Jawaban Mke Uts NUGROHO
3
Nama : NUGROHO TOHARI S N B I : 421103985 Kelas : A IKUT KLAS R state Temperature (F) Pressure (Psia) Entropy Btu/lbm.R Enthalpy (Btu/lbm) 1 1050 2000 1.5811 1505,65 2 640 500 1.5811 1322.2 3 1050 500 1.7536 1547,6 4 162 5 1.7371 1063 5 162 5 0.2336 129.5 6 162 2000 0.2336 129.5 JAWABAN ETS MKE KELAS R – 421103985 State 1 P = 2000 Psia T = 1050 (interpolasi) P=2000 Psia T 1000 F T 1060 F T 1100 S (Btu/lbm.R) 1.5603 1.5811 1.6019 h (Btu/lbm) 1474.1 1505,65 1537.2 S = 1.5603 + {(1050-1000) (1.6019 – 1.5603 / 1100 – 1000)} = 1.5811 Btu/lbm h = 1474.1 + {(1050-1000) (1537.2-1474.1 / 1100 – 1000)} = 1505,65 Btu/lbm.R State 2 S2 = S1 pada reheat P=500 Psia (interpolasi) S (Btu/lbm.R Pada 500 Psia 1.5810 1.5811 1.5915 Temperature 640 640 660 h(Btu/lbm) 1322 1322,2 1334.2 T = 640 + {(1.5811-1.5810) (660 – 640) / (1.5915-1.5810)} = 640 f h= 1322 + {(1.5811-1.5810) (1334,2 – 1322 ) / 1.5915 – 1.5810 ) = 1322,2 Btu/lbm.R State 3 Reheat konstan S4=S3 , T3=T1 Pada P = 500 Psia T = 1050 F (interpolasi ) S= 1.7363 + {(1050- 1000) (1.8056-1.7363) / (1200-1000)}=1,7536 Btu/lbm.R h= 1519 + {(1050-1000) / (1627.6-1519.6) / (1200-1000) = 1546,6 Btu/lbm P = 500 Psia T 1000 F T 1060 F T 1200 S (Btu/lbm.R) 1.7363 1,7536 1.8056 h (Btu/lbm) 1519.6 1546,6 1627.6
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Transcript of Jawaban Mke Uts NUGROHO
Nama : NUGROHO TOHARI SN B I : 421103985Kelas : A IKUT KLAS R
stateTemperature
(F)Pressure
(Psia)Entropy
Btu/lbm.REnthalpy(Btu/lbm)
1 1050 2000 1.5811 1505,652 640 500 1.5811 1322.23 1050 500 1.7536 1547,64 162 5 1.7371 10635 162 5 0.2336 129.56 162 2000 0.2336 129.5
JAWABAN ETS MKE KELAS R – 421103985
State 1 P = 2000 Psia T = 1050 (interpolasi)
P=2000 Psia T 1000 F T 1060 F T 1100S (Btu/lbm.R) 1.5603 1.5811 1.6019
h (Btu/lbm) 1474.1 1505,65 1537.2
S = 1.5603 + {(1050-1000) (1.6019 – 1.5603 / 1100 – 1000)} = 1.5811 Btu/lbm
h = 1474.1 + {(1050-1000) (1537.2-1474.1 / 1100 – 1000)} = 1505,65 Btu/lbm.R
State 2 S2 = S1 pada reheat P=500 Psia (interpolasi)
S (Btu/lbm.RPada 500 Psia
1.5810 1.5811 1.5915
Temperature 640 640 660h(Btu/lbm) 1322 1322,2 1334.2
T = 640 + {(1.5811-1.5810) (660 – 640) / (1.5915-1.5810)} = 640 f
h= 1322 + {(1.5811-1.5810) (1334,2 – 1322 ) / 1.5915 – 1.5810 ) = 1322,2 Btu/lbm.R
State 3 Reheat konstan S4=S3 , T3=T1 Pada P = 500 Psia T = 1050 F (interpolasi )
S= 1.7363 + {(1050-1000) (1.8056-1.7363) / (1200-1000)}=1,7536 Btu/lbm.R
h= 1519 + {(1050-1000) / (1627.6-1519.6) / (1200-1000) = 1546,6 Btu/lbm
State 4 Turbin kedua S4=S3 pada P= 5 Psia
X= (1,7536 - 0.2336) / 1.6124 = 0,942
h = 129.45 + (0.942 * 1001.3) = 1072,6 Btu/lbm.R
P = 500 Psia T 1000 F T 1060 F T 1200S (Btu/lbm.R) 1.7363 1,7536 1.8056
h (Btu/lbm) 1519.6 1546,6 1627.6
P = 5 Psia
Sf = 0.2336 Btu/lbm.RSfg = 1.6124 Btu/lbm.R
Hf = 129.45Btu/lbmHfg = 1001.3Btu/lbmV = 0.016445 Ft ³/lbm
Nama : NUGROHO TOHARI SN B I : 421103985Kelas : A IKUT KLAS R
State 5 keluar kondenssor kondisi saturate pada 5 Psia
P = 5 PsiaSf = 0.2336
Hf = 129.45Btu/lbmT = 162.24 F
State 6 Kerja pompa h6 =h5 T6 = T5
JAWABAN ;
a) Q = (15O5,65– 129.5 ) + (1546,6 – 106.3) =2816,45 Btu/lbm
b) Wt1 = (1511.96 – 106.3) = 448.96 Btu/lbm
Wt2 = 1372 - 1327.24 = 44.76 Btu/lbm
Wtotal = 448.96 + 44.76 = 493.72 Btu/lbm
Wbersih = 493.72 * 85% =419.66 Btu/lbm
c) Efisiensi thermal (Ƞ) = 493.72/1691.46 =0.29 = 29 %
d) Kualitas pintu keluar turbin (X)= uap / campuran = (S - Sf)/Sfg
= ( 1.7336-0.2336) / 1.6124 = 0.942 =94.2 %
e) Gambar skema diagram aliran dan diagram T –s
