Question

before compression, an otto cycle with 0.005 kg of air is at 100 kpa, 300 k. the volume is then decreased by a factor of 10. the cycle reaches its maximum temperature at 1800 k. on both an air-standard and a cold air-standard basis, determine: a. the heat addition, in kj

Answer 1

The heat addition in an Otto cycle can be determined using the first law of thermodynamics and the ideal gas law. By calculating the initial and final temperatures of the air and using the specific heat at constant volume for air, we can find the heat addition. In this case, the heat addition is equal to 1080 kJ.

Step-by-step explanation:

The heat addition in an Otto cycle can be determined using the first law of thermodynamics which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In an air-standard analysis, we assume the air behaves as an ideal gas and follows the ideal gas law.

Here, the initial volume of the air is decreased by a factor of 10, so the final volume is 0.001 times the initial volume. Based on the given information, we can calculate the initial and final temperatures of the air using the ideal gas law:

Initial volume: V1 = 0.005 kg

Initial pressure: P1 = 100 kPa

Initial temperature: T1 = 300 K

Final volume: V2 = 0.001 * V1 = 0.001 * 0.005 = 0.000005 kg

Using the ideal gas law, P1 * V1 / T1 = P2 * V2 / T2, we can calculate the final temperature:

T2 = (P2 * V2 * T1) / (P1 * V1) = (100 kPa * 0.000005 kg) / (100 kPa * 0.005 kg) * 300 K = 0.006 K

Since the cycle reaches its maximum temperature at 1800 K, we can conclude that 1800 K is the temperature after the compression. Now, to calculate the heat addition, we can use the first law of thermodynamics:

Heat addition = Change in internal energy + Work done

Since the cycle is adiabatic, the change in internal energy is equal to zero. Therefore, the heat addition is equal to the work done. To calculate the work done, we can use the formula:

Work done = Specific heat at constant volume * Change in temperature

Using the specific heat at constant volume for air, which is approximately 0.72 kJ/(kg·K), and the change in temperature, we can calculate the heat addition:

Heat addition = Work done = Specific heat at constant volume * Change in temperature = 0.72 kJ/(kg·K) * (1800 K - 300 K) = 0.72 kJ/(kg·K) * 1500 K = 1080 kJ