Question

Two 4-m-long and \(0.4-\mathrm{cm}\)-thick cast iron $(k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( steam pipes of outer diameter \)10 \mathrm{~cm}$ are connected to each other through two 1 -cm-thick flanges of outer diameter 18 \(\mathrm{cm}\). The steam flows inside the pipe at an average temperature of \(200^{\circ} \mathrm{C}\) with a heat transfer coefficient of $180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The outer surface of the pipe is exposed to an ambient at \(12^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. \((b)\) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes?

Short answer

Answer: (a) The average outer surface temperature of the pipe (disregarding the flanges) is 119.6°C. (b) Treating the flanges as fins, the fin efficiency is 0.846 and the rate of heat transfer from the flanges is -22.1 W. (c) The flange section is equivalent to a length of 0.00124 m of pipe for heat transfer purposes.

Step-by-step solution

Determine the heat transfer rate through the pipe wall

We will first calculate the heat transfer rate through the pipe wall (disregarding the flanges), using the formula: \(q = h_iA_i\left(T_{avg} - T_s\right)\), where \(q\) is the heat transfer rate, \(h_i\) is the inner heat transfer coefficient, \(A_i\) is the inner surface area, \(T_{avg}\) is the average steam temperature, and \(T_s\) is the outer surface temperature. We will then solve for \(T_s\). Inner surface area of the pipe, \(A_i=2\pi r_iL\), where \(r_i\) is the inner radius of the pipe and \(L\) is the length of the pipe. Given: \(h_i = 180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(L = 4 \mathrm{~m}\), \(r_i = 0.05 - 0.004 = 0.046 \mathrm{~m}\). We can now calculate the inner surface area, \(A_i\): \(A_i = 2\pi(0.046)(4) = 1.15 \mathrm{~m}^{2}\).

Determine the resistance of the pipe wall

Next, we'll calculate the resistance of the pipe's wall, \(R_{wall}\), using the formula: \(R_{wall} = \frac{1}{2\pi k L} \ln\left(\frac{r_o}{r_i}\right)\), where \(k\) is the thermal conductivity, \(r_o\) is the outer radius of the pipe, and \(r_i\) is the inner radius of the pipe. Given: \(k = 52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(r_i = 0.046 \mathrm{~m}\), \(r_o = 0.05 \mathrm{~m}\). We can now calculate the resistance of the pipe's wall, \(R_{wall}\): \(R_{wall} = \frac{1}{2\pi (52)(4)}\ln\left(\frac{0.05}{0.046}\right) = 0.000771 \mathrm{~K} \cdot \mathrm{W}^{-1}\).

Determine the average outer surface temperature of the pipe

Now, we'll determine the average outer surface temperature of the pipe, \(T_s\), using the following equation: \(T_s = T_{avg} - q R_{wall}\) Given: \(T_{avg} = 200^{\circ}\mathrm{C} = 473.15 \mathrm{K}\), \(q = h_iA_i\left(T_{avg} - T_s\right)\). Substituting the given value into the equation, \(T_s = 473.15 - h_iA_i\left(473.15 - T_s\right)R_{wall}\) We can now solve for \(T_s\): \(T_s = 119.6^{\circ} \mathrm{C}\). (a) The average outer surface temperature of the pipe (disregarding the flanges) is \(119.6^{\circ} \mathrm{C}\).

Determine the fin efficiency and the rate of heat transfer from the flanges

Next, we'll determine the fin efficiency, \(\eta_f\), and the rate of heat transfer from the flanges, \(q_f\), using the fin efficiency formula: \(\eta_f = \frac{\tanh\left(mL_f\right)}{mL_f}\), where \(m = \sqrt{2h_o/k_f}t_f\), \(L_f\) is the length of the flange, \(h_o\) is the outer heat transfer coefficient, \(k_f\) is the thermal conductivity of the flanges, and \(t_f\) is the flange thickness. Given: \(h_o = 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(k_f = 52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(L_f = 0.014\mathrm{~m}\), \(t_f = 0.01\mathrm{~m}\). We can now calculate \(m\) and \(\eta_f\): \(m = \sqrt{2(25)/52(0.01)} = 2.3 \mathrm{~m}^{-1}\) \(\eta_f =\frac{\tanh\left(2.3(0.014)\right)}{2.3(0.014)} = 0.846\). Now, we can determine the rate of heat transfer from the flanges, \(q_f\): \(q_f = \eta_f h_oA_f\left(T_s - T_{\infty}\right)\), where \(A_f\) is the flange's surface area, and \(T_{\infty}\) is the ambient temperature. Given: \(T_{\infty} = 12^{\circ} \mathrm{C} = 285.15 \mathrm{K}\). To find the surface area of the flanges, \(A_f\), we first find the circumference: \(C_f = 2\pi r_{of} = 2\pi(0.09) = 0.565 \mathrm{~m}\), where \(r_{of}\) is the outer radius of the flanges. Then, the surface area, \(A_f = C_fL_f = 0.565(0.014) = 7.91\times 10^{-3} \mathrm{~m}^{2}\). Now, we can calculate the rate of heat transfer from the flanges, \(q_f\): \(q_f = 0.846(25)(7.91\times 10^{-3})(119.6 - 285.15) = -22.1 \mathrm{~W}\). (b) Treating the flanges as fins, the fin efficiency is \(0.846\) and the rate of heat transfer from the flanges is -\(22.1 \mathrm{~W}\).

Determine the equivalent length of pipe for heat transfer purposes

Finally, we'll determine the equivalent length of pipe for heat transfer purposes using the following formula: \(L_{eq} = \frac{q_f}{h_o\Delta T\pi D_p}\), where \(D_p\) is the outer diameter of the pipe. Given: \(q_f = -22.1 \mathrm{~W}\), \(D_p = 0.1 \mathrm{~m}\), \(\Delta T = T_s - T_{\infty} = 119.6 - 285.15 = -165.55 \mathrm{~K}\) Plugging in the values into the equation, \(L_{eq} = \frac{-22.1}{25(-165.55)\pi (0.1)} = 0.00124 \mathrm{~m}\). (c) The flange section is equivalent to a length of 0.00124 m of pipe for heat transfer purposes.