6.0 Efficiency of cyclic process

  First Law of Thermodynamics

In a cyclic process. $$\Delta U = 0\quad {\text{and}}\quad {{\text{Q}}_{net}} = {W_{net}}$$ (from first law of thermodynamics)

First we see what is the meaning of efficiency of a cycle. Suppose $100$ $J$ of heat of supplied to a system (in our case it is an ideal gas) and the system does $60$ $J$ of work. Then efficiency of the cycle is $60\%.$ Thus efficiency $\left(\eta \right)$ of a cycle can be defined as $$\eta = \left( {\frac{{\frac{{{\text{Work done by the working substance}}}}{{{\text{(an ideal gas in our case)during a cycle}}}}}}{{{\text{Heat supplied to the gas during the cycle}}}}} \right) \times 100$$ $$ = \frac{{{W_{Total}}}}{{\left| {{Q_{ + ve}}} \right|}} \times 100$$ $$ = \frac{{\left| {{Q_{ + ve}}} \right| - \left| {{Q_{ - ve}}} \right|}}{{\left| {{Q_{ + ve}}} \right|}} \times 100$$ $$ = \left\{ {1 - \left| {\frac{{{Q_{ - ve}}}}{{{Q_{ + ve}}}}} \right|} \right\} \times 100$$ Thus, $$\eta = \frac{{{W_{Total}}}}{{\left| {{Q_{ + ve}}} \right|}} \times 100 = \left\{ {1 - \left| {\frac{{{Q_{ - ve}}}}{{{Q_{ + ve}}}}} \right|} \right\} \times 100$$

Note: (i) There can't be a cycle whose efficiency is $100\%.$ Hence, $\eta $ is always less then $100\%.$

Thus, $${W_{Total}} \ne {Q_{ + ve}}$$ (ii) It is just like a shopkeeper. He takes some money from you. $($suppose he takes $Rs. 100/-$ from you$).$ In lieu of this he provides services to you $($ Suppose he provides services of worth $Rs.180/-).$ Then the efficiency of the shopkeeper is $80\%.$ There can't be a shopkeeper whose efficiency is $100\%.$ Otherwise what will he save $?$

Question 17. Find the efficiency of the given cycle.

Slution. From table $18.4$ we can see that ${Q_{ + ve}}$ during the cycle is $700$ $J,$ while the total work done in the cycle is $600$ $J.$ $$\eta = \frac{{{W_{Total}}}}{{\left| {{Q_{ + ve}}} \right|}} \times 100 = \left( {\frac{{600}}{{700}}} \right) \times 100$$ $$=85.71\%$$