Question

A heat exchanger is used to condense steam coming off the turbine of a steam power plant by cold water from a nearby lake. The cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the condenser at \(16^{\circ} \mathrm{C}\) at a rate of \(20 \mathrm{~kg} / \mathrm{s}\) and leaves at \(25^{\circ} \mathrm{C}\), while the steam condenses at \(45^{\circ} \mathrm{C}\). The condenser is not insulated, and it is estimated that heat at a rate \(8 \mathrm{~kW}\) is lost from the condenser to the surrounding air. The rate at which the steam condenses is (a) \(0.282 \mathrm{~kg} / \mathrm{s}\) (b) \(0.290 \mathrm{~kg} / \mathrm{s}\) (c) \(0.305 \mathrm{~kg} / \mathrm{s}\) (d) \(0.314 \mathrm{~kg} / \mathrm{s}\) (e) \(0.318 \mathrm{~kg} / \mathrm{s}\)

Short answer

Answer: The rate at which steam condenses in the heat exchanger is \(0.290 \: \mathrm{kg/s}\).

Step-by-step solution

Calculate the Enthalpy Difference for the Water Stream

First, we need to find the enthalpy difference (∆H) for the water entering and leaving the condenser. To do this, we can use the formula: \(\Delta H = m_w \cdot c_p \cdot (T_{out} - T_{in})\) Where \(m_w\) is the water mass flow rate, \(c_p\) is the specific heat capacity of the water, and \(T_{out}\) and \(T_{in}\) are the outlet and inlet water temperatures, respectively. Using the given values, \(\Delta H = 20 \: kg/s \cdot 4.18 \: kJ/kg \cdot K \cdot (25^{\circ}C - 16^{\circ}C)\) \(\Delta H = 1502.8 \: \mathrm{kJ/s}\)

Apply the Conservation of Energy Equation

Now we will apply the conservation of energy equation to the problem. The sum of the heat inputs must equal the sum of the heat outputs. In this case, the heat input is from the condensing steam, while the heat outputs are due to the cooling water and the heat loss to surrounding air. The equation for this is: \(q_{steam} = \Delta H + q_{loss}\) Where \(q_{steam}\) is the heat transfer from the condensing steam, \(\Delta H\) is the enthalpy difference calculated in step 1, and \(q_{loss}\) is the heat loss to the surroundings. Using the values calculated above as well as the given 8 kW heat loss, \(q_{steam} = 1502.8 \: \mathrm{kJ/s} + 8 \: \mathrm{kW}\) \(q_{steam} = 1510.8 \: \mathrm{kJ/s}\)

Calculate the Condensation Rate of Steam

The final step is to calculate the condensation rate (\(m_s\)) of the steam, which could be found using the formula: \(m_s = \frac{q_{steam}}{c_p \cdot (T_s - T_{in})}\) Using the values calculated above, \(m_s = \frac{1510.8 \: \mathrm{kJ/s}}{4.18 \: \mathrm{kJ/kg} \cdot K \cdot (45^{\circ}C-16^{\circ}C)}\) \(m_s = 0.290 \: \mathrm{kg/s}\) Therefore, the correct answer is (b) \(0.290 \: \mathrm{kg/s}\).

Key concepts

Enthalpy Difference Calculation

Understanding the concept of enthalpy difference is crucial when studying heat exchangers and other thermodynamic systems. Enthalpy, often denoted as 'H', is a measure of total energy in a thermodynamic system. It includes both internal energy and the energy required to make space for it in the environment (pressure-volume work).

When cold water flows through a heat exchanger to absorb heat from steam, its temperature increases. This process changes the water's enthalpy. The enthalpy difference calculation is a way to quantify the energy exchanged via heat between two substances at different temperatures.

In the given exercise, we established the enthalpy difference by multiplying the mass flow rate of water (\(m_w\text{ in kg/s}\)), the specific heat capacity of water (\(c_p\text{ in kJ/kg·K}\)), and the temperature change of water (\(T_{out} - T_{in}\text{ in K}\)). Specific heat capacity is an intensive property that represents the amount of heat required to raise the temperature of a unit mass of a substance by one degree.

Conservation of Energy Equation

The conservation of energy principle, also known as the first law of thermodynamics, asserts that energy cannot be created or destroyed in an isolated system. In the context of the exercise, we're dealing with a heat exchanger where energy is transferred, not created or lost. The energy that goes into the system has to equal the energy that leaves it. This is crucial for calculating the efficiency of the heat exchanger.

The equation applied in the exercise puts this principle into practice. It sets the heat absorbed by the cold water (\(ΔH\text{ in kJ/s}\)) plus the heat loss to the surroundings (\(q_{loss}\text{ in kW}\)) equal to the heat released by the steam (\(q_{steam}\text{ in kJ/s}\)). Heat loss can occur due to lack of insulation or other inefficiencies.

An important takeaway is that in real-world scenarios, not all the heat from the steam is transferred to the cold water. Some energy is always lost to the surroundings, and an effective heat exchanger design aims to minimize such losses to increase efficiency.

Condensation Rate of Steam

The condensation rate of steam is a term that describes how quickly steam turns into liquid water when it loses energy. This is an essential variable in many industrial applications, including power plants and HVAC systems.

In the exercise, we calculated the condensation rate by utilizing the conservation of energy equation and the concept of enthalpy. By dividing the heat transfer from the condensing steam (\(q_{steam}\text{ in kJ/s}\)) by the product of the specific heat capacity of steam at constant pressure (\(c_p\text{ in kJ/kg·K}\)) and the temperature difference between the steam and entering water (\(T_s - T_{in}\text{ in K}\)), we determine the mass of steam that can condense per second.

The calculated condensation rate connects the thermal energy transfer with the phase change of the steam. Phase change, in this case, the process of condensation, is heavily influenced by both temperature and pressure. The specific enthalpy of condensation is a property that is unique to the substance undergoing the phase change and the conditions under which it condenses.