This question was previously asked in

ESE Mechanical 2014 Official Paper - 1

Option 2 : Lower the sink temperature to 30°c

CT 3: Building Materials

2962

10 Questions
20 Marks
12 Mins

__Explanation:__

Improved performance means improved efficiency. We know that the efficiency of a reversible Carnot engine is given by-

\(\eta = 1 - \frac{{{T_2}}}{{{T_1}}}\)

[Where T_{1} is the higher temperature or temperature of the source and T_{2} is the lower temperature or temperature of the sink]

It T_{2} is constant, then-

\({\left( {\frac{{\partial \eta }}{{\partial {T_1}}}} \right)_{{\rm{T}}2}} = \frac{{{T_2}}}{{T_1^2}}\)

Now, as T_{1} increases, η increases, and the slope \({\left( {\frac{{\partial \eta }}{{\partial {T_1}}}} \right)_{{\rm{T}}2}}\) decreases. If T_{1} is constant, \({\left( {\frac{{\partial \eta }}{{\partial {T_2}}}} \right)_{{\rm{T}}1}} = - \frac{1}{{{T_1}}}\)

As T_{2} decreases, η increases, but the slope \({\left( {\frac{{\partial \eta }}{{\partial {T_2}}}} \right)_{{\rm{T}}1}}\) remains constant.

Also,

\({\left( {\frac{{\partial \eta }}{{\partial {T_1}}}} \right)_{{\rm{T}}2}} = \frac{{{T_2}}}{{T_1^2\;}}{\rm{\;and}}\;{\left( {\frac{{\partial \eta }}{{\partial {T_2}}}} \right)_{{\rm{T}}1}} = - \frac{{{T_1}}}{{T_1^2}}\)

Since T_{1 }> T_{2},

\(\therefore \;{\left( {\frac{{\partial \eta }}{{\partial {T_2}}}} \right)_{{\rm{T}}1}} > {\left( {\frac{{\partial \eta }}{{\partial {T_1}}}} \right)_{{\rm{T}}2}}\)

So, the more effective way to increase the efficiency is to decrease T_{2}, though in both ways the efficiency of the reversible cycle will increase. **So, options (1) and (2) both are correct but option (2) is better. So, we will choose option (2).**

__Alternative way __

To solve this question is to calculate the efficiency values-

The efficiency of a reversible engine-

\({\eta _1} = 1 - \frac{{{T_2}}}{{{T_1}}} = 1 - \left( {\frac{{60 + 273}}{{260 + 273}}} \right) = 1 - \frac{{333}}{{533}}\)

**∴**** η _{1} = 0.3752**

**Now,**

When source temperature T_{1} is 300°C

New efficiency

\({{\rm{\eta }}_2} = 1 - \frac{{333}}{{\left( {300 + 273} \right)}} = 1 - \frac{{333}}{{573}}\)

**∴**** η _{2 }= 0.41885**

When sink temperature T_{2} is 30°C

New efficiency

\({{\rm{\eta }}_3} = 1 - \frac{{\left( {273 + 30} \right)}}{{533}}\)

**∴**** η _{3 }= 0.43152**

**Hence, η _{3} > η_{2} > η_{1}**

**So, options (1) and (2) both are correct but option (2) is better. So, we will choose option (2).**