Chapter 9Natural Convection Chapter 9 NATURAL CONVECTION Physical Mechanisms of Natural Convection 9-1C Natural convection is the mode of heat transfer that occurs between a solid and a fluid which moves under the influence of natural means. Natural convection differs from forced convection in that fluid motion in natural convection is caused by natural effects such as buoyancy. 9-2C The convection heat transfer coefficient is usually higher in forced convection because of the higherfluid velocities involved. 9-3C The hot boiled egg in a spacecraft will cool faster when the spacecraft is on the ground since there is no gravity in space, and thus there will be no natural convection currents which is due to the buoyancy force. 9-4C The upward force exerted by a fluid on a body completely or partially immersed in it is called the buoyancy or “lifting” force. The buoyancy force is proportional to the density of the medium. Therefore, the buoyancy force is the largest in mercury, followed by in water, air, and the evacuated chamber. Note that in an evacuated chamber there will be no buoyancy force because of absence of any fluid in the medium. 9-5C The buoyancy force is proportional to the density of the medium, and thus is larger in sea water than it is in fresh water. Therefore, the hull of a ship will sink deeper in fresh water because of the smallerbuoyancy force acting upwards. 9-6C A spring scale measures the “weight” force acting on it, and the person will weigh less in waterbecause of the upward buoyancy force acting on the person’s body. 9-7C The greater the volume expans ion coefficient, the greater the cha nge in density with temperature, the greater the buoyancy force, and thus the greater the natural convection currents. 9-8C There cannot be any natural convection heat transfer in a medium that experiences no change in volume with temperature. 9-9C The lines on an interferometer photograph represent isotherms (constant temperature lines) for a gas, which correspond to the lines of constant density. Closely packed lines on a photograph represent a large temperature gradient. 9-10C The Grashof number represents the ratio of the buoyancy force to the viscous force acting on a fluid. The inertial forces in Reynolds number i s replaced by the buoyancy forces in Grashof number. 9-1
9-1C Natural convection is the mode of heat transfer that occurs between a solid and a fluid which movesunder the influence of natural means. Natural convection differs from forced convection in that fluidmotion in natural convection is caused by natural effects such as buoyancy.
9-2C The convection heat transfer coefficient is usually higher in forced convection because of the higher fluid velocities involved.
9-3C The hot boiled egg in a spacecraft will cool faster when the spacecraft is on the ground since there isno gravity in space, and thus there will be no natural convection currents which is due to the buoyancyforce.
9-4C The upward force exerted by a fluid on a body completely or partially immersed in it is called the buoyancy or “lifting” force. The buoyancy force is proportional to the density of the medium. Therefore,the buoyancy force is the largest in mercury, followed by in water, air, and the evacuated chamber. Notethat in an evacuated chamber there will be no buoyancy force because of absence of any fluid in themedium.
9-5C The buoyancy force is proportional to the density of the medium, and thus is larger in sea water thanit is in fresh water. Therefore, the hull of a ship will sink deeper in fresh water because of the smaller buoyancy force acting upwards.
9-6C A spring scale measures the “weight” force acting on it, and the person will weigh less in water because of the upward buoyancy force acting on the person’s body.
9-7C The greater the volume expansion coefficient, the greater the change in density with temperature, thegreater the buoyancy force, and thus the greater the natural convection currents.
9-8C There cannot be any natural convection heat transfer in a medium that experiences no change involume with temperature.
9-9C The lines on an interferometer photograph represent isotherms (constant temperature lines) for a gas,which correspond to the lines of constant density. Closely packed lines on a photograph represent a largetemperature gradient.
9-10C The Grashof number represents the ratio of the buoyancy force to the viscous force acting on a fluid.The inertial forces in Reynolds number is replaced by the buoyancy forces in Grashof number.
9-16 A horizontal hot water pipe passes through a large room. The rate of heat loss from the pipe by naturalconvection and radiation is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 The temperature of the outer surface of the pipe is constant.
Properties The properties of air at 1 atm and the film temperature
of (T s+T ∞)/2 = (65+22)/2 = 43.5°C are (Table A-15)
1-
25
K 00316.0K )2735.43(
11
7245.0Pr
/sm10735.1
CW/m.02688.0
=+
==
=
×=
°=−
f T
k
β
υ
Analysis (a) The characteristic length in this case is the outer diameter of the pipe, m.06.0== D Lc
Then,
805,692)7245.0()/sm10735.1(
)m06.0)(K 2265)(K 00316.0)(m/s81.9(Pr
)(225
3-12
2
3
=×
−=
−=
−∞
υ
β DT T g Ra s
( )[ ] ( )[ ]15.13
7245.0/559.01
)805,692(387.06.0
Pr /559.01
387.06.0
2
27/816/9
6/12
27/816/9
6/1
=
++=
++=
Ra Nu
2
2
m885.1)m10)(m06.0(
C.W/m893.5)15.13(m06.0
CW/m.02688.0
===
°=°
==
π π DL A
Nu D
k h
s
W477.6=°−°=−= ∞ C)2265)(m885.1)(C.W/m893.5()( 22T T hAQ s s
(b) The radiation heat loss from the pipe is
[ +−+×=−=
− 44428244
)K 27322()K 27365().K W/m1067.5)(m885.1)(8.0()( surr s srad T T AQ σ ε
9-17 A power transistor mounted on the wall dissipates 0.18 W. The surface temperature of the transistor isto be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 Any heattransfer from the base surface is disregarded. 4 The local atmospheric pressure is 1 atm. 5 Air properties are
evaluated at 100°C.
Properties The properties of air at 1 atm and the given film
temperature of 100°C are (Table A-15)
1-
25
K 00268.0K )273100(
11
7111.0Pr
/sm10306.2
CW/m.03095.0
=+
==β
=
×=υ
°=
−
f T
k
Analysis The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by “guessing” the surface temperature to be 165°C for the evaluation of h. This is
the surface temperature that will give a film temperature of 100°C. We will check the accuracy of this
guess later and repeat the calculations if necessary.
The transistor loses heat through its cylindrical surface as well as its top surface. For convenience,we take the heat transfer coefficient at the top surface of the transistor to be the same as that of its sidesurface. (The alternative is to treat the top surface as a vertical plate, but this will double the amount of calculations without providing much improvement in accuracy since the area of the top surface is muchsmaller and it is circular in shape instead of being rectangular). The characteristic length in this case is the
outer diameter of the transistor, m.004.0== D Lc Then,
6.292)7111.0()/sm10306.2(
)m004.0)(K 35165)(K 00268.0)(m/s81.9(Pr
)(225
3-12
2
3
=×
−=
−=
−∞
υ
β DT T g Ra s
( )[ ] ( )[ ]039.2
7111.0/559.01)6.292(387.06.0
Pr /559.01387.06.0
2
27/816/9
6/12
27/816/9
6/1
=
++=
++= Ra Nu
222
2
m0000691.04/m)004.0()m0045.0)(m004.0(4/
C.W/m78.15)039.2(m004.0
CW/m.03095.0
=+=+=
°=°
==
π π π π D DL A
Nu D
k h
s
and
[ ]C187
)K 27325()273()1067.5)(m0000691.0)(1.0(
C)35)(m0000691.0)(C.W/m8.15(W18.0
)()(
4482
22
44
°= →
+−+×+
°−°=
−+−=
−
∞
s
s
s
surr s s s s
T T
T
T T AT T hAQ σ ε
which is relatively close to the assumed value of 165°C. To improve the accuracy of the result, we repeat
the Rayleigh number calculation at new surface temperature of 187°C and determine the surface
temperature to be
T s = 183°C Discussion W evaluated the air properties again at 100°C when repeating the calculation at the new surface
temperature. It can be shown that the effect of this on the calculated surface temperature is less than 1°C.
delta_a=LRa_a=(g*beta*(T_s-T_infinity)*delta_a^3)/nu^2*PrNusselt_a=0.59*Ra_a^0.25h_a=k/delta_a*Nusselt_aA=L^2Q_dot_a=h_a*A*(T_s-T_infinity)"(b), plate is horizontal with hot surface facing up"delta_b=A/pp=4*LRa_b=(g*beta*(T_s-T_infinity)*delta_b^3)/nu^2*PrNusselt_b=0.54*Ra_b^0.25h_b=k/delta_b*Nusselt_bQ_dot_b=h_b*A*(T_s-T_infinity)
"(c), plate is horizontal with hot surface facing down"delta_c=delta_bRa_c=Ra_bNusselt_c=0.27*Ra_c^0.25h_c=k/delta_c*Nusselt_cQ_dot_c=h_c*A*(T_s-T_infinity)
9-21 A cylindrical resistance heater is placed horizontally in a fluid. The outer surface temperature of theresistance wire is to be determined for two different fluids.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 4 Any heat transfer by radiation is ignored. 5 Properties are evaluated at
500°C for air and 40°C for water.
Properties The properties of air at 1 atm and 500°C are (Table A-15)
1-
25
K 001294.0K )273500(
11
6986.0Pr
/sm10804.7
CW/m.05572.0
=+
==β
=
×=υ
°=−
f T
k
The properties of water at 40°C are
1-
26
K 000377.0
32.4Pr
/sm106582.0/
CW/m.631.0
=β
=
×=ρµ=υ
°=
−
k
Analysis (a) The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by “guessing” the surface temperature to be 1200 °C for the calculation of h. We
will check the accuracy of this guess later and repeat the calculations if necessary. The characteristic length
in this case is the outer diameter of the wire, m.005.0== D Lc Then,
)6986.0()/sm10804.7(
)m005.0(C)201200)(K 001294.0)(m/s81.9(Pr
)(
225
3-12
2
3
=×
°−=
−=
−∞
υ
β DT T g Ra s
( )[ ] ( )[ ] 919.16986.0/559.01
)7.214(387.06.0
Pr /559.01
387.06.0
2
27/816/9
6/12
27/816/9
6/1
=
++=
++=
Ra Nu
2
2
m01571.0)m1)(m005.0(
C.W/m38.21)919.1(m005.0
CW/m.05572.0
===
°=°
==
π π DL A
Nu D
k h
s
and
C1211°=
°−°=
−= ∞
s
s
s s
T
T
T T hAQ
C)20)(m01571.0)(C.W/m38.21(W400
)(
22
which is sufficiently close to the assumed value of 1200°C used in the evaluation of
h, and thus it is notnecessary to repeat calculations.
(b) For the case of water, we “guess” the surface temperature to be 40°C. The characteristic length in this
case is the outer diameter of the wire, m.005.0== D Lc Then,
9-22 Water is boiling in a pan that is placed on top of a stove. The rate of heat loss from the cylindrical sidesurface of the pan by natural convection and radiation and the ratio of heat lost from the side surfaces of the pan to that by the evaporation of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gaswith constant properties. 3 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of
(T s+T ∞)/2 = (98+25)/2 = 61.5°C are (Table A-15)
1-
25
K 00299.0K )2735.61(
11
7198.0Pr
/sm10910.1
CW/m.02819.0
=+
==
=
×=
°=
−
f T
k
β
υ
Analysis (a) The characteristic length in this case is the height of the pan, m.12.0== L Lc Then,
6
225
3-12
2
3
10299.7)7198.0()/sm10910.1(
)m12.0)(K 2598)(K 00299.0)(m/s81.9(Pr
)(×=
×
−=
−=
−
∞
υ
β LT T g Ra s
We can treat this vertical cylinder as a vertical plate since
4/14/164/1
35 and thus 0.25<07443.0
)7198.0/10299.7(
)12.0(3535
Gr
L D
Gr
L≥=
×
=
Therefore,
60.28
7198.0
492.01
)10299.7(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/16
2
27/816/9
6/1
=
+
×+=
+
+= Nu
2
2
m09425.0)m12.0)(m25.0(
C.W/m720.6)60.28(m12.0
CW/m.02819.0
===
°=°
==
π π DL A
Nu L
k h
s
and
W46.2=°−°=−= ∞ C)2598)(m09425.0)(C.W/m720.6()( 22T T hAQ s s
(b) The radiation heat loss from the pan is
[ ] W56.1=+−+×=
−=− 444282
44
)K 27325()K 27398().K W/m1067.5)(m09425.0)(95.0(
)( surr s srad T T AQ σ ε
(c) The heat loss by the evaporation of water is
W1254kW254.1)kJ/kg2257)(kg/s3600/2( ==== fg hmQ
Then the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water then becomes
9-23 Water is boiling in a pan that is placed on top of a stove. The rate of heat loss from the cylindrical sidesurface of the pan by natural convection and radiation and the ratio of heat lost from the side surfaces of the pan to that by the evaporation of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gaswith constant properties. 3 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of
(T s+T ∞)/2 = (98+25)/2 = 61.5°C are (Table A-15)
1-
25
K 00299.0K )2735.61(
11
7198.0Pr
/sm10910.1
CW/m.02819.0
=+
==
=
×=
°=
−
f T
k
β
υ
Analysis (a) The characteristic length in this case is the height of the pan, m.12.0== L Lc Then,
6
225
3-12
2
3
10299.7)7198.0()/sm10910.1(
)m12.0)(K 2598)(K 00299.0)(m/s81.9(Pr
)(×=
×
−=
−=
−
∞
υ
β LT T g Ra s
We can treat this vertical cylinder as a vertical plate since
4/14/164/1
35 and thus 0.25<07443.0
)7198.0/10299.7(
)12.0(3535
Gr
L D
Gr
L≥=
×
=
Therefore,
60.28
7198.0
492.01
)10299.7(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/16
2
27/816/9
6/1
=
+
×+=
+
+= Nu
2
2
m09425.0)m12.0)(m25.0(
C.W/m720.6)60.28(m12.0
CW/m.02819.0
===
°=°
==
π π DL A
Nu L
k h
s
and
W46.2=°−°=−= ∞ C)2598)(m09425.0)(C.W/m720.6()( 22T T hAQ s s
(b) The radiation heat loss from the pan is
[ ] W5.9=+−+×=
−=− 444282
44
)K 27325()K 27398().K W/m1067.5)(m09425.0)(10.0(
)( surr s srad T T AQ σ ε
(c) The heat loss by the evaporation of water is
W1254kW254.1)kJ/kg2257)(kg/s3600/2( ==== fg hmQ
Then the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water then becomes
9-24 Some cans move slowly in a hot water container made of sheet metal. The rate of heat loss from thefour side surfaces of the container and the annual cost of those heat losses are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 3 Heat loss from the top surface is disregarded.
Properties The properties of air at 1 atm and the film temperature of
(T s+T ∞)/2 = (55+20)/2 = 37.5°C are (Table A-15)
1-
25
K 003221.0K )2735.37(
11
7261.0Pr
/sm10678.1
CW/m.02644.0
=+
==
=
×=
°=−
f T
k
β
υ
Analysis The characteristic length in this case is the height of the bath,
m.5.0== L Lc Then,
8
225
3-12
2
3
10565.3)7261.0()/sm10678.1(
)m5.0)(K 2055)(K 003221.0)(m/s81.9(Pr
)(×=
×
−=
−=
−∞
υ
β LT T g Ra s
84.89
7261.0
492.01
)10565.3(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/18
2
27/816/9
6/1
=
+
×+=
+
+= Nu
[ ] 2
2
m5.4)m5.3)(m5.0()m1)(m5.0(2
C.W/m75.4)84.89(m5.0
CW/m.02644.0
=+=
°=°
==
s A
Nu L
k h
and
W1.748C)2055)(m5.4)(C.W/m75.4()( 22 =°−°=−= ∞T T hAQ s s
9-25 Some cans move slowly in a hot water container made of sheet metal. It is proposed to insulate theside and bottom surfaces of the container for $350. The simple payback period of the insulation to pay for itself from the energy it saves is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm. 3 Heat loss from the top surface is disregarded.
Properties Insulation will drop the outer surface temperature to a value close to the ambient temperature.The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh
number and thus the Nusselt number depends on the surface temperature, which is unknown. We assume
the surface temperature to be 26°C. The properties of air at the anticipated film temperature of
(26+20)/2=23°C are (Table A-15)
1-
25
K 00338.0K )27323(
11
7301.0Pr
/sm10543.1
CW/m.02536.0
=+
==
=
×=
°=
−
f T
k
β
υ
Analysis We start the solution process by “guessing” the outer surface temperature to be 26 °C . We willcheck the accuracy of this guess later and repeat the calculations if necessary with a better guess based on
the results obtained. The characteristic length in this case is the height of the tank, m.5.0== L Lc
Then,
7
225
3-12
2
3
10622.7)7301.0()/sm10543.1(
)m5.0)(K 2026)(K 00338.0)(m/s81.9(Pr
)(×=
×
−=
−=
−∞
υ
β LT T g Ra s
53.56
7301.0
492.01
)10622.7(387.0825.0
Pr
492.01
Ra387.0825.0 Nu
2
27/816/9
6/17
2
27/816/9
6/1
=
+
×+=
+
+=
[ ] 2
2
m7.4)m60.3)(m5.0()m10.1)(m5.0(2
C.W/m868.2)53.56(m5.0
CW/m.02536.0
=+=
°=°
==
s A
Nu L
k h
Then the total rate of heat loss from the outer surface of the insulated tank by convection and radiation becomes
W5.97
])K 27320()K 27326)[(.K W/m1067.5)(m7.4)(1.0(+
C)2026)(m7.4)(C.W/m868.2(
)()(
444282
22
44
=+−+×
°−°=
−+−=+=
−
∞ surr s s s srad conv T T AT T hAQQQ σ ε
In steady operation, the heat lost by the side surfaces of the tank must be equal to the heat lost from theexposed surface of the insulation by convection and radiation, which must be equal to the heat conductedthrough the insulation. The second conditions requires the surface temperature to be
m05.0
C)(55)mC)(4.7W/m.035.0(W97.5 2tank °−
°=→−
==s s
sinsulation
T
L
T T kAQQ
It gives T s = 25.38°C, which is very close to the assumed temperature, 26°C. Therefore, there is no need to
repeat the calculations.
The total amount of heat loss and its cost during one year are
9-26 A printed circuit board (PCB) is placed in a room. The average temperature of the hot surface of the board is to be determined for different orientations.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas withconstant properties. 3 The local atmospheric pressure is 1 atm. 3 The heatloss from the back surface of the board is negligible.
Properties The properties of air at 1 atm and the anticipated film
temperature of (T s+T ∞)/2 = (45+20)/2 = 32.5°C are (Table A-15)
1-
25
K 003273.0K )2735.32(
11
7275.0Pr
/sm10631.1
CW/m.02607.0
=+
==
=
×=
°=
−
f T
k
β
υ
Analysis The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown
(a) Vertical PCB . We start the solution process by “guessing” the surface temperature to be 45°C for the
evaluation of the properties and h. We will check the accuracy of this guess later and repeat the calculations
if necessary. The characteristic length in this case is the height of the PCB, m.2.0== L Lc Then,
225
3-12
2
3
10756.1)7275.0()/sm10631.1(
)m2.0)(K 2045)(K 003273.0)(m/s81.9(Pr
)(×=
×
−=
−= −
∞
υ
β LT T g Ra s
78.36
7275.0
492.01
)10756.1(387.0825.0
Pr
492.01
Ra387.0825.0 Nu
2
27/816/9
6/17
2
27/816/9
6/1
=
+
×+=
+
+=
2
2
m03.0)m2.0)(m15.0(
C.W/m794.4)78.36(m2.0
CW/m.02607.0
==
°=°
==
s A
Nu L
k h
Heat loss by both natural convection and radiation heat can be expressed as
which is sufficiently close to the assumed value of 45°C for the evaluation of the properties and h.
(b) Horizontal, hot surface facing up Again we assume the surface temperature to be 45 ° C and use the properties evaluated above. The characteristic length in this case is
which is sufficiently close to the assumed value of 45°C in the evaluation of the properties and h.
(c) Horizontal, hot surface facing down This time we expect the surface temperature to be higher, andassume the surface temperature to be 50 °C . We will check this assumption after obtaining result andrepeat calculations with a better assumption, if necessary. The properties of air at the film temperature of
C352
2050
2°=
+=
+= ∞T T
T s f are (Table A-15)
1-
25
K 003247.0K )27335(
11
7268.0Pr /sm10655.1
CW/m.02625.0
=+
==
=×=
°=
−
f T
k
β
υ
The characteristic length in this case is, from part (b), Lc = 0.0429 m. Then,
37,166)7268.0()/sm10655.1(
)m0429.0)(K 2050)(K 003247.0)(m/s81.9(Pr
)(225
3-12
2
3
=×
−=
−=
−∞
υ
β c s LT T g Ra
453.5)379,166(27.027.0 4/14/1=== Ra Nu
C.W/m340.3)453.5(m0429.0
CW/m.02625.0 2
°=
°
== Nu L
k
h c
Considering both natural convection and radiation heat loses
"ANALYSIS""(a), plate is vertical"delta_a=LRa_a=(g*beta*(T_s_a-T_infinity)*delta_a^3)/nu^2*PrNusselt_a=0.59*Ra_a^0.25h_a=k/delta_a*Nusselt_aA=w*LQ_dot=h_a*A*(T_s_a-T_infinity)+epsilon*A*sigma*((T_s_a+273)^4-(T_surr+273)^4)"(b), plate is horizontal with hot surface facing up"delta_b=A/p
p=2*(w+L)Ra_b=(g*beta*(T_s_b-T_infinity)*delta_b^3)/nu^2*PrNusselt_b=0.54*Ra_b^0.25h_b=k/delta_b*Nusselt_bQ_dot=h_b*A*(T_s_b-T_infinity)+epsilon*A*sigma*((T_s_b+273)^4-(T_surr+273)^4)"(c), plate is horizontal with hot surface facing down"delta_c=delta_bRa_c=Ra_bNusselt_c=0.27*Ra_c^0.25h_c=k/delta_c*Nusselt_cQ_dot=h_c*A*(T_s_c-T_infinity)+epsilon*A*sigma*((T_s_c+273)̂ 4-(T_surr+273)^4)
9-28 Absorber plates whose back side is heavily insulated is placed horizontally outdoors. Solar radiation isincident on the plate. The equilibrium temperature of the plate is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the anticipated film
temperature of (T s+T ∞)/2 = (115+25)/2 = 70°C are (Table A-15)
1-
25
K 002915.0K )27370(
11
7177.0Pr
/sm10995.1
CW/m.02881.0
=+
==
=
×=
°=−
f T
k
β
υ
Analysis The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by “guessing” the surface temperature to be 115 °C for the evaluation of the
properties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.
The characteristic length in this case is m.24.0)m8.0m2.1(2
)m8.0)(m2.1(=
+==
p
A L sc Then,
225
3-12
2
3
1414.6)7177.0()/sm10995.1(
)m24.0)(K 25115)(K 002915.0)(m/s81.9(Pr
)(×=
×
−=
−=
−∞
υ
β c s LT T g Ra
33.48)10414.6(54.054.0 4/174/1=×== Ra Nu
C.W/m801.5)33.48(m24.0
CW/m.02881.0 2 °=°
== Nu L
k h
c
2m96.0)m2.1)(m8.0( == s A
In steady operation, the heat gain by the plate by absorption of solar radiationmust be equal to the heat loss by natural convection and radiation. Therefore,
which is identical to the assumed value. Therefore there is no need to repeat calculations.
If the absorber plate is made of ordinary aluminum which has a solar absorptivity of 0.28 and anemissivity of 0.07, the rate of solar gain becomes
W2.188)m96.0)(W/m700)(28.0( 22=== s AqQ α
Again noting that in steady operation the heat gain by the plate by absorption of solar radiation must beequal to the heat loss by natural convection and radiation, and using the convection coefficient determinedabove for convenience,
9-29 An absorber plate whose back side is heavily insulated is placed horizontally outdoors. Solar radiationis incident on the plate. The equilibrium temperature of the plate is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The localatmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the anticipated film
temperature of (T s+T ∞)/2 = (70+25)/2 = 47.5°C are (Table A-15)
1-
25
K 00312.0K )2735.47(
11
7235.0Pr
/sm10774.1
CW/m.02717.0
=+
==
=
×=
°=−
f T
k
β
υ
Analysis The solution of this problem requires a trial-and-error approach since the determination of theRayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We
start the solution process by “guessing” the surface temperature to be 70 °C for the evaluation of the
properties and h. We will check the accuracy of this guess later and repeat the calculations if necessary.
The characteristic length in this case is m.24.0)m8.0m2.1(2
)m8.0)(m2.1(=
+==
p
A L sc Then,
225
3-12
2
3
10379.4)7235.0()/sm10774.1(
)m24.0)(K 2570)(K 00312.0)(m/s81.9(Pr
)(×=
×
−=
−=
−∞
υ
β c s LT T g Ra
93.43)10379.4(54.054.0 4/174/1=×== Ra Nu
C.W/m973.4)93.43(m24.0
CW/m.02717.0 2 °=°
== Nu L
k h
c
2m96.0)m2.1)(m8.0( == s A
In steady operation, the heat gain by the plate by absorption of solar radiation must be equal to the heat loss by natural convection and radiation. Therefore,
which is close to the assumed value. Therefore there is no need to repeat calculations.
For a white painted absorber plate, the solar absorptivity is 0.26 and the emissivity is 0.90. Thenthe rate of solar gain becomes
W7.174)m96.0)(W/m700)(26.0( 22 === s AqQ α
Again noting that in steady operation the heat gain by the plate by absorption of solar radiation must beequal to the heat loss by natural convection and radiation, and using the convection coefficient determined
above for convenience (actually, we should calculate the new h using data at a lower temperature, anditerating if necessary for better accuracy),
9-30 A resistance heater is placed along the centerline of a horizontal cylinder whose two circular sidesurfaces are well insulated. The natural convection heat transfer coefficient and whether the radiation effectis negligible are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gaswith constant properties. 3 The local atmospheric pressure is 1 atm.
Analysis The heat transfer surface area of the cylinder is
2
m05027.0)m8.0)(m02.0( === π π DL A Noting that in steady operation the heat dissipated from the outer surface must equal to the electric power consumed, and radiationis negligible, the convection heat transfer is determined to be
![Chap09[2] LS Modif](https://static.documents.pub/doc/80x56/577ce4bb1a28abf1038f0760/chap092-ls-modif.jpg)
![chap09 [호환 모드]parkjonghyuk.net/.../programming1/Korean/chap09.pdf · 2011. 3. 7. · • 카드는4개의무늬와각무늬당13개의숫자로이 루어짐 • 카드를비트필드로표현하면메모리를절약할수](https://static.documents.pub/doc/80x56/60fef2bf45e60b55476f5622/chap09-eeoe-2011-3-7-a-eoee4eoeeeeeee13eoeeoe.jpg)
