What is the efficiency of a Carnot engine operating between higher temperature TH and lower temperature TL?
TH−TLTH
The work done by the gas over the entire cycle divided by the heat consumed from the hot reservoir is the efficiency of the Carnot engine.
Let us consider each isothermal process and find the work done in each of them.
Process a - b (isothermal)
Work done in an isothermal process is given by the formula
W=nRTℓn(v2v1)
In this case
W1=nRTℓn(vnvb). . . . (1)
Similarly the work done in process c - d is
W3=nRTℓn(vcvd). . . . (2)
Since both of these process are isothermal
QL=W3...(4) and QH=W1...(5)
The Carnot engine is represented below
Applying law of conservation of energy we get
QH−QL=W
Dividing throughout by QH we get
1−QLQH=WQH=η. . . . .(6)
Substituting for QH and QL in(6)
We get
1−nRTLln(vcvd)nRTHln(vbva)=η
⇒η=1−TLTHln(vcvd)(vbva)
In process b - c since it is adiabatic the state variables T and V are connected using the relation
T(1γ−1)b×Vb=T(1γ−1)c×Vc ......(8)
and similarly
T(1γ−1)d×Vd=T(1γ−1)a×Va .......(9)
From (8) and we get
vcvd=vbva
⇒ln(vcvd)ln(vbva) .......(10)
Substituting (10) in (9) we get
eta=1−TL
η=1−TLTH