Second Law of Thermodynamics

Although the net heat supplied in a cycle is equal to the net work done, the gross heat supplied must be greater than the net work done. That is,

Qj > W

And hence the thermal efficiency of a heat engine will be less than 100%.

Consider the Otto cycle. By calculating the heat supplied and rejected in the cycle, its thermal efficiency can be found. The heat flow in a reversible constant volume process is given by

Q = mCv(AT)

Where

M = mass

AT = temperature difference

Cv = specific heat at a constant volume (heat required to raise the unit mass through one degree temperature rise) Figure 13.2 Basic pres­sure volume curve.

Thus, referring to the Otto cycle PV diagram again, the heat supplied Q2 through to 3 at constant volume between temperatures T2 and T3 (per kilogram of air) is given by

Q(2 to 3) = Cv(T3 — T2) (13.3)

Similarly, the heat rejected Q4 through to 1 at constant volume between temperatures T4 and Tj (per kilogram of air) is given by

Q(4to 1) = Cv(T4 -Tj) (13.4)

Because the compression and expansion phases are adiabatic (i. e., involve no heat transfer), Eqs. 13.3 and 13.4 contain the required terms to substitute into Eq. 13.2 as

Cv(T4 — Tt) Cv(T3 — T2)

(T4- Tt) (T3-T2)

 T1 = 1 — Cp Cv Also, because the compression and expansion phases are isentropic processes (i. e., revers­ible and adiabatic), the following relationships apply

 T2 _ "V Y-l And T3 _ "v4~ Ti _V2_ T4 _V3_
 Y-l

Where y

Because V = V, and V„ = V,

4 1 2 3’

 Y-l

 J2 Ti

 H T4

 V,

Now,

Vi Swept volume + Clearance volume ^ . .

—- =—————————————— = Compression ratio

V2 Clearance volume

Thus, it can be seen that the thermal efficiency of the Otto cycle depends on only the compression ratio.