Draw a t-s diagram of a simple rankine diagram witht his data
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
Related questions
Question
Draw a t-s diagram of a simple rankine diagram witht his data
![State 4s
P4s PL = 30 kPa
=
S4s S3=6.8826 kJ/kg.K
According to Table A5, when pressure 30 kPa
Sf= 0.9441kJ/kg.K
S.fg
= 6.8234 kJ/kg.K
Sg= 7.7675 kJ/kg.K
hf= 191.8 kJ/kg
hfg=2392.2 kJ/kg
h4s=hf + x4s x hg= 28.27+0.87 ×2335.3 =2320.981 kJ/kg
Since S4s is smaller than Sg and larger than Sf at the pressure of 30 kPa, the status at 4s is a saturated
mixture.
S4S-Sf
Sfg
Wturn isen = h3 h4s=3423.1-2320.981-1102.119
Wturb*act h3h4s 0.86x1102.119=947.82 kJ/kg
X4s
=
Wout
State 4a
= W
hsa
h4a
6.826-0.9441/6.8234-0.87
P4a=PL = 30kPa
turb*act=947.82g
kj
= h3 - Wturb act =3423.1 -947.82
= 2475.28 kJ/kg](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9fd72755-fbcd-4bf5-b9e3-e08fe5879c40%2F674d6165-56a3-478e-9439-fbb43105b7cc%2Ffgy5v_processed.png&w=3840&q=75)
Transcribed Image Text:State 4s
P4s PL = 30 kPa
=
S4s S3=6.8826 kJ/kg.K
According to Table A5, when pressure 30 kPa
Sf= 0.9441kJ/kg.K
S.fg
= 6.8234 kJ/kg.K
Sg= 7.7675 kJ/kg.K
hf= 191.8 kJ/kg
hfg=2392.2 kJ/kg
h4s=hf + x4s x hg= 28.27+0.87 ×2335.3 =2320.981 kJ/kg
Since S4s is smaller than Sg and larger than Sf at the pressure of 30 kPa, the status at 4s is a saturated
mixture.
S4S-Sf
Sfg
Wturn isen = h3 h4s=3423.1-2320.981-1102.119
Wturb*act h3h4s 0.86x1102.119=947.82 kJ/kg
X4s
=
Wout
State 4a
= W
hsa
h4a
6.826-0.9441/6.8234-0.87
P4a=PL = 30kPa
turb*act=947.82g
kj
= h3 - Wturb act =3423.1 -947.82
= 2475.28 kJ/kg
![State 1
Status: Saturated Liquid
P₁ = PL= 30 kPa
From Table A5 when P₁ = 30 kPa
v1 vf = 0.001022 m³/kg
h1 hf 289.27 kJ/kg
State 2
Status: Compressed liquid
P2s PH=6000 kPa
h2a= 296.04 kJ/kg _[h₂a = h1 + Wpact = 289.27 + 6.77 = 296.04
WP, ise V₁ (P2 - P₁) = 0.001022 (6000 - 30) = 6.101
kg
Wpinsen 6.101
0.9
Wpact nisen pum
Win = W
State 3
pact=6.77Kg
Status: Superheated Steam
P3-PH=6000kpa
= 6.77
h3-3423.1 kJ/Kg
S3-6.8826 kJ/Kg.K
kJ
kg
T3 Tmax= 500°C
From Table 6A, When P3= 6000 kPa
Heat In (Boiler)
Qin=h3 - h₂a = 3423.1 - 296.04 = 3127.06 kJ/Kg
Kg](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9fd72755-fbcd-4bf5-b9e3-e08fe5879c40%2F674d6165-56a3-478e-9439-fbb43105b7cc%2Fkuuc2l_processed.png&w=3840&q=75)
Transcribed Image Text:State 1
Status: Saturated Liquid
P₁ = PL= 30 kPa
From Table A5 when P₁ = 30 kPa
v1 vf = 0.001022 m³/kg
h1 hf 289.27 kJ/kg
State 2
Status: Compressed liquid
P2s PH=6000 kPa
h2a= 296.04 kJ/kg _[h₂a = h1 + Wpact = 289.27 + 6.77 = 296.04
WP, ise V₁ (P2 - P₁) = 0.001022 (6000 - 30) = 6.101
kg
Wpinsen 6.101
0.9
Wpact nisen pum
Win = W
State 3
pact=6.77Kg
Status: Superheated Steam
P3-PH=6000kpa
= 6.77
h3-3423.1 kJ/Kg
S3-6.8826 kJ/Kg.K
kJ
kg
T3 Tmax= 500°C
From Table 6A, When P3= 6000 kPa
Heat In (Boiler)
Qin=h3 - h₂a = 3423.1 - 296.04 = 3127.06 kJ/Kg
Kg
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