Question

Engineers testing the efficiency of an electric generator gradually vary the temperature of the hot steam used to power it while leaving the temperature of the cooling water at a constant $20°C$. They find that the generator’s efficiency increases at a rate $3.5\times {10}^{-4}{\mathrm{K}}^{-1}$at steam temperatures near $300°C$. What is the ratio of the generator’s efficiency to the efficiency of a Carnot engine?

Short answer

Ratio of generator’s efficiency to the efficiency of a Carnot engine$=0.392$

Step-by-step solution

Introduction 

Carnot engine:

It is a heat engine having maximum efficiency which converts heat energy into mechanical energy by extracting heat from hot source and dumping some heat energy into cold sink. The maximum efficiency is given by the following expression.

$\eta =\left(1-\frac{T_{c_{}}}{T_h}\right)$

$\eta$ is efficiency of engine , T_c is temperature of cold sink and T_h is temperature of hot source.

Derive expression

We may deduce from the problem that our generator's efficiency can be expressed as the efficiency of a Carnot engine multiplied by a constant, which will almost certainly be between 0 and 1. .

$\eta =c\left(1-\frac{T_c}{T_h}\right)$

Differentiate it with respect to T_h.

$$\begin{gathered} \frac{d\eta }{d{T}_h}=-c{T}_c\frac{d}{d{T}_h}\frac{1}{T_h} \\ =\frac{c{T}_c}{T_h^2} \end{gathered}$$

Ratio of generator’s efficiency to the efficiency of a Carnot engine 

We can solve for this constant as follows:

$c=\frac{T_h^2}{T_c}\frac{d\eta }{d{T}_h}$

Put the values in equation,

localid="1649653195006" $$\begin{gathered} c=\frac{{573}^2}{293}\times 3.5\times {10}^{-4} \\ =0.392 \end{gathered}$$