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

[mclame22]

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

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?

 

[Andrew Mason]

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, W = -\Delta U = -nC_v\Delta T

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)

= -nC_v(T_b - T_a)

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

= -nC_v(T_d - T_c)

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

 

[mclame22]

I'm sorry, but I still can't see how to find Td... Your equations seem essentially the same as mine.

 

[Andrew Mason]

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

 

[Andrew Mason]

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, \Delta Q_h = nC_v\Delta T_{b-c} \text{ and } \Delta Q_c = nC_v\Delta T_{d-a}.

AM