Question

Consider a modern extra-supercritical pressure steam power plant that operates Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater, as shown below. Steam enters the high-pressure turbine at 32 MPa and 700 °C and is condensed in the condenser at a pressure of 10 kPa. Steam exits the high-pressure turbine at 4 MPa and is routed into a reheater (inside the boiler) to be reheated to 640 °C. Steam at P-8 MPa is extracted from the high-pressure turbine to the open feedwater heater and steam at P, 1 MPa is extracted from the low-pressure turbine to the closed feedwater heater (see below). The extracted steams are all completely condensed into saturated liquid the given pressures, respectively. For the open feedwater heater, all steams entering the device have the same pressure so there is no free expansion in the device. For the closed-feedwater heater, the feedwater is to be heated to To,-180 °C. The total mass flow rate entering the high-pressure turbine is rm,-64.4 kg/s All turbines and pumps operate isentropically. Neglect stray heat transfer, pressure drops along the connecting pipes, and the kinetic and potential effect, you are asked to do the following: 1) Determine the properties of given primary states, and fill in the blank boxes in the Table on nex on a reheat-regenerative at page. Plot the cycle on the given T-s diagram, clearly show all principle states. 2) Determine the extracted steam fractions, y and z, for two heaters. 3) (Note that y m, / m and z Determine the thermal efficiency of the cycle, m, / m) 4) 5) Determine the total net power output from the system, net Reheater 1 Low-P High-P Turbine Turbine Boiler 2 5 Condenser Closed FWH Open FWH 8 -13 +11 10 12 Pump l Pump II Pump Il

Answer 1

The following pressures will be equal: P = Po = 10 kPa Pz = P12 = 1000 kPa P2 = P4 = 4000 kPa Po = P = P10 = P13 = P2 = 8000 kPa Puu = P= 32000 kPa Since all pumps and turbines are isentropic: S2 = Si = S3 S4 = S5 = So S10 = $11 S12 = S13 S7 = Sg

(1) s (kJ/kg-K) By using EES and above conditions: State P (kPa) T(°C 32000 700 2 8000 4000 4 4000 640 1000 -- 10 45.8 10 45.8 8000 46.1 8000 180 8000 295 11 | 32000 304.4 1000 179. 9 13 8000 h (kJ/kg) 3726.8 3242.5 3050.6 3766.4 3268.6 2367.9 191.7 199.8 766.8 1316.6 1349.4 762.8 770.8 6.514 6.514 6.514 7.472 7.472 7.472 0.649 0.649 0.91 0 3.207 3.207 2.139 2.139 0 The T-s plot is: 700........ 640 .... 32 MPa 11 1078 MPa 74 MPa .....1 MPa 1801 10 kPa

For open feed water heater: 1-Y-> Open FWH 713 +10 The energy balance for open feed water heater: Em = EW yh, + (1 - y, z)h, +(1- y) h3 = ho y(3242.5) + (1 - y - z)(766.8) + z(770.8) = 1316.6 z=137.45 - 618.925 y ...... (1) Closed feed water heater: Closed FWH 1 -1 1-v-zl V -12 7 +12 L The energy balance for closed feed water heater: Em = EU zhz + (1 - y–z) ha = (1 - y-z)h, + zh2 z (3268.6) + (1 - y-2)(199.8) =(1- y - 2)(766.8)+z(762.8) z = 0.1845 -0.1845 y ......(2) Solving (1) and (2) equations: 137.45 - 618.925 y = 0.1845 -0.1845 y y = 0.2218 z = 0.1845 -0.1845y z = 0.1845 – 0.1845(0.2218) z = 0.1436

The extracted steam functions are: At state 2: m2 = ym m2 = 0.221864.4 m2 = 14.28 kg/s At state 5: ,_ms m mg = zm mg = 0.1436 x 64.4 mg = 9.25 kg/s Heat input: Pin = h - 1 +(1-y)(h4 - hz) = (3726.8 - 1349.4) + (1 -0.2218) (3766.4 - 3050.6) = 2934.43 kJ/kg Heat input: yout = (1-y-z) (h6 -h,) = (1 -0.2218 - 0.1436) (2367.9-191.7) = 1381 kJ/kg The thermal efficiency of the cycle: Tak = 1 - You ein I- 1381 2934.43 = 0.5294 Then = 52.94% (5) The total net power output from the system: W net = 17.hin = 0.5294 x 2934.43 W = 1553.49 kJ/kg