Question

If the compression ratio of an engine working on Otto cycle is increased from $ 5 $ to $ 7 $ . The percentage increase in efficiency will be
(A) $ 2\% $
(B) $ 4\% $
(C) $ 8\% $
(D) $ 14\% $

Answer 1

Hint: To solve this question, we need to use the formula for the efficiency of the otto cycle. Putting the values of the initial and the final compression ratios, we can calculate the initial and the final efficiencies. Using these we can calculate the required percentage increase in efficiency.

Formula used: The formula used to solve this question is given by
 $ \eta = 1 - \dfrac{1}{{{r^{\gamma - 1}}}} $ , here $ \eta $ is the efficiency of an engine working on an otto cycle having a compression ratio of $ r $ and the gas used in the engine having the specific heat ratio of $ \gamma $ .

Complete step-by-step solution
We know that the efficiency of an engine working in an otto cycle is given by
 $ \eta = 1 - \dfrac{1}{{{r^{\gamma - 1}}}} $
In the given question, we are not given any information regarding the type of gas used in the engine. So we assume that the gas used in the given engine is air. We know that the heat capacity ratio for the air is approximately $ 1.4 $ . So we substitute $ \gamma = 1.4 $ in the above expression to get
 $ \eta = 1 - \dfrac{1}{{{r^{1.4 - 1}}}} $
 $ \Rightarrow \eta = 1 - \dfrac{1}{{{r^{0.4}}}} $ ..........................(1)
Now, according to the question, the engine has an initial compression ratio of the engine is equal to $ 5 $ . So substituting $ r = 5 $ in (1), we get the initial efficiency as
 $ {\eta _1} = 1 - \dfrac{1}{{{5^{0.4}}}} $
 $ \Rightarrow {\eta _1} = 0.47 $ ..........................(2)
Now, the engine has a final compression of $ 7 $ . Therefore substituting $ r = 7 $ in (1) we get
 $ {\eta _2} = 1 - \dfrac{1}{{{7^{0.4}}}} $
 $ \Rightarrow {\eta _2} = 0.54 $ ..........................(3)
Now, the increase in the efficiency can be given by
 $ \Delta \eta = {\eta _2} - {\eta _1} $
The fractional change in efficiency can be given by
 $ \dfrac{{\Delta \eta }}{{{\eta _1}}} = \dfrac{{{\eta _2} - {\eta _1}}}{{{\eta _1}}} $
And the percentage change in efficiency can be written as
 $ \dfrac{{\Delta \eta }}{{{\eta _1}}} \times 100 = \dfrac{{{\eta _2} - {\eta _1}}}{{{\eta _1}}} \times 100 $
Putting (2) and (3) in the above equation, we get
 $ \dfrac{{\Delta \eta }}{{{\eta _1}}} \times 100 = \dfrac{{0.54 - 0.47}}{{0.47}} \times 100 $
 $ \Rightarrow \dfrac{{\Delta \eta }}{{{\eta _1}}} \times 100 = 14.9\% \approx 14\% $
Thus, the percentage increase in the efficiency is equal to $ 14\% $ .
Hence, the correct answer is option D.
Note
An otto cycle is used in the operation of various automobiles. It describes hw an engine turns the chemical energy of the fuel into the useful work. The fuel used is a mixture of gasoline and air.