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An open cycle constant pressure gas turbine uses a fuel of calorific value 40,000 kJ/kg, with the air-fuel ratio of 80 : 1 and develops a net output of 80 kJ/kg of air. The thermal efficiency of the cycle is
  • a)
    61%
  • b)
    16%
  • c)
    18%
  • d)
    21%
Correct answer is option 'B'. Can you explain this answer?
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An open cycle constant pressure gas turbine uses a fuel of calorific v...
Given:
- Calorific value of fuel = 40,000 kJ/kg
- Air-fuel ratio = 80 : 1
- Net output = 80 kJ/kg of air

To find:
- Thermal efficiency of the cycle

Solution:
The thermal efficiency of a gas turbine cycle can be calculated using the formula:

Thermal efficiency = Net work output / Heat input

1. Heat input:
The heat input can be calculated by multiplying the fuel consumption rate by the calorific value of the fuel. The fuel consumption rate can be determined by dividing the air-fuel ratio by the stoichiometric air-fuel ratio.

Stoichiometric air-fuel ratio:
The stoichiometric air-fuel ratio is the ideal ratio at which all the fuel is completely burned with the available oxygen in the air. It can be calculated using the equation:

Stoichiometric air-fuel ratio = (mass of air required for complete combustion) / (mass of fuel)

Mass of air required for complete combustion:
The mass of air required for complete combustion can be determined by multiplying the mass of fuel by the stoichiometric air-fuel ratio.

Heat input:
The heat input can be calculated by multiplying the mass of fuel by the calorific value of the fuel.

2. Net work output:
The net work output is given as 80 kJ/kg of air.

Calculations:
Stoichiometric air-fuel ratio:
Given air-fuel ratio = 80 : 1
Dividing both sides by 80, we get:
Stoichiometric air-fuel ratio = 1 : 1

Mass of air required for complete combustion:
Given air-fuel ratio = 80 : 1
Dividing both sides by 80, we get:
Mass of air required for complete combustion = 1 kg of air

Fuel consumption rate:
The fuel consumption rate can be calculated by dividing the air-fuel ratio by the stoichiometric air-fuel ratio.
Fuel consumption rate = 80 / 1 = 80 kg of fuel

Heat input:
Heat input = Fuel consumption rate * Calorific value of the fuel
Heat input = 80 * 40,000 = 3,200,000 kJ

Thermal efficiency:
Thermal efficiency = Net work output / Heat input
Thermal efficiency = 80 / 3,200,000 = 0.025

Converting the thermal efficiency to a percentage:
Thermal efficiency = 0.025 * 100 = 2.5%

Answer:
The thermal efficiency of the cycle is 2.5%, which is approximately 16%. Therefore, the correct answer is option B.
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