The brake fuel conversion efficiency of an engine is 0.3. The mechanical efficiency is 0.8. The combustion efficiency is 0.94. The heat losses to the coolant and oil are 60 kW. The fuel chemical energy entering the engine per unit time, , is 190 kW. What percentage of this energy becomes (a) brake work; (b) friction work; (c) heat losses; (d) exhaust chemical energy; and (e) exhaust sensible energy

Explanation:

The combustion efficiency is 0.94 which means 6 % ( 1 - 0.94) is wasted as exhaust chemical energy

exhaust chemical energy = 0.06 × 190 kW = 11.4 kW

heat loss to the coolant and oil = 60 kW

The chemical energy remaining = 190 kW - 60 kW - 11.4kW = 118.6 kW

The brake conversion efficiency = 0.3 yielding 0.3 ×  118.6 kW = 35.58 kW

with ( 118.6 - 35.58 ) been exhaust sensible energy = 83.02 kW

only 0.8 of this 35.58 kW is useful which equals = 0.8 × 35.53 kW = 28.464 kW

amount lost to friction = 35.58 kW - 28.464 kW = 7.116 kW

a) percentage brake work = 28.464 kW / 190 kW × 100 = 14.98%

b) percentage friction work = 7.116 kW / 190kW × 100 = 3.745%

c) heat losses = 60 kW / 190 kW × 100 = 31.579 %

d) percentage exhaust chemical energy = 6%

d) percentage exhaust sensible energy = 83.02 kW / 190 kW × 100 = 43.694 %

fanswer:

explanation:

f

answer:

put more,