Question

A four-cylinder gasoline engine has an efficiency of 28 % and delivers 280 J of work per cycle per cylinder. A) If the engine runs at 25 cycles per second (1500 rpm), determine the work done per second. B)What is the total heat input per second from the gasoline. C) If the energy content of gasoline is 130 MJ per gallon, how long does one gallon last?

Answer 1

Answer:

A). 28 $\frac{kJ}{s}$ at 25 cycles per second

B). 100 kJ

C). 6500s or 180.3 minutes or 1.8 hr

Explanation:

Four-cylinder= 280J cycleper cylinder

Efficiency =28%

A).

$W=280J (\frac{1}{cycle}*\frac{1}{cylinder} *\frac{4cylinder}{1} *\frac{25cycle}{s} )\\W=280J*100\frac{1}{s}\\ W=28000\\W=20x10^{3} = 28kJ$

W=28$\frac{kJ}{s}$

B).

The efficiency is 28% and have a work of 28k0J

so:

If 20kJ⇒28%

  xkJ  ⇒ 100%

$100%*28kJ=xkJ*28%\\xkJ=\frac{100*28}{28}\\ x=100kJ$

Total heat is 100kJ

C).

130MJ per gallon

$t_{gallon}=\frac{130MJ}{20\frac{kJ}{s} }= \frac{130x10^{6} }{20x10^{3}}=6500s\\ 6500s*\frac{1minute}{60s}*\frac{1hr}{60minute}=1.8hr$