Answered

Get comprehensive solutions to your questions with the help of IDNLearn.com's experts. Discover detailed answers to your questions with our extensive database of expert knowledge.

p5.7 during a test, the four-cycle, 15.649 l (955 in3) engine in a large tractor consumes 73 l/h (19.3 gal/h) of diesel fuel while running at 2100 r/min (rpm) and producing 1160 nm (856 lb-ft) of torque. the imep during the test is 1110 kpa (161 psi). the fuel density is 0.835 kg/l (7.0 lb/gal), and its heating value is 45,434 kj/kg (19,533 btu/lb). calculate:

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

#SPJ4

Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.