We have ɳ _{HE } = W _{net }/ Q_{1} = Q_{1} –Q_{2} / Q_{1 }= 1 – Q_{2} / Q_{1 }

as t_{1} > t_{2}

so t_{1} – t_{2} > 0 ( must be for heat flow)

We know reversible heat engine’s efficiency depends on source and sink temperatures.

Thus ɳ _{rev} = /1 (t_{1}, t_{2}) where f _{1 }is a function of t_{1} and t_{2}

From(7.7)

1 –Q_{2} / Q_{1} = f_{1} ( t_{1} , t_{2})

or Q_{2} / Q_{1} = 1 – F_{1} ( T_{1}, T_{2})

or Q_{1} / Q_{2} = 1 / 1- f ( t_{1} , t_{2}) = f_{2 }( t_{1},t_{2})

Let us now take three heat engines arranged as shown in Fig. 7.5

Q_{2 }/ Q_{3} = f_{3} ( t_{2} ,t_{3})

And Q_{1 }/ Q_{3} = f_{4} ( t_{1} , t_{3})

Again

Q_{1} / Q_{2 }= Q_{1} / Q_{3 }/ Q_{2}/Q_{3}

= f_{4}(t_{1},t_{3}) / f_{3} (t_{2},t_{3})

or f_{2} ( t_{1},t_{2}) = f_{4} ( t_{1},t_{3} / f_{3} ( t_{2} , t_{3})

Now t _{1}, t_{2}, t_{3} are independent and t_{3} does not affect the operation of HE1 which is

operating between t_{1} and t_{2}.

To satisfy equation (7.11), we must have

and f_{4} ( t_{1},t_{3}) = ϕ (t_{1}) / ϕ(t_{3})

and f_{3}(t_{2},t_{3}) = ϕ( t_{1}) / ϕ(t_{2}) = Q_{1} / Q_{2}

Since ϕ (t) is an arbitrary function, the simpliest possible way to define absolute

thermodynamic temperature Tis to let <j>(t) = T, proposed by Kelvin.

Thus from equation (7.12), we have

Q_{1 }/ Q_{2} = T_{1}/T_{2}

This scale is also known as Kelvin scale. It is independent of characteristic of any substance. In Kelvin scale, the triple point of water is taken as the standard reference point. For a Carnot engine operating between T and T _{tr} (triple point of water), we have

Q /Q _{tr} = T / T _{tr}

T _{tr} = 273.16 k (arbitrary assigned value)

T=273.16 Q / Q _{tr}

If T changes, Q also changes. Thus Q plays an important role of thermodynamic property.

Fig. 7.7 shows a series of heat engines.

T_{1} / T_{2} = Q_{1 }/ Q_{2}

T_{1}-T_{2} / T_{2} = Q_{1} – Q_{2} / Q_{2}

T_{1} –T_{2} / T_{2} = Q_{1 }– Q_{2} / Q_{2}

T_{1} –T_{2} = T_{2} / Q_{2} ( Q_{1} –Q_{2})

Similarly T_{2 }– T_{3} = T_{3 }/ Q_{3} ( Q_{2} –Q_{3})

= T_{2} / Q_{2} ( Q_{2} – Q_{3})

T_{3} – T_{4 }= T_{2} / Q_{2} ( Q3 –Q_{4})

And so on.

Let W_{1} = W_{2} = W_{3} =… = W

Q_{1} –Q_{2} = Q_{2} – Q_{3} = Q_{3} –Q_{4}

Thus T_{1} – T_{2} = ( T_{2} –T_{3}) = ( T_{3} –T_{4}) =….(7.16)

Also if it is assumed Q_{1 }– Q_{2} = Q_{2} – Q_{3} = Q_{3} – Q_{4} = …

Then also W_{1} = W_{2} = W_{3} = … = W

Equation (7.16) shows equal temperature intervals usually steam point and ice point are taken as fixed reference temperatures and the difference between the two is arbitrary taken as 100°C.

T_{S }– T_{i} =100 , [T_{S }= steam point T_{i} = ice point]

ɳ _{rev} = 1 – T_{i} / T_{S}= 0.268 [ ɳ _{rev} measured and found to be 0.268]

T_{i} / T_{S} = 0.732

(7.17) => T_{S} – 0.732 ts = 100

T_{s} = 100 / 0.268 = 373.15 ( steam point)

T_{i }= 273.15 (ice point)

Triple point of water is at 0.01°C. For this in Kelvin scale 273.15 + 0.01 = 273.16 K.