BecomeMyFan=D
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.
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BecomeMyFan=D
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i dont even know where to start
BecomeMyFan=D
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ANY help, even a formula or a rule or a hint, anything will help
BecomeMyFan=D
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i just know efficiency = input/output
lokisan
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Yep, calm down.,,
BecomeMyFan=D
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oh, sorry, it is output/input
BecomeMyFan=D
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ok,.... i am calm
lokisan
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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.
lokisan
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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.
BecomeMyFan=D
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so, can i say that efficiancy is 20.9%?
lokisan
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Yes
BecomeMyFan=D
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ok, thank you, i already faned you long time ago, so cant fan again :(
lokisan
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Yeah I know - no worries :)
lokisan
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The answer makes sense - I've seen values between 10 and 50% for internal combustion engines.
BecomeMyFan=D
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then it is probably correct, i also know that cobustion engines are very unefficient
BecomeMyFan=D
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hey anwar
AnwarA
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hi :)
BecomeMyFan=D
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want a fan?
AnwarA
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would you be one if I say yes?
BecomeMyFan=D
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...
AnwarA
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lol
BecomeMyFan=D
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you got it
AnwarA
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thanks!!
BecomeMyFan=D
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anwar, can you please fan lokisan, because he helped me alot and i can only fan him once
AnwarA
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lol you really care about others :P.. I already did anyways
AnwarA
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so you faned me just as an exchange for me to fan the guy huh? :(
BecomeMyFan=D
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nah, would have faned you anyway
sstarica
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lol andy, you and your fanning job~ ^_^