Forced convection lab report

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

1

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 =

𝐶𝑃 𝜇 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 =

𝐶𝑃 𝜇 1007 × 184.6 × 10−7 = = 0.7068 𝑘 26.3 × 10−3

b. 𝑃𝑟 =

𝐶𝑃 𝜇 1004 × 230.1 × 10−7 = = 0.6903 𝑘 33.8 × 10−3

c. 𝑅𝑒 =

𝑈𝐷𝜌 5 × 0.02 × 1.1614 = = 6291 𝜇 184.6 × 10−7

d. 2

1

1

𝑃𝑟 4 0.7068 4 ̅𝑢𝐷 = 𝐶𝑅𝑒𝐷𝑚 𝑃𝑟 𝑛 ( ) = 0.26 × 62910.6 × 0.70680.37 × ( 𝑁 ) = 43.75 𝑃𝑟𝑆 0.6903 ℎ̅ =

̅𝑢𝐷 0.0263 × 43.75 𝑘𝑁 = = 72.8 𝐷 0.0158

5). From equation 15 we have 1 2 2∆𝑝𝐻2𝑂 𝜌𝑈 = ∆𝑝𝐻2𝑂 => 𝑈 = √ 2 𝜌 With the correction constant we have: 2∆𝑝𝐻2𝑂 2∆𝑝𝐻2𝑂 𝑅𝑇∞ 𝑈 = 𝐶𝐷 √ = 𝐶𝐷 √ 𝜌 𝑝𝑎𝑖𝑟 𝑀𝑎𝑖𝑟 Since 1mmH2O = 9.81 Pa we have 2 × 9.81 × 8.314 × ∆𝑝𝐻2𝑂 𝑇∞ ∆𝑝𝐻2𝑂 𝑇∞ 𝑈 = 0.98√ = 73.48√ 0.029 × 𝑝𝑎𝑖𝑟 𝑝𝑎𝑖𝑟

LOG SHEET Forced Convection Experiment Atmospheric pressure pair = 101000 Pa

25V Speed (Hz) Power P (W) Air temperature T∞ (oC) Surface temperature TS (oC) ΔpH2O (mmH2O) U h

20 8.93 22.3 48.4 18 16.86 137.83

25 8.93 22.6 46 27 20.66 153.73

30 8.93 22.8 44.1 41 25.47 168.89

35 8.93 22.9 42.6 51 28.41 182.61

40 8.93 23.1 41.6 65 32.08 194.45

35V Speed (Hz) Power P (W) Air temperature T∞ (oC) Surface temperature TS (oC) ΔpH2O (mmH2O) U h

20 17.50 22.9 74.4 17 16.40 136.91

25 17.50 23.1 66.8 28 21.05 161.34

30 17.50 23.2 64.4 40 25.17 171.14

35 17.50 23.4 60.3 50 28.15 191.08

40 17.50 23.6 58.2 64 31.85 203.78

3

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 16.86 × 15.8 × 10−3 × 1.18 = = 17192.8 𝜇 182.5 × 10−7

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 =

𝐶𝑃 𝜇 1006.875 × 182.5 × 10−7 = = 0.709 𝑘 25.9 × 10−3

Prandtl number at surface: 𝑃𝑟 =

𝐶𝑃 𝜇 1007.625 × 195 × 10−7 = = 0.701 𝑘 27.875 × 10−3

Calculated Nusselt number: 1

1

𝑃𝑟 4 0.709 4 𝑚 𝑛 ̅̅̅̅̅̅ 𝑁𝑢 ) = 0.26 × 17192.80.6 × 0.7090.37 × ( ) = 79.816 𝐷 = 𝐶𝑅𝑒𝐷 𝑃𝑟 ( 𝑃𝑟𝑆 0.701 Experimental Nusselt number at T∞: ̅̅̅̅̅̅ 𝑁𝑢 𝐷 =

ℎ̅𝐷 137.83 × 0.0158 = = 83.96 𝑘 25.936 × 10−3

4

25V 2.100 2.080 2.060

log10(Nu)

2.040 2.020 2.000 1.980

Theoretical Nu

1.960

Experimental Nu

1.940 1.920 1.900 1.880 4.200

4.250

4.300

4.350

4.400

4.450

4.500

4.550

log10(Re)

35V 2.120 2.100 2.080 2.060

log10(Nu)

2.040 2.020 2.000

Experimental Nu

1.980

Theoretical Nu

1.960 1.940 1.920 1.900 1.880 4.200

4.250

4.300

4.350

4.400

4.450

4.500

4.550

log10(Re)

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 5

̅𝑢. However the consistently higher than its actual value and hence higher experimental 𝑁 experimental model to calculate average Nusselt number is still within good range of agreement with the theoretical model.

6