Question

1/4 Thermo I-Take-Home Final Summer 2019 Thermodynamics I: Final, Summer 2019 (7/30, Tuesday) (Total 200 points) Name: Consider a modern extra-supercritical pressure steam power plant that operates on a reheat-regenerative 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 Ps- 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 at 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 T,-180 °C. The total mass flow rate entering the high-pressure turbine is m,-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 next page. 2) Plot the cycle on the given T-s diagram, clearly show all principle states. 3) Determine the extracted steam fractions, y and z, for two heaters. (Note that y m,/m, and z=ms / m,). Determine the thermal efficiency of the cycle, h Determine the total net power output from the system, Wrt 4) 5) ner Reheater High-P Turbine Low-P Turbine Boiler 2+ 6 Condenser Closed Open FWH 9 FWH 8 13 -11 10 12 Pump I Pump IIl Pump II

please show all work with full steps.

thanks

Answer 1

The pressures will be г. f = ofо. P, Pa = дело Рf%. f, a n = 100 К% % = f – Р. = f = f = 800 Kfm for = f = 32 000 Efo Ptr%) T®сс) Пcro 32 050 9 orov 250 Qovo 050 6 до 3269- 6 10 5-8 (кг/к) (кг/кух) 31 С-8 6S 14 32 Ч2. S 6.519 3o SD-6 6.79 316 :y 17:472 7. V72 23 C7,6 7.972 191:1 0-60 192.8 о. С чо 1 с. 8 3.2.1 2.2017 13 99.9 • 139 To 8 2.190 о 4% . 18 о 295 304: о go 8 сю denro 32 слvО Но 8 слю 1366 79. 74 2:0

640 32 MB 10 137 - ampa - - - - Pa Impa. LokPa 32 for over feed water heater Min Ein = Cout 9h. + (1-1-2) Ag + (-4) bs = bo y (3242.5) + (1-4-2) (7668) +(1+y)(770.8) = 1316-6 Z= 2.22344 + 0.2882 0 for closed seed water heater Ein = Cout 2 hg + (1-Y-2) ho= CI-Y-2) hg 4 (ty) hiz z (3268.6) +(1--296199.89 = (1-Y-7) 1766.9)+114)762.8 zs 0.347- 0.347y -

from © ¢ ® TY= 0.02287 di z = 0.389 Now at state 2 Y = Me » inga yom, Ma = 0.522872 64:Y Me = 1.473 kg 2 atstede 5 Z= ms MG= 0.3390 64.4 IMG = 21.03 Kglb. y Meat imput = (3726.6-1349.4)+(1-0-82284) * (3766-4-3050.6) = 3076 kilky Heat output = (1-4-2) Cho-hy) = 1388.7 KS1189 Nens me = 1- grond gin = I 1388-7 3076.83 1015486 no = 54.86