Question

Steam in a heating system flows through tubes whose outer diameter is $3 \mathrm{~cm}\( and whose walls are maintained at a temperature of \)120^{\circ} \mathrm{C}\(. Circular aluminum alloy fins \)(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( of outer diameter \)6 \mathrm{~cm}$ and constant thickness \(t=2 \mathrm{~mm}\) are attached to the tube, as shown in Fig. P3-201. The space between the fins is \(3 \mathrm{~mm}\), and thus there are 200 fins per meter length of the tube. Heat is transferred to the surrounding air at \(25^{\circ} \mathrm{C}\), with a combined heat transfer coefficient of $60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.

Short answer

Answer: The increase in heat transfer from the tube per meter of its length as a result of adding fins is 129.45 W/m.

Step-by-step solution

Calculate the surface area of the bare tube per meter of its length

To calculate the surface area of the bare tube, we can use the formula for the surface area of a cylinder per unit length: \(A_{tube} = 2\pi r L\) where \(r\) is the radius of the tube and \(L\) = 1m (since we're calculating per meter of length) Outer diameter = 3 cm, so the radius of the tube is 1.5 cm (divide by 2). Convert it to meters, 0.015 m. \(A_{tube} = 2\pi(0.015)(1)= 0.094 m^2/m\)

Calculate the heat transfer rate from the bare tube

Using the heat transfer coefficient given, we can find the rate of heat transfer from the bare tube with the following formula: \(q_{tube} = h A_{tube} \Delta{T}\) \(h\) = 60 W/m²·K (heat transfer coefficient), \(\Delta{T}\) = \(120^{\circ} - 25^{\circ}\) = 95 K (Temperature difference between the tube wall and surrounding air) \(q_{tube} = 60(0.094)(95) = 536.6 \mathrm{~W/m}\)

Calculate the fin efficiency

The efficiency of a single fin is given by the formula: \(\eta_{fin} = \tanh(mL)/ mL\) where, \(m = \sqrt{2h/(kt)}\) and \(L = r_2 - r_1\) In our case: \(k\) = 180 W/m·K (thermal conductivity of aluminum alloy), \(t\) = 2 mm = 0.002 m (thickness of the fins), \(h\) = 60 W/m²·K (heat transfer coefficient), \(r_1\) = 1.5 cm = 0.015 m (radius of tube), \(r_2\) = 3 cm = 0.03 m (outer radius of fins), \(L\)= \(0.03-0.015\)= 0.015 m (length of the fin) Calculate \(m\): \(m = \sqrt{2(60)/(180*0.002)}=8.66 \mathrm{m^{-1}}\) Now, calculate the fin efficiency: \(\eta_{fin} = \tanh(8.66*0.015)/ (8.66*0.015) = 0.850\)

Calculate the effective surface area of the fins

The effective area of a single fin is given by the formula: \(A_{fin} = 2 \pi (r_2 - r_1)L\) There are 200 fins per meter of tube length, so the total fin area per meter of tube length is: \(A_{fins} = 200 A_{fin}\) Now, calculate the effective fin area per meter of the tube: \(A_{fins} = 200(2\pi(0.03-0.015)(0.015))=0.0283 \mathrm{~m^2/m}\)

Calculate the heat transfer rate from the fins

Using the fin efficiency, the heat transfer rate from the fins per meter of the tube length can be calculated by: \(q_{fins} = h \eta_{fin} A_{fins} \Delta{T}\) Substitute the values and calculate: \(q_{fins} = 60(0.850)(0.0283)(95) \approx 129.45 \mathrm{~W/m}\)

Calculate the total heat transfer rate with fins

The total heat transfer rate with fins is the sum of the heat transfer rates from the bare tube and fins. \(q_{total} = q_{tube} + q_{fins}\) \(q_{total} = 536.6 + 129.45 = 666.05 \mathrm{~W/m}\)

Calculate the increase in heat transfer as a result of adding fins

The increase in heat transfer as a result of adding fins can be found by subtracting the heat transfer rate without fins from the total heat transfer rate with fins. \(\Delta{q} = q_{total} - q_{tube}\) \(\Delta{q} = 666.05 - 536.6 = 129.45 \mathrm{~W/m}\) The increase in heat transfer from the tube per meter of its length as a result of adding fins is 129.45 W/m.