Question

If compression ratio of an engine working on Otto cycle is increased from 5 to 6, its air standard efficiency will increase by
(a) 1%
(6) 20%
(c) 16.67%
(d) 8%
(e) 25%.​

Answer 1

Answer:

8 % is the correct answer

Answer 2

Answer:

The air standard efficiency, $\eta_{otto}$, will increase by $8\%$.

Therefore, the correct option is option d) $8 \%$.

Explanation:

Given,

The compression ratio, r = $\frac{6}{5}$

As we know,

The ratio of specific heat, i.e., $\frac{C_{p} }{C_{v} }$ for an ideal gas, γ = $1.5$

The standard efficiency of air will increase by, $\eta_{otto}$ =?

As we know,

• The standard efficiency of an air-Otto cycle can be calculated by the equation given below:
• $\eta_{otto} = 1 - \frac{1}{r^{\gamma - 1} }$    -------equation (1)

Firstly, we have to calculate the value of ${r^{\gamma - 1} }$

After putting the values in ${r^{\gamma - 1} }$, we get:

• ${r^{\gamma - 1} }$ =  ${\frac{6}{5} ^{1.5 - 1} }$
• ${r^{\gamma - 1} }$ = $[{\frac{6}{5} ]^{\frac{1}{2} } }$
• ${r^{\gamma - 1} }$ = $\sqrt{1.2}$
• ${r^{\gamma - 1} }$ = $1.09$

Now,

• $\frac{1}{{r^{\gamma - 1} }}$ = $\frac{1}{1.09}$ = $0.917$

After putting the value of $\frac{1}{{r^{\gamma - 1} }}$ in equation (1), we get:

• $\eta_{otto} = 1 - 0.917$
• $\eta_{otto} = 0.08$

To change it in percentage, multiply it by $100$

• $\eta_{otto} = 0.08 \times 100$
• $\eta_{otto} = 8 \%$

Hence, standard efficiency will increase by $8\%$.