Objectives:To determine the convection heat transfer coefficient and to investigate for the

variation of Nusselt and Reynolds numbers with air flow past cylindrical tubes.

Data & Result:-

Data & Result for throttling opining

100%

Throttle Opening 100%P (mmH2O) =18.5 Ta=17 C+273=290 K

Time Temperature (T-Ta) Log10 (T-Ta)0 80 63 1.7995 76 59 1.77110 71 54 1.73215 66 49 1.69020 61 44 1.64325 56 39 1.59130 52 35 1.54435 49 32 1.50540 46 29 1.46245 42 25 1.39850 40 23 1.36255 37 20 1.30160 35 18 1.25570 31 14 1.14680 28 11 1.04190 26 9 0.954100 24 7 0.845110 22 5 0.69897120 21 4 0.60205999130 20 3 0.47712125140 19 2 0.30103150 19 2 0.30103160 18 1 0170 18 1 0

opening 100 %y = -0.0104x + 1.8558

0.000

0.200

0.400

0.600

0.800

1.000

1.200

1.400

1.600

1.800

2.000

0 20 40 60 80 100 120 140 160

time

log

(T-T

i)

Calculation:-

Slope=M =-0.0104 (from graph)

h = -2.3026(mc/A1) M = -2.3026*(0.1093*3800/0.00404)*-0.0104 = 246.19 W/m2.k

Nu=h d / k = 246.19*0.01242/25.6*10-3 = 122.3 V1=237.3(H1TA/PA) 0.5 = 237.3* (1.85*(17+273.15)/91.17*103)0.5

= 18.2 m/s

V=2V1 = 36.4 m/s

Re = ρ V d/μ = 1.2*36.4*0.01242 / (180.6*10-7) = 30039.1 => Turbulent

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Data & Result for throttling opining 70%

Throttle Opening 70%P (mmH2O) =11.5 Ta=17 C+273=290 K

Time Temperature (T-Ta) Log10 (T-Ta)0 80 63 1.7995 78 61 1.78510 75 58 1.76315 71 54 1.73220 67 50 1.69925 63 46 1.66330 59 42 1.62335 56 39 1.59140 53 36 1.55645 50 33 1.51950 47 30 1.47755 44 27 1.43160 42 25 1.39870 37 20 1.30180 34 17 1.23090 33 16 1.204100 29 12 1.079110 26 9 0.954120 25 8 0.903130 23 6 0.778140 22 5 0.69897150 20 3 0.477160 20 3 0.477170 19 2 0.301180 18 1 0.000

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Throttling 70 % y = -0.0091x + 1.9063

0.000

0.500

1.000

1.500

2.000

2.500

0 50 100 150 200

Time

log

(T

-Ti)

Calculation:-

Slope=M =-0.0091 (from graph)

h = -2.3026(mc/A1) M = -2.3026*(0.1093*380/0.00404 m2)*-0.0091 = 215.4 W/m2.k

Nu=h d / k = 215.4*0.01242/25.6*10-3 = 104.5

V1=237.3(H1TA/PA) 0.5 = 237.3* (1.15*(17+273.15)/91.17*103)0.5 = 14.3 m/s

V=2V1 = 28.6 m/s

Re = ρ V d/μ = 1.2*28.6*0.01242 / (180.6*10-7) = 23602.126 => Turbulent

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Data & Result for throttling opining 30%

Throttle Opening 30%P (mmH2O) =3 mm Ta=17 C+273=288 KTime Temperature (T-Ta) Log10 (T-Ta)

0 80 63 1.7995 79 62 1.79210 78 61 1.78515 76 59 1.77120 74 57 1.75625 71 54 1.73230 69 52 1.71635 66 49 1.69040 64 47 1.67245 61 44 1.64350 59 42 1.62355 57 40 1.60260 55 38 1.58070 51 34 1.53180 48 31 1.49190 45 28 1.447100 41 24 1.380120 39 22 1.342130 37 20 1.30103140 35 18 1.255150 33 16 1.204160 31 14 1.146170 29 12 1.079180 28 11 1.041190 27 10 1.000200 26 9 0.954210 25 8 0.903220 24 7 0.845

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throttling 30%y = -0.0044x + 1.8381

0.000

0.200

0.400

0.600

0.800

1.000

1.200

1.400

1.600

1.800

2.000

0 50 100 150 200 250

Time

log

(T-T

i)

Calculation:-

Slope=M =-0.0044 (from graph)

h = -2.3026(mc/A1) M = -2.3026*(0.1093*380/0.00404 m2)*-0.0044 = 104.15 W/m2.k

Nu=h d / k = 104.15*0.01242/25.6*10-3 = 50.53

V1=237.3(H1TA/PA) 0.5 = 237.3* (0.3*(17+273.15)/91.17*103)0.5 = 7.33 m/s

V=2V1 = 14.66 m/s

Re = ρ V d/μ = 1.2*14.66*0.01242 / (180.6*10-7) = 12098.15 => Turbulent

Throttle Opening % H1 (cmH2O) V1 (m/s) V (m/s) M*102 (Nu) (Re)100 1.85 18.2 36.4 -0.00104 122.3 30039.170 1.15 14.3 26.6 -0.0091 104.5 23602.12630 0.3 7.33 14.66 -0.0044 50.53 12098.15

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Nu Vs Re

0

20

40

60

80

100

120

140

0 5000 10000 15000 20000 25000 30000 35000

Re

Nu

Discussion & conclusion:-

From the results obtained we can conclude the following: As the throttle opening decreases, the static pressure increases thus the

stream velocity decreases. This yields to a decrease in Reynolds number.

As a result, it can be seen from the graph that Nusselt number is proportional to Reynolds number. .So any decrease in Reynolds number will cause a decrease in Nusselt number. Then the convection heat transfer coefficient will decrease. . h depends on both material physical properties and its geometry. This yield to decrease heat transfer.

In our experiment, the transient cooling conduction of the copper element was approximated using the lumped method.

Thus when plotting a log scale with time, the resulting curve would be linear:

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