T: What is the net work done by an Otto engine?

In summary: For this cycle, the work can be calculated easily using W = Qh-Qc. Since a-b and c-d are adiabatic, heat flows only from b-c and d-a. Since it is isochoric, \Delta Q_h = nC_v\Delta T_{b-c} \text{ and } \Delta Q_c = nC_v\Delta T_{d-a}.
  • #1
mclame22
13
0

Homework Statement


Show that the net work done by the otto engine per cycle is
Cv(Tc - Tb)(1 - Ta/Tb)

PV diagram:
[PLAIN]http://www.antonine-education.co.uk/physics_a2/options/module_7/topic_4/Otto_indic.gif [Broken]

Homework Equations


Work done by a kmole of gas expanding irreversibly and adiabatically from
P1 V1 T1 to P2 V2 T2:
= (P1V1 - P2V2)/(γ - 1) = R(T1 - T2)/(γ - 1)

The Attempt at a Solution


The only work done by the otto engine is during the adiabatic expansions/compressions. So the work done by the gas from point A to point B is
= R(Ta - Tb)/(γ - 1)

Likewise, the work done by the gas from point C to point D is
= R(Tc - Td)/(γ - 1)

So the net work done by the gas will be
= R(Ta - Tb)/(γ - 1) + R(Tc - Td)/(γ - 1)
= (R/(γ - 1))(Ta - Tb + Tc - Td)
= Cv[(Tc - Tb) + Ta - Td]

Here I get stuck. I don't see why the answer doesn't have a Td term. Have I gone wrong somewhere?
 
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  • #2
mclame22 said:

Homework Equations


Work done by a kmole of gas expanding irreversibly and adiabatically from
P1 V1 T1 to P2 V2 T2:
= (P1V1 - P2V2)/(γ - 1) = R(T1 - T2)/(γ - 1)
Since it is adiabatic, [itex]W = -\Delta U = -nC_v\Delta T[/itex]

The Attempt at a Solution


The only work done by the otto engine is during the adiabatic expansions/compressions. So the work done by the gas from point A to point B is
= R(Ta - Tb)/(γ - 1)
[itex] = -nC_v(T_b - T_a)[/itex]

Likewise, the work done by the gas from point C to point D is
= R(Tc - Td)/(γ - 1)
[itex] = -nC_v(T_d - T_c)[/itex]
So the net work done by the gas will be
= R(Ta - Tb)/(γ - 1) + R(Tc - Td)/(γ - 1)
= (R/(γ - 1))(Ta - Tb + Tc - Td)
= Cv[(Tc - Tb) + Ta - Td]

Here I get stuck. I don't see why the answer doesn't have a Td term. Have I gone wrong somewhere?
Your work is correct. Can you express Td in terms of the other temperatures?

AM
 
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  • #3
I'm sorry, but I still can't see how to find Td... Your equations seem essentially the same as mine.
 
  • #4
mclame22 said:
I'm sorry, but I still can't see how to find Td... Your equations seem essentially the same as mine.
Think of W as Qin-Qout or Qh-Qc. What are Qh and Qc in terms of Ta, Tb, Tc and Td?

AM
 
  • #5
What you have to do here is express Td in terms of Ta, Tb and Tc

To do that you must first determine the ratio of Ta to Tb (it depends on the ratio of V1/V2 and gamma). Then determine the ratio of Td to Tc (try to express it in a similar way). Once you do that you will be able to express Td in terms of the other temperatures.

For this cycle, the work can be calculated easily using W = Qh-Qc. Since a-b and c-d are adiabatic, heat flows only from b-c and d-a. Since it is isochoric, [itex]\Delta Q_h = nC_v\Delta T_{b-c} \text{ and } \Delta Q_c = nC_v\Delta T_{d-a}[/itex].

AM
 

1. What is the Otto Cycle?

The Otto Cycle is a thermodynamic cycle that is commonly used in spark-ignition internal combustion engines. It describes the process of converting heat energy into mechanical work, which is then used to power the engine.

2. How does the Otto Cycle work?

The Otto Cycle consists of four processes: intake, compression, power, and exhaust. In the intake process, the fuel-air mixture is drawn into the cylinder. In the compression process, the mixture is compressed by the piston. In the power process, the spark plug ignites the mixture, causing it to expand and push the piston down, creating mechanical work. In the exhaust process, the exhaust gases are expelled from the cylinder.

3. What is the net work done in an Otto Cycle?

The net work done in an Otto Cycle is the difference between the work done in the power process and the work done in the compression process. This work is measured in joules (J) or kilojoules (kJ) and represents the amount of energy that is converted into mechanical work.

4. What factors affect the net work done in an Otto Cycle?

The net work done in an Otto Cycle is affected by several factors, including the compression ratio, air-fuel ratio, engine speed, and efficiency of the engine. A higher compression ratio and air-fuel ratio typically result in a higher net work done, while a lower engine speed and lower efficiency can decrease the net work done.

5. How is the net work done in an Otto Cycle calculated?

The net work done in an Otto Cycle is calculated by subtracting the work done in the compression process from the work done in the power process. This can be represented by the equation Wnet = Wpower - Wcompression, where Wnet is the net work done, Wpower is the work done in the power process, and Wcompression is the work done in the compression process.

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