A compressor is bring where the actual exit temperature is 650°C. Find the isentropic compressor efficiency and the entropy generation. Solution AT T=650°C e P. = 17.5 MPa es Vactual P; 1MPA h; - hes h; – h. W. Ncompressor W. a 59 2ND Law Analysis for Control Volume Efficiency: Worked Example No. 3 Solution Inlet State: P=1MPA, T=179.91°C, s=6.5864 kJ/kg-K, h=2778.08 kJ/kg Ideal Exit State: Pes=17.5MPa, ses=S;=6.5864 kJ/kg-K, hs=3560.1 kJ/ky Actual Exit State: P=17.5MPA, T,=650°C, s̟= 6.7357 kJ/kg-K, h,=3693.9 kJ/kg Therefore ideal work w, = hes - h, = (3560.1 – 2778.08) kJ/kg = 782.02 kJ/kg And the actual work w = h. - h, = (3693.9 – 2778.08) kJ/kg = 915.82 kJ/kg Thus compressor efficiency W. 782.02 Ncompressor = 85.4% W 915.82 a II

A compressor is bring where the actual exit temperature is 650°C. Find the isentropic compressor efficiency and the entropy generation. Solution AT T=650°C e P. = 17.5 MPa es Vactual P; 1MPA h; - hes h; – h. W. Ncompressor W. a 59 2ND Law Analysis for Control Volume Efficiency: Worked Example No. 3 Solution Inlet State: P=1MPA, T=179.91°C, s=6.5864 kJ/kg-K, h=2778.08 kJ/kg Ideal Exit State: Pes=17.5MPa, ses=S;=6.5864 kJ/kg-K, hs=3560.1 kJ/ky Actual Exit State: P=17.5MPA, T,=650°C, s̟= 6.7357 kJ/kg-K, h,=3693.9 kJ/kg Therefore ideal work w, = hes - h, = (3560.1 – 2778.08) kJ/kg = 782.02 kJ/kg And the actual work w = h. - h, = (3693.9 – 2778.08) kJ/kg = 915.82 kJ/kg Thus compressor efficiency W. 782.02 Ncompressor = 85.4% W 915.82 a II

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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