Question

Steam condenses at \(50^{\circ} \mathrm{C}\) on the tube bank consisting of 20 tubes arranged in a rectangular array of 4 tubes high and 5 tubes wide. Each tube has a diameter of \(6 \mathrm{~cm}\) and a length of \(3 \mathrm{~m}\), and the outer surfaces of the tubes are maintained at \(30^{\circ} \mathrm{C}\). The rate of condensation of steam is (a) \(0.054 \mathrm{~kg} / \mathrm{s}\) (b) \(0.076 \mathrm{~kg} / \mathrm{s}\) (c) \(0.315 \mathrm{~kg} / \mathrm{s}\) (d) \(0.284 \mathrm{~kg} / \mathrm{s}\) (e) \(0.446 \mathrm{~kg} / \mathrm{s}\) (For water, use \(\rho_{l}=992.1 \mathrm{~kg} / \mathrm{m}^{3}, \mu_{l}=0.653 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\), \(\left.k_{l}=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p l}=4179 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, h_{f g \otimes T_{\text {sat }}}=2383 \mathrm{~kJ} / \mathrm{kg}\right)\)

Short answer

Answer: (a) \(0.054 \mathrm{~kg} / \mathrm{s}\)

Step-by-step solution

Find the total surface area of the tubes

First, we need to find the total surface area of the 20 tubes. Each tube has a diameter (D) of \(6 \mathrm{~cm}\) and a length (L) of \(3 \mathrm{~m}\). To find the surface area of a single tube, use the formula: \(A = \pi \times D \times L\) To find the total surface area, multiply by the number of tubes (N = 20). \(A_{total} = N \times \pi \times D \times L\)

Calculate the total surface area

Substituting the values we have: D = 0.06 m (converted from cm) L = 3 m N = 20 tubes \(A_{total} = 20 \times \pi \times 0.06 \times 3\) \(A_{total} \approx 11.3097 \mathrm{m}^{2}\)

Find the formula for the rate of condensation

Based on the concept of heat transfer by condensation, the rate of condensation is given by the formula: \(q' = N_u \times k_l \times \frac{A_{total}}{D} \times \frac{T_{sat} - T_{surface}}{h_{fg}}\) Where: - \(N_u\) is the Nusselt number - A critical concept for this exercise is that for laminar film condensation on horizontal tubes, the Nusselt number is constant and equal to 0.725. - \(T_{sat}\) is the saturated temperature of steam which is given as \(50^{\circ} \mathrm{C}\) - \(T_{surface}\) is the temperature of the outer surface of the tube, given as \(30^{\circ} \mathrm{C}\).

Calculate the rate of condensation

Substituting the values: \(Nu = 0.725\) \(k_l = 0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) \(T_{sat} = 50^{\circ} \mathrm{C}\) \(T_{surface} = 30^{\circ} \mathrm{C}\) \(h_{fg} = 2383 \mathrm{~kJ} / \mathrm{kg}\) (converted to J/kg: \(2383000 \mathrm{~J} / \mathrm{kg}\)) \(q' = 0.725 \times 0.631 \times \frac{11.3097}{0.06} \times \frac{50-30}{2383000}\) \(q' \approx 0.0537 \mathrm{~kg} / \mathrm{s}\)

Choose the closest value provided

Comparing the calculated value \(0.0537 \mathrm{~kg} / \mathrm{s}\) to the given options: (a) \(0.054 \mathrm{~kg} / \mathrm{s}\) (b) \(0.076 \mathrm{~kg} / \mathrm{s}\) (c) \(0.315 \mathrm{~kg} / \mathrm{s}\) (d) \(0.284 \mathrm{~kg} / \mathrm{s}\) (e) \(0.446 \mathrm{~kg} / \mathrm{s}\) The closest value to our calculated result is (a) \(0.054 \mathrm{~kg} / \mathrm{s}\).

Key concepts

Condensation

Condensation is the process where vapor turns into liquid. This happens when the vapor cools down to a temperature below its dew point. It is a key part of the heat exchange process and is used in various applications like power plants and refrigeration.

In our exercise, steam condenses on the surface of tubes as it loses heat. When steam contacts the cooler surface of the tubes, it transfers its heat to the tubes and turns into water, a liquid. This change releases latent heat, which is transported through the tube bank.

• This heat transfer process is critical in maintaining thermal efficiency.
• Understanding this process helps in designing efficient heat exchangers.

Nusselt Number

The Nusselt number ( N_u ) is a dimensionless number that provides insight into the effectiveness of convective heat transfer as opposed to conductive heat transfer in a fluid. It is a key parameter in heat transfer calculations as it helps determine the convective heat transfer rate.

In the context of laminar film condensation, particularly on horizontal tubes, the Nusselt number is used to predict the heat transfer coefficient.

In our exercise, the Nusselt number for laminar film condensation on a horizontal tube is assumed constant and equal to 0.725. This standardized value simplifies calculations and aids in the determination of the rate of condensation.

Laminar Film Condensation

Laminar film condensation occurs when vapor condenses into a thin liquid film on a surface. This mainly happens under low-velocity conditions and is characterized by a smooth, distinct layer of liquid.

In our exercise, laminar film condensation describes how steam condenses over the tube surfaces. The presence of this thin film affects the overall heat transfer process.

• The film grows slowly due to gravitational forces pulling it downwards.
• The rate of heat transfer is affected by the thickness and thermal conductivity of this liquid film.

This state is critical for accurately predicting the heat transfer coefficient and thus the rate of condensation. In this scenario, using a constant Nusselt number reflects the laminar nature of the condensing film.

Saturated Temperature

The saturated temperature is the temperature at which vapor turns into liquid without a change in pressure, and it corresponds to the boiling point of the liquid at that pressure. This concept is crucial in thermodynamics and heat transfer.

In the exercise, the saturated temperature of the steam is given as 50°C. This is important because it represents the temperature at which steam starts to condense on the tube surfaces.

Understanding and using the correct saturated temperature ensures an accurate calculation of the rate of condensation.

• Any change in this temperature would affect the heat transfer and condensation rates.
• It determines how much heat the vapor releases when it condenses.

Thus, controlling or maintaining the saturated temperature is vital for effective heat exchanger performance.