Can you please draw the TS Diagram of the problem below. I already have a solution but I badly need the TS Diagram.Problem:A 25,000 kW turbo-generator is supplied with 128,000 kg/h of steam at 2.50 MP and 400°C when developing it rated load. There are actually extracted 10,400 kg/h at 0.3 MPa and 8300 kg/h at 0.06 MPa. The condenser pressure is 0.007 MPa and actual feedwater temperature is 127°C. Again, please only draw the TS Diagram. Thanks!Please refer to the image below. Reference solution for making the TS Diagram@ Point 1 (Superheated)@ Point 5 (Saturated Liquid)P₁ = 2.5 MPa > T₁ = 400h₁ = 32 39.3 kJ /kgP5 = P4 = 0.007 MPahshes = 163.40 kJ/kgV₁ = Vf5 = 0.0010074 m³/kgS₁ = 7.0148 kJ/kg@ Point 2S₂P₂ = 0.3 MPaSgS, 7.0148 kJ/kg= 6.99 19 kJ /kg k < 5₂ (suSh6.97267.01487.0263SigSf = 0.5592Sfg=7.716727 39.4h₂ = 2734.710056 kJ/kgh₁ = 163.40hfg = 2409.1@ Point 3 (Wet)S3 S17.0148 kJ/kgkP3 = 0.00 MPaSg= 7.5320 S3 (wet)S = 11453= 6.38672717.5h₂x = 0.9190192118h3 = 359.86 + x(2293.6)h3 =2467.722464 kJ/kgWpi= has hs0.0533922hf 359.86hfg = 2293.6=has 163.40has 163.4533922 kJ/kg@ Point 4 (Wet)SA S₁ = 7.048 kJ/kgkP4 = 0.007 MPaS4= 5 + XSpg7.0148 0.5592 + X(7.7167)h4 = hf + xhfgX= 0.83657522163.40+ x(24091)2178.793362@ Point B, (Subcooled)WP₁ = V₂ (Po-Ps)-0.001 0074 (60-7)= 0.0533922 kJ/kg@ Point 6 (Saturated Liquid)P6 = P3 = 0.06 MPaho hr6 = 359.86 kJ/kgV6 Vf6. 0.0010 331 m²/kg@ Point B (Subcooled)WP2 V6 (PP)-0.00160331 (300- 60)= 0.247944Wp₂= hp 359.86he6 = 360. 107944@ Point 7 (Saturated Liquid)P7= P₂ = 0.3 MPahz = hf7 561.47 kJ/kgV7 V7 = 0.0010732 m³/kg@ Point B, (Subcooled)V7(P₁-P+)WP3= 0.0010732( 2500-300)=2.36 104hez-haWp32. 36104 = hs7 -561.47h87 563. 83104

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Answered: Can you please draw the TS Diagram of…

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 please draw the TS Diagram of the problem below. I already have a solution but I badly need the TS Diagram.

Problem:

A 25,000 kW turbo-generator is supplied with 128,000 kg/h of steam at 2.50 MP and 400°C when developing it rated load. There are actually extracted 10,400 kg/h at 0.3 MPa and 8300 kg/h at 0.06 MPa. The condenser pressure is 0.007 MPa and actual feedwater temperature is 127°C.

Again, please only draw the TS Diagram. Thanks!

Please refer to the image below.

Reference solution for making the TS Diagram
@ Point 1 (Superheated)
@ Point 5 (Saturated Liquid)
P₁ = 2.5 MPa > T₁ = 400
h₁ = 32 39.3 kJ /kg
P5 = P4 = 0.007 MPa
hshes = 163.40 kJ/kg
V₁ = Vf5 = 0.0010074 m³/kg
S₁ = 7.0148 kJ/kg
@ Point 2
S₂
P₂ = 0.3 MPa
Sg
S, 7.0148 kJ/kg
= 6.99 19 kJ /kg k < 5₂ (su
S
h
6.9726
7.0148
7.0263
Sig
Sf = 0.5592
Sfg=
7.7167
27 39.4
h₂ = 2734.710056 kJ/kg
h₁ = 163.40
hfg = 2409.1
@ Point 3 (Wet)
S3 S17.0148 kJ/kgk
P3 = 0.00 MPa
Sg= 7.5320 S3 (wet)
S = 11453
= 6.3867
2717.5
h₂
x = 0.9190192118
h3 = 359.86 + x(2293.6)
h3 =
2467.722464 kJ/kg
Wpi= has hs
0.0533922
hf 359.86
hfg = 2293.6
=
has 163.40
has 163.4533922 kJ/kg
@ Point 4 (Wet)
SA S₁ = 7.048 kJ/kgk
P4 = 0.007 MPa
S4= 5 + XSpg
7.0148 0.5592 + X(7.7167)
h4 = hf + xhfg
X= 0.83657522
163.40+ x(24091)
2178.793362
@ Point B, (Subcooled)
WP₁ = V₂ (Po-Ps)
-0.001 0074 (60-7)
= 0.0533922 kJ/kg
@ Point 6 (Saturated Liquid)
P6 = P3 = 0.06 MPa
ho hr6 = 359.86 kJ/kg
V6 Vf6. 0.0010 331 m²/kg
@ Point B (Subcooled)
WP2 V6 (PP)
-0.00160331 (300- 60)
= 0.247944
Wp₂= hp 359.86
he6 = 360. 107944
@ Point 7 (Saturated Liquid)
P7= P₂ = 0.3 MPa
hz = hf7 561.47 kJ/kg
V7 V7 = 0.0010732 m³/kg
@ Point B, (Subcooled)
V7(P₁-P+)
WP3
= 0.0010732( 2500-300)
=
2.36 104
hez-ha
Wp3
2. 36104 = hs7 -561.47
h87 563. 83104

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