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The efficiency of an otto cycle is 60%. Compression begins at 1 bar and 27 0C. The maximum pressure is 81 bar. The maximum temperature of the cycle is ______________ 0C.
Take Cv = 0.7 kJ/kg , R = 0.3 kJ/kg/K
  • a)
    2590
  • b)
    2592
Correct answer is between '2590,2592'. Can you explain this answer?
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The efficiency of an otto cycle is 60%. Compression begins at 1 bar a...
Given data:
Efficiency of the Otto cycle = 60%
Initial pressure (P1) = 1 bar
Initial temperature (T1) = 27°C
Maximum pressure (P3) = 81 bar
Specific heat at constant volume (Cv) = 0.7 kJ/kg
Gas constant (R) = 0.3 kJ/kg/K

Efficiency of the Otto cycle:
The efficiency of the Otto cycle is given by the formula:
η = 1 - (1 / compression ratio)^(γ-1)
where γ is the ratio of specific heats (Cp/Cv).

We are given the efficiency as 60%. Let's substitute the values and solve for the compression ratio.

0.6 = 1 - (1 / compression ratio)^(γ-1)

Solving for compression ratio:
(1 / compression ratio)^(γ-1) = 0.4

Taking the logarithm of both sides:
(γ-1) * ln(1 / compression ratio) = ln(0.4)

ln(1 / compression ratio) = ln(0.4) / (γ-1)

1 / compression ratio = e^(ln(0.4) / (γ-1))

compression ratio = 1 / (e^(ln(0.4) / (γ-1)))

Calculating the maximum temperature (T3):
The maximum temperature (T3) can be calculated using the ideal gas equation:

P1 * V1 / T1 = P3 * V3 / T3

Since the volume remains constant in the Otto cycle, we can simplify the equation to:

P1 / T1 = P3 / T3

Solving for T3:
T3 = (P3 * T1) / P1

Calculating the maximum temperature in °C:
To convert the temperature from Kelvin to Celsius, we use the formula:
T(°C) = T(K) - 273.15

Now let's substitute the given values and calculate the maximum temperature.

T3 = (81 * (27 + 273.15)) / 1

T3 = 21987.15 K

T3(°C) = 21987.15 - 273.15

T3(°C) ≈ 21714

Conclusion:
The maximum temperature of the Otto cycle is approximately 21714°C.
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The efficiency of an otto cycle is 60%. Compression begins at 1 bar and 27 0C. The maximum pressure is 81 bar. The maximum temperature of the cycle is ______________ 0C.Take Cv = 0.7 kJ/kg , R = 0.3 kJ/kg/Ka)2590b)2592Correct answer is between '2590,2592'. Can you explain this answer?
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