Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
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.

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.

Elements Of Electromagnetics
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
See similar textbooks
icon
Related questions
Question

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
Transcribed Image Text: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
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Power Plant Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY
SEE MORE TEXTBOOKS