Question

In large steam power plants, the feedwater is frequently heated in closed feedwater heaters, which are basically heat exchangers, by steam extracted from the turbine at some stage. Steam enters the feedwater heater at \(1.6 \mathrm{MPa}\) and \(250^{\circ} \mathrm{C}\) and leaves as saturated liquid at the same pressure. Feedwater enters the heater at \(4 \mathrm{MPa}\) and \(30^{\circ} \mathrm{C}\) and leaves \(10^{\circ} \mathrm{C}\) below the exit temperature of the steam. Neglecting any heat losses from the outer surfaces of the heater, determine (a) the ratio of the mass flow rates of the extracted steam and the feedwater heater and \((b)\) the reversible work for this process per unit mass of the feedwater. Assume the surroundings to be at \(25^{\circ} \mathrm{C}\).

Short answer

The ratio of mass flow rates of the extracted steam and the feedwater is 0.277, and the reversible work for this process per unit mass of the feedwater is -424 kJ/kg.

Step-by-step solution

Find the enthalpies of steam and feedwater entering the feedwater heater

To find the enthalpies, we need to look up the properties of steam and water in a steam table. For steam entering the heater: - Pressure: \(P_1 = 1.6 \,\mathrm{MPa}\) - Temperature: \(T_1 = 250^{\circ} \mathrm{C}\) We can look these up in a steam table and find that the specific enthalpy for steam at these conditions is: \(h_1 = 2878 \, \mathrm{kJ/kg}\) For feedwater entering the heater: - Pressure: \(P_2 = 4 \, \mathrm{MPa}\) - Temperature: \(T_2 = 30^{\circ} \mathrm{C}\) We can look these up in a water table and find that the specific enthalpy for water at these conditions is: \(h_2 = 127 \, \mathrm{kJ/kg}\)

Find the enthalpies of steam and feedwater leaving the feedwater heater

For steam leaving the heater, it is given that it leaves as a saturated liquid at the same pressure as the entering steam. - Pressure: \(P_3 = 1.6 \, \mathrm{MPa}\) We can look up the properties of saturated liquid water at this pressure in a water table and find that the specific enthalpy for saturated liquid water at this pressure is: \(h_3 = 844 \, \mathrm{kJ/kg}\) For feedwater leaving the heater, the temperature is \(10^{\circ} \mathrm{C}\) below the exit temperature of the steam, which means: - Temperature: \(T_4 = 844 \,\mathrm{kJ/kg} - 10^{\circ} \mathrm{C}\) Using the steam table again, we find the specific enthalpy for water at this condition to be: \(h_4 = 700 \, \mathrm{kJ/kg}\)

Calculate the mass flow rate ratio

Conservation of energy for the feedwater heater can be written as: \(m_s h_1 + m_f h_2 = m_s h_3 + m_f h_4\) Where \(m_s\) is the mass flow rate of the extracted steam and \(m_f\) is the mass flow rate of the feedwater. The problem asks for the ratio \(\frac{m_s}{m_f}\), so we can rewrite the equation as: \(\frac{m_s}{m_f} = \frac{h_4 - h_2}{h_1 - h_3}\) Now we can plug in the values we found earlier: \(\frac{m_s}{m_f} = \frac{700 \,\mathrm{kJ/kg} - 127 \, \mathrm{kJ/kg}}{2878 \, \mathrm{kJ/kg} - 844 \, \mathrm{kJ/kg}} = 0.277\) So the ratio of the mass flow rates of the extracted steam and the feedwater is \(0.277\).

Calculate the reversible work per unit mass of the feedwater

The reversible work per unit mass of the feedwater can be determined using the second law of thermodynamics. The equation for the reversible work per unit mass is: \(W_{rev} = T_0 \Delta S\) Where \(T_0\) is the surrounding temperature and \(\Delta S\) is the change in entropy of the combined steam and feedwater system. The entropy change can be calculated using the mass flow rates and the specific entropies: \(\Delta S = m_s (s_3-s_1) + m_f (s_4 - s_2)\) Using the steam tables, we can find the specific entropies at the given conditions: - \(s_1 = 6.269 \, \mathrm{kJ/kg \cdot K}\) - \(s_2 = 0.439 \, \mathrm{kJ/kg \cdot K}\) - \(s_3 = 2.320 \, \mathrm{kJ/kg \cdot K}\) - \(s_4 = 2.102 \, \mathrm{kJ/kg \cdot K}\) Plugging in the values and the mass flow rate ratio, we get: \(\Delta S = 0.277 (2.320 \,\mathrm{kJ/kg \cdot K} - 6.269 \, \mathrm{kJ/kg \cdot K}) + (2.102 \, \mathrm{kJ/kg \cdot K} - 0.439 \, \mathrm{kJ/kg \cdot K}) = -1.422 \, \mathrm{kJ/kg \cdot K}\) Now we can calculate the reversible work per unit mass using the surrounding temperature \(T_0 = 25^{\circ} \mathrm{C} = 298 \, \mathrm{K}\): \(W_{rev} = 298 \, \mathrm{K} \times (-1.422 \, \mathrm{kJ/kg \cdot K}) = -424 \, \mathrm{kJ/kg}\) Hence, the reversible work for this process per unit mass of the feedwater is \(-424 \, \mathrm{kJ/kg}\).