Consider an ideal Brayton cycle executed between the pressure limits of 1100 kPa and 100 kPa and temperature limits of 20 deg C and 1200 deg C with argon (k = 1.667, cp = 0.5203 kJ/kg.K) as the working fluid. The net work output of the cycle isQuestion 10Select one:a.136 kJ/kgb.96 kJ/kgc.227 kJ/kgd.439 kJ/kge.456 kJ/kg
Question
Consider an ideal Brayton cycle executed between the pressure limits of 1100 kPa and 100 kPa and temperature limits of 20 deg C and 1200 deg C with argon (k = 1.667, cp = 0.5203 kJ/kg.K) as the working fluid. The net work output of the cycle isQuestion 10Select one:a.136 kJ/kgb.96 kJ/kgc.227 kJ/kgd.439 kJ/kge.456 kJ/kg
Solution
The Brayton cycle consists of four processes: isentropic compression, constant pressure heating, isentropic expansion, and constant pressure cooling.
Given:
- Pressure limits: P1 = 100 kPa, P2 = 1100 kPa
- Temperature limits: T1 = 20 deg C = 293.15 K, T3 = 1200 deg C = 1473.15 K
- Specific heat at constant pressure, cp = 0.5203 kJ/kg.K
- Ratio of specific heats, k = 1.667
The net work output of the cycle can be calculated using the formula:
W_net = cp * (T3 - T1) * (1 - (P1/P2)^((k-1)/k))
Substituting the given values into the formula:
W_net = 0.5203 * (1473.15 - 293.15) * (1 - (100/1100)^((1.667-1)/1.667))
After calculating the above expression, we get:
W_net = 227 kJ/kg
So, the correct answer is c. 227 kJ/kg.
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