Question

find efficiency of an otto cycle engine relative to the carnot cycle using tge same maximum pressure 21bar and temperature 1650°c and same minimum pressure 1.05bar and temperature 38°c. assume working fluid as air​

Answer 1

Answer:

9-2C It is less than the thermal efficiency of a Carnot cycle. ... produce the same amount of net work as that produced during the actual cycle. ... minimum volume formed in the cylinder.

Answer 2

The efficiency of the otto cycle is 0.374 and of the Carnot cycle is 0.84.

Given:

Maximum Pressure = 21 bar

Maximum Temperature = 1650 + 273 = 1923 K

Minimum Pressure = 1.05 bar

Minimum Temperature = 38 + 273 = 311 K

To find:

The efficiency of the otto cycle and Carnot cycle.

Solution:

The efficiency of the Otto cycle is given as

$\eta = 1 - \frac{T_{4} - T_{1} }{T_{3} - T_{2} } = 1 - \frac{V_{1}}{V_{3}}^{1-\gamma}$

We have been given

$T_{3} = 1923\\P_{3} = 21\\T_{1} = 311\\P_{1} = 1.05$

We know that

$\frac{V_{1}}{V_{3}} = \frac{T_{1}P_{3}}{T_{3}P_{1}}$

$\frac{V_{1}}{V_{3}} = \frac{311*21}{1923*1.05} = 3.235$

Therefore

$\eta = 1 - 3.235^{1-1.4} =0.374$

The efficiency of the Carnot cycle is given as

$\eta = 1 - \frac{T_{1}}{T_{3}}$

$\eta = 1 - \frac{311}{1923} = 0.84$

Hence the Carnot cycle is much more efficient than the otto cycle.

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