forced 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: Forced Heat Convection
2. Experiment Aim:
1. Calculating the forced heat convection coefficient (ℎ 𝐹𝐶) for a heated cylinder
2. Find the relations between Re N, and NuNo. for fluid flow around a cylinder
3. Introduction:
 Convection; is the mode of energy transfer between a solid surface and
the adjacent liquid or gas that is in motion, and it involves the
combined effects of conduction and fluid motion, convection is divided
into two types:
1. Forced Convection; the fluid is forced to flow (due to pressure drop)
over the surface by external means such as a fan, pump, or the wind,
2. Free Convection: the fluid motion is caused by buoyancy forces that
are induced by density differences due to the variation of temperature
in the fluid,
Note: There are three types of fluid flow: laminar, transition, and
turbulent, the fluid flow type can be known from a dimensionless Value
that reflects the value of the average velocity, fluid density, fluid dynamic
viscosity, and the characteristic length for the geometry used, this
dimensionless number is called Reynolds number and it is given by:
𝑅𝑒 ≡
𝜌 𝑉𝐷
𝜇
= 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟
For circular pipe, the Reynolds number for the three flow types is:
Re <2300 Laminar Flow
4000<Re< 2300 Transition Flow
Re <4000 Turbulent Flow
* 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)
*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

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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

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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, simultaneously, the gate is
opened 100% and the fan is switched on. 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) Q% t(s)
V
m/s
Q% t(s) V m/s Q% t(s) V m/s Q% t(s) V m/s
90
100
0
17 80
0
15 60
0
12 40
0
9.5
80 10 11 12 14
70 20 21 25 29
60 33 35 40 47
50 49 51 60 71

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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
ln
𝑇 − Ta
𝑇1 − Ta
Ta = 18 ċ 𝑇1 = 90 ċ
1. ln
90−18
90−18
=0
2. ln
80−18
90−18
=-0.14963
3. ln
70−18
90−18
=-0.325422
4. ln
60−18
90−18
=-0.53899
5. ln
50−18
90−18
=-0.81099

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100%; slope =
−0.4 + 0.3
24.4 − 18.4
= −0.01666
80%; slope =
−0.64 + 0.556
41 − 36
= −0.0168
60%; slope =
−0.7 + 0.64
52.1 − 47.6
= −0.01333
40%; slope =
−0.78 + 0.7
69 − 61
= −0.01
ℎ =
𝑚∗𝑐∗𝑠𝑙𝑜𝑝𝑒
п𝑑𝑙
,,,,, 𝑁𝑢 =
ℎ𝑑
𝑘
1. h =
𝟎.𝟏𝟎𝟗𝟑∗𝟑𝟖𝟎 ∗ 𝟎.𝟎𝟏𝟔𝟔𝟔
𝟑.𝟏𝟒∗𝟎.𝟎𝟗𝟓∗𝟎.𝟎𝟏𝟐𝟒
= 186.369 w/m2
𝑐0
𝑁𝑢 =
186.369 ∗ 𝟎. 𝟎𝟏𝟐𝟒
𝟎. 𝟎𝟐𝟔
= 88.88
Log (88.88) =1.9488
2. h =
𝟎.𝟏𝟎𝟗𝟑∗𝟑𝟖𝟎∗ 𝟎.𝟎𝟏𝟔𝟖
𝟑.𝟏𝟒∗𝟎.𝟎𝟗𝟓∗𝟎.𝟎𝟏𝟐𝟒
= 188.64 w/m
2
𝑐
0
𝑁𝑢 =
188.64 ∗ 0.0124
0.026
= 89.9
Log (89.9) =1.9537
3. h =
𝟎.𝟏𝟎𝟗𝟑∗𝟑𝟖𝟎∗ 𝟎.𝟎𝟏𝟑𝟑𝟑
𝟑.𝟏𝟒∗𝟎.𝟎𝟗𝟓∗𝟎.𝟎𝟏𝟐𝟒
= 149.34 w/m
2
𝑐
0
𝑁𝑢 =
149.34 ∗ 0.0124
0.026
= 71.22
Log (71.22) =1.8526
4. h =
𝟎.𝟏𝟎𝟗𝟑∗𝟑𝟖𝟎∗ 𝟎.𝟎𝟏
𝟑.𝟏𝟒∗𝟎.𝟎𝟗𝟓∗𝟎.𝟎𝟏𝟐𝟒
= 112.28 w/m
2
𝑐
0
𝑁𝑢 =
112.28 ∗ 0.0124
0.026
= 53.54
Log (53.54) =1.7286
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
0 10 20 30 40 50 60 70 80
ln
t(s)
100%
80%
60%
40%

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𝑅𝑒 =
𝜌 𝑉𝐷
𝜇
Q=100%
1. 𝑅𝑒 =
1.19796∗17∗0.0124
18.1328∗10−6
=13926.694
Log (13926.694) =4.1438
Q=80%
2. 𝑅𝑒 =
1.19796∗15∗0.0124
18.1328∗10−6
=12288.25
Log (12288.25) =4.0894
Q=60%
3. 𝑅𝑒 =
1.19796∗12∗0.0124
18.1328∗10−6
=9830.6
Log (9830.6) =3.9925
Q=40%
4. 𝑅𝑒 =
1.19796∗9.5∗0.0124
18.1328∗10−6
=7782.56
Log (7782.56) =3.8911
1. Q=100%
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
Log(Nu)
log(Re)
Log(c)
n = (1.4-1)/ (3.5-3.09) =0.9756
Log (c) = -1.8

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T(ċ) 90 80 70 60 50
t(s) 0 10 20 33 49
𝑅𝑒 13926.694
Log( 𝑅𝑒) 4.1438
h w/m
2
𝑐
0
186.369
𝑁𝑢 88.88
Log( 𝑁𝑢) 1.9488
2. Q=80%
T(ċ) 90 80 70 60 50
t(s) 0 11 21 35 51
𝑅𝑒 12288.25
Log( 𝑅𝑒) 4.0894
h w/m
2
𝑐
0
188.64
𝑁𝑢 89.9
Log( 𝑁𝑢) 1.9537
3. Q=60%
T(ċ) 90 80 70 60 50
t(s) 0 12 25 40 60
𝑅𝑒 9830.6
Log( 𝑅𝑒) 3.9925
h w/m
2
𝑐
0
149.34
𝑁𝑢 71.22
Log( 𝑁𝑢) 1.8526
4. Q=60%
T(ċ) 90 80 70 60 50
t(s) 0 12 25 40 60
𝑅𝑒 7782.56
Log( 𝑅𝑒) 3.8911
h w/m
2
𝑐
0 112.28
𝑁𝑢 53.54
Log( 𝑁𝑢) 1.7286

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6. DISCUSSION:
1. Draw the relation between the forced convection and the
increasing velocity. discuss the diagram
The heat transfer coefficient increase with
an increase in the velocity of the fluid the
heat transfer co efficient will increase.
Because, increasing the velocity will increase
the Reynolds number value. The amount of
increase in heat transfer co efficient will
depend upon the flow conditions.
0
100
200
300
400
500
600
0 2 4 6 8 10 12 14 16 18
QW
V m/s
T=90C
T=80C
T=70C
T=60C
T=50

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2. Was the experiment done in steady state situation? Why?
𝑄𝑠𝑡𝑜𝑟𝑒 = Qin − Qout
Qout = ℎ𝐴(T − Ta)
𝑄𝑠𝑡𝑜𝑟𝑒 = 𝑚𝑐
𝑑𝑇
dt
The condition is unstable at the other end, the temperature changes with time
"Steady state" is a relative term. Unsteadiness can even be caused by, say voltage
fluctuations, which may cause a change in flow rate. You might be able to set an
arbitrary band (say 1% of the measured value about the mean), so that if the
fluctuations are less than that, you may take it as 'nominally steady'. That should do
for your purpose.
3. What is the experimental principle that was depended on in this experiment?
Convective heat transfer, often referred to simply as convection, is the transfer of
heat from one place to another by the movement of fluids. Convection is usually the
dominant form of heat transfer in liquids and gases. Although often discussed as a
distinct method of heat transfer, convective heat transfer involves the combined
processes of unknown conduction (heat diffusion) and advection (heat transfer by
bulk fluid flow).
Forced convection is a mechanism, or type of transport in which fluid motion is
generated by an external source (like a pump, fan, suction device, etc.). It should be
considered as one of the main methods of useful heat transfer as significant
amounts of heat energy can be transported very efficiently.

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4. There are some practical applications for the forced convection, mention
number of them?
Cooling Towers
5. What are the main results that you have deducted from this experiment?
Calculating the forced heat convection coefficient (ℎ 𝐹𝐶) for a heated cylinder
Find the relations between Re N, and NuNo. For fluid flow around a cylinder

forced heat convection | HEAT TRANSFER Laboratory

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