Question

An engine burns 1.4 kg of gasoline per hour and yields 3.5 kW of power.
Calculate the efficiency of the engine when the heating value of gasoline is
43 MJ/kg.

Answer 1

i dont even know where to start

Answer 2

ANY help, even a formula or a rule or a hint, anything will help

Answer 3

i just know efficiency = input/output

Answer 4

Yep, calm down.,,

Answer 5

oh, sorry, it is output/input

Answer 6

ok,.... i am calm

Answer 7

The heating value is the amount of heat energy produced by complete combustion in a given quantity.

Take a basis of 1 hour. Then your engine burns 1.4kg of gasoline, which yields a maximal amount of thermal energy to use for work as\[43MJ/kg \times 1.4 kg=60.2MJ\]In one hour, your engine produces from this energy, \[\frac{3500J}{s}\times \frac{3600s}{h}=12.6MJ\]The efficiency is the amount of work per unit energy input, which is \[\eta = \frac{12.6MJ}{60.2MJ}=0.209\]to 3 significant figures.

Answer 8

This is the first time I've used 'heating value' for anything like this, so if the answer's wrong, I couldn't find an appropriate definition.

Answer 9

so, can i say that efficiancy is 20.9%?

Answer 10

Yes

Answer 11

ok, thank you, i already faned you long time ago, so cant fan again :(

Answer 12

Yeah I know - no worries :)

Answer 13

The answer makes sense - I've seen values between 10 and 50% for internal combustion engines.

Answer 14

then it is probably correct, i also know that cobustion engines are very unefficient

Answer 15

hey anwar

Answer 16

hi :)

Answer 17

want a fan?

Answer 18

would you be one if I say yes?

Answer 19

...

Answer 20

lol

Answer 21

you got it

Answer 22

thanks!!

Answer 23

anwar, can you please fan lokisan, because he helped me alot and i can only fan him once

Answer 24

lol you really care about others :P.. I already did anyways

Answer 25

so you faned me just as an exchange for me to fan the guy huh? :(

Answer 26

nah, would have faned you anyway

Answer 27

lol andy, you and your fanning job~ ^_^