Question

Steam is to be condensed on the shell side of a 1 -shellpass and 8-tube-passes condenser, with 50 tubes in each pass, at \(30^{\circ} \mathrm{C}\left(h_{f g}=2431 \mathrm{~kJ} / \mathrm{kg}\right)\). Cooling water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(15^{\circ} \mathrm{C}\) at a rate of \(1800 \mathrm{~kg} / \mathrm{h}\). The tubes are thin-walled, and have a diameter of \(1.5 \mathrm{~cm}\) and length of \(2 \mathrm{~m}\) per pass. If the overall heat transfer coefficient is \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine \((a)\) the rate of heat transfer and \((b)\) the rate of condensation of steam.

Short answer

Question: Calculate (a) the rate of heat transfer and (b) the rate of condensation of steam in a condenser. Solution: (a) The rate of heat transfer is 169113.3 W. (b) The rate of condensation of steam is 250.2 kg/h.

Step-by-step solution

Calculate the rate of heat transfer

The rate of heat transfer (Q) can be found using the following formula: \[ Q = U \cdot A \cdot \Delta T \] Where: - Q is the rate of heat transfer - U is the overall heat transfer coefficient - A is the surface area of the tubes - ΔT is the temperature difference between the cooling water and the steam First, let's calculate the total surface area of the tubes (A) using the given diameter (D) and length (L) per pass, as well as the total number of tubes (N_tubes): \[ A = N_{tubes} \cdot \pi D \cdot L \] Using the given values, we have: \[ A = 8 \times 50 \times \pi \times 0.015 \mathrm{m} \times 2\mathrm{m} = 3.7699\mathrm{m}^2 \] Next, we need to calculate the temperature difference (ΔT) between the water and the steam. As the initial temperature of the cooling water is given, and we know the condensing temperature of the steam, we can use these directly: \[ \Delta T = 30^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C} = 15 \mathrm{K} \] Now, we can substitute these values into the heat transfer equation to obtain the rate of heat transfer (Q): \[ Q = 3000 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}} \cdot 3.7699 \mathrm{m}^2 \cdot 15\mathrm{K} = 169113.3\mathrm{W} \] So, \((a)\) the rate of heat transfer is 169113.3 W.

Calculate the rate of condensation of steam

The rate of condensation of steam (m) can be found by dividing the rate of heat transfer by the specific enthalpy of vaporization (h_fg): \[ m = \frac{Q}{h_{fg}} \] We are given that the specific enthalpy of vaporization is 2431 kJ/kg, which we can convert to W/kg: \[ h_{fg} = 2431 \frac{\mathrm{kJ}}{\mathrm{kg}} \cdot \frac{1000 \mathrm{W}}{1 \mathrm{kJ}} = 2431000 \frac{\mathrm{W}}{\mathrm{kg}} \] Now, we can substitute these values into the above formula to obtain the rate of condensation of steam (m): \[ m = \frac{169113.3\mathrm{W}}{2431000 \frac{\mathrm{W}}{\mathrm{kg}}} = 0.0695 \frac{\mathrm{kg}}{\mathrm{s}} \] To convert the rate of condensation to kg/h, we multiply by the conversion factor between seconds and hours: \[ m_{\mathrm{h}} = 0.0695 \frac{\mathrm{kg}}{\mathrm{s}} \cdot \frac{3600 \mathrm{s}}{1 \mathrm{h}} = 250.2 \frac{\mathrm{kg}}{\mathrm{h}} \] So, \((b)\) the rate of condensation of steam is 250.2 kg/h.

Key concepts

Condenser Design

In a thermal system, a condenser plays a vital role by converting vapor into liquid, primarily by heat removal through a cooling medium. A well-designed condenser, such as the one in this exercise, maximizes heat exchangement effectiveness. This particular design involves a 1-shell pass and 8-tube passes configuration, which means the steam flows once through the shell, while the cooling water passes through the tubes eight times.
This setup enhances heat transfer efficiency by maximizing contact time and area between the steam and cooling water. With 50 tubes per pass, a total of 400 tubes facilitate efficient heat dissipation.
Key design parameters include tube diameter, length, and material, all contributing to effective heat transfer. Proper selection of these parameters ensures efficient energy transfer and optimal condensation performance.

Rate of Condensation

The rate of condensation is a measure of how quickly vapor, like steam, is converted into liquid. This process is heavily reliant on the rate of heat transfer within the system.
To calculate the condensation rate, the heat removed from the steam must be known, as this determines how much steam is condensed. Dividing this rate of heat transfer by the latent heat of vaporization provides the amount of steam converted to liquid per unit time.
This exercise showed a condensation rate of 0.0695 kg/s, equating to 250.2 kg/h. By understanding this rate, engineers can optimize condenser operations, ensuring that the cooling system meets the required performance without exceeding water capacity or energy costs.

Overall Heat Transfer Coefficient

In a heat exchanger, the overall heat transfer coefficient, symbolized as 'U', is crucial for determining how effectively heat is conducted through the materials separating hot and cold fluids.
This coefficient is affected by factors like material conductivity, surface area, and temperature differences. In the exercise, a U-value of 3000 W/m²•K signifies efficient thermal conductance, indicating that the condenser's design supports rapid heat exchange.
An accurate determination of 'U' helps in calculating the heat transfer rate, crucial for sizing components and optimizing thermal efficiency. Such precision allows for predictability in system performance, ensuring that heat exchangers provide the desired cooling effects.

Latent Heat of Vaporization

Latent heat of vaporization represents the amount of energy required to convert a unit mass of a liquid into vapor without changing its temperature. This property is fundamental in processes involving phase change, such as condensation.
In the given exercise, the latent heat of vaporization for steam at 30°C is 2431 kJ/kg. This value informs the heat energy necessary to convert each kilogram of steam to water.
Understanding this concept allows engineers to calculate exactly how much heat needs to be removed in the process of condensation, which in turn dictates the condenser capacity and performance requirements, balancing energy efficiency with system demands.