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In the diesel engine experiment, if: engine speed= 1302 Torque (N.m) = 9.9 Power(W)= 1344 Air mass flow rate (Kg/sec) = 0.002178 fuel mass flow rate (Kg/sec) = 0.000140 Tamb (C)= 20.1 Texh (C)= 292 Pamb (mbar)= 929 Air box Pressure (Pa)= -82 The Volumetric efficiency will be:

In the diesel engine experiment, if: engine speed= 1302 Torque (N.m) = 9.9 Power(W)= 1344 Air mass flow rate (Kg/sec) = 0.002178 fuel mass flow rate (Kg/sec) = 0.000140 Tamb (C)= 20.1 Texh (C)= 292 Pamb (mbar)= 929 Air box Pressure (Pa)= -82 The Volumetric efficiency will be:

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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In the diesel engine experiment, if: engine speed= 1302 Torque (N.m) = 9.9 Power(W)= 1344 Air mass flow rate (Kg/sec) = 0.002178 fuel mass flow rate (Kg/sec) = 0.000140 Tamb (C)= 20.1 Texh (C)= 292 Pamb (mbar)= 929 Air box Pressure (Pa)= -82 The Volumetric efficiency will be: Select one: a. 84.89% O b. 78.33% O c. 65.27% O d. 71.86%
Transcribed Image Text:In the diesel engine experiment, if: engine speed= 1302 Torque (N.m) = 9.9 Power(W)= 1344 Air mass flow rate (Kg/sec) = 0.002178 fuel mass flow rate (Kg/sec) = 0.000140 Tamb (C)= 20.1 Texh (C)= 292 Pamb (mbar)= 929 Air box Pressure (Pa)= -82 The Volumetric efficiency will be: Select one: a. 84.89% O b. 78.33% O c. 65.27% O d. 71.86%
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