Combined Convection and Radiation Heat

Transfer (H2)

King Fahd University of Petroleum and

Minerals
Chemical Engineering Department
Chemical Engineering Laboratory (CHE309)

Instructor: Mohammed Alkhater

Title
A

Experiment Date:11 /4/2016

Submission Date:25/4/2016
ABSTRACT

The combined convection and radiation system has been designed to demonstrate
heat transfer from a solid surface to its surroundings.
The objectives of this experiment are:
1- To determine the combined heat transfer (Qrad +Qconv) from a horizontal
cylinder in natural convection over a wide range of power inputs and
corresponding surface temperature. Also, to demonstrate the relationship
between power input and surface temperature in free convection.
2- To compare the contribution of heat transfer by convection with heat transfer
by radiation and from the measurements to show the domination of the
convective heat transfer coefficient (Hc) at low surface temperatures and the
domination of the radiation heat transfer coefficient (H r) at high surface
temperatures.
3- To determine the effect of forced convection on heat transfer from the surface
of a cylinder at varying air velocities and the surface temperatures. Also, to
demonstrate the relationship between air velocity and surface temperature for
a cylinder subjected to forced convection.
4- To demonstrate that the local heat transfer coefficient varies around the
circumference of a horizontal cylinder when subjected to forced convection.

In this experiment, the surface temperature of the cylinder was studied at different air
velocities and different positions. The experiment consists of three parts, the first part
was free convection and radiation heat transfer determination and their contribution
to the total heat transfer at different voltages. The second part was studying the
effect of forced convection. The third part was studying the variation of local heat
transfer coefficient around aa cylinder at different positions. The Q tot in the
experiment 1 was 2.684 at 5 volts. The Nusselt number was calculated to be 9.81 at
20 volts and 1 m/s air velocity and 60o degree.

INTRODUCTION

Combined convection and radiation heat transfer is a particular case of simultaneous

radiative, convective and conductive heat transfer. Combined radiation and
convection occurs between the boundary surfaces and the surrounding medium,
between the surfaces separated by the moving medium and inside the moving
medium. The experiment was done by physical process by studying the surface
temperature of the barriers.
. In this experiment, there is a Heat Transfer Service Unit which is connected to the
Combined Convection and Radiation Heat Transfer Accessory, in the first
experiment the voltage was changing while in the second and third experiments the
voltage was kept constant, but the air velocity was changing only in the third
experiment.

Theoretical Background

The following theoretical analysis has used an empirical relationship for the heat
transfer due to natural convection proposed by VT Morgan [1].

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The total heat loss from the cylinder:
Qtot = Qc + Qr (1)
The heat loss due to natural convection:
Qc = Hcm As (Ts – Ta) (2)
The heat loss due to radiation:
Qr = Hrm As (Ts – Ta) (3)
Where the heat transfer area = πDL
The average heat transfer coefficient for radiation (Hrm) in (W/m2.K) can be calculated
from the following equation:
4 4
(T s −T a )
H rm=σξF (4)
(T s−T a )
Where,
σ = Stefan Boltzmann constant (56.7 ×10−9 W /m2 K 4)
ξ = Emissivity of surface
F=1= View factor
The average heat transfer coefficient for natural convection Hcm can be calculated
using the following equation:
k Num
H cm = (5)
D
Where,
k = Thermal conductivity of air
Num = average Nusselt number = c (RaD)n
C and n are constant can be taken from table 1 below
RaD = Rayleigh number
3
G β ( T s−T a ) D
RaD = (6)
v2
1
β= (7)
T film

T s +T a
T film= (8)
2
Table 1: Contact c and Exponent n for natural convection on a horizontal cylinder

RaD c n
10 to 10-2
-9
0.675 0.058
10-2 to 102 1.020 0.148
102 to 104 0.850 0.188
104 to 107 0.480 0.250
107 to 1012 0.125 0.333

Alternatively a simplified equation may be used to calculate the heat transfer

coefficient for free convection.
0.25
T s−T a
H cm =1.32( ) (9)
D
The heat transfer coefficients Hfm due to forced convection can be calculated using
the following relationship:
k
H fm= Num (10)
D

3
An empirical formula [3] can be used to calculate the value for the average Nusselt
number Num as follows:
( 0.62 ℜ0.5 Pr 0.33 ) 0.5
Num=0.3+ ( (1+ )ℜ
)

(( ( ) ))
0.4
0.66 0.25 282000 (11)
1+
Pr

Where,
Re= Reynolds number = (Uc D/ v)
Pr = Prandtl number for air
Uc = corrected air velocity = 1.22 Ua

Procedure

Experiment1 1: free convection and radiation heat transfer determination and their
contribution to the total heat transfer
1- The front main switch was switched on.
2- The heater voltage was set to 5 volts.
3- The accessory was allowed to stabilize and the temperature of the cylinder
was monitored.
4- When the temperature reached equilibrium, the icon (Go) was selected on
thee software menu bar to record (T9 , T10 , V, I).
5- The heater voltage was set to 8 volts.
6- The accessory was allowed to stabilize and then the above readings were
repeated.
7- The heater voltage was set to 12, 15 and 20 volts.
8- The accessory was allowed to stabilize and then the above readings were
repeated.
Experiment 2: Effect of forced convection
1- The fan was set to give an air flow rate of 0.5 m/s.
2- The heater voltage was set to 20 volts.
3- The accessory was allowed to stabilize.
4- When the temperature reached equilibrium, the icon (Go) was selected on
thee software menu bar to record (T9 , T10 , V, I, Ua).
5- The fan was adjusted to give an air flow rate of 1 m/s.
6- The accessory was allowed to stabilize and then the above readings were
repeated.
7- The above procedure was repeated with changing the air velocity in steps of 1
m/s until 7m/s.
Experiment 3: variation of local heat transfer coefficient around a cylinder
1- The fan was set to give an air flow rate of 1 m/s.
2- The heater voltage was set to 20 volts.
3- When the temperature reached equilibrium, the icon (Go) was selected on
thee software menu bar to record (T9 , T10 , V, I, position).
4- The accessory was allowed to stabilize and then the above readings were
repeated.
5- The cylinder was rotated in steps of 30o until the thermocouple is located on
the top of the cylinder.

4
 6- The fan was adjusted to give an air flow rate of 5 m/s then The above
procedure was repeated.

Results
Table 1
V(volt I(A) T9(◦C T10(◦C Ts(K) Ta(K) As(m^2 Tfilm Rad Nu Hcm Qc(W) Hrm Qr(W) Qtot(W Qin(W
) ) ) ) (K) (W/m^2K (W/m^2K ) )
) )
5 0.77 21.9 74.1 347. 295.0 2.20E- 321.2 1832.3 3.49 16.195 1.859 7.184 0.825 2.684 3.8
2 3 5 03 0
8 1.25 22.2 132.8 406. 295.3 2.20E- 350.7 3555.5 3.95 19.175 4.663 9.520 2.315 6.979 10.1
5 0 5 03 4
12 1.85 22.7 212.2 485. 295.8 2.20E- 390.6 5468.9 4.28 21.006 8.753 13.595 5.665 14.419 22.2
1 4 5 03 7
15 2.30 23.2 269.9 543. 296.3 2.20E- 419.7 6626.1 4.44 22.222 12.05 17.305 9.388 21.443 34.4
5 1 5 03 4 5
20 3.07 23.7 361 634. 296.8 2.20E- 465.5 8168.1 4.62 23.807 17.65 24.586 18.23 35.894 61.7
6 2 5 03 3 8 6

Table 2

V(volt Ua T9(◦C T10(◦C Ts(K) Ta(K) I(A) Hrm Qr(W Hcm Qfc Qtot Uc(m/
) (m/s) ) ) ) s)
0.5 24.5 333.4 606. 297.6 3.0 22.2 15.10 17.5 11.8 26.9 0.61
6 5 8 3 0 9 9
1 24.8 255.1 528. 297.9 3.0 16.3 8.29 16.2 8.24 16.5 1.22
3 5 8 7 6 2
2 25.4 172.1 445. 298.5 3.0 11.5 3.71 14.5 4.69 8.40 2.44
3 5 9 1 3
3 25.6 133.1 406. 298.7 3.1 9.66 2.28 13.4 3.18 5.46 3.66
3 5 0 4
20
4 25.7 109.7 382. 298.8 3.1 8.66 1.60 12.6 2.33 3.93 4.88
9 5 3 4
5 25.7 92.8 366. 298.8 3.1 7.99 1.18 11.9 1.76 2.94 6.1
0 5 3 5
6 25.7 81.7 354. 298.8 3.1 7.58 0.93 11.4 1.41 2.34 7.32
9 5 2 2
7 25.7 73.2 346. 298.8 3.1 7.27 0.76 10.9 1.14 1.90 8.54
4 5 3 6

Table 3

Ua=1m/s

5
 Position T10( C Kinemitic
V(volt) T9( C) I(A) Tfilm Pr Re Nu
( Degree) ) vescosity
135.3 10.013
0 26.9 243.8 3.0859 2.65E-05 0.695 3.78E+02
5 9
9.9610
30 27.5 247.9 3.0908 137.7 2.67E-05 0.695 3.74E+02
9
9.8112
60 27.5 252.7 3.0762 140.1 2.76E-05 0.695 3.63E+02
1
115.4 10.409
20 90 27.7 203.2 3.0811 2.47E-05 0.704 4.04E+02
5 7
11.162
120 27.5 144.4 3.0811 85.95 2.16E-05 0.706 4.63E+02
5
11.602
150 27.5 114.3 3.0859 70.9 2.00E-05 0.708 4.99E+02
5
11.760
180 27.8 103.8 3.01201 65.8 1.95E-05 0.708 5.13E+02
1

Table 4

Ua=5 m/s
Position T9 T10 Kinemitic
V(volt) I(A) Tfilm Pr Re Nu
( Degree) ( C) ( C) vescosity
2.64E+0
0 26.3 93.4 3.125 59.85 1.89E-05 0.709 27.69267
3
2.66E+0
30 26.5 91.4 3.125 58.95 1.88E-05 0.709 27.76306
3
2.67E+0
60 26.5 89.5 3.130 58 1.87E-05 0.709 27.83798
3
2.88E+0
20 90 26.6 61.5 3.125 44.05 1.74E-05 0.711 29.02773
3
2.95E+0
120 26.7 53.1 3.120 39.9 1.70E-05 0.712 29.40994
3
2.98E+0
150 26.9 48.8 3.120 37.85 1.68E-05 0.712 29.59436
3
3.01E+0
180 26.9 44.8 3.120 35.85 1.66E-05 0.713 29.79377
3

Graph 1

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 T10 vs Qin
400
350 f(x) = 4.81422621344643 x + 82.7118589164764
300 R² = 0.959573397081289
250
Qin(W)
T10(◦C)

200
Linear (Qin(W))
150
100
50
0
0 10 20 30 40 50 60 70
Qin(W)

Graph 2

25.000 Hcm vs Ts
f(x) = 0.0254019840789541 x + 8.20797017194875 Hcm
20.000 R² = 0.952034793705384 (W/
m^...
Hcm (W/m^2K)

15.000

10.000

5.000

0.000
300.0 350.0 400.0 450.0 500.0 550.0 600.0 650.0 700.0
Ts(K)

Graph 3

Hrm vs Ts
30.000

25.000
f(x) = 0.0603611521963112 x − 14.725411598303
Hrm (W/m^2K)

20.000 R² = 0.981205635850657
Hrm (W/m^2K)
15.000
Linear (Hrm (W/m^2K))
10.000

5.000

0.000
300.0 400.0 500.0 600.0 700.0
Ts(K)

7
 Graph 4
T10 vs Uc
400
350
300
250
T10(◦C)

200 Uc(m/s)
150
100
50
0
0 1 2 3 4 5 6 7 8 9
Uc(m/s)

Graph 5
Hcm & Hrm vs Ts
700.0
600.0
500.0
400.0 Hcm (W/m^2K)
Ts(K)

300.0 Hrm (W/m^2K)

200.0
100.0
0.0
5.000 10.000 15.000 20.000 25.000 30.000
Hcm & Hrm (W/m^2K

Graph 6
T10 vs Position at Ua=1m/s
300
250
200
T10( C)

150 Position ( Degree)

100
50
0
0 20 40 60 80 100 120 140 160 180 200
Position ( Degree)

Graph 7

8
 T10 vs Position at Ua=5m/s
100
80
60
T10( C)
Position ( Degree)
40
20
0
0 20 40 60 80 100 120 140 160 180 200
Position ( Degree)

Graph 8
Nu vs Position at Ua=1m/s
12
11.5
11
10.5
Position ( Degree)
Nu

10
9.5
9
8.5
0 20 40 60 80 100 120 140 160 180 200
Position ( Degree)

Graph 9

Nu vs Position at Ua=5m/s
30
29.5
29
28.5
Position ( Degree)
Nu

28
27.5
27
26.5
0 20 40 60 80 100 120 140 160 180 200
Position ( Degree)

9
Discussion of Results

Experiment 1:

The comparison of Qc and Qr is in table 1 for all readings. The main difference
between Qinput and Qtot is that Qin is higher than Qtot. While increasing the voltage
and the temperature of the film, the input heat transfer is increasing. The value of H cm
obtained from the simplified empirical equation was less than from that of full
empirical equation with 25% difference. As it can be seen from graph1 the relation
between T10 and Qin is linearly independent.

Experiment 2:

The main reason behind the difference between Qinput and Qtot was that some of the
temperature lost to the surrounding giving low experimental heat transfer value.
From graph 4, it is clear that T10 is directly proportional to the velocity of the air Uc .

Experiment 3:

Graphs 6-9: Shows the relationship between the surface temperature T10 and the
angular position for each setting of air velocity Ua =1 m/s and 5 m/s and also shows
the relationship between the local Nusselt number and the angular position for each
setting of air velocity Ua =1 m/s and 5 m/s.Ts & Nu for both 1&5 m/s are randomly
change versus angular position so, we cannot determine any relation from the
graphs. When the volumetric flowrate increases, the inlet temperature of the hot fluid
slightly increases. However, there is a slight decrease in the outlet temperature of
the hot fluid as the flowrate increases.

Conclusions and Recommendations

Having concluded the experiment, it was noticed that there are two factors that can
affect the heat transfer rate. The difficulty in controlling the air velocity and the
temperature. Also the angular position has great effect on the heat transfer rate.

From an engineering point of view, it’s recommended that ability to control the air
velocity and temperature will give accurate results. It would be also suggested that
using digital rotator in order to give precise angel would minimize the error in the
heat transfer rate.

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Literature Cited

1. “Chemical Engineering Laboratory-I Manual”, 2nd Edition, Dhahran:

King Fahd University of Petroleum & Minerals, 2014. Pages M5-1 – M5-
7.

2. Perry, Robert H and Green, Don W. “Perry’s Chemical Engineers’

Handbook, 7th Edition, New York: McGrow-Hill, 1998.

Nomenclature
V Voltage to heated cylinder V
A
I Current to heated cylinder
Qin Power supplied to heated cylinder W
D Diameter of heated cylinder M
L Heated length of cylinder M
As Heat transfer area M2
Ua Air velocity in duct m/s
Uc Corrected air velocity m/s
Qc Heat loss due to natural convection W
Qf Heat loss due to forced convection W
Qr Heat loss due to radiation W
Qtot Total heat loss from cylinder W
Hc Heat transfer coefficient for natural convection W/m2K
Hf Heat transfer coefficient for forced convection W/m2K
Hr Heat transfer coefficient for radiation W/m2K
σ Stefan Boltzman constant W/m2K4

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 ξ Emissivity of cylinder -
F Area factor -
v Kinematic viscosity of air M2/s
K Thermal conductivity of air W/mK
Re Reynolds number -
Nu Nusselt number -
Pr Prandtl number -
θ Angular position of thermocouples Degrees
o
T10 Surface temperature of heated cylinder C
Ts Surface temperature of heated cylinder k
o
T9 Temperature of ambient/surroundings C
Ta Temperature of ambient/surroundings K
Tfilm Film temperature of air K

Appendices
A1. Raw Data
Table1: Log sheet for Experiment 1

12
Table2: Log sheet for Experiment 2

13
Table3: Log sheet for Experiment 3

14
A2. Analysis of Data and Sample Calculations

15
Sample Calculation:

Experimental 1:

Qc =H cm A s (T S −T a )
A s=πDL

k Num
H cm =
D
Num = c (RaD)n
G β ( T s−T a ) D3
RaD =
v2

T s +T a 347.25+294.95
T film= = =321.1 K
2 2

1 1 −3 −1
β= = =3.114 ×10 K
T film 321.1
3
G β ( T s−T a ) D (9.81)(3.114 × 10−3 )(347.25−294.95)(0.013)
RaD = = =1835.88
v2 (2.95 ×10−5)2

c= 0.850 & n = 0.188 from table 1

Num = c (RaD)n = 3.492
k= 0.0464 W/m.K

W
H cm =16.20 2
.K
m
−3
Qc =( 16.20 ) (2.199× 10 ) ( 347.25−294.95 ) =1.863W

Qr = Hrm As (Ts – Ta)

(T s4−T a4 ) ( 347.254 −294.954 )

H rm=σξF =( 56.7 × 10−9 ) ( 0.95 ) =7.18 W /m2 K
(T s−T a ) ( 347.25−294.95 )
Qr = (7.18)( 2.199 ×10−3 ) ( 347.25−294.95 )=0.809 W
Qtot= 0.809+1.863= 2.672 W
0.25
T s−T a
H cm =1.32( ) =20.31 W /m2 K
D

Experiment 2:

Qtot = Qc + Qr

Qr = Hrm As (Ts – Ta)

16
 4
(T s −T a )
4
( 606.554 −297.654 )
=( 56.7 × 10 ) ( 0.95 )
−9 2
H rm=σξF =22.23W /m K
(T s−T a ) ( 606.55−296.65 )

Qr = (22.23)( 2.199 ×10−3 ) ( 505.55−297.65 )=15.1W

Qfc = Hcm As (Ts – Ta)
0.25
T s−T a 2
H cm =1.32( ) =17.49 23 W /m K
D

Qfc = 11.88 W
Qfc exp= 15.6 W
Qtot= 26.98 W
Qtotexp= 30.711 W

Experiment 3:

( 0.62 ℜ0.5 Pr 0.33 ) ℜ

0.5
Num=0.3+ (
(1+ ) )

(( ( ) ))
0.4
0.66 0.25
282000
1+
Pr

for position 0 at Ua= 1 m/s

Tfilm = 408.5 K
Pr= 0.695
K= 0.0515 W/m.k
Ua= 1.06 m/s
U= 62.53×10−6 m2/s
Uad
ℜ= =3291.62
kv
Num = 30.937

17