Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, \(1.7\)-m-long pipe with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). A refrigerant at \(-15^{\circ} \mathrm{C}\) flows inside the pipe and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. Air properties at the average temperature are \(k=0.0240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.735\), \(\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The rate of heat transfer to the pipe is (a) \(343 \mathrm{~W}\) (b) \(419 \mathrm{~W}\) (c) \(485 \mathrm{~W}\) (d) \(547 \mathrm{~W}\) (e) \(610 \mathrm{~W}\)

First, we need to compute the Reynolds number (\(Re\)) of the airflow around the pipe: $$Re = \frac{Vd}{\nu}$$ where \(V = 4 \mathrm{~m} / \mathrm{s}\) is the airflow velocity, \(d = 0.05 \mathrm{~m}\) is the diameter of the pipe, and \(\nu = 1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) is the kinematic viscosity. Plugging in the values, we get: $$Re = \frac{4 \times 0.05}{1.382 \times 10^{-5}}$$

Now that we have the Reynolds number, we can determine if the airflow around the pipe is laminar or turbulent. The critical Reynolds number for pipes is \(Re_{crit} = 2300\). If \(Re < 2300\), the flow is laminar, and if \(Re > 2300\), the flow is turbulent. In our case: $$Re > 2300$$ so the flow is turbulent.

For turbulent flow over an isothermal cylinder, the appropriate correlation to determine the Nusselt number (\(Nu\)) is: $$Nu = 0.3 + \frac{0.62Re^{0.5} Pr^{1/3}}{[1 + (0.4 / Pr)^{2/3}]^{1/4}}$$ Next, plug in the given values for \(Re\) and \(Pr\), and calculate the Nusselt number: $$Nu = 0.3 + \frac{0.62Re^{0.5} Pr^{1/3}}{[1 + (0.4 / Pr)^{2/3}]^{1/4}}$$

Now we need to calculate the convective heat transfer coefficient (\(h\)) from the Nusselt number using the formula: $$h = \frac{Nu \times k}{d}$$ where \(k = 0.0240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is the thermal conductivity of the air and \(d\) is the diameter of the pipe. Plug in the values and solve for \(h\): $$h = \frac{Nu \times 0.0240}{0.05}$$

Finally, we can determine the heat transfer rate from the air to the refrigerant inside the pipe using the formula: $$Q = hA(T_{air} - T_{refrigerant})$$ where \(A = \pi dL\) is the surface area of the pipe, with \(L = 1.7 \mathrm{~m}\) being the length of the pipe, and \(T_{air} = 25^{\circ}\mathrm{C}\) and \(T_{refrigerant} = -15^{\circ}\mathrm{C}\) are the temperatures of the air and refrigerant, respectively. Plug in the values and solve for \(Q\): $$Q = h(\pi \times 0.05 \times 1.7)(25 - (-15)) \mathrm{~W}$$ Compare the calculated heat transfer rate \(Q\) with the options given and choose the closest value. (a) \(343 \mathrm{~W}\) (b) \(419 \mathrm{~W}\) (c) \(485 \mathrm{~W}\) (d) \(547 \mathrm{~W}\) (e) \(610 \mathrm{~W}\)