Free heat convection |HEAT TRANSFER Laboratory

SAIF AL-DIN ALI MADI
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
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[HEAT TRANSFER Laboratory]
University of Baghdad
Name: - Saif Al-din Ali -B-

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TABLE OF CONTENTS
Experiment Name................................................................I
Experiment Aim..............................................................II
Introduction...................................................................III
THEORY...........................................................................V
Calculations and results.................................................VI
DISCUSSION……………….................................................VII

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1. Experiment Name: Free Heat Convection from a Horizontal Heated
Cylinder
2. Experiment Aim:
1. Calculating the free heat convection coefficient (ℎ 𝑁𝐶) for a
horizontal heated solid cylinder.
2. Find the relationship between RaNo. And NuNo. for fluid flow
around a cylinder
3. Introduction:
Free Convection: is the heat transfer between a solid surface and the
adjunct fluid as a result of the temperature difference between them. This
temperature difference leads to a density difference near the surface and
causes buoyancy forces that give the fluid a natural motion which leads to
a heat transfer between the fluid and the solid surface.
 Grashof Number: It is a dimensionless number that expresses
the fluid motion that results from the buoyancy the solid surface
as shown in following formula:
𝐺𝑟 =
𝐵⋅𝑔⋅𝜌2⋅𝛥𝑇⋅𝐿 𝐶
3
𝑢2 (Diffusivity for momentum) / (Diffusivity for heat)
Where:
𝜌 ; is the fluid density at the film temperature
𝑢 ; is the fluid dynamic viscosity at the film temperature,
𝐿 𝐶 ; is the characteristic length of the geometry ( 𝐿 𝐶 =d for horizontal
cylinder and𝐿 𝐶 =L for vertical cylinder
 Prandtle Number: It is a dimensionless number that relates the
thermo-physical properties of the fluid, The Prandtle numbers
for gases are about 1.
𝑝𝑟 =
𝜇𝑐 𝑝
𝑘
=
𝑣
𝑎
= (Diffusivity for momentum) / (Diffusivity for heat)

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 Rayleigh Number: It is the multiplication of Gr and Pr:
Ra=Gr.Pr
 Nusslet Number: It is a dimensionless group that represents the ratio
of the heat transfer by convection to the heat transfer by construction,
𝑁𝑢 =
ℎ𝑑
𝑘
4. Experiment Theory:-
1. The Apparatus used: the apparatus used in this experiment is
called Cross Flow Heat Exchanger. Draw a schematic shape for the
apparatus demonstrating its parts clearly, the apparatus parts are
as:
 Horizontal wind tunnel with specified length and has multiple holes
 Electric heater used for heating the copper specimen.
 Alcohol manometer for measuring pressure
 Fan + Gate for changing the amount of the air drawing opening
 Digital Thermometer for measuring specimen temperature.
 Digital Timer for measuring required time to reach a specified temperature
 Pitot - Static Tube for measuring static pressure
SAIF AL-DIN ALI

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2. The Experiment Procedure:
The specimen is heated by using the electric heater until reaching a specified
temperature Value (for example 85 "C or 90 ° C), Then the specimen is put
horizontally into one of the holes in the wind tunnel, Consequently, the forced air
attaches the copper specimen Surface as a result the specimen temperature is
decreased with increasing time etc....
5. Calculations and results
Data
T(ċ) 100 90 80 70 60 50
t(sec) 0 74 172 295 454 676
𝐓𝐚( ċ ) 18
dr (m) Lr (m) Cpr(J/kg.c) Ar (m^2) mr (kg)
0.0124 0.095 380 0.00404 0.1093
Qout = ℎ𝐴(T − Ta)
Qin = 0
𝑄𝑠𝑡𝑜𝑟𝑒 = 𝑚𝑐
𝑑𝑇
dt
𝑄𝑠𝑡𝑜𝑟𝑒 = Qin − Qout → 𝑚𝑐
𝑑𝑇
dt
= −ℎ𝐴(T − Ta)
1 = −
ℎ𝐴
𝑚𝑐
×
(T − Ta)
𝑑𝑇
dt
∫
𝑑𝑇
(T − Ta)
𝑇
𝑇1
= − ∫
ℎ𝐴
𝑚𝑐
1
0
dt
ln
𝑇 − Ta
𝑇1 − Ta
= −
ℎ𝐴
𝑚𝑐
𝑡

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𝐥𝐧
𝑻 − 𝐓𝐚
𝑻𝟏 − 𝐓𝐚
Ta = 18 ċ 𝑇1 = 100 ċ
1. ln
100−18
100−18
= 0
2. ln
90−18
100−18
= - 0.13005
3. ln
80−18
100−18
= - 0.2795
4. ln
70−18
100−18
= - 0.4554
5. ln
60−18
100−18
= - 0.669
6. ln
50−18
100−18
= - 0.9409
𝐬𝐥𝐨𝐩𝐞 =
−𝟎. 𝟔 + 𝟎. 𝟒
𝟒𝟏𝟎 − 𝟐𝟔𝟎
= − 𝟎. 𝟎𝟎𝟏𝟑𝟑𝟑
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
0 100 200 300 400 500 600 700 800
ln〖(𝑇−T_a)/(𝑇1−T_a)〗
t (sec)
SAIF AL-DIN ALI

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𝒉 =
𝒎∗𝒄∗𝒔𝒍𝒐𝒑𝒆
𝑨𝒓
& 𝑵𝒖 =
𝒉𝒅
𝒌
& 𝑻 𝒇 =
𝑻+𝑻∞
𝟐
& 𝑮 𝒓 =
𝑩⋅𝒈⋅𝝆 𝟐
⋅𝜟𝑻⋅𝑳 𝑪
𝟑
𝒖 𝟐
&
Ra=Gr.Pr & 𝒑 𝒓 =
𝝁𝒄 𝒑
𝒌
=
𝒗
𝒂
h =
0.1093 ∗ 380 ∗ 0.001333
0.00404
= 13.7041 w/m2
𝑐0
1. 𝑻 𝒇 =
𝟏𝟎𝟎+𝟏𝟖
𝟐
= 𝟓𝟗 𝑐0
Property; at 𝑻 𝒇
Density (𝝆 ) 1.04992
Coefficient of
expansion (𝑩)
3.02036 * 10-3
Kinematic
viscosity (v)
19.1290*10-6
Gravity(g) 9.81
Dynamic
viscosity (u)
20.084*10-6
Prandtl-
Number (pr)
0.710172
Heat
conductance
(k)
28.5208* 10-3
1. 𝑻 𝒔 = 100 𝑐
𝑮 𝒓 =
𝑩 ⋅ 𝒈(𝑻 𝒔 − 𝑻∞)𝑫 𝟑
𝒗 𝟐
𝑮𝒓 =
3.02036 ∗ 10−3
∗ 9.81(100 − 18)(0.0124)3
(19.1290 ∗ 10−6
) 𝟐
= 𝟏𝟐𝟔𝟓𝟗. 𝟔𝟔
Ra=Gr.Pr
Ra= 𝟏𝟐𝟔𝟓𝟗. 𝟔𝟔 *0.710172=8990.537
𝑵𝒖 =
𝒉𝒅
𝒌
𝑵𝒖 =
13.7041 ∗ 0.0124
28.5208 ∗ 10−3
= 𝟓. 𝟗𝟓𝟖𝟏
Log ( 𝑵𝒖) =0.7751
Log (Ra) =3.95378

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2. 𝑻 𝒇 =
𝟗𝟎+𝟏𝟖
𝟐
= 𝟓𝟒 𝑐0
Coefficient of
expansion (𝑩)
3.06516* 10−3
Kinematic
viscosity (v)
18.63315*10-6
Gravity(g) 9.81
Dynamic
viscosity (u)
19.854*10-6
Prandtl-
Number (pr)
0.710632
Heat
conductance
(k)
28.1648* 10-3
2. 𝑻 𝒔 = 90 𝑐
𝑮 𝒓 =
𝑩 ⋅ 𝒈(𝑻 𝒔 − 𝑻∞)𝑫 𝟑
𝒗 𝟐
𝑮𝒓 =
3.06516 ∗ 10−3
∗ 9.81(90 − 18)(0.0124)3
(18.63315 ∗ 10−6
) 𝟐
= 𝟏𝟏𝟖𝟗𝟏. 𝟏𝟓𝟓
Ra=Gr.Pr
Ra= 𝟏𝟏𝟖𝟗𝟏. 𝟏𝟓𝟓 *0.710632=8450.235
𝑵𝒖 =
𝒉𝒅
𝒌
𝑵𝒖 =
13.7041 ∗ 0.0124
28.1648 ∗ 10−3
= 𝟔. 𝟎𝟑𝟑𝟒𝟒
Log ( 𝑵𝒖) =0.78056
Log (Ra) =3.92686

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3. 𝑻 𝒇 =
𝟖𝟎+𝟏𝟖
𝟐
= 𝟒𝟗 𝑐0
Coefficient of
expansion (𝑩)
3.11148* 10−3
Kinematic
viscosity (v)
18.14201*10-6
Gravity(g) 9.81
Dynamic
viscosity (u)
19.6224*10-6
Prandtl-
Number (pr)
0.711124
Heat
conductance
(k)
27.8072* 10-3
3. 𝑻 𝒔 = 80 𝑐
𝑮 𝒓 =
𝑩 ⋅ 𝒈(𝑻 𝒔 − 𝑻∞)𝑫 𝟑
𝒗 𝟐
𝑮𝒓 =
3.11148 ∗ 10−3
∗ 9.81(80 − 18)(0.0124)3
(18.14201 ∗ 10−6
) 𝟐
= 𝟏𝟎𝟗𝟔𝟐. 𝟖𝟏𝟏
Ra=Gr.Pr
Ra= 𝟏𝟎𝟗𝟔𝟐. 𝟖𝟏𝟏 *0.711124=7795.9186
𝑵𝒖 =
𝒉𝒅
𝒌
𝑵𝒖 =
13.7041 ∗ 0.0124
27.8072 ∗ 10−3
= 𝟔. 𝟏𝟏𝟏𝟎𝟑
Log ( 𝑵𝒖) =0.786114
Log (Ra) =3.889857

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4. 𝑻 𝒇 =
𝟕𝟎+𝟏𝟖
𝟐
= 𝟒𝟒 𝑐0
Coefficient of
expansion (𝑩)
3.16388 * 10−3
Kinematic
viscosity (v)
17.6285 *10-6
Gravity(g) 9.81
Dynamic
viscosity (u)
19.3844 *10-6
Prandtl-
Number (pr)
0.711744
Heat
conductance
(k)
27.4432* 10−3
4. 𝑻 𝒔 = 70 𝑐
𝑮 𝒓 =
𝑩 ⋅ 𝒈(𝑻 𝒔 − 𝑻∞)𝑫 𝟑
𝒗 𝟐
𝑮𝒓 =
3.16388 ∗ 10−3
∗ 9.81(70 − 18)(0.0124)3
(17.6285 ∗ 10−6
) 𝟐
= 𝟗𝟗𝟎𝟐. 𝟎𝟖𝟓𝟗
Ra=Gr.Pr
Ra= 𝟗𝟗𝟎𝟐. 𝟎𝟖𝟓𝟗 *0.711744=7047.75023
𝑵𝒖 =
𝒉𝒅
𝒌
𝑵𝒖 =
13.7041 ∗ 0.0124
27.4432 ∗ 10−3
= 𝟔. 𝟏𝟗𝟐𝟎𝟗𝟐
Log ( 𝑵𝒖) =0.79183
Log (Ra) =3.84805

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5. 𝑻 𝒇 =
𝟔𝟎+𝟏𝟖
𝟐
= 𝟑𝟗 𝑐0
Coefficient of
expansion (𝑩)
3.21628* 10−3
Kinematic
viscosity (v)
17.13171*10-6
Gravity(g) 9.81
Dynamic
viscosity (u)
19.1464*10-6
Prandtl-
Number (pr)
0.712364
Heat
conductance
(k)
27.0792* 10−3
5. 𝑻 𝒔 = 60 𝑐
𝑮 𝒓 =
𝑩 ⋅ 𝒈(𝑻 𝒔 − 𝑻∞)𝑫 𝟑
𝒗 𝟐
𝑮𝒓 =
3.21628 ∗ 10−3
∗ 9.81 ∗ (60 − 18)(0.0124)3
(17.13171 ∗ 10−6
) 𝟐
= 𝟖𝟔𝟎𝟖. 𝟔𝟔
Ra=Gr.Pr
Ra= 𝟖𝟔𝟎𝟖. 𝟔𝟔 *0.712364=6132.50257
𝑵𝒖 =
𝒉𝒅
𝒌
𝑵𝒖 =
13.7041 ∗ 0.0124
27.0792 ∗ 10−3
= 𝟔. 𝟐𝟕𝟓𝟑𝟐
Log ( 𝑵𝒖) =0.79763
Log (Ra) =3.78763

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6. 𝑻 𝒇 =
𝟓𝟎+𝟏𝟖
𝟐
= 𝟑𝟒 𝑐0
Coefficient of
expansion (𝑩)
3.26868 * 10−3
Kinematic
viscosity (v)
16.6505 *10-6
Gravity(g) 9.81
Dynamic
viscosity (u)
18.9084 *10-6
Prandtl-
Number (pr)
0.712984
Heat
conductance
(k)
26.7152* 10−3
6. 𝑻 𝒔 = 50 𝑐
𝑮 𝒓 =
𝑩 ⋅ 𝒈(𝑻 𝒔 − 𝑻∞)𝑫 𝟑
𝒗 𝟐
𝑮𝒓
=
3.26868 ∗ 10−3
∗ 9.81 ∗ (50 − 18)(0.0124)3
(16.6505 ∗ 10−6) 𝟐
= 𝟕𝟎𝟓𝟔. 𝟕𝟎𝟑
Ra=Gr.Pr
Ra=𝟕𝟎𝟓𝟔. 𝟕𝟎𝟑 *0.712984=5031.316
𝑵𝒖 =
𝒉𝒅
𝒌
𝑵𝒖 =
13.7041 ∗ 0.0124
26.7152 ∗ 10−3
= 𝟔. 𝟑𝟔𝟎𝟖𝟐
Log ( 𝑵𝒖) =0.803513
Log (Ra) =3.70168

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T
(C) t(sec) Ra Log(Ra)
h
(W/m^2*c)
𝑵𝒖 Log (𝑵𝒖
100 0 8990.537 3.95378
13.7041
𝟓. 𝟗𝟓𝟖𝟏 0.7751
90 74 8450.235 3.92686 𝟔. 𝟎𝟑𝟑𝟒𝟒 0.78056
80 174 7795.9186 3.889857 𝟔. 𝟏𝟏𝟏𝟎𝟑 0.786114
70 295 7047.75023 3.84805 𝟔. 𝟏𝟗𝟐𝟎𝟗𝟐 0.79183
60 454 6132.5025 3.78763 𝟔. 𝟐𝟕𝟓𝟑𝟐 0.79763
50 676 5031.316 3.70168 𝟔. 𝟑𝟔𝟎𝟖𝟐 0.803513
Log (NU) = Log(c) +m Log (Ra)
m= (0.47-0.4)/ (3-2.45) = 0.12727
Log(c) =-0.22
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
Log(𝑵𝒖)
Log (Ra)
Log(c)
SAIF AL-DIN ALI
SAIF AL-DIN ALI

14 | P a g e
6. DISCUSSION:
1. Discuss the results that you have found and compare your formula
with the formula found in your textbook for the same case. give the
reasons for this differences.
Because of the free transition we note a few comparison with the
previous experience if we compare our work with the theoreticians in the
book note the book closer to the ideal case. In the practical there is a loss
in time and speed of the student's response to the experience and when
drawing relations draw drawing to a linear relationship and this is not
effective, which caused the inaccuracy -
There is no speed so the Re does not exist but a new description appears,
which is Gr, and another expression Ra
2. Can you obtain the same formula for the same cylinder but in
vertical position? Why?
In vertical mode the characteristic
length varies if the cylinder is
horizontal and the boundary layer
is changed as in the picture

15 | P a g e
3. Was the experiment done in steady state situation? why?
The existence of time and reliability to calculate heat transfer coefficient
makes the process not state situation
This is a state situation ; a system or a process is in a steady state if the
variables (called state variables) which define the behavior of the system
or the process are unchanging in time. In continuous time, this means
that for those properties p of the system, the partial derivative with
respect to time is zero and remains so:
4. What is the experimental principle that was depended on in this
experiment?
In free convection fluid motion is due to buoyancy forces within the
fluid, while in forced convection it is externally imposed. Buoyancy is due
to the combined presence of a fluid density gradient and a body force
that is proportional to density. In practice, the body force is usually
gravitational, although it may be a centrifugal force in rotating fluid
machinery or a Coriolis force in atmospheric and oceanic rotational
motions. There are also several ways in which a mass density gradient
may arise in a fluid, but for the most common situation it is due to the
presence of a temperature gradient. We know that the density of gases
and liquids depends on temperature, generally decreasing (due to fluid
expansion) with increasing temperature

16 | P a g e
5. There are several practical applications for the free convection,
mention number of them?

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