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ch14 - Basis for Problems 2 through 10 100 kg dry coal/min...

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Unformatted text preview: Basis for Problems 2 through 10: 100 kg dry coal/min Problem 14.2 104.8 kg wet coal/min ' 100.00 kg dry coal/min H2O entering with coal 4.8000 kg/min 0.2665 kmol/min Molar Flow Rates: Component kg/min 75.20 5.00 .60 3.50 7.50 Ash 7.20 100.00 11.920 0(0sz H Problem 14.3 02 req'd = (molar flow rate of C) x (1 kmol O2/kmol C) + (molar flow rate of H) x (1 kmol H2/2 kmol H) x 0.5 kmol OZ/kmol H2) + (molar flow rate of S) x (1 kmol OZ/kmol S) - (molar flow rate of O) x (l kmol 02/2 kmol 0) Problem 14.4 (a) for 15% excess 02 02 ed = 8.485 kmo /min = 1.15 x 02 required N2 fed = 31.918 kmol/min = (79 kmol N2/21 kmol 02) x 02 fed (b) P = 1 atm = 760.0 mm Hg r.h. = (pH20/p*H20 @ 25°C) x 100% 50% p*H20 = 23.756 mm Hg from Table B.3 pH20 = 11.878 mm Hg yH O = pH 0 P = I.I . for each mole of dry air (79 mole% N2, 21 mole% 02): component moles MW mass mole frac N2 0.7900 28 22.12 0.7777 02 0.2100 32 6.72 0.2067 Water 0.0159 18 0.29 0.0156 Total 1.0159 29.13 1.0000 AVG MW = = . g 0 the dew point is the temperature at which water has a vapor pressure equal to its partial pressure, 11.8780 mm Hg from Table B.3, the dew point is about the degrees of superheat = T - T dew point =_l_ 14—2 Problem 14.4 (cont'd) flow rate of water in air = 02 flow rate x (mole fraction H20/mole fraction 02) (c) air feed rate = 02 fed + N2 fed + H20 fed 41.04 kmol/min 919.4 std cubic meters/mi 1,003.6 cubic meters/min Problem 14.5 composition of furnace flue gas (ignoring fly ash) C02 flow rate = flow rate of carbon in coal x (1 kmol C02/1 kmol C) H20 flow rate = [flow rate of water in air + flow rate of H in coal x (1 kmol H20/2 kmol H)] + flow rate of water in wet coal N2 flow rate = flow rate of N2 in air + flow rate of N in coal x (l kmol N2/2 kmol N) 802 flow rate = flow rate of S in coal x (1 kmol SOZ/kmol S) 02 flow rate 02 fed with air — 02 required for complete combustion Component kmol/minmo e/ rac g/min C02 6.261 0.1461 275.50 H20 3.391 .0792 61.04 N2 31.975 .7463 895.31 802 0.109 .0025 6.99 . .0258 35.41 total 42.844 1.0000 1274.26 OOOO fly ash in the flue gas prior to the electrostatic precipitator = 5.760 kg/min fly ash in the flue gas after the the electrostatic precipitator = 0.006 kg/min or 0.0001 kg ash/kmol off—gas so that the fly ash removed by the 14-3 Problem 14.6 Feed to each scrubber is 1/2 total flue gas flow rate (a) The only constituents of the feed gas'whose flow rates change (i.e. rather than remain constant as they flow through the scrubber) are water, 502, C02 (because of the formation of C02 upon reaction of $02 with CaCO3), and fly ash. 90% of the inlet 802 is absorbed and reacted with CaCO3 and all of the entering fly ash is removed from the gas stream. Since the gas stream leaving the absorber is saturated with water, pHZO in that stream = vapor pressure of water at 53C = mole fraction water x Pressure. p* H20(53°C) yHZO 0.14305 bar (interpolated from Table B.5) p*H20/P 0.1412 n_H20/n_total Gas entering scrubber Gas leaving the scrubber mole mole comp kmol/min frac kg/min75 MOI/min frac kg/min 0.00 21.422 1.0000— 22.969 1.0000 663.99 0.0001 = Conv function* = 100*(difference) *Use Goal Seek tool to adjust amount of water in gas stream leaving the scrubber so that the H20 mole fraction is that calculated from the vapor pressure. (b) slurry feed to scrubber liquid = 15.2 x gas inlet flow rate (kg/min) 9684 kg/min solids = liquid flow rate/9 (kg/min) 1076 kg/min total 10760 kg/min (c) each kmol of CaSOB is formed by reacting 1 kmol of 802 with 1 kmol CaCO3 SO2 absorbed = 0.049 kmol/min 4.73 kg/min CaSO3 formed = _ 0.049 kmol/min 5.91 kg/min CaC03 consumed = 0.049 kmol/min 4.92 kg/min Therefore, assuming that the solution concentrations of CaCO3 and CaSO3 remain constant (the liquid is saturated), the slurry leaving the absorber must contain additional solids in the amount of 0.99 kg/min = CaSO3 formed - CaCO3 consumed Furthermore, the flow rate of water in the slurry has also changed. Water has been consumed in reaction 14.3 and a water balance around the absorber reads 14-4 Problem 14.6 (cont'd) (d) (6) water fed with gas + water generated + water entering with slurry = water leaving in slurry + water leaving in gas water fed with gas = 1.696 kmol/min 0. water generated 025 kmol/min water leaving in gas water entering with slurry water leaving in slurry 3.243 kmol/min 537.997 kmol/min (from below) 536.474 kmol/min (from mass balance) entering the scrubber liquid in slurry CaCO3 in liquid CaSO3 in liquid 9684 kg/min 0.002 kg/lOO kg water 0.003 kg/lOO kg water water in liquid = 9683.9 kg/min (determined using Goal Seek) total liquid flow = 9684.4 kg/min (CaCO3+CaSO3+H20) Convergence Function* = 0 (difference) x 100 *Balance is solved by using the Goal Seek tool to adjust the water flow rate until the convergence function (liquid in slurry entering the scrubber — total liquid flow) = 0. CaSO3 dissolved in liquid leaving absorber = CaSO3 concentration x water flow rate = 0.2897 kg/min CaCO3 dissolved in liquid leaving absorber = CaCO3 concentration x water flow rate = 0.1931 kg/min Therefore, the ratio of solid to liquid in the slurry leaving the absorber is 0.1115 kg solid/kg liquid m_limestone = (nSO2 fed x 0.90) x (1.052 kmole CaCO3/kmol 802) x (1 kg limestone/0.921 kg CaCO3) (83 kg CaCO3/kmole CaCO3) 4.66 kg limestone/min Since inerts leave the system only in the wet solids: (m_inerts)_WS = m_limestone x (0.079 kg inerts/0.921 kg CaCO3) = 0.400 g inerts/min The flow rate of CaCO3 in the wet solids can be determined from a balance around a system comprised of the scrubber, blending tank, and filter. The flow rate of CaCO3 in the wet solids = flow rate of CaCO3 in the limestone - that reacted with $02 in the scrubber. (n_CaCO3)_WS = (n_CaCO3)_fed - (n_SOZ)_reacted x {(n_CaCO3)_reacted/(n_802)_reacted} = [(1.052*0.9(n_502)_fed - 0.90 (n_SO2)_fed )1 x (l kmol CaCO3/kmol SOZ) (n_CaCO3)_WS 0.0026 kmol CaCO3/min 14-5 Problem 14.6 (cont'd) Likewise, the flow rate of CaSO3 leaving with the wet filter cake is given by a balance on CaSO3 around the system defined above: the flow rate of CaSO3 = rate at which it is generated by the reaction of 802: (n_CaSO3)_WS 0.9(n_802)_fed x (lmol CaSOB/mol 502) 0.0491 kmol CaSO3/min 5.060 kg CaSO3/min As all of the fly ash in the entering gas is removed in the scrubber, (m_fly ash). = 0.003 kg 1y ash/min Summary for the Wet Solids from the filter mass frac (n_CaSO3)_WS (m_CaSO3)_WS inerts 0.3995 0. CaCO3(s)* 0.2119 0.0194 CaCO3(aq)** 0.0001 0.0000 CaSO3(s) 5.0599 0.4639 CaSO3(aq)** 0.0002 0.0000 fly ash 0.0029 0.0003 H2O 5.2324 0.4797 total 10.9069 1.0000 * CaCO3(s) and CaSO3 (s) are total amounts calculated above — the amounts dissolved: i.e., CaCO3(aq) and CaSOB(aq) ** CaCO3(aq) and CaSO3(aq) are adjusted by Goal Seek so that the ratio of each to water corresponds to the solubility of each; i.e. the converge l and converge 2 values below are forced to zero solubilities: CaCO3 0.002 kg/lOO kg water converge 1 —8.65983E—05 CaSO3 0.003 kg/lOO kg water converge 2 8.82333E—06 in the wet solids comprising the filter cake: liquid = 5.233 kg/min = CaSO3(aq) + CaCO3(aq) + H2O solids = 5.674 kg/min inerts + CaC03(s) + CaSO3(s) + fly ash 14—6 Problem 14.6 (cont'd) (f) From the solid—to—liquid ratio calculated in part (e), the slurry leaving the absorber is 10.0% wt% solids.' The total solids in the slurry sent to the filter is given in the above table, and he mass of liquid in this slurry is (mass of solids)x(l—fraction solids)(fraction solids) m_liquid in slurry to filter = 50.9 kg/min The recycle (filtrate) flow rate, m_recycle, is the flow rate of liquid in the slurry to the filter — the flow rate of liquid in the wet solids. The volumetric flow rate of recycle is 0.988 x m_recycle m_recycle = 45.6 kg recycle/min V_recycle = 45.1 L recycle/min A water balance around a system composed of the absorber, the filter and the blending tank gives the rate of make—up water addition. make—up water liquid water leaving with wet solids + water leaving as CaSO3 hemihydrate + water leaving with cleaned off-gas - water entering absorber with off—gas from furnace = 33.52 kg make-up water/min 33.52 L make—up water/min 14-7 Problem 14.7 Basis: 100 kg dry coal/min The rate at which heat is removed from the furnace can be determined from an energy balance around the furnace. Use reference conditions as follows coa1(s), H20 (1), all other species (g), all at 25°C. Tref (0C) 25 Energy Balance: Q = AH = Z nAern/v + Z(nH)out — Z(nH)in = (mcoal)(—HHV)+ 2(nH)out — 2(nH)in (mcoal)(—HHV) = —3,078,000 kJ/min kJ/kmol kJ/kmol C02 H20 (v) H20 (1) . N2 . , 273,570 502 . 0 02 . , 75,936 ash* bottom ash* 12,982 81,284 54,623 185,247 0 9,024 288,548 13,635 1,489 9,434 10,440 281 1,618 OJ I—‘UIHOI—‘OWOWO *kg/min, kJ/kg gas 42.8441 If none of heat transferred from the furnace is lost to the surrounding but all is used to generate steam in the power cycle, then the flow rate of steam is given as the heat transferred to the steam divided by the change in specific enthalpy of the steam. heat to steam = 2,892,410 kJ/min steam enthalpy change =H_out - H_in = H(vapor, T=540C, P=24.l MPa) — H(liq, T=38C, P=24.l MPa) H(liq, T=38°C, P=24.l MPa) = H_ref steam tables + H(sat liq, 38°C) + V(24.l—0.00655)MPa H_ref steam tables = 0 H(sat liq, 38°C) = 159.1 kJ/kg =1 L/kg V(24.1-0.00655)MPa = (1 L/kg)(l mA3/1000L)(24.1—.00655)MPa x (10 6Pa/MPa)(1 N/mAZ/Pa) x (lJ/N m)(l kJ/lOOOJ) = 24.1 kJ/kg 14—8 Problem 14.7 (cont'd) H(liq, T=38°C, P=24.1 MPa) 24.1 MPa = 235 bar H(T=540°C, P=24.1 MPa) 183.2 kJ/kg 3,321.1 kJ/kg found by interpolation of values from steam tables interpolation to evaluate P = 24.1 MPa = 235 3,188.9 ‘3,321.1 3,354.2 250 3166 3337 so that AH_Steam 3137.9 kJ/kg m_steam x AH_steam = m_steam = 921.8 kg/min Problem 14.8 for 5% excess 02 02 fed 7.747 kmol/min = 1.05 x 02 required N2 fed 29.143 kmol/min = (79 kmol N2/21 kmol 02)x02 fed H20 vapor fed 0.5857 kmol/min = 02 fed x y_H20/y_02 IN kJ/kmol kmol/min Coal* kJ/min kmol/min kJ/kmol kJ/min 0 0.0000 0 C02 0.000 O 0 6.2614 12,982 81,284 H20 (v) 0.586 54,078 31,673 3.3356 54,623 182,200 H20 (1) 0.267 0 0 0.0000 0 N2 29.143 8,571 249,781 29.1999 9,024 263,502 802 0.000 0 0 0.1092 13,635 1,489 02 7.747 8,950 69,333 0.1345 9,434 1,269 ash* 5.7600 281 1,618 grate ash* 1 14-9 Problem 14.8 (cont'd) for 25% excess 02 02 fed N2 fed H20 vapor fed 9.222 kmol/min 34.694 kmol/min 0.6973 kmol/min 1.05 x 02 required (79 kmol N2/21 kmol 02) x 02 fed 02 fed x y_H20/y_02 om: kmol/min kJ/kmol kJ/min kmol/min kJ/kmol kJ/min Component Coal* . 0 C02 H20 (V) H20 (1) N2 802 02 ash* grate ash* *kg/min, kJ/kg % excess 02 Q 5 -2,896,266 15 —2,892,410 25 —2,892,976 37,706 0 297,359 0 82,539 (kJ/mln) m_steam (kg/mln) 923.0 921.8 921.9 81,284 188,294 0 313,595 1,489 15,189 1,618 1,160 Would not reduce the excess 02 feed to 0% excess because of concerns regarding unburned carbon (soot formation). 14—10 Problem 14.9 What is the temperature of the flue gas after it has exchanged heat with the feed air? 567, 008 81,284 185,247 288,548 182, 812 5,784 152,040 24,048 1,489 . 109 10,440 . 832 0 * kg/min, kJ/kg 384,196 34,690 273,570 75,936 Convergence O . 00 *Use Goal Seek tool to adjust the temperature of the flue gas until the energy balance is satisfied; i.e. (H_in)_f1ue gas + (H_in)_air — (H_out)_flue gas — (H_out)_air = 0 sea excess air _— kmol/min kJ/kmol kJ/min kmo319/min kJ/kmol kJ/min 39.041 529, 745 .041 178, 957 6.261 12,982 81,284 36.261 1,012 6,338 3.336 54,623 182,200 3.336 44,871 149,671 29.200 9,024 263,502 29.200 778 22,730 .109 13,635 1,489 0.109 1,033 113 1,269 0.135 778 105 0 O kmol/min kJI/kmol kJ/min kmol7/min kJ/kmol kJ/min 350, 788 7.586 54,078 31,673 .143 8,571 249,781 .747 8,950 69,333 315 °C Convergence 0.00 14-11 Problem 14.9 (cont'd) 599,851 . 182,247 81,284 . 5,108 188,294 . 154,076 313,595 . -21,936 1,489 . 95 15,189 . 1,031 417,604 54,078 37,706 8,571 297,359 8,950 82,539 Convergence 0.00 THERMODYNAMIC DATA USED IN CALCULATIONS AH , (2500mm) Cp(kJ /kmol°C)=a-1>bT(°C)+cT2+1le3 AH__vap(25C) kJ/mol a bxlOA2 cxlO“5 dx10A9 kJ/kmol Coal* C02 . . . H2O (V) . . . 43,965.0 H20 (1) N2 802 02 ash* grate ash* , H=AHW + [0, dt=a(T —T,,,) TN] NTo— (T2 _TI:I)+§(T3 —TI:;f)+%(T4 _TI:I) 14-12 Problem 14.10 Given the performances of the scrubber in Problem 6 and the furnace in Problem 7, has the second EPA requirement of no more than 520 ng SOZ/J been met? ' with a basis of 100 kg dry coal/min — Q = 2,892,410 kJ/min 802 from each scrubber = 0.0055 kmol/min 802 from both scrubbers = 0.0109 kmol/min = 0.6987 kg/min = 6.987E+11 ng/min S02 from the scrubber/Q = 241,560 ng/kJ = 242 ng/J ere ore, " s anoaro 13 me w1 'IZ remova . Problem 14.11 — Q from basis = 2,892,410 kJ/min for a furnace with 1 MW of heat transferred equivalent to 0.39 Mwe Power Output = 500 MWe 1 MW per 0.39 MWe or Q_act = 500 MWe*(l MW/O.39 MWe)*(lOA6 W/MW)(1 MW/lOOO kJ/s)(3600 S/h) Q_act 4.6154E+09 kJ/h scale factor = Q_act/Q_basis = 1,596 min/h Applying the scale factor, the following flow rates are determined: (a)Coal feed rate 159,569 kg dry coal/h 167,228 kg wet coal/h 65,494 kmol/h 1,467,067 std cubic meters/min 1,601,414 cubic meters/min (b)Air feed rate (c)F1ue Gas Flow Rates basis component kmol/min 9,991 439,618 5,412 97,408 51,023 1,428,639 174 11,149 1,766 56,510 9,191 42.844 68,366 2,042,515 * (kg/mm) (d)Steam Generation: basis actual (kg/min) (kg/h) 921.8 1,470,836 14—13 Problem 14.12 For each scrubber Gas entering scrubber o / 34,183 36,652 flow feed flow flow flow m_recycle V_recycle fresh water flow of slurry to each scrubber of slurry from each scrubber of limestone to each blending tan of wet solids from each filter of filtrate from each filter = rate of flue gas to each scrubber rate of gas from each scrubber mo e rac 1.0000 1.0000 14—14 1,016,666 223,258 93,139 714,319 557 1,059,529 17,170,364 kg/h 17,171,940 kg/h 7,432 kg/h 17,404 kg/h 72,833 kg/h 34,183 kmol/h 36,652 kmol/h 908,593 mA3/h 980,381 mA3/h 72,833 kg recycle/min 71,959 L recycle/min 53,490 kg make-up water/h 52,848 L make—up water/h Problem 14.13 The flue gas is reheated so that its den51ty will be reduced and the gas will rise through the stack and into the atmosphere. Problem 14.14 (a) (b) Reject Alternative 1 because it calls for direct release of particulates and 502. Moreover, it would require an accurate on—line analysis of $02 in order to adjust the fraction of the flue gas that bypasses the scrubber. Such variations would also make it difficult to control the temperature of the gas sent to the stack. CH4 + 202 ———> C02 + 2H20 Basis: 71 kmol CH4/r1Value is assumed and then adjusted using the Goal Seek tool to bring the temperature of the stack gas to the specified value (80°C) for 10% excess 02 02 fed = 155.425 kmol 02/mi:= 1.10 x 02 required N2 fed = 584.693 kmol N2/mi = (79 kmol N2/21 kmol 02)XO2 fed H20 vapor fed = 11.7509 kmolH20/mi:= 02 fed x y_H20/y_02 T(°C) air 315 reference conditions are CH4 25 elements at 25°C and 1 atm. Comb Gases* 2,097 Reference 25 17 =AH ; + :[Cp dt=a(T _r,,,)._’2’(rz —T,ff)-o—:(T3 4,1, d(T‘ —T,:,) T u; 70.6 —74,850 ~5.288E+06 0.0 -38l,223 0.000E+00 11.8 —231,717 -2.723E+06 584.7 8,463 4.948E+06 0.0 0.000E+00 70.6 —278,305 —l.966E+07 153.0 -151,391 —2.317E+07 584.7 68,622 4.012E+07 155.4 8,950 1.391E+06 14.1 73,399 1.037E+06 —1.672E+06 *Converge O = enthalpy of inlet gas - enthalpy of outlet gas * Goal Seek tool is used to adjust the temperature of the combusion gases until enthalpies of inlet and outlet gas streams are the same. 14—15 Problem 14.14 (cont'd) The flow rate of gases entering the stack is twice that of the gas leaving each scrubber. Therefore the flows below are twice those from Problem 12. Ref Temp (°C) 25 Off Gas Temp (°C) 53 Stack Gas Temp (°C) =80 Scrubbed Flue Gas Stack Gas component 10,148 -392,444 —3.983E+09 10,349 —240,885 —2.493E+09 10,219 —391,396 —4.000E+09 10,502 —239,969 —2.520E+09 51,023 815 4.157E+07 51,608 1,602 8.269E+07 17 —392,369 —6.835E+O6 17 —391,252 —6.816E+06 1,766 827 1.461E+06 1,780 1,633 2.907E+06 -6.439E+09 74,126 —6.441E+09 Converge** 2.278E-06 ** Use Goal Seek to adjust flow rate of methane (basis in the above calculations) so that enthalpy of stack gas — enthalpy of scrubbed flue gas — enthalpy of combustion gases = O (c) Ref Temp (°C) 25 Off Gas Temp (°C) 53 Stack Gas Temp (°C) =80 Scrubbed Flue Gas Stack Gas —WW 10,148 —392,444 —3.983E+09 10,148 -391,396 -3.972E+09 10,349 —240,885 —2.493E+09 10,349 -239,969 —2.483E+09 51,023 815 4.157E+07 51,023 1,602 8.175E+07 17 —392,369 —6.835E+06 17 -391,252 —6.816E+06 1,766 827 1.461E+06 1,766 1,633 2.884E+06 73,303 -6.439E+09 73,303 —6.377E+09 Q = 6.175E+07 kJ/h = H_stackgas — H_scrubbedfluegas = 17,152 kW 16,270 Btu/s The heat released from burning coal was determined to be 28,924 kJ/kg dry coal This means that to provide the additional heating of the scrubbed flue gas, the following addition coal would have to be burned: 135 kg coal/h = Q/28,924 kJ/kg dry coal The coal useage is thereby increased 1.3% The heat required could be transferred to the flue gas (1) heating ambient air that is fed to the stack with the flue gas (this actually requires a somewhat greater amount of coal) or (2) by heating a portion of the flue gas and then injecting the heated flue gas into the unheated portion. In either case, the heating could be done by passing the appropriate stream through tubes in the furnace. by 14-16 Problem 14.14 (cont'd) THERMODYNAMIC DATA USED IN CALCULATIONS o 2 3 Cp(kJ/km01°C)=a+bT(CFcT +dT AH_vap(25C) kJ/mol a bx1002 cx1005 dX10“9 kJ/kmol Coal* 1.046 AH,@5°chm) C02 —393.5 36.110 4.233 —2.887 7.464 H20 (v) -241.83 33.460 0.688 0.760 -3.593 43965 H20 (1) 75.400 N2 0 29.000 0.220 0.572 —2.871 802 —393.5 38.910 3.904 —3.104 8.606 02 0 29.100 1.158 —0.608 1.311 CH4 —74.85 34.310 5.469 0.366 —11.000 ash* 0.921 grate ash* .921 *kJ/kg Problem 14.15 Flow rate of stack ga 73,303 kmol/h V = 2.133E+06 m‘3/h v*nd“2/4 d = (4*V/Vfl)“(0.5) v (m/s) d (m) d (ft) _—‘_5 "—12—.29" " ‘40".3 6 11.21 36.8 7 10.38 34.1 8 9.71 31.9 9 9.16 30.0 10 8.69 28.5 11 8.28 27.2 12 7.93 26.0 13 7.62 25.0 14 7.34 24.1 15 7.09 23.3 Effect of Gas Velocity on Stack Diameter 14.00 A1200 E 7:10.00 E Q E, 8.00 '5 6.00 15 :9 4.00 0’ 2.00 0.00 o 5 10 15 20 Gas Velocity (mls) 14-17 Problem 14.16 From Problem 11(d): Steam flow rate in power cycle 1,470,836 kg/h Cooling Water In, 25°C Wet Steam 27.5% liquid, 6-55 kPa Cooling Water Out, 28°C Steam Condensate 6.55 kPa Evaluate enthalpies of liquid and vapor by interpolation from steam tables, pressure table (Table B.6): 6.55 kPa —-—> .0655 bar Enthalpy of H20 (sat liq, 6.55 kPa) 158.0 kJ/kg Enthalpy of H20 (sat vap, 6.55 kPa) 2,570.3 kJ/kg Enthalpy of wet steam fed to condenser = frac vapor x H_vapor + frac liq x H_liq = 1906.9 kJ/kg Enthalpy of condensate = 158.0 kJ/kg Heat removal required for condensation = steam flow rate x (H_vap — H_liq) = 2.572E+09 kJ/h Water flow rate to condenser = heat removal rate/Cp(T_out — T_in)_water = 2.046E+08 kg/h Problem 14.17 From a mechanical energy balance: eff*W_s/m_condensate = A P/p = [(24.1 — 0.0066) MPa/lOOO kg/m3] x [1.01325 x 10A5 N/m“2/101.325 kPa] x [1000 kPa/MPa] x [1 kJ/lOOO N m] x [1 kW s/kJ] = 24.09 kW 5 W_s = 12,305 kW 16,501 hp 14-18 Problem 14.18 Enthalpy of steam leaving the furnace* 3321 kJ/kg Enthalpy of wet steam entering the condens« 1907 kJ/kg A H_steam for process 1414 kJ/kg steam flow rate 1,470,836 kg/h Total enthalpy change of steam 2.080E+O9 kJ/h 578 MW Steam cycle efficiency 8..5% * from Problem 14.7 (or steam tables) Problem 14.19 The use of two criteria 18 to ensure that both low—sulfur em1351ons (520 ng SOZ/J) are achieved and that appropriate technology is used (90% removal of 802). If these standards were not applied together, then it would be possible for users of low-sulfur coal to forego use of flue—gas scrubbing. Likewise, users of high—sulfur coal could emit significant quantities of 802 and still be in compliance without the 520 ng/J...
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