Water at free stream temperature of T∞ = 60 °C flows past a bank of tubes at a free stream velocity of U = 0.5m/s. The tubes in the bank are of diameter D = 0.018m each and arranged in a staggered arrangement with ST = 2D, SL = 2D. Determine the mean heat transfer coefficient if the tubes are

Question

Water at free stream temperature of [latex]T_{\infty}=60^{\circ} C[/latex] flows past a bank of tubes at a free stream velocity of U = 0.5m/s. The tubes in the bank are of diameter D = 0.018m each and arranged in a staggered arrangement with [latex]S_{T}[/latex] = 2D, [latex]S_{L}[/latex] = 2D. Determine the mean heat transfer coefficient if the tubes are maintained at a mean temperature of [latex]T_{w}=20^{\circ} C[/latex].

Accepted Answer

Step 1 The fluid properties, as usual, are evaluated at the mean temperature given by [latex]T_{m}=\frac{T_{\infty}+T_{w}}{2}=\frac{60+20}{2}=40^{\circ} C[/latex].

Density: [latex] ho_{m}=992.3 kg / m ^{3}[/latex]

Kinematic viscosity: [latex] u_{m}=6.564 imes 10^{-7} m ^{2} / s[/latex]

Thermal conductivity: [latex]k_{m}=0.630 W / m ^{\circ} C[/latex]

Prandtl number: [latex]P r_{m}=4.32[/latex]

The Prandtl numbers at the wall and free stream temperatures are

[latex]P r_{\infty}=2.967, \quad P r_{w}=6.957[/latex]

Step 2 The Reynolds number may be determined as

[latex]\operatorname{Re}_{D}=\frac{U D}{ u_{m}}=\frac{0.5 imes 0.018}{6.564 imes 10^{-7}}=13711[/latex]

The ratio of transverse to longitudinal pitch is given by

[latex]\frac{S_{T}}{S_{L}}=\frac{2 D}{2 D}=1[/latex]

Step 3 The constants in the Nusselt number correlation are chosen fromTable 14.2 as C = 0.35, m = 0.6. The Nusselt number is obtained using Eq. 14.55 as

Table 14.2 Constants for use with correlating Eq. 14.55

	Aligned		Staggered
[latex]R e_{D} ext { range }[/latex]	C	m	C	m
[latex]10-10^{2}[/latex]	0.8	0.4	0.9	0.4
[latex]10^{2}-10^{3}[/latex]	Use single tube formula
[latex]10^{3}-2 imes 10^{5}[/latex]	[latex]\frac{S_{T}}{S_{L}}<0.7[/latex]		[latex]\frac{S_{T}}{S_{L}}<2[/latex]
	Do not use		0.35	0.6
	[latex]\frac{S_{T}}{S_{L}}>0.7[/latex]		[latex]\frac{S_{T}}{S_{L}}>0.7[/latex]
	0.27	0.63	0.4	0.6
[latex]2 imes 10^{5}-10^{6}[/latex]	0.021	0.84	0.022	0.84

[latex]\overline{N u}_{D}=C R e_{D}^{m} \operatorname{Pr}^{0.36}\left(\frac{P r_{\infty}}{P r_{w}} ight)^{\frac{1}{4}}[/latex] (14.55)

[latex]\overline{N u}_{D}=0.35 imes 13711^{0.6} imes 4.32^{0.36}\left(\frac{2.967}{6.957} ight)^{0.25}=145.4[/latex]

Step 4 The mean heat transfer coefficient may then be obtained as

[latex]\bar{h}=\frac{\overline{N u}_{D} k_{m}}{D}=\frac{145.4 imes 0.630}{0.018}=5089 W / m ^{2 \circ} C[/latex]

Question 14.8: Water at free stream temperature of T∞ = 60 °C flows past a ...

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