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 \(42 \mathrm{~kg} / \mathrm{s}\) 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 of \(8 \mathrm{~kW}\) is lost from the condenser to the surrounding air. The rate at which the steam condenses is (a) \(0.228 \mathrm{~kg} / \mathrm{s}\) (b) \(0.318 \mathrm{~kg} / \mathrm{s}\) (c) \(0.426 \mathrm{~kg} / \mathrm{s}\) (d) \(0.525 \mathrm{~kg} / \mathrm{s}\) (e) \(0.663 \mathrm{~kg} / \mathrm{s}\)

Short answer

Answer: (e) 0.663 kg/s

Step-by-step solution

Calculate the energy gained by the cold water

First, we need to calculate the energy gained by the cold water as it passes through the heat exchanger. To do this, we will use the formula: \(Q_\text{coldwater} = m_\text{coldwater} \cdot c_p \cdot \Delta T\) where \(Q_\text{coldwater}\) is the energy gained by the cold water, \(m_\text{coldwater}\) is the mass flow rate of cold water, \(c_p\) is the specific heat capacity of the water, and \(\Delta T\) is the temperature change of the cold water. Given: \(m_\text{coldwater} = 42 \, \frac{\text{kg}}{\text{s}}\) \(c_p = 4.18 \, \frac{\text{kJ}}{\text{kg} \cdot \text{K}}\) \(\Delta T = 25^{\circ} \mathrm{C} - 16^{\circ} \mathrm{C} = 9^{\circ} \mathrm{C}\) Calculate the energy gained by the cold water: \(Q_\text{coldwater} = (42 \, \frac{\text{kg}}{\text{s}}) \cdot (4.18 \, \frac{\text{kJ}}{\text{kg} \cdot \text{K}}) \cdot (9^{\circ} \mathrm{C}) = 1580.04 \, \mathrm{kJ/s}\)

Calculate the total energy entering the condenser

Now, we need to calculate the total energy entering the condenser. The energy entering the condenser is the sum of the energy gained by the cold water plus the heat loss to the surrounding air. Given: \(Q_\text{loss} = 8 \, \mathrm{kW} = 8 \, \mathrm{kJ/s}\) Calculate the total energy entering the condenser: \(Q_\text{total} = Q_\text{coldwater} + Q_\text{loss} = 1580.04 \, \mathrm{kJ/s} + 8 \, \mathrm{kJ/s} = 1588.04 \, \mathrm{kJ/s}\)

Calculate the rate at which the steam condenses

The total energy entering the condenser is equal to the rate at which the steam condenses times the energy gained by the steam as it condenses, which is given by: \(Q_\text{total} = m_\text{steam} \cdot h_\text{steam-condensation}\) where \(m_\text{steam}\) is the mass flow rate of the steam and \(h_\text{steam-condensation}\) is the heat of condensation of the steam. Since the steam condenses at \(45^{\circ} \mathrm{C}\), the heat of condensation can be found in steam tables (approximately 2420 \(\frac{\text{kJ}}{\text{kg}}\)). Rearrange the equation to solve for the mass flow rate of the steam: \(m_\text{steam} = \frac{Q_\text{total}}{h_\text{steam-condensation}}\) Calculate the mass flow rate of the steam: \(m_\text{steam} = \frac{1588.04 \, \mathrm{kJ/s}}{2420 \, \frac{\text{kJ}}{\text{kg}}} \approx 0.656 \, \frac{\text{kg}}{\text{s}}\) Since the value of 0.656 is closest to the given option (e) \(0.663 \, \frac{\text{kg}}{\text{s}}\), we conclude that the correct answer is (e) \(0.663 \, \frac{\text{kg}}{\text{s}}\).