Water at an average temperature of \(180^{\circ} \mathrm{F}\) and an average velocity of \(4 \mathrm{ft} / \mathrm{s}\) flows through a thin-walled \(\frac{3}{4}\)-indiameter tube. The water is cooled by air that flows across the tube with a velocity of \(12 \mathrm{ft} / \mathrm{s}\) at an average temperature of \(80^{\circ} \mathrm{F}\). Determine the overall heat transfer coefficient.

To calculate the heat transfer coefficient for water, we will use the following equation: \(h_{water} = k \frac{Nu}{D}\) Here, \(Nu\) is the Nusselt number, \(k\) is the fluid's thermal conductivity, and \(D\) is the tube's diameter. To find the Nusselt number, we will use the Sieder-Tate correlation: \(Nu = 0.027 \cdot Re^{0.8} \cdot Pr^{0.33}\) Reynolds Number (Re) and Prandtl Number (Pr) can be found using the given fluid properties (temperature and velocity). Please note that the properties of water may change slightly due to differing temperatures, so be prepared to adjust calculations accordingly. In this case, we assume constant properties for simplicity.

The process is similar to finding the heat transfer coefficient for water. We will use the Colburn analogy to find the Nusselt number for the air-cooled tubes: \(Nu = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3}\) You can then find the Reynolds and Prandtl numbers for air, as we did for water, using the given temperature and velocity data.

To find the surface areas, we need the tube's diameter and length. The diameter is given as \(\frac{3}{4}\) inches, which needs converting to feet: \(D_{inner} = \frac{3}{4} \cdot \frac{1 \mathrm{ft}}{12 \mathrm{in}}\) The outer diameter (D_outer) can be determined by assuming negligible wall thickness, so it is the same as D_inner. We will need to know the length (L) of the tube to calculate the surface area, which is not provided in the problem. We will use L as a variable in the next steps and find the heat transfer coefficients per unit length.

The Logarithmic Mean Temperature Difference (LMTD) is necessary to find the heat transfer rate. In this problem, we are given the average water and air temperatures, so we can use them directly to derive LMTD: \(LMTD = \frac{T_{water} - T_{air}}{\ln{\left(\frac{T_{water}}{T_{air}}\right)}}\)

We can now use the heat transfer coefficients, surface areas, and LMTD to find the overall heat transfer coefficient (U) using the following heat transfer rate equation: \(Q = U \cdot A \cdot LMTD\) Where Q is the heat transfer rate, A is the surface area, and LMTD is the logarithmic mean temperature difference. However, we will not find Q explicitly because it is not required in this problem. Instead, we will express U for per unit length of the tube to solve the problem. The overall heat transfer coefficient (U) can be related to the convection heat transfer coefficients for water (inside) and air (outside) as: \(\frac{1}{U \cdot A} = \frac{1}{h_{water} \cdot A_{inner}} + \frac{1}{h_{air} \cdot A_{outer}}\) Assuming that the inner and outer surface areas are equal (since the tube is thin-walled): \(U = \frac{1}{\frac{1}{h_{water}} + \frac{1}{h_{air}}}\) By substituting the convection heat transfer coefficients from steps 1 and 2 into this equation, we can find the overall heat transfer coefficient per unit length (U) for the given conditions.