the determine and pipe entire the in flow turbulent developed fully
assume can we Therefore, pipe. theof length total the than shorter much is which
m 0.3 0.03 10 10D L L
roughly is length
entry the and turbulent is flow the Therefore, 10,000. than greater is which
10,760/sm 10 0.658
m) m/s)(0.03 (0.236DVRe
m/s 0.236 m/min 14.15 m 10 7.069
/minm 0.010
A
VV
: numberReynolds the and waterof
velocity mean the find to needfirst we t,coefficien transfer heatthe determine To
0.40.80.40.8
th
26-
m
24-
3
c
m
C115C · W/m 1462
W/m 73,460 C65
h
qT T
becomes exit the at pipe theof etemperatur surface the and
C W/m 1462 (69.5)m 0.03
C · W/m 0.631Nu
D
kh
Then;
2
2
s
ms
2
Discussion Note that the inner surface temperature of the pipe will be 50°C higher than the mean water temperature at the pipe exit. This temperature difference of 50°C between the water and the surface will remain constant throughout the fully developed flow region.
Example 8 Heat Loss from the Ducts of a Heating System Hot air at atmospheric pressure and 80°C enters an 8–m-long uninsulated square duct of cross section 0.2 m 0.2 m that passes through the attic of a house at a rate of 0.15 m3/s. The duct is observed to be nearly isothermal at 60°C. Determine the exit temperature of the air and the rate of heat loss from the duct to the attic space.
We do not know the exit temperature of the air in the duct, and thus we cannot determine the bulk mean temperature of air, which is the temperature at which the properties are to be determined. The temperature of air at the inlet is 80°C and we expect this temperature to drop somewhat as a result of heat loss through the duct whose surface is at 60°C. At 80°C and 1 atm we read;
C J/kg1008 c
0.7154 Pr /sm 10 2.097
C0.02953W/m k kg/m 0.99994
p
25-
3
The characteristic length (which is the hydraulic diameter), the mean velocity, and the Reynolds number in this case are
91.4 (0.7154) 65)0.023(35,7 Pr Re 0.023k
hD Nu
from numberNusselt
the determine and duct entire the in flow turbulent developed fully assume
can we Therefore, duct. theof length total the than shorter much is which
m 2 m 0.2 10 10D L L
roughly is length
entry the and turbulent is flow the Therefore, 10,000. than greater is which
35,765/sm 10 2.097
m) m/s)(0.2 (3.75DVRe
m/s 3.75 m) (0.2
/minm 0.15
A
VV
m 0.2 a4a
4a
p
4AD
0.30.80.30.8h
th
25-
m
2
3
c
m
2
ch
Then,
C3.71C) · J/kg kg/s)(1008 (0.151
)m C)(6.4 · W/m (13.5exp C]80) - [(60 - C06
cmhA-)expT -(TTT
from oilof etemperatur exit the determine we Next
kg/s 0.151 /s)m )(0.15kg/m (1.009 V m
m 6.4 m) m)(8 (0.2 4 4aLLp A
,Also
C.W/m5.1391.40.2m
C.0.02953W/mNu
D
kh
22
psisse
33
2
s
2
h
W 1313 - C))(-15.2m C)(6.4W/m (13.5T hAQ
C2.15
80 - 60
71.3 - 60ln
3.1708
T - T
T - Tln
TTT
become air the from
loss of heat rate the and difference etemperatur mean clogarithmi The
22
lns
e
is
es
eiln
Therefore, air will lose heat at a rate of 1313 W as it flows through the duct in the attic. Discussion The average fluid temperature is (80 + 71.3)/2 = 75.7°C, which is sufficiently close to 80°C at which we evaluated the properties of air. Therefore, it is not necessary to re-evaluate the properties at this temperature and to repeat the calculations.
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