FUND OF ENG THERMODYN(LLF)+WILEYPLUS
FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
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Steady-state operating data are shown in the figure below for an open feedwater heater. Heat transfer from the feedwater heater to its surroundings occurs at an average outer surface temperature of 50°C at a rate of 100 kW. Ignore the effects of motion and gravity and let To = 25°C, po = 1 bar. Determine (a) the ratio of the incoming mass flow rates, m/ṁ2. (b) the rate of exergy destruction, in kW. P2 = 1 bar Tz = 400°C 1 ṁy = 0.7 kg/s Pi = 1 bar T, = 40°C Feedwater heater X3 = 25% P3 = 1 bar Tp = 50°C %3D 2)
7.58 Figure PZ.58 shows a gas turbine power plant using air as the working fluid. The accompanying table gives steady-state operating data. Air can be modeled as an ideal gas. Stray heat transfer and the effects of motion and gravity can be ignored Let To 290 K, po = 100 kPa. Determine, each in kJ per kg of air flowing, (a) the net power developed, (b) the net exergy increase of the air passing through the heat exchanger, (eg- e), and (c) a full exergy accounting based on the exergy supplied to the plant found in part (b). Comment. State p(kPa) T(K) h(kJ/kg) s° (kJ/kg K) 1100 290 290.16 1.6680 500 505 508.17 2 2.2297 3 500 875 904.99 2.8170 4 100 635 643.93 2.4688 a o is the variable appearing in Eq. 6.20a and Table A-22. Heat exchanger Compressor Turbine FIGURE P7.58
A domestic water heater holds 189 L of water at 60°C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in m, would a 1000-kg mass have to be raised from zero elevation for its exergy to equal that of the hot water? Let T0 = 298 K, p0 = 1 atm, g = 9.81 m/s2 .
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