1
BENHA UNIVERSITY/ FACULTY OF ENGINEERING MECHANICAL ENGINEERING DEPARTMENT FOURTH YEAR MECHANICAL POWER FINAL EXAMINATION TIME ALLOWED: 4 HRS DATE: May 4th 2009 MAXIMUM MARKS: 100 TRADITIONAL POWER STATIONS
QUESTION ONE 25 MARKS A power station works at the critical state of pressure and temperature, has an installed rated capacity of 400 MW. The ambient temperature in the o boiler house is 31 C, air humidity is 0.0204 kg/kg dry air, velocity of wind around the boiler =3 m/s and the ratio of bottom ash to fly ash = 90:10. Assume barometer reading is 760 mm of mercury. Percentage composition of coal fuel input to the power station is taken as: C=88%, H2=4.3%, N2=0.7%, O2=4%, S=1%, Ash=2% Calculate I. The gross calorific value of the fuel. solution CVH= [4.18/100][8080 C% + 34500 (H%-O%/8)+ 2220S%] = [4.18/100][8080*8 + 34500 (4.3-4/8)+ 2220*1] = 35294.24kJ/kg CVL = CVH – mw*4.18*588.76 = 35294.24 – (0.043x18/2)*4.18*588.76 (0.043x18/2)*4.18*588.76 = 34341.83 kJ/kg =GCV II. The minimum air/fuel ratio. product of combustion kg /kg i f
kgO2 /kgf
Mi
C=0.88
2.346667 CO2=44/12
yi=xi*Mi
%
Mi
= 3.226667 0.243535 44
10.71555
0.55804 55.80401 24.55377
O2=0.04 -0.04
O2=0.25x2.66
=0.665167 =0.665167 0.050204 32
1.606527
0.083664 8.366409 2.677251
N2=0.007
N2=2.66x77/23 =8.950449 =8.950449 0.241904 28
6.773313
0.352738 35.27379 9.876662
H2=0.043 0.344
H2O=0.043x18/2=0.387 O=0.043x18/2=0.387
0.029209 2
0.058418
0.003042 0.304228 0.006085
S=0.01
SO2=0.01x64/2 = 0.02
0.00151 32
0.048305
0.002516 0.251558 0.080499
Sum
0.01
2.660667 Sum
Ash=0.02 (a/f) min=
13.24928
138 19.20211
1
2.6601x100/23= 11.57
Minimum mass of air= [2.66066x100/23] [2.66066x100/23] =11.57 kg a/kgf (A/f)min = 11.57 kga/kgf
100
37.19426
Rg=
0.223637
2
III. If 25% excess air is supplied compute the percentage composition by volume of the dry flue gases. Compute the gas constant of the flue gases. solution Product of combustion CO2 0.243535 =23.4% O2 0.050204 =5.02% N2 0.241904 =24.19% H2O 0.029209 =2.92% SO2 0.001510 =0.151% R g = 0.223637= 0.223637 kJ/kgK
IV. If a chain stoker is used to burn this fuel having grate width 5 m find the grate length. You may assume any reasonable with justification. Draw the chain grate stoker used showing dimension calculated. solution Consider the overall efficiency is being=33%=1/3 Thus =400MW/Qf = 1/3 Qf = 3x400MW =1200 MW=CVLxmf =34341.83xmf mf = 1200x1000/(34341.83)= 34.94281kg/s= 1174078 kg/h 5
2
Astoker =mf xCVL/HA = 1174078x34341.83/(15x10 x4.18)= 643.0622 m Lstoker = Astoker /W=643.0622 /5=128.6124m about 130 m
V. The steam produced from this boiler is used to run a Regenerative steam power cycle. Assume cycle with suitable assumption for number of feed water heaters. Calculate: performance on a) overall cycle efficiency, b) specific steam and c) specific fuel consumption. Solution o
At T =374.14 C, P =220.9 bar: c ritical c ritical o
h1= 2138.2 kJ/kg, T1= 374.14 C , P1 =220.9 bar o o Assume the economizer temperature rise = 40 C of range (30 to 50 C) o TfB= 374.14 – 40 = 334.14 C o Pcondenser = 0.07 bar, Tf, condenser =39.024 C, hcondenser =163.32 kJ/kg Temperature rise in the feed water heaters = 334.14 – 39.024
3
o
=295.116 C Number of feed water heaters = 295.116/50 = 5.90232 = 6 o Temperature per FWH = 295.116/6 = 49.186 C hB=h1- 4.18 *TfB = 2138.2 – 4.18 * 334.14 = 741.4948 kJ/kg Enthalpies:
h2'=1391.6 kJ/kg, x2'=0.51 h2 = h1-T(h1-h2') = 2138.2-o.80*(2138.2-1391.6) = 1540.92 kJ/kg Water températures: o T3 =39.024 C =T4 o T5 = 39.024+49.186 = 88.21 C o Tf,pf =88.21+3=91.21 C Pf =0.734bar hh'= 1579.7kJ/kg hh=2138.2-o.80*(2138.2-1579.7)=1691.4kJ/kg o T6 = 88.21+49.186=127.234 C o Te,pf =127.234+3=130.234 C Pe=2.72bar he'=1699.9kJ/kg, he=2138.2-o.80*(2138.2-1699.9)= 1787.16kJ/kg o T7 = 127.234+49.186 =176.42 C o Td,pf =176.42+3=179.42 C Pd=9.894bar hd'=1829.2kJ/kg o T8=176.42+49.186=225.606 C o o Tc,pf =225.606 +3=228.606 C Pc=27.279bar, hc'=1935.2kJ/kg hc=2138.2-o.80*(2138.2-1935.2)= 1975.8kJ/kg o T9 = 225.606+ 49.186 = 274.792 C o T b,pf =274.792 +3=277.792 C P b=62.072bar, h b'=2040.7kJ/kg h b=2138.2-o.80*(2138.2-2040.7)=2060 kg o T10 = 274.792+ 49.186 = 323.978 C o o T b,pf =(274.792 +324)/2 C= 299.396 C Pa= 85.158 bars ha= 2063.5 kJ/kg
4
o
Teconomizer = 323.978+ 50 = 374 C heconomizer = 374*4.18 = 1563.32kJ/kg Heat Balance of FWHs: Assume sixth FWH is open feed water heater (1-ma)(c p*T9) +ma ha=c pT10 ma= 4.18*49.186/(2063.5-4.18*274.792)= 0.224729 The fifth FWH Heat balance: m b*(h b – cp*T9) = (1-ma)*c p*(T9 – T8) m b=(1-ma)*c p*(T9 – T8)/ (h b – cp*T9) = (1-0.0545)*4.18*(49.186)/ (2060 – 4.18*274.792) = 0.174226 mc=(1-ma-m b)*c p*(T8 – T7)/ (hc – cp*T8) =(1-0.0479-0.21)*4.18*(49.186)/ (1975.8 – 4.18x225.606) = 0.142677 Mass fraction: ma=0.224729 mb=0.174226 mc=0.142677 md=0.0761 me=0.004914 mh=0.054681 Wnet=h1-ha+(1-ma)(ha-h b)+(1-ma-m b)(h b-hc)+(1-ma-m b-mc)(hc-hd)+ (1-m=a-m b-mc-md)(hd-he)+(1-ma-m b-mc-md-me)(he-hh)+ (1-ma-m b-mc-md-me-mh)(hh-h2) = 342.5379kJ/kg ms=400X1000/(0.9x342.5379) =1297.504kg/sec = 4671.016 t/h Heat addition = h1-h9=783.972kJ/kg Thermal efficiency=wnet/qadd=342.5379/783.97= 0.436926=43% SSC=3600/ mwnet=11.67754kg/kwh SFC=3600/ (0.8x0.436926x35294.24)=0.29181kg/kWh
5
Chinemy
1
Turbine
superheater
2 Condenser
Coal Fuel 3
4
5
130m
11 6
Economizer
7 10 8
9
6
VI. Calculate: The boiler capacity in tons/hour, and the firing fuel rate in o tons/h, boiler feed water temperature in C, and % CO2 in flue gases, % CO in flue gases. Solution ms= 400X1000/(0.9x342.5379) =1297.504kg/sec = 4671.016 t/h choice of water tube mf = sfc *400=116.7241t/h < 117.4078 t/h of stoker design. C moles N 2 moles CO 2th
88 100 * 12 77 * Airth
0.073333 %N 2fuel
100 * 28 100 * C moles N 2moles
C moles
0.809
0.149
Execss Air Supplied,% EAS
7900 * (CO 2th - %CO 2flue ) %CO 2flue * (100
CO 2th )
25%
VII. Show the method of firing this fuel and give description for it. i) Hand firing. ii) Semi mechanized firing iii) Stoker firing. VIII. Describe the types of draught and compute the height of the chimney to get static draught of 22 mm of water if the mean flue gas temperature in o the chimney is measured as 190 C. Solution P= (air -gas) H=P[(1/R aTa)-(1/R gTg)]H 5 =1.013x10 [(1/ (287x304)-1/(224x463)] -3 = w hw =1000x22x10 =22 5 H=22/ {1.013x10 [(1/(287x304)-1/(224x463)]}=120.9127 m
IX. Compute: all losses in the boiler and its efficiency if the surface area of 2 the boiler of heat transfer equals 600 m . AAS
(1
EAS 100
) * Airth =1.25*11.52=14.4
7
L1
100 * 14.38273 * 0.23 * (190
31)
34341.83
100 * 9 (
%H 2fuel 100
L2
) 584
=1.531592
0.45(Tflue
Tatm ) * 4.18
CV 100 * 9 (
4.3 100
) 584
0.45(190
31) * 4.18
34341.83 8.208255
L3
L4
L5
M moisture
584
0.45 T f
T a
4.18
CV
0 100 * 14.4 * 0.0204 * 0.45 * (190 34341.83 12.54 %[ 14.9 88] 5744 %[ 14.9
88]34341.83
31)
* 4.18
4.18
8.908
L6= 0.548 x [[(70+273)/55.55] – [(31+273)/55.55] +1.957 x(70-31) 0.5 2 x[(196.85x3 +68.9)/68.9] ]*4.18= 628.405 W/m L 6 (%)
L7
100
4
4
1.25
100 * 628.405 * 600x3600 3
116 .724 x 10 33.86175098
x 34341.83
Total Ash Collected/ kg of Fuel Burnt * CVflyash CV
0.058 L8
100
Total Ash Collected/ kg of Fuel Burnt * CVbottomash CV
5.24 L1
L2
L3
L4
L5
L6
L6%
L7
L8
1.531 0.00446
8.208 0.023
0 0
12.54 0.0365
8.908 0.0259
628.405
33.86 0.098
0.058 0.00017
5.24 0.00017
Efficiency of the boiler = 100-20.5=79.5%
Sum 70.35 0.205
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X. \Find: the ratio of circulating water to steam flows if the condenser pressure is 0.07bar and the cooling tower cools the water to 25 oC. Design and draw the condenser & feed water heaters : that would handle and drains water to the boiler. Solution
Q AUΔUm ΔTm
(Ts i
Tw i ) (Ts o Tw o ) Ts i Tw i ln Ts o Tw o
U C1 C 2 C3 C 4
, Ts i
Ts o
V
A Total Surface Area of Tubes .
Q m C w Tw o
Tw i
Water inlet velocity is limited to a maximum 2.5 m/s to minimize erosion, and a minimum 1.5 to 1.8 m/s for good heat transfer. Values between 2.1 to 2.5 m/s are most common. C1: Dimensional factor depending upon tube outer diameter : Tube outer diameter- inch C1 3/4 = 1.905 cm 2777 7/8 = 2.2225 cm 2705 1 = 2.54 cm 2582
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C2: Dimensionless correction factor for circulating water inlet temperature: Twi, o F
35
40
45
50
55
60
70
80
90
100
C2
0.57
0.64
0.72
0.79
0.86
0.92
1
1.04
1.08
1.1
C3: Dimensionless correction factor for tube material: Tube Material
304 Stainless Steel
Admiralty arsenic copper
Aluminum brass Muntz Metal
Aluminum Bronze 90-10 Cu-Ni
70-30 Cu-Ni
18 gauge
0.58
1
0.96
0.9
0.83
17 gauge
0.56
0.98
0.94
0.87
0.8
16 gauge
0.54
0.96
0.91
0.84
0.76
C4: Dimensionless cleanliness factor C4 = 0.85 for clean tubes, less for algae covered or sludge tubes Design of the condenser & Heaters Item
U
mw/ms
DTm
A
N
Ncorrect
A
condenser
2.785566
33.15677
18.4429
26977.76
2260.958
2261
2697.825
H
3.722131
0.452109
39.65229
629.7971
52.78219
53
63.2396
E
3.722131
0.037237
39.65229
51.87227
4.347324
5
5.966
D
3.722131
0.527587
39.65229
734.94
61.59403
62
73.9784
C
3.722131
0.880384
39.65229
1226.394
102.7819
103
122.8996
B
3.722131
0.941797
39.65229
1311.943
109.9517
110
131.252
mcwxC pxTcw=msX(h2-h3)(1-ma) mcw/ms=(h2-h3)(1-ma)/C pxDTcw = 33.155677
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QUESTION TWO 25 MARKS The daily load curve for a power station is given by the following equation: 2 L=350 + 10t -t , where t is the time in hours from 0 to 24 hours and L is in MW. Calculate: I. Draw the daily load curve and daily duration curve. t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
L 359 366 371 374 375 374 371 366 359 350 339 326 311 294 275 254 231 206 179 150 119 86 51 14
I 62.36838667 64.77420889 66.52052 67.57945333 67.93428889 67.57945333 66.52052 64.77420889 62.36838667 59.34206667 55.74540889 51.63972 47.09745333 42.20220889 37.04873333 31.74292 26.40180889 21.15358667 16.13758667 11.50428889 7.41532 4.043453333 1.572608889 0.197853333
0.173728 0.176979 0.179301 0.180694 0.181158 0.180694 0.179301 0.176979 0.173728 0.169549 0.164441 0.158404 0.151439 0.143545 0.134723 0.124972 0.114294 0.102687 0.090154 0.076695 0.062314 0.047017 0.030835 0.014132
Hr 5.756121 5.650397 5.577226 5.534226 5.52004 5.534226 5.577226 5.650397 5.756121 5.898008 6.081218 6.31297 6.603329 6.96646 7.422656 8.001784 8.749401 9.738301 11.09212 13.03862 16.04786 21.26895 32.43019 70.75949
11
400 350 300 L I Efficiency HR
250 W200 M 150 100 50 0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 time, hour
Variation of L, I versus t
II.
III.
IV. V.
VI.
VII.
Peak load and its time: 2 L=350+10t-t dL/dt=10-2t*= 0 , thus t*=5 h, 2 L p=350+10x5-(5) = 375 MW Average load: 24 24 2 2 3 Lav = [1/24]∫ (L=350+10t -t )dt =[1/24][350t+5t -t /3] 0 0 = 274 MW Load factor: L.f = Lav/L p = 274/375 = 0.74 Reserve factor of that station: R.f = LR /LP = (5x80)/(375) = 1.067 Average heat rate and efficiency for a day: 6 2 av=Lav/Iav =[274]/[4.18x10 (8+6x274+0.4x(274) )] = 0.180694 HR= Iav/Lav = 5.534226 Input-time curve for a day: see Figure.
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QUESTION THREE 25 MARKS Consider a combined reheat and regenerative cycle utilizing steam as the working fluid. The net power output of the turbine is 400MW. Steam enters the high-pressure turbine at 8 MPa, 550oC, and is extracted for purpose of feed-water heating in closed heater at 1.2 MPa. The remainder of the steam o is reheated at this pressure to 550 C and fed to the low-pressure turbine. Steam is extracted from the low-pressure turbine at 0.4 MPa for feed-water heating in an open heater. The condenser pressure is 10 kPa. The condensate from the closed heater is throttled into the open heater. The isentropic efficiencies of high and low pressure turbines are 90% and 85% respectively, and the pump isentropic efficiency is 80%. I. Draw a line diagram of the unit and show the state points on a T-s and h-s diagrams.
II.
What is the steam flow rate to the high-pressure turbine?
III.
What power motor is required to drive each of the pumps?
IV.
o If there is a 10 C rise in the temperature of the cooling water, what is the rate of flow of cooling water through the condenser?
V.
If the velocity of the steam flowing from the turbine to the condenser is limited to a maximum of 120 m/s. What is the diameter of this connecting pipe?
Solution
h1= 3520 kJ/kg, h2’=2960 kJ/kg, h2 = 3016 kJ/kg, h 3 = 3582 kJ/kg, h 4’ = 2480 kJ/kg, h4 = 3274.3 kJ/kg, h5 = 2645.3 kJ/kg, h 6 = 191.81 kJ/kg, h 7’ = 191.81+0.001(400-10)=192.2 kJ/kg, h7 = 192.3 kJ/kg, h 8 = 604.73 kJ/kg, h9’ = 604.73 +0.001(8000-400)=614.33 kJ/kg, h10 = 798.641 kJ/kg, h 11 = h10 = 798.641 kJ/kg, h 12 = 798.641 kJ/kg. m1= 0.0832, m2 =0.117, wnet= 1279.18 kJ/kg, ms= 312.7 kg/s, W p1=122.54 kW, W p2 =2970.65 kW, mcw= 1467.1 kg/s, D=6.515 m. h1= 3520 kJ/kg, h2’=2960 kJ/kg, h2 = 3016 kJ/kg, h 3 = 3582 kJ/kg, h 4’ = 2480 kJ/kg, h4 = 3274.3 kJ/kg, h5 = 2645.3 kJ/kg, h 6 = 191.81 kJ/kg, h 7’ = 191.81+0.001(400-10)=192.2 kJ/kg, h7 = 192.3 kJ/kg, h 8 = 604.73 kJ/kg, h9’ = 604.73 +0.001(8000-400)=614.33 kJ/kg, h10 = 798.641 kJ/kg, h 11 = h10 = 798.641 kJ/kg, h 12 = 798.641 kJ/kg. m1= 0.0832, m2 =0.117, wnet= 1279.18 kJ/kg, ms= 312.7 kg/s, W p1=122.54 kW, W p2 =2970.65 kW, mcw= 1467.1 kg/s, D=6.515 m.
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QUESTION FOUR
25 MARKS
A combined-cycle unit combines the Rankine (steam turbine) and Brayton (gas turbine) thermodynamic cycles by using heat recovery boilers to capture the energy in the gas turbine exhaust gases for steam production to supply a steam turbine as shown in the Figure "Combined-Cycle Cogeneration Unit". The Process steam can be also provided for industrial o purposes for 20% of steam produced. Take Tamb=25 C, relative air humidity, =50%, maximum temperature after the combustion chamber = 1100 K , and pressure ratio of the compressor = 30:1. Assume reasonable assumption with justification to design the combined cycle shown for each pressure and temperature and mass flow rate of steam per kg of circulating air in the compressor. Calculate: a) the thermal efficiency and b) SFC if the fuel Calorific value is 6x104kJ/kg.
Solution Gas Turbine Cycle T1, K
Steam Cycle 298
T2, K
787.5006
T3, K
1100
T4, K
50% DBT
25
WBT
18
416.2536
Tcooling Tower
376
Tcondenser
T5, K
h8 h9
With Best and Great Success
19.4 32.4
135.432
292.6
Dr Atef A Aly
