This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 143 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 144 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 145 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 offgas  water entering absorber with off—gas from furnace = 33.52 kg makeup water/min
33.52 L make—up water/min 147 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(24.10.00655)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* 1 149 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 1411 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)=a1>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) 1412 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 makeup 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** 2.278E06 ** 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 1416 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) 1417 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, 655 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 1418 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 lowsulfur 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...
View
Full Document
 Spring '08
 wetzel

Click to edit the document details