can you draw and schematic and t-s fiagram for this actual reheat rankine cycle
can you draw and schematic and t-s fiagram for this actual reheat rankine cycle
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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can you draw and schematic and t-s fiagram for this actual reheat rankine cycle
State 1:
Status: Saturated Liquid
30 kPa
From Table A5 when
ʋ1=ʋf = 0.001022 m3/kg
h1=hf = 289.27 kJ/kg
State 2a:
Status: Compressed liquid
P2s=PH= 6000 kPa
h2a= 296.04 kJ/kg [
WP, ise=
Wpact =
State 3
Status: Superheated Steam
P3=PH=6000kpa
T3=Tmax= 550°C
From Table 6A, When = 6000 kPa and =550
=3423.1 kJ/Kg
=6.8826 kJ/Kg⸼K

Transcribed Image Text:State 4s
P4s = P₁= 2000 kPa
S4s S3=6.8826 kJ/kg.K
According to Table A5, when pressure 2000 kPa
Sf= 2.4467 kJ/kg.K
Sfg=3.8923 kJ/kg.K
Sg= 6.3390 kJ/kg.K
hf=908.47 kJ/kg
hfg= 1889.8 kJ/kg
hg=2798.3 kJ/kg
Since S4s is larger than Sg and larger than Sf at the pressure of 2000 kPa, the status at 4s is
superheated steam.
By interpolation at S4s 6.8826 kJ/kg from table A6
6.8826-6.7684
6.9583-6.784
Wturn*isen = h3 h4s=3423.1-3098.56 =324.54 kJ/kg
Wturb*act = WHPTX - nisenxturb= 0.86x324.54=279.1044 kJ/kg
h4s
=
State 4a
P4a = PI
h4a
h4a
=
=
(3137.7 3024.2) (3024.2) 3098.56 kj/kg
2000kPa
h3 -Wturb act =3423.1 -279.1044 =
3143.99 kPa
Since ha is larger than hg = 2798.3 kJ/ kg at the pressure of 2000 kpa, the status at 4a is superheated
steam.
State 5
Status: Superheated Steam
P5= P₁= 2000 kPa
Tr=Tr = 400
From Table A6 when P5 = 2000 kPa and T5 = 400 °C
h5 = 3248.4 kJ/kg
S5 7.1292 kJ/kg x k
=
Qin2
=
h5 -h4a = 3248.4-3143.99 = 104.41 kJ/Kg
State 6s
Status: Superheated Steam
Pos PL= 30 kPa
S6S = S5 = 7.1292 kJ/kg
According to Table A5, when pressure of 30 kPa, the status at 6, is saturated mixture.
Sf= 0.9441 kj/kg k
Sfg= 6.8234 kj/kg k
Sg- 7.7675 kj/kg k
hf 289.27 kj/kg
hfg=2335.3 kj/kg
hg=2624.6 kj/kg
X6s
=
Since S6s is smaller than Sf at the pressure of 30 kPa, the status at 6s is saturated mixture.
-
S6s-sf 7.4337-0.9441
Sfg
6.8234
hos = 2405.0518 kj/kg
WLpt isen=
= 0.906
n=h5-h6s =3248.4-2405.0518=843.3482 kJ/kg

Transcribed Image Text:WLpt act=WLpt isen xnisenxturb =843.3482 x 0.86 = 725.279 kJ/kg
State 6a
h6a-h5-WLpt act= 3248.4 - 725.279= 2523.121 kj/kg
Since ha is larger than hg at the pressure of 30 kPa, the status at 6a is superheated steam
Qout-hoa-h₁-2523.121 — 289.27 = 2233.851 kJ/kg
Checking if W,
Qin-lin1+Qin2 = 3127.06+109.659= 3236.711 kJ/kg
Onet-lin-Qout = 1002.860 kJ/kg
Wout WHPT act + WLPT act = 725.279+284.359= 1009.6384 kJ/kg
Wnet Wout + Win =13009.63+6.77= 1002.860 kJ/kg
Since Wnet-Qnet, the Rankine cycle obey the 1st law of thermodynamics and the calculation is correct.
net=Qout
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