Calculating Gasoline Needed for 50.0hp Heat Engine Output

In summary, the conversation is discussing a physics problem involving the efficiency of a car's engine when burning gasoline. The formula P=W/t is used to determine the time needed to develop a power output of 50.0hp, but the answer is not correct because the car's engine is only 25% efficient. To account for this, the energy output must be multiplied by 0.25.
  • #1
ChunkymonkeyI
35
0

Homework Statement


When gasoline(density=.7297 g/cm^3) is burned, it gives off 5.00 times 10^4 J/g(its heat of combustion). If a car's engine is 25% efficient, how much gasoline per hour must it burn in order to develop an output of 50.0hp. 1hp=746 W


Homework Equations


P=W/t
e=W/Qh

The Attempt at a Solution


I believed I was solving for time so what I did was I used P=W/t
then I rearranged it to t=W/P and plugged 5.00 times 10^4 for w and 746 times 50 for P and it didn't get me the right answer because the answer is 10.7 kg/h
 
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  • #2
Hi ChunkymonkeyI! :smile:
ChunkymonkeyI said:
… I used P=W/t
then I rearranged it to t=W/P and plugged 5.00 times 10^4 for w and 746 times 50 for P and it didn't get me the right answer because the answer is 10.7 kg/h

erm :redface: … what about your 25% efficiency? :wink:
 
  • #3
tiny-tim said:
Hi ChunkymonkeyI! :smile:erm :redface: … what about your 25% efficiency? :wink:[/QUOTE

Do I use the 25 percent efficiency to find W and if so what do I do after that lol?
 
Last edited:
  • #4
ChunkymonkeyI said:
Idk what to do with the 25 percent efficiency please help me!

aha!

everything in an exam question is there for a reason!

if you're totally stuck, just try either multiplying the answer by 0.25 (ie 25%), or dividing by it …

50% chance you'll get full marks! :biggrin:

in fact, the "car's engine is 25% efficient" means that the power output (to the wheels) is only 25% of the power input (from the gasoline) :wink:
 
  • #5
tiny-tim said:
aha!

everything in an exam question is there for a reason!

if you're totally stuck, just try either multiplying the answer by 0.25 (ie 25%), or dividing by it …

50% chance you'll get full marks! :biggrin:

in fact, the "car's engine is 25% efficient" means that the power output (to the wheels) is only 25% of the power input (from the gasoline) :wink:

Lol but I want to learn how to do the steps can u please show me them it's bothering me that I can't solve this physics heat engine problem pleaaaaaaaaaaase
 
  • #6
the power output (to the wheels) is only 25% of the power input (from the gasoline) …

so put that into the equation :smile:
 
  • #7
tiny-tim said:
the power output (to the wheels) is only 25% of the power input (from the gasoline) …

so put that into the equation :smile:

What equation
 
  • #8
(just got up :zzz: …)
ChunkymonkeyI said:
What equation

this one :smile:
ChunkymonkeyI said:
… I used P=W/t
then I rearranged it to t=W/P and plugged 5.00 times 10^4 for w and 746 times 50 for P …
 
  • #9
tiny-tim said:
(just got up :zzz: …)


this one :smile: …[/QUOT

But I don't know the values that I need to solve this problem can u please just show me the steps I'm not trying 2 cheat or anything I just want 2 learn how 2 do this
 
  • #10
find the energy from the gasoline,

then multiply by 0.25 to get the energy output :smile:
 

What is a heat engine?

A heat engine is a device that converts thermal energy into mechanical energy, typically used to produce electricity or power machinery.

How is horsepower (hp) related to heat engine output?

Horsepower is a unit of power that measures the rate at which work is done. In the context of a heat engine, it is used to express the amount of mechanical energy produced per unit of time.

What factors affect the amount of gasoline needed for a 50.0hp heat engine output?

The amount of gasoline needed for a 50.0hp heat engine output depends on several factors, including the efficiency of the engine, the type of fuel being used, and the load on the engine. Other factors such as temperature, altitude, and maintenance of the engine can also play a role.

How can I calculate the amount of gasoline needed for a 50.0hp heat engine output?

To calculate the amount of gasoline needed for a 50.0hp heat engine output, you will need to know the specific fuel consumption (SFC) of the engine. This is the amount of fuel consumed per unit of power produced. You can then use the formula Gasoline Needed (in gallons) = (50.0hp/SFC) * (1/number of hours of operation).

Are there any other methods for determining the amount of gasoline needed for a 50.0hp heat engine output?

Yes, there are other methods for determining the amount of gasoline needed for a 50.0hp heat engine output. One method is to use a dynamometer, which measures the torque and rotational speed of the engine to calculate its power output. Another method is to use a flow meter, which measures the amount of fuel being consumed by the engine in real-time.

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