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SCHOOL OF CHEMICAL AND BIOMEDICAL

ENGINEERING (Division of Chemical & Biomolecular Engineering)

Nanyang Technological

University

Yr 2 / SEMESTER 2

N1.2-B4-16

CH2702

Experiment C5

Forced Convection

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Name: Le Vu Anh Phuong Student ID: U1320848B Group: 14 Date: 3/2/15

Experiment description

The experiments aim to calculate the heat transfer coefficients h of forced convection for a

heated cylinder under cross air flow, and compare with that obtained from theoretical formula

with empirical corrections. Experimental h can be computed based on power delivered to the

heater, area of heat transfer and the temperature difference. To compare that with the

theoretical model, Nusselt numbers using experimental data and theoretical model are

calculated and plotted as a function of Reynolds numbers. From here h from both models can be

obtained and compared at each Re number.

Pre-laboratory problems:

1).

Forced convection: fluid movement caused by external forces such as a fan, pump, wind, ect.

Natural convection: fluid movement caused by its own density differences within the fluid body,

leading to buoyancy forces acting on fluid elements.

2).

𝑁𝑢 =ℎ𝐿

𝑘

𝑅𝑒 =𝑈𝑅𝜌

𝜇

𝑃𝑟 =𝐶𝑃𝜇

𝑘

All are dimentionless

3).

Nusselt number: the measure of convection heat transfer

Reynolds number: the ratio of inertial force to viscous force in fluid

Prandtl number: ratio of momentum diffusivity to thermal diffusivity.

4).

a.

𝑃𝑟 =𝐶𝑃𝜇

𝑘=

1007 × 184.6 × 10−7

26.3 × 10−3= 0.7068

b.

𝑃𝑟 =𝐶𝑃𝜇

𝑘=

1004 × 230.1 × 10−7

33.8 × 10−3= 0.6903

c.

𝑅𝑒 =𝑈𝐷𝜌

𝜇=

5 × 0.02 × 1.1614

184.6 × 10−7= 6291

d.

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�̅�𝑢𝐷 = 𝐶𝑅𝑒𝐷𝑚𝑃𝑟𝑛 (

𝑃𝑟

𝑃𝑟𝑆)

14

= 0.26 × 62910.6 × 0.70680.37 × (0.7068

0.6903)

14

= 43.75

ℎ̅ =𝑘�̅�𝑢𝐷

𝐷=

0.0263 × 43.75

0.0158= 72.8

5).

From equation 15 we have

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2𝜌𝑈2 = ∆𝑝𝐻2𝑂 => 𝑈 = √

2∆𝑝𝐻2𝑂

𝜌

With the correction constant we have:

𝑈 = 𝐶𝐷√2∆𝑝𝐻2𝑂

𝜌= 𝐶𝐷√

2∆𝑝𝐻2𝑂𝑅𝑇∞

𝑝𝑎𝑖𝑟𝑀𝑎𝑖𝑟

Since 1mmH2O = 9.81 Pa we have

𝑈 = 0.98√2 × 9.81 × 8.314 × ∆𝑝𝐻2𝑂𝑇∞

0.029 × 𝑝𝑎𝑖𝑟= 73.48√

∆𝑝𝐻2𝑂𝑇∞

𝑝𝑎𝑖𝑟

LOG SHEET Forced Convection Experiment

Atmospheric pressure pair = 101000 Pa

25V Speed (Hz) 20 25 30 35 40

Power P (W) 8.93 8.93 8.93 8.93 8.93

Air temperature T∞ (oC) 22.3 22.6 22.8 22.9 23.1

Surface temperature TS (oC) 48.4 46 44.1 42.6 41.6

ΔpH2O (mmH2O) 18 27 41 51 65

U 16.86 20.66 25.47 28.41 32.08

h 137.83 153.73 168.89 182.61 194.45

35V Speed (Hz) 20 25 30 35 40

Power P (W) 17.50 17.50 17.50 17.50 17.50

Air temperature T∞ (oC) 22.9 23.1 23.2 23.4 23.6

Surface temperature TS (oC) 74.4 66.8 64.4 60.3 58.2

ΔpH2O (mmH2O) 17 28 40 50 64

U 16.40 21.05 25.17 28.15 31.85

h 136.91 161.34 171.14 191.08 203.78

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Sample calculation: 20 Hz, 25 V

Air temperature T∞(oC): 22.3

Surface temperature TS (oC): 48.4

Duct air velocity U (m/s): 16.86

Mass density of air ρ at T∞ (kg/m3): 1.18

µ viscosity of air at T∞ (kg/s.m): 182.5x10-7

µ viscosity of air at TS (kg/s.m): 195x10-7

Reynolds number

𝑅𝑒𝐷 =𝑈𝐷𝜌

𝜇=

16.86 × 15.8 × 10−3 × 1.18

182.5 × 10−7= 17192.8

Air thermal conductivity k in flow T∞ (W/m.K): 25.9x10-3

Air thermal conductivity k at surface TS (W/m.K): 27.875x10-3

Specific heat CP of air in flow T∞ (J/kg.K): 1006.875

Specific heat CP of air at surface TS (J/kg.K): 1007.625

Prandtl number in fluid:

𝑃𝑟 =𝐶𝑃𝜇

𝑘=

1006.875 × 182.5 × 10−7

25.9 × 10−3= 0.709

Prandtl number at surface:

𝑃𝑟 =𝐶𝑃𝜇

𝑘=

1007.625 × 195 × 10−7

27.875 × 10−3= 0.701

Calculated Nusselt number:

𝑁𝑢𝐷̅̅ ̅̅ ̅̅ = 𝐶𝑅𝑒𝐷

𝑚𝑃𝑟𝑛 (𝑃𝑟

𝑃𝑟𝑆)

14

= 0.26 × 17192.80.6 × 0.7090.37 × (0.709

0.701)

14

= 79.816

Experimental Nusselt number at T∞:

𝑁𝑢𝐷̅̅ ̅̅ ̅̅ =

ℎ̅𝐷

𝑘=

137.83 × 0.0158

25.936 × 10−3= 83.96

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Discussion and conclusion

From both sets of experiments, both experimental and theoretical Nusselt number follow a

linear relation with the Reynolds number. The experimental Nu graph is consistently higher but

closed to the theoretical values. The discrepancy between them could have been due to

experimental errors. For instance the power delivered by the electrical source to the cylindrical

heater may be less than what indicated from the voltmeter, possibly due to internal resistance

of the instrument causing heat loss. This makes the calculated heat transfer coefficient

1.880

1.900

1.920

1.940

1.960

1.980

2.000

2.020

2.040

2.060

2.080

2.100

4.200 4.250 4.300 4.350 4.400 4.450 4.500 4.550

log1

0(N

u)

log10(Re)

25V

Theoretical Nu

Experimental Nu

1.880

1.900

1.920

1.940

1.960

1.980

2.000

2.020

2.040

2.060

2.080

2.100

2.120

4.200 4.250 4.300 4.350 4.400 4.450 4.500 4.550

log1

0(N

u)

log10(Re)

35V

Experimental Nu

Theoretical Nu

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consistently higher than its actual value and hence higher experimental �̅�𝑢. However the

experimental model to calculate average Nusselt number is still within good range of agreement

with the theoretical model.