Understanding Otto Cycle: Calculating Cycles, Heat Input, and Efficiency

  • Thread starter Thread starter kris24tf
  • Start date Start date
  • Tags Tags
    Cycle Otto
Click For Summary
The discussion revolves around calculating various aspects of the Otto cycle for a four-stroke engine delivering 150 hp at 3600 rpm. To determine the number of cycles per minute, it is established that there are 1800 cycles, as the engine completes two cycles per revolution. For calculating heat input, the thermal efficiency of 20% is noted, and the formula E=W/Q is referenced, with the need to find work done using the engine's horsepower. The conversion of horsepower to kilowatts is highlighted, emphasizing that 1 hp equals 0.7457 kW. Understanding these calculations is essential for determining both heat input and wasted heat in the engine's operation.
kris24tf
Messages
35
Reaction score
0
Otto Cycle?? Please help ASAP!

I have a question that my entire class is confused over. I need help on it for a test, but I have no idea where to go with it. It doesn't seem to fit any of the equations or situations I have been using.

The question is: A four-stroke engine runs on the otto cycle. It delivers 150 hp at 3600 rpm. a) how many cycles are there in one minute?
I know that this is just 3600/2 = 1800 cycles

b) If the thermal efficiency of the engine is 20% what is the heat input per minute?

I don't know how to get this one. I know the equation I want is E=W/Q, but I don't know how to find the work done. do I use the 150 hp, 3600 rpm, or combine them somehow?

c)How much heat is wasted to the environment / minute.

I need the answer to part b I think for this right?

Any help at all would be great. Thanks for looking.
 
Physics news on Phys.org
For (b), the 150 horse power is the clue.

1 hp = 0.7457 kW

What is the energy delivered in 1 minute.
 

Thread 'Magnitude of buoyant force in fluids of different densities'

The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?
View full post »

Similar threads

Calculating Efficiency and Work in Otto Cycle Engine
  • · Replies 1 ·
Replies
1
Views
1K
Efficiency of Stirling Cycle - Is it the same as the Carnot Engine?
  • · Replies 14 ·
Replies
14
Views
2K
Engineering Reviewing Efficiency of Otto & Diesel Cycle Heat Engines
  • · Replies 8 ·
Replies
8
Views
4K
Calculating work of Otto cycle stages - Thermodynamics.
  • · Replies 1 ·
Replies
1
Views
3K
Calculating Otto Engine Clearance w/ Efficiency 46.4% & k=1.3
  • · Replies 1 ·
Replies
1
Views
2K
Otto Cycle Engine Homework: Total Work & Efficiency Calculation
  • · Replies 1 ·
Replies
1
Views
8K
Thermodynamic Cycle -- Work done as a function of Heat absorbed
  • · Replies 3 ·
Replies
3
Views
2K
How Do You Calculate Thermal Efficiency in a Diatomic Gas Heat Engine Cycle?
  • · Replies 2 ·
Replies
2
Views
4K
Closed cycle of an ideal gas
  • · Replies 3 ·
Replies
3
Views
1K
Calculate minimum work input of refrigeration cycle
  • · Replies 6 ·
Replies
6
Views
6K