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Sagot :
During a test, the four-cycle FEP=1051.44 hp, IP=407.43 HP and brake power BP= 342.26 hr
Given : Displacemen Volume Vs = 955 in³
density S: 7 lbigat
ΔV = volumetric fuel consumption = 19.3 gal/h
N = 2100 pm
Ti ( Torque ) = 856 lb-it
Calorific value ( C . V ) = 19 ,533 BTU /lb
( After cut off ) 7F = 163 lb- ft .
Fuel consumption ( mf ) = δx ΔV
= 7 x 19 .3
mf = 135 . 1 lb/ br
Now Brake power :
BP = TixN/5252(hp )
B. P =
85 6X 2100/5252
= 342 . 26 hp
B . P = 342 . 26 hp
Now Fuel equivalent power
fEr :my x C . V
135. 1 ( eb / hr ) x 19, 533 ( BTU/ RD )
135. 1 X19533/60 x 41.83
FEP
= 1051.44 hp
Learn more about Brake power here:
https://brainly.com/question/13279330
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