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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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Solution 1.07 x 10 Pac = 1.20 kg/m 287 x 310 V x Pac me 0.5 x 0.0011 x 1.20 = 0.00066 kg/cycle %3! 4. %3! 2800 x 0.0011 x 1.20 = 1.08 0.5 = 0.46 (quite low) R 1.08 bp 60 644 IC Engities Mass flow rate of air supplied by the blower 1.2 x 0.0992 x 1.21 = 0.144 kg/cycle ATisen ATactuol TMower %3D ()* (1.15) a286 = 312.2 K 300 x 312.2-300 0.75 ATtual = 16.3 K 300+ 16.3 = 316.3 K %3D W. rmC,AT =0.102 x 1.005 x 16.3 %3D 1.6709 kJ/cycle 720 = 20 kW 60 1.6709 x %3D d21 = 6 x x 02 x 0.5 0.0942 m/cycle by V,N 0.0942 x 720 x 105 120 x 6 bmep x 60 x 1000 %3D Ans 6.37 bar Total mechanical loss %3D 1.5 x 20 = 30 kW ip = bp + fp= 6 x 120 + 30 = 750 kW %3D Fuel consumed per hour bsfex bp = 0.21 x 720 = 151.2 h ip x 100 = TinyCV 750 x 3600 151.2 x 44000 Tith !! 0.406 = 40.6% Ans %3D 151.2 60 x 720 x 0.0992 Fuel-air ratio Ane 0.0353
Transcribed Image Text:Solution 1.07 x 10 Pac = 1.20 kg/m 287 x 310 V x Pac me 0.5 x 0.0011 x 1.20 = 0.00066 kg/cycle %3! 4. %3! 2800 x 0.0011 x 1.20 = 1.08 0.5 = 0.46 (quite low) R 1.08 bp 60 644 IC Engities Mass flow rate of air supplied by the blower 1.2 x 0.0992 x 1.21 = 0.144 kg/cycle ATisen ATactuol TMower %3D ()* (1.15) a286 = 312.2 K 300 x 312.2-300 0.75 ATtual = 16.3 K 300+ 16.3 = 316.3 K %3D W. rmC,AT =0.102 x 1.005 x 16.3 %3D 1.6709 kJ/cycle 720 = 20 kW 60 1.6709 x %3D d21 = 6 x x 02 x 0.5 0.0942 m/cycle by V,N 0.0942 x 720 x 105 120 x 6 bmep x 60 x 1000 %3D Ans 6.37 bar Total mechanical loss %3D 1.5 x 20 = 30 kW ip = bp + fp= 6 x 120 + 30 = 750 kW %3D Fuel consumed per hour bsfex bp = 0.21 x 720 = 151.2 h ip x 100 = TinyCV 750 x 3600 151.2 x 44000 Tith !! 0.406 = 40.6% Ans %3D 151.2 60 x 720 x 0.0992 Fuel-air ratio Ane 0.0353
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