How Does the Otto Cycle Affect Internal Energy Change?

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In summary: Then if deltaU = Q - W, deltaU = 0, Q = 0, then -W will be left alone, but my question is are there other ways to find what work is other than using pdeltaV?Yes. If U = 0, \Delta Q = W. In other words, W = Qh-Qc.
  • #1
doggieslover
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The idealized cycle shown is known as the Otto cycle. (Intro 1 figure) Suppose an engine is executing this Otto cycle, using a gas (not necessarily ideal) as its working substance. From state A to state B, the gas is allowed to expand adiabatically. (An adiabatic process is one in which no heat is added to, or given off by, the working gas.) The gas is then cooled at constant volume until it reaches state C, at which point it is adiabatically compressed to state D. Finally, it is heated at constant volume until it returns to state A.

The pressure and volume of the gas in state A are p_A and V_A respectively. The pressure and volume of the gas in state C are p_C and V_C respectively.

http://session.masteringphysics.com/problemAsset/1011140/12/STH_tc_2.jpg

Part C
What is DeltaU, the change in the gas's internal energy after a complete cycle?
Express your answer in terms of any needed variables from the problem introduction.

Okay I know that Q is zero, and deltaU = Q - W, so deltaU = -W, and W = p*deltaV, and a complete cycle means it returns to its original state.

So my answer should be p_A*V_A, but it's incorrect, it says the answer does not depend on those variables, what variables are they looking for then?

Help?
 
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  • #2
doggieslover said:
The idealized cycle shown is known as the Otto cycle. (Intro 1 figure) Suppose an engine is executing this Otto cycle, using a gas (not necessarily ideal) as its working substance. From state A to state B, the gas is allowed to expand adiabatically. (An adiabatic process is one in which no heat is added to, or given off by, the working gas.) The gas is then cooled at constant volume until it reaches state C, at which point it is adiabatically compressed to state D. Finally, it is heated at constant volume until it returns to state A.

The pressure and volume of the gas in state A are p_A and V_A respectively. The pressure and volume of the gas in state C are p_C and V_C respectively.

http://session.masteringphysics.com/problemAsset/1011140/12/STH_tc_2.jpg

Part C
What is DeltaU, the change in the gas's internal energy after a complete cycle?
Express your answer in terms of any needed variables from the problem introduction.

Okay I know that Q is zero,

No, I don't think that is correct. The heat flow is zero for the adiabatic parts of the cycle, but for the two times when it is heated and cooled at constant volume there is a heat flow. Do you see what to do now?
 
  • #3
doggieslover said:
What is DeltaU, the change in the gas's internal energy after a complete cycle?
Express your answer in terms of any needed variables from the problem introduction.

Okay I know that Q is zero, and deltaU = Q - W, so deltaU = -W, and W = p*deltaV, and a complete cycle means it returns to its original state.

So my answer should be p_A*V_A, but it's incorrect, it says the answer does not depend on those variables, what variables are they looking for then?

Help?
It's a trick question. If the gas returns to its original state, has there been any change to its internal energy?

AM
 
  • #4
I know that Q is zero already, so now deltaU is zero too since there's no change in internal energy?
 
  • #5
doggieslover said:
I know that Q is zero already, so now deltaU is zero too since there's no change in internal energy?
The work done per cycle and the heat flow from the hot to cold reservoir is not zero. But the change in internal energy of the gas is zero over one complete cycle.

AM
 
  • #6
Then if deltaU = Q - W, deltaU = 0, Q = 0, then -W will be left alone, but my question is are there other ways to find what work is other than using pdeltaV?
 
  • #7
doggieslover said:
Then if deltaU = Q - W, deltaU = 0, Q = 0, then -W will be left alone, but my question is are there other ways to find what work is other than using pdeltaV?

Q is not zero; it is only zero for two parts of this four-part cycle.
 
  • #8
doggieslover said:
Then if deltaU = Q - W, deltaU = 0, Q = 0, then -W will be left alone, but my question is are there other ways to find what work is other than using pdeltaV?
Yes. If U = 0, [itex]\Delta Q = W[/itex]. In other words, W = Qh-Qc.

You could work out Qh and Qc from the BC and DA parts of the path (constant volume). For example, if this was an ideal gas, from the ideal gas equation: [itex]Q = nC_v\Delta T = VC_v\Delta P/R[/itex]

AM
 

1. What is a Simplified Otto Cycle?

A Simplified Otto Cycle is a theoretical thermodynamic cycle used to represent the operation of a four-stroke spark ignition engine. It consists of four processes: intake, compression, power, and exhaust.

2. How does a Simplified Otto Cycle work?

During the intake process, fuel and air are mixed and drawn into the cylinder. The mixture is then compressed during the compression process. Next, a spark ignites the compressed mixture, causing a rapid expansion and producing power. Finally, the exhaust process releases the remaining exhaust gases from the cylinder.

3. What are the major assumptions of the Simplified Otto Cycle?

The Simplified Otto Cycle assumes that the working fluid (air-fuel mixture) is an ideal gas, the combustion process is constant volume, and there are no heat losses to the surroundings. It also assumes that all processes are reversible and adiabatic.

4. What is the purpose of a Simplified Otto Cycle?

The Simplified Otto Cycle is used to analyze and understand the thermodynamic behavior of spark ignition engines. It helps in the design and optimization of engine components and the evaluation of engine performance.

5. Are there any limitations to the Simplified Otto Cycle?

Yes, the Simplified Otto Cycle is a simplified model and does not take into account real-world factors such as friction, heat transfer, and combustion inefficiencies. It also does not account for the variation in fuel-air mixture during the intake process.

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