4150-02-B-E4
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Transcript of 4150-02-B-E4
MATAKULIAH : TERMODINAMIKA IIMODEL TUGAS/QUIZ : QUIZ- ETS-04MATERI : SIKLUS RANKINE DG REHEATAWAL TUGAS : 12 MEI 2015AKHIR TUGAS : 13 MEI 2015, ,JAM 10.00 PAGI
Melalui email : [email protected]
NAMA : Much. Shochib ChabibiNBI/KLAS : 421204150-B-Q-ETS-04email :
Perhatikan gambar dibawah ini :
Diketahui :
P1 = 344 + 50 = 394 Psia
P3 = 7000 + 50 = 7050 Psia
T3 = 1570 ℉
P5 = 2600 + 50 = 2650 Psia
T = 1700 ℉
P=7000 psia + 2 NBI dan T=1570 F
P=2600 psia + 2 NBI dan T=1700 F
P=344 psia + 2 NBI
Tentukan : Effisiensi siklus tsb
Untuk mencari h3 dan s3, dengan menggunakan tabel b2, dengan menginterpelasi 1570 F
dengan P = 7050 psia
P(psia)Temperatur
1500 1570 1600
7000h (Btu/lbm) 1714,4 1783,5
s (Btu/lbm 0R) 1,5751 1,6096
7050h (Btu/lbm) 1713,765 1762,174 1782,97
s (Btu/lbm 0R) 1,57397 1,598246 1,60865
8000h (Btu/lbm) 1701,7 1772,9
s (Btu/lbm 0R) 1,5552 1,5906
h3= 1762,174 s3=1,598246
P(psia)Entalpi Volume
hf Vf
300 394,1 0,01890394 422,394 0,01931400 424.2 0,01934
hfp1=422,394Btu/lbmVfp1=0,01931ft3/lbm
P(psia)Entropi
Sf Sfg
2600 0,9237 0,29712650 0,9348 0,2725
3203.8 1,0582 0,0000
Sfp4 = 0,9348 Btu/lbm0R
Sfgp4 = 0,2725 Btu/lbm0R
X4 = S3−S fp4
S fgp 4 =
1,598246−0,93480,2725 = 2,43
Untuk mencari h4, dengan uji coba interpelasi pada P = psia 2650
P(psia)Temperatur1100 1200
2000h (Btu/lbm) 1537,2 1598,6
s (Btu/lbm 0R) 1,6019 1,6400
2650h (Btu/lbm) 1519,52 1584,3
s (Btu/lbm 0R) 1,5640 1,60425
3000h (Btu/lbm) 1510,0 1576,6
s (Btu/lbm 0R) 1.5436 1.5850
S3 = S4
P = 2650 psia 1200 F 1300 Fh (Btu/lbm) 1519,52 h4 = 1574,64 1584,3s (Btu/lbm 0R) 1,5640 s4 = 1,598246 1,60425
P(psia)Temperatur
1700
2000h (Btu/lbm) 1898.1
s (Btu/lbm 0R) 1.7978
2650h (Btu/lbm) 1891,8
s (Btu/lbm 0R) 1,7663
3000h (Btu/lbm) 1888.4
s (Btu/lbm 0R) 1.7494
h5 =1891.8 Btu/lbm 0R.
S5 =1.7663 Btu/lbm 0R.
P(psia)Entropi
Sf Sfg
300 0,5885 0,9232394 0,6199 0,8674400 0,6219 0,8639
Sfp6 = 0,6199 Btu/lbm0R
Sfgp6 =0,8674 Btu/lbm0R
X6 = S5−S fp6
S fgp 6 =
1.7663−0,61990,8674 = 1,32
Untuk mencari h6, dengan uji coba interpelasi pada P = 394 psia.
P(psia)Temperatur1000 1100
300h (Btu/lbm) 1526,4 1579,8
s (Btu/lbm 0R) 1,7966 1,8319
394h (Btu/lbm) 1523,8 1577,5
s (Btu/lbm 0R) 1,7654 1,7823
400h (Btu/lbm) 1523.6 1577.4
s (Btu/lbm 0R) 1.7634 1.7991
S5 = S6
P = 435 psia 1000 F 1100 Fh (Btu/lbm) 1523,8 h6 = 1526,66 1577,5s (Btu/lbm 0R) 1,7654 S6 = 1.7663 1,7823
h2= h1+V1(P3–P1)
h2= 422,394 + 0,01931 ft3/lbm (7050 – 394) lbf/in2 x 144 in2/ft2 x Btu/778lbf.ft
= 422,394 + 23,79 = 446,184 Btu/lbm
η = W t1+W t 2−W p
QB+QR=
(h3−h4 )+(h5−h6 )−(h2−h1)
(h3−h2 )+(h5−h4)
=(1762,174 –1574,64 )+(1891,8– 1526,66 )−(446,184−422,394)
(1762,174−446,184 )+(1891,8−1574,64 )
η = 187,534+365,14−23,79
1315,99−3 17,16=
528,884998,83
= 0,5295 = 52,95 %