Question

Consider an ideal reheat-regenerative Rankine cycle with one open feedwater heater. The boiler pressure is \(10 \mathrm{MPa}\), the condenser pressure is \(15 \mathrm{kPa}\), the reheater pressure is \(1 \mathrm{MPa}\), and the feedwater pressure is \(0.6 \mathrm{MPa}\). Steam enters both the high- and low- pressure turbines at \(500^{\circ} \mathrm{C} .\) Show the cycle on a \(T\) -s diagram with respect to saturation lines, and determine (a) the fraction of steam extracted for regeneration and \((b)\) the thermal efficiency of the cycle.

Short answer

Based on the provided information and calculations, the fraction of steam extracted for regeneration in the ideal reheat-regenerative Rankine cycle with one open feedwater heater is 0.179, and the thermal efficiency of the cycle is 45.05%.

Step-by-step solution

Determine Cycle States

Using the given pressure and temperature values, we can determine the following cycle states: 1. State 1: Steam enters the high-pressure turbine at \(10 \,\text{MPa}\) and \(500^{\circ}\mathrm{C}\). 2. State 2: Steam is partially expanded in the high-pressure turbine and reaches the reheater pressure, \(1\,\text{MPa}\). 3. State 3: Steam leaves the reheater at \(1\,\text{MPa}\) and \(500^{\circ}\mathrm{C}\) and enters the low-pressure turbine. 4. State 4: Steam is further expanded in the low-pressure turbine and exits at the condenser pressure, \(15\,\text{kPa}\). 5. State 5: Saturated liquid exits condenser at \(15\,\text{kPa}\). 6. State 6: Subcooled liquid leaves the open feedwater heater at \(0.6\,\text{MPa}\). 7. State 7: Compressed liquid enters the boiler at \(10\,\text{MPa}\). Now, we need to calculate the enthalpies for each state considering the ideal Rankine cycle.

Calculate enthalpies

We get the enthalpies, considering an ideal Rankine cycle (using steam tables) for each state as follows: 1. State 1: \(h_1 = 3373.7\,\text{kJ/kg}\) 2. State 2: \(h_2 = h_1 - \eta_T (h_1 - h_2')\), with \(\eta_T = 1\, (\text{for an ideal turbine}) \implies h_2 = 2787.4\,\text{kJ/kg}\) 3. State 3: \(h_3 = 3373.2\,\text{kJ/kg}\) 4. State 4: \(h_4 = h_3 - \eta_T (h_3 - h_4')\), with \(\eta_T = 1\, (\text{for an ideal turbine}) \implies h_4 = 2262.4\,\text{kJ/kg}\) 5. State 5: \(h_5 = 225.9\,\text{kJ/kg}\) 6. State 6: \(h_6 = 372.5\,\text{kJ/kg}\) 7. State 7: \(h_7 = h_6 + v_6(h_{1f} - h_{6f})\), with \(\eta_p = 1\, (\text{for an ideal pump}) \implies h_7 = 391.9\,\text{kJ/kg}\)

Calculate the extracted steam fraction

Now, we need to find the mass fraction of the extracted steam (\(y\)) for the open feedwater heater. Using energy balance for the open feedwater heater: \(1\,\text{kg}\cdot h_2 = (1-y)\,\text{kg}\cdot h_3 +y\,\text{kg}\cdot h_6\) Solving for \(y\), we get: \(y = \frac{h_2 - h_3}{h_6 - h_3} = \frac{2787.4 - 3373.2}{372.5 - 3373.2} = 0.179\).

Calculate the thermal efficiency

Finally, we will calculate the thermal efficiency of the cycle using the first law of thermodynamics and the energy balance as follows: \(\eta_{th} = \frac{W_{net,out}}{Q_{in}} = \frac{W_{out} - W_{in}}{Q_{in}}\) For the ideal Rankine cycle, we have \(W_{turbine,out} = w_{turbine, HP} + w_{turbine, LP} = (h_1 - h_2) + (1-y)(h_3 - h_4)\) \(W_{pump,in} = h_7 - h_6\) \(Q_{in} = q_{boiler} + q_{reheater} = (h_1 - h_7) + (h_3 - h_2)\) Substituting the values for the enthalpies obtained in Step 2 and the extracted steam fraction we obtained in Step 3, we can calculate the thermal efficiency: \(\eta_{th} = \frac{(3373.7 - 2787.4) + (1 - 0.179)(3373.2 - 2262.4)}{(3373.7 - 391.9) + (3373.2 - 2787.4)} = 0.4505 = 45.05\%\) Now, we have found both the fraction of steam extracted for regeneration and the thermal efficiency of the reheat-regenerative Rankine cycle. The results are: (a) The fraction of steam extracted for regeneration is \(y=0.179\). (b) The thermal efficiency of the cycle is \(\eta_{th}=45.05\%\).