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BecomeMyFan=D
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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.
 3 years ago
 3 years ago
BecomeMyFan=D Group Title
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.
 3 years ago
 3 years ago

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BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
i dont even know where to start
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
ANY help, even a formula or a rule or a hint, anything will help
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
i just know efficiency = input/output
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Yep, calm down.,,
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
oh, sorry, it is output/input
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
ok,.... i am calm
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
so, can i say that efficiancy is 20.9%?
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
ok, thank you, i already faned you long time ago, so cant fan again :(
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Yeah I know  no worries :)
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
The answer makes sense  I've seen values between 10 and 50% for internal combustion engines.
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
then it is probably correct, i also know that cobustion engines are very unefficient
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
hey anwar
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
want a fan?
 3 years ago

AnwarA Group TitleBest ResponseYou've already chosen the best response.0
would you be one if I say yes?
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
you got it
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
anwar, can you please fan lokisan, because he helped me alot and i can only fan him once
 3 years ago

AnwarA Group TitleBest ResponseYou've already chosen the best response.0
lol you really care about others :P.. I already did anyways
 3 years ago

AnwarA Group TitleBest ResponseYou've already chosen the best response.0
so you faned me just as an exchange for me to fan the guy huh? :(
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
nah, would have faned you anyway
 3 years ago

sstarica Group TitleBest ResponseYou've already chosen the best response.0
lol andy, you and your fanning job~ ^_^
 3 years ago
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