Question

1. The geothermal power plant at Chena Hot Springs operates on a simple Rankine cycle with R-134e as the working fluid. For more details, see the suggested reading The cycle operates between pressure limits of 232 psia and approximately 63.6 psia. So that we can actually do this problem by hand, let's estimate that the cycle operates between pressures of approximately opsie and 232 pse. The maximum temperature reached by the cycle is equivalent to the temperature of the geothermal source, and is 164". The mass flowrate of the refrigerant is 28.6 lbm/sec. Assuming a perfectly ideal Rankine cycle, sketch this cycle on a T-s diagram and calculate the following W d. for the cycle . for a power cycle operating between the same temperature limits e. for the cycle & Holdmann quotes the actual steady state pump power for this cycle as 40 kW. What is the approximate isentropic efficiency of the pump? h. Holdmann quotes the actual steady state turbine power for this cycle as 250 kW. What is the approximate isentropic efficiency of the turbine? Mint: If Table A-13E had entries for P = 232 psia, they would have the values on the back. You're welcome. Table A-130, P=232 psia u, Btu/ibm h, Btu/bm Btu/ TF 160 v ft/ m 0.22650 117.12 126 37 0.22862 180 0.24352 122 13 132.12 0.23774

Answer 1

Subdivisions a, b, c, d are solved here

Given:

• Operating pressures are P_{​​​​​​1} = 232 psia and P_{​​​​​2} = 60 psia.
• Maximum temperature T_{​​​​​​1} = 164^$\degree$F
• Mass flow rate, $\dot{m}$ = 28.6 lbm/sec

Step by step process for finding first four subdivisions are done below.

Let Let Let Let STEP 1: turbine inlet is represented by 1 (suffix) turbine outlet is represented by 2 (Suffix) pump ineket is represented by 3 (suffix) pump outlet is represented by 4 suffix) Find hi and si at 164°F and 232 psia by interpolation. b- 126.37 S,-0.22862 = 164-160 180 - 160 164 -160 180-160 132. 12 - 126.37 0.23774 - 0-22862 MF 127.52 Btu/ebm 51 = 0.230444 Btu/iom-R STEP 2: Find type of state at turbine outlet and hz. Process 1-2 → Isentropic process in turbine. Hence Si=S2 → 52 = 0.230 444 Btuksm-R NOTE : Below tables are taken from steam tables. © At P= 60 psia, Specific volume bat. temp Enthalpy (Btu/lam) Enthropy (Btu/iom ) T, OF ' t HtFlam) hf hfg | hg Sf Spg Sg 49.84 0.01270 0.79361 28.08 82.03 110.11 0.06029 0.16098 0.22127 i At P=60psia, TOF h (Btu/elom) s (Bruxlbm-R) 60 112.39 0.22570 80 116.82 0.23 407

From Note 4 & SQ = 0.22127 Blu Ron-R. S2 7 Sg state 2 is Hence in superheated state. T-S diagram: By interpolation, ha - 112.39 116.82 - 112.39 - 0.230444 - 0.22570 0.23 407 - 0.22570 h 2 = 114.901 Btu/lbim Step 3: Find hz and ha. h3= 28.08 Btullbm ruump ( h z=hg at P=60 psia) (From Note ) Pump work Wp i ha ha vf (P1-P2) = hq-h3. of Ve = 0.01270 ftylom P.-P2 = (232-60) psia & P-P2 = 172 ebbline (:poia= ebeline) 7-9 = 1+2 x(2° lbf (inin ta *) -Pa = 24768 lbk/etz If (P.-P2) = 0. 01270 ft x 24768 lbl/442 libro + V (P-P2) = 314.5536 lbf-t = 0.4058 Btu/lom Ilom

Werm> lis 01-Pa) = 0.4058 Btwem eft = 0.00129 Blu] [t lbf-ft = 0.00129 Batu] in Wp= h 4-h3 Vf (P1-P2) = (ha-ha) 0.4058 = hq - 28.08 ha = 28.4858 Btu/ Step 4: Find Whemp. pump work in Btu/sec ³ W pump = in Wouma = 28.6 x (ha - h) - 28.6 x (28.4859 - 28.08) - 11. 606 Btu/sec Steps Find turbine Turbine work in Blu W turbine in (hi-he) - 28.6 (127.52 - 14.901 Wurw 360.9034 Btu Step 6: Find din Bin = in (hi-h4) = 29.6 ( 127.52 - 28.4968) din - 2832. 378 Btu/sec Find th of the cople: M = 1- hr-h3 steps. hi ha = 1- 114.901 - 28.08 127.52 - 28.4858 = 0.1233 M = 12.33% |

al pump = 11.606 Btu/sec b] Wturbine = 360.9034 Btysec o Qin = 2832.376 Btly sec d] MER = 12.33% [Note: 1th can also be find by using below formula. th= net = Wturbine-Wpump pump Qin Qin = 360, 9034 – 11.606 2832. 378 - On 1233