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Determine the Efficiency, in %, of a simple rankine cycle. Use: 8 MPa and 480 C (Table 3. Vapor) h1 = 3348.4 kJ/kg s1 = 6.6586 kJ/kg – K 2 MPa (Table 3. Vapor; note: s1 = s2) h2 = 2963.145 kJ/kg 2 MPa and 460 C (Table 3. Vapor; note: s3 = s4) h3 = 3379.5 kJ/kg s3 = 7.3147 kJ/kg – K 60C (Table 1. Saturation Temperatures) P5 = 0.019940 MPa = 19.94

Determine the Efficiency, in %, of a simple rankine cycle. Use: 8 MPa and 480 C (Table 3. Vapor) h1 = 3348.4 kJ/kg s1 = 6.6586 kJ/kg – K 2 MPa (Table 3. Vapor; note: s1 = s2) h2 = 2963.145 kJ/kg 2 MPa and 460 C (Table 3. Vapor; note: s3 = s4) h3 = 3379.5 kJ/kg s3 = 7.3147 kJ/kg – K 60C (Table 1. Saturation Temperatures) P5 = 0.019940 MPa = 19.94

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
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Determine the Efficiency, in %, of a simple rankine cycle.

Use:

8 MPa and 480 C (Table 3. Vapor)

  • h1 = 3348.4 kJ/kg
  • s1 = 6.6586 kJ/kg – K

2 MPa (Table 3. Vapor; note: s1 = s2)

  • h2 = 2963.145 kJ/kg

2 MPa and 460 C (Table 3. Vapor; note: s3 = s4)

  • h3 = 3379.5 kJ/kg
  • s3 = 7.3147 kJ/kg – K

60C (Table 1. Saturation Temperatures)

  • P5 = 0.019940 MPa = 19.94 kPa
  • hf = 251.13 kJ/kg = h5
  • hfg = 2358.5 kJ/kg
  • sf = 0.8312 kJ/kg – K
  • sfg = 7.0784 kJ/kg – K
  • vf = 0.0010172 m3/kg = v5

Solving for h4

s3 = s4 = sf + xsfg

7.3147 kJ/kg – K = 0.8312 kJ/kg – K + x (7.0784 kJ/kg – K)

x = 0.9159555832

h4 = hf + x(hfg)

h4 = 251.13 kJ/kg + 0.9159555832 (2358.5 kJ/kg)

  • h4 = 2411.411243 kJ/kg

Solving for h6

h6 = v5 (P6 – P5) + h5

h6 = 0.0010172 m3/kg (8000 kN/m2 – 19.94 kN/m2) + 251.13 kJ/kg

h6 = 8.117317032 kJ/kg + 251.13 kJ/kg

  • h6 = 259.247317 kJ/kg
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