Question

In large steam power plants, the feedwater is frequently heated in a closed feedwater heater by using steam extracted from the turbine at some stage. Steam enters the feedwater heater at \(1 \mathrm{MPa}\) and \(200^{\circ} \mathrm{C}\) and leaves as saturated liquid at the same pressure. Feedwater enters the heater at \(2.5 \mathrm{MPa}\) and \(50^{\circ} \mathrm{C}\) and leaves at \(10^{\circ} \mathrm{C}\) below the exit temperature of the steam. Determine the ratio of the mass flow rates of the extracted steam and the feedwater.

Short answer

The ratio of the mass flow rates of the extracted steam and the feedwater in the feedwater heater is approximately 0.238.

Step-by-step solution

Find the Enthalpy of the Inlet Steam

To find the enthalpy of the inlet steam, we can use the steam tables to look up the enthalpy at the given pressure and temperature: $$ P_{steam\_in} = 1\,\text{MPa}\\ T_{steam\_in} = 200^\circ\,\text{C} $$ Using the steam tables, we find that the enthalpy of the inlet steam is: $$ h_{steam\_in} = 2825.8\;\frac{\text{kJ}}{\text{kg}} $$

Find the Enthalpy of the Outlet Steam

We are told that the outlet steam is saturated liquid at the same pressure as the inlet steam, so we can look up the enthalpy of saturated liquid at 1 MPa using the steam tables: $$ h_{steam\_out} = 762.81\;\frac{\text{kJ}}{\text{kg}} $$

Find the Enthalpy of the Inlet Feedwater

Using the steam tables, we can find the enthalpy of the inlet feedwater at the given pressure and temperature: $$ P_{feedwater\_in} = 2.5\,\text{MPa}\\ T_{feedwater\_in} = 50^\circ \text{C} $$ The enthalpy of the inlet feedwater is: $$ h_{feedwater\_in} = 209.33\;\frac{\text{kJ}}{\text{kg}} $$

Find the Enthalpy of the Outlet Feedwater

The outlet feedwater temperature is 10 degrees Celsius below the exit temperature of the steam. Since the steam leaves as saturated liquid, we can find the saturation temperature of the steam at 1 MPa and subtract 10 degrees Celsius to find the outlet feedwater temperature: $$ T_{feedwater\_out} = T_{sat\_1MPa} - 10^\circ \text{C} $$ Using the steam tables, we find the saturation temperature at 1 MPa is: $$ T_{sat\_1MPa} = 179.9^\circ\,\text{C} $$ So, $$ T_{feedwater\_out} = 179.9 - 10 = 169.9^\circ\,\text{C} $$ Now we can use the steam tables to find the enthalpy of the outlet feedwater at the given pressure and temperature: $$ h_{feedwater\_out} = 721.11\;\frac{\text{kJ}}{\text{kg}} $$

Determine the Ratio of the Mass Flow Rates

Using the principle of conservation of energy, we can equate the energy taken away from the steam to the energy added to the feedwater. The energy is equal to the mass flow rate times the change in enthalpy: $$ m_{steam} \times (h_{steam\_in} - h_{steam\_out}) = m_{feedwater} \times (h_{feedwater\_out} - h_{feedwater\_in}) $$ Now we need to find the ratio between the mass flow rates of the extracted steam and the feedwater: $$ \frac{m_{steam}}{m_{feedwater}} = \frac{h_{feedwater\_out} - h_{feedwater\_in}}{h_{steam\_in} - h_{steam\_out}} = \frac{721.11 - 209.33}{2825.8 - 762.81} $$ Calculating the ratio, we get: $$ \frac{m_{steam}}{m_{feedwater}} = 0.238 $$ So, the ratio of the mass flow rates of the extracted steam and the feedwater is approximately 0.238.