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Heat transferMass transferTwo approachesExperimental or empirical approachTheoretical approach (solving the boundary layer equations for a particular geometry

Experiment for measuring the average convection heat transfer coefficient(5.1)

Dimensionless representation of convection heat transfer measurements(5.2)Film temperature(5.3)

(5.4)(5.5)(5.6)(5.7)

(5.8)(5.9)(5.10)(5.11)For fluid of small Prandtl number, namely liquid metal:Single correlating equation, which applies for all Prandtl number:(5.12)

(5.13)(5.15)(5.16)Expressions for the average coefficients may now be determined. However, since the turbulent boundary layer is generally preceded by a laminar boundary layer, we first consider mixed boundary layer conditions.(5.14)

(5.17)where it is assumed that transition occurs abruptly at x = xc.(5.18)(5.19)For a completely turbulent boundary layer (Rex = 0), A = 0.For a transition Reynolds number of Rex,c = 5x105, A =871.

(5.20)For laminar flow(5.21)For turbulent flow

Analogous mass transfer results are obtained by replacing (Nux, Pr) with (Shx, Sc).(5.22)where p = 2 for laminar flow and p = 8 for turbulent flow.

(5.23)For laminar flow(5.25)For turbulent flow(5.24)If the heat flux is known, the convection coefficient may be used to determine the local surface temperatureAverage surface temperaturewhere Nux is obtained from appropriate convection correlation. From eq. (5.23)(5.26)(5.27)

Friction DragReD < 2Separation effects negligibleForm or Pressure Drag

Re > 1x105Two minima

C and m from TablesProperties evaluated at film temperature

Overall AverageStagnation Point(5.29)Table 5.1(5.28)

Sheet1

ReDCm

0.4 to 40.9890.33

4 to 400.9110.385

40 to 4,0000.6830.466

4000 to 40,0000.1930.618

40,000 to 400,0000.0270.805

Table 5.2 Constant of Eq. 5.29 for noncircular cylinders in cross flow of a gas

C and m from TablesProperties evaluated at free stream temperature except PrsPrs evaluated at the surfacePr10 n=0.36(5.30)Table 5.3

Sheet1

ReDCm

1 to 400.750.4

40-10000.510.5

103-2x1050.260.6

2x105-1060.0760.7

Recommended for ReD Pr > 0.2Properties evaluated at film temperatureNo Tables

(5.31)Finally, we note that by invoking the heat and mass transfer analogy, Equations 5.28 through 5.31 may be applied to problems involving convection mass transfer from a cylinder in cross flow. It is simply a matter of replacing

Air at 100 kPa and 35oC flows across a 50-mm-diameter cylinder at a velocity of 50 m/s. The cylinder surface is maintained at a temperature of 150oC. Calculate the heat loss per unit length of the cylinder.We first calculate the Reynolds number and then find the applicable constants from Table 5.1 for use with Eq. 5.29.

From Table 5.1From Eq. 5.29The heat transfer per unit length is therefore

(5.32)

Air at 100 kPa and 27oC blows across a 12-mm-diameter sphere at a free stream velocity of 4 m/s. A small heater inside the sphere maintains the surface temperature at 77oC. Calculate the heat lost by the sphere.

From Eq. 5.32The heat transfer is then

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Note:AlignedStaggeredTransverse Pitch STLongitudinal Pitch SL

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(5.33)Table 5.4

Sheet1

ST/D

1.251.523

SL/DC1mC1mC1mC1m

Aligned

1.250.3480.5920.2750.6080.10.7040.06330.752

1.50.3670.5860.250.620.1010.7020.06780.744

20.4180.570.2990.6020.2290.6320.1980.648

30.290.6010.3570.5840.3740.5810.2860.608

Staggered

0.6------0.2130.636

0.9----0.4460.5710.4010.581

1--0.4970.558----

1.125----0.4780.5650.5180.56

1.250.5180.5560.5050.5540.5190.5560.5220.562

1.50.4510.5680.460.5620.4520.5680.4880.568

20.4040.5720.4160.5680.4820.5560.4490.57

30.310.5920.3560.580.440.5620.4280.574

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Note:Factor Airflow by 1.13 Pr1/3Properties evaluated at film temperature

(5.34)

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(5.35)Table 5.5

Sheet1

Correction Factor C2

NL123456789

Aligned0.640.80.870.90.920.940.960.980.99

Staggered0.680.750.830.890.920.950.970.980.99

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Note:Maximum velocity occurs in the plane of the tubesIt differs for aligned or staggered tubes

Mass conservationStaggeredAt A2At A1(5.36)(5.37)

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Note:Use Table for C and mProperties evaluated at inlet fluid temperature except PrsIt differs for aligned or staggered tubes

(5.38)Table 5.6

Sheet1

ConfigurationReD,maxCm

Aligned10-1020.80.4

Staggered10-1020.90.4

Aligned102-103appr as single cylinder

Staggered102-103appr as single cylinder

Aligned (ST/SL>0.7)103-2x1050.270.63

Staggered (ST/SL2)103-2x1050.40.6

Aligned2x105-2x1060.0210.84

Staggered2x105-2x1060.0220.84

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If NL < 20Use correction to previous relation

(5.39)Table 5.7

Sheet1

Correction Factor C2

NL123457101316

Aligned0.70.80.860.90.920.950.970.980.99

Staggered0.640.760.840.890.920.950.970.980.99

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In general heat transfer is enhanced by a more tortuous flow

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Log Mean Temperature DifferenceN is the total number of tubes in the bank.NT is the number of tubes in the transverse plane.To outlet temperature estimated from(5.40)(5.41)

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N is the total number of tubes in the bankFriction factor fCorrection factor c(5.42)(5.43)

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Friction factor fCorrection factor cFriction factor fCorrection factor c103106

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Friction factor fCorrection factor c

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A pressurized water unit is used as a space heater. It consists of a tube bundle with hot water flowing through the tubes. Air is blown over the tubes to extract the energy. The tubes are arranged in a staggered array and the exterior tube diameters are 16.4 mm. The longitudinal and transverse pitches are SL= 34.3 mm and ST=31.3 mm, respectively. There are seven rows with eight tubes in each as shown. The temperature of the tube surface is 70oC and the inlet air temperature is 15oC. The air is moving with a velocity of 6 m/s. Determine the air side convection coefficient, the rate of heat transfer and the pressure drop through the bundle.

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From Eq. 5.38, the air-side Nusselt number isSinceis greater thanThe maximum velocityoccurs on the transverse plane, A1. Hence from Eq.5.36

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it follows from Table 5.6 and 5.7 that From Eq. 5.41 and

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Hence from Eq.5.40andThe pressure drop may be obtained from Eq.5.43

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Hence with NL = 7Comments:1. With properties evaluated at Tf, it is readily found that ReD,max = 11,876. With ST/D = 2 and SL/D = 2, it follows from Tables 5.4 and 5.5 that C1 = 0.482, m = 0.556, and C2 = 0.97. From Eqs. 5.34 and 5.35, NuD = 86.7, and h = 144.8 W/m2 K. Values of h obtained from Eqs. 5.34 and 5.38 therefore agree to within 7%, which is well within their uncertainties.

2. Had DT = Ts - Ti been used in lieu of DTlm in Eq. 5.42, the heat rate would have been overpredicted by 11%.

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Gas flow in a bed of spheres(5.44)are the Colburn j factors defined by andNote:Properties evaluated at arithmetic mean of the fluid temperatures entering and leaving the bed

If the particles are at a uniform temperature Ts where Ap,t is the total surface area of the particles

(5.45)To outlet temperature estimated from(5.46)where Ac,b is the bed cross-sectional area

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