A brass tube of 6mm ID and 7mm OD has short integral fins of height 7mm and 0.3mm thickness arranged with a spacing of 3mm. The heat transfer coefficient to the hot fluid at 75^◦C flowing inside the tube is 270W/m^2^◦C. The fins and the external surface of the tube are exposed to air at 20^◦C with

Question

A brass tube of 6mm ID and 7mm OD has short integral fins of height 7mm and 0.3mm thickness arranged with a spacing of 3mm. The heat transfer coefficient to the hot fluid at 75°C flowing inside the tube is [latex] 270W/m^{2}°C[/latex] . The fins and the external surface of the tube are exposed to air at 20°C with a heat transfer coefficient of [latex] 45W/m^{2}°C[/latex] . Determine the heat transfer from a meter length of tube.

Accepted Answer

Step 1 The nomenclature for this problem is based on Fig. 4.20. The given data is written down.

Tube inner radius: [latex]r_{i}= 0.003 m[/latex]
Tube outer radius:[latex] r_{o} = 0.0035 m[/latex]
Fin length: [latex] L = 0.007 m[/latex]
Fin tip radius:[latex]r_{2} = r_{o} + L[/latex]
[latex]= 0.0035 + 0.007 = 0.0105 m[/latex]
Fin semi-thickness: [latex] t = 0.00015 m[/latex]
Fin spacing: [latex] S = 0.003 m[/latex]
Thermal conductivity of
tube and fin material:[latex] k = 111 W/m°C[/latex]
Tube side heat transfer coefficient: [latex]h_{1} = 270 W/m^{2} °C[/latex]
Fin side heat transfer coefficient: [latex]h_{2} = 45 W/m^{2} °C[/latex]
Tube side fluid temperature: [latex]T_{1} = 75 °C[/latex]
Fin side fluid temperature: [latex]T_{2} = 20 °C[/latex]

Step 2 Non-dimensional fin parameter is calculated as

[latex]\mu =\sqrt{\frac{h_{2} }{kt} } \cdot r_{2} =\sqrt{\frac{45}{111 imes 0.00015} }\cdot 0.0105=0.546[/latex]

Radius ratio that characterizes the fin is

[latex]\alpha =\frac{r_{2}}{r_{0}} =\frac{0.0105}{0.0035}=3[/latex]

Step 3 Fin efficiency is read off in Fig. 4.13 as η = 0.92. The various areas needed are calculated per meter length of tube basis. Since the fin spacing is S = 3mm or 0.003 m, the center to center distance between fins [latex]\acute{S}[/latex] is

[latex]\acute{S} = S + 2t = 0.003 + 2 × 0.00015 = 0.0033 m[/latex]

The number of fins per m length of tube is determined as

[latex]n=\frac{1}{\acute{S}} =\frac{1}{0.0033} \approx 303[/latex]

The above is obtained by rounding the number to an integer. The parent area per m of tube length is obtained as

[latex]A_{p}= 2πr_{o} = 2 × π × 0.0035 = 0.02199m^{2}[/latex]

Fin area per m of tube length is

[latex]A_{f} = 2π(r^{2}_{2} − r^{2}_{o} )n = 2 × π × (0.0105^{2} − 0.0035^{2}) × 303 = 0.1866m^{2}[/latex]

The total exposed area on the outside is

[latex]A_{t} = A_{f} + 2πr_{o}(n − 1)S[/latex]

[latex]= 0.1866 + 2 × π×0.0035 × (303 − 1) × 0.003 = 0.2065m^{2}[/latex]

Step 4 The overall surface efficiency may now be calculated, using Eq. 4.70 as

[latex]\eta _{0} =\frac{Q_{t} }{Q_{max}}=\left[1-\frac{A_{f} }{A_{t}}(1-\eta ) ight][/latex] (4.70)

[latex]\eta _{0} =\left[1-\frac{0.1866 }{0.2065}(1-0.92 ) ight] =0.928[/latex]

The three resistances that correspond to those shown in Fig. 4.20 are

[latex]R_{1} =\frac{1}{2\pi r_{i}h_{1}} =\frac{1}{2 imes \pi imes 0.003 imes 270} =0.1965m°C/W[/latex]

[latex]R_{2} =\frac{1}{2\pi k}\ln \left(\frac{r_{0} }{r_{2}} ight) =\frac{1}{2 \pi imes 111}\ln \left(\frac{0.0035}{0.003} ight) =0.1965m2.2103 imes 10^{-4}m°C/W[/latex]

[latex]R_{3} =\frac{1}{h_{2}A_{t}\eta _{0} } =\frac{1}{45 imes 0.2065 imes 0.928}=0.1160m°C/W[/latex]

The overall thermal resistance is then given by

[latex]R_{tot} = 0.1965 + 2.2103 × 10^{-4} + 0.1160 = 0.3127m°C/W[/latex]

Step 5 Tube heat loss per m of tube length is then given by

[latex]Q=\frac{T_{1}-T_{2}}{R_{tot}} =\frac{75-20}{0.3127}=175.9W/m[/latex]

Question 4.11: A brass tube of 6mm ID and 7mm OD has short integral fins of...

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