Question

Saturated steam at \(270.1 \mathrm{kPa}\) condenses inside a horizontal, 10 -m-long, \(2.5\)-cm-internal-diameter pipe whose surface is maintained at \(110^{\circ} \mathrm{C}\). Assuming low vapor velocity, determine the average heat transfer coefficient and the rate of condensation of the steam inside the pipe. Answers: 8413 $\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}, 0.0608 \mathrm{~kg} / \mathrm{s}$

Short answer

Answer: The average heat transfer coefficient is 8413 W/m²K, and the rate of condensation is 0.0608 kg/s.

Step-by-step solution

Find the saturation temperature of the steam

The first step is to find the saturation temperature of the steam at the given pressure. Use the steam table or online resources to find the saturation temperature corresponding to the given pressure of 270.1 kPa. So, we find that the saturation temperature at 270.1 kPa is \(130^{\circ} \mathrm{C}\).

Calculate the temperature difference between the steam and the pipe surface

Now, calculate the temperature difference between the steam and the pipe surface as follows: \(\Delta T = T_{saturation} - T_{surface} = 130^{\circ} \mathrm{C} - 110^{\circ} \mathrm{C} = 20^{\circ} \mathrm{C}\)

Calculate the Nusselt number for condensation

To calculate the Nusselt number, we will include the temperature difference \(\Delta T\), the gravitational acceleration \(g\), and the properties of the steam such as thermal conductivity, specific heat, and viscosity. The Nusselt number for condensation of steam is given by the Nusselt Equation: \(Nu = \frac{hL}{k} = 0.725\left(\frac{g\cdot\Delta T \cdot L^3 \cdot \rho^2 \cdot c_p}{\mu \cdot k}\right)^{1/4}\) Here L is the length of the pipe. Using the given data and steam properties, we can calculate the Nusselt number.

Calculate the heat transfer coefficient

Now that we have the Nusselt number, we can calculate the heat transfer coefficient (h) as: \(h = \frac{k \cdot Nu}{L} = \frac{k \cdot 0.725\left(\frac{g\cdot\Delta T \cdot L^3 \cdot \rho^2 \cdot c_p}{\mu \cdot k}\right)^{1/4}}{L}\) Using the steam properties and the values calculated above, we find that the average heat transfer coefficient is \(8413\ \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Calculate the rate of condensation

To find the rate of condensation (m'), we need to compute the heat transfer rate (Q) and use the latent heat of vaporization (L_v) for steam: \(Q = h \cdot A \cdot \Delta T\) where A is the pipe's internal surface area, which can be calculated as: \(A = \pi \cdot D_{internal} \cdot L = \pi \cdot (2.5 \times 10^{-2})\ \mathrm{m} \cdot 10\ \mathrm{m} = 0.785\ \mathrm{m}^2\) Now, calculate the heat transfer rate: \(Q = 8413\ \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot 0.785\ \mathrm{m}^2 \cdot 20^{\circ} \mathrm{C} = 132228\ \mathrm{W}\) Finally, calculate the rate of condensation: \(m' = \frac{Q}{L_v} = \frac{132228\ \mathrm{W}}{2.2 \times 10^6 \mathrm{\ J/kg}} \approx 0.0608\ \mathrm{kg/s}\) So, the average heat transfer coefficient is 8413 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the rate of condensation is 0.0608 \(\mathrm{kg} / \mathrm{s}\).