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Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.
What will be the back work ratio? (Answer up to two decimal places)
Correct answer is '0.58'. Can you explain this answer?
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Most Upvoted Answer
Consider a gas turbine power plant operates on the simple Brayton cyc...
The temperature at the exit of combustion chamber is
8555 = 12.81 x 1.093 x (T3 – 533)
T3 = 1144 K
The temperature at the turbine exit is determined using isentropic efficiency relation.
The back work ratio is
= 12.81 x 1.093 x (1144 – 754.4) = 5455 Kw
≈ 0.58
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Community Answer
Consider a gas turbine power plant operates on the simple Brayton cyc...
Given Data:
Pressure limits: P1 = 100 kPa, P2 = 700 kPa
Inlet temperature: T1 = 30°C = 303 K
Outlet temperature: T2 = 260°C = 533 K
Inlet mass flow rate: m = 12.6 kg/s
Air-fuel ratio: A/F = 60
Combustion efficiency: η_comb = 97%
Heating value of diesel: HHV = 42000 kJ/kg
Isentropic efficiency of turbine: η_turbine = 85%
Specific heat capacity of combustion gases: cp = 1.093 kJ/kg−K
Specific heat capacity of combustion gases at constant volume: cv = 0.806 kJ/kg−K
Ratio of specific heats: k = 1.357
Solution:
1. Air Compressor:
The air enters the compressor at state 1 and leaves at state 2.
Using the ideal gas law, we can calculate the density of air at state 1 and state 2:
ρ1 = P1 / (RT1) and ρ2 = P2 / (RT2)
where R is the specific gas constant for air.
2. Combustion Chamber:
The air and diesel fuel are burned in the combustion chamber with an air-fuel ratio of 60.
The mass of air is given by ma = m / (A/F + 1)
The mass of fuel is given by mf = m - ma
The energy released from the combustion of fuel is given by Qr = mf * HHV
The actual energy released in the combustion chamber is given by Qa = η_comb * Qr
3. Turbine:
The combustion gases leave the combustion chamber and enter the turbine at state 3.
The isentropic efficiency of the turbine is given by η_turbine = (h3s - h2) / (h3 - h2)
where h2 is the enthalpy at state 2 and h3s is the isentropic enthalpy at state 3.
We can calculate h2 using the specific heat capacity at constant volume cv = R / (k - 1) and temperature T2:
h2 = cv * T2
4. Back Work Ratio:
The back work ratio (BWR) is given by BWR = Wc / Qa
where Wc is the work done by the compressor and Qa is the actual energy released in the combustion chamber.
Calculations:
1. Air Compressor:
Using the ideal gas law, we can calculate the density of air at state 1 and state 2:
ρ1 = P1 / (RT1) = (100 * 10^3) / (R * 303)
ρ2 = P2 / (RT2) = (700 * 10^3) / (R * 533)
2. Combustion Chamber:
The mass of air is given by ma = m / (A/F + 1) = 12.6 / (60 + 1)
The mass of fuel is given by mf = m - ma
The energy released from the combustion of fuel is given by Qr = mf * HHV = mf
Pressure limits: P1 = 100 kPa, P2 = 700 kPa
Inlet temperature: T1 = 30°C = 303 K
Outlet temperature: T2 = 260°C = 533 K
Inlet mass flow rate: m = 12.6 kg/s
Air-fuel ratio: A/F = 60
Combustion efficiency: η_comb = 97%
Heating value of diesel: HHV = 42000 kJ/kg
Isentropic efficiency of turbine: η_turbine = 85%
Specific heat capacity of combustion gases: cp = 1.093 kJ/kg−K
Specific heat capacity of combustion gases at constant volume: cv = 0.806 kJ/kg−K
Ratio of specific heats: k = 1.357
Solution:
1. Air Compressor:
The air enters the compressor at state 1 and leaves at state 2.
Using the ideal gas law, we can calculate the density of air at state 1 and state 2:
ρ1 = P1 / (RT1) and ρ2 = P2 / (RT2)
where R is the specific gas constant for air.
2. Combustion Chamber:
The air and diesel fuel are burned in the combustion chamber with an air-fuel ratio of 60.
The mass of air is given by ma = m / (A/F + 1)
The mass of fuel is given by mf = m - ma
The energy released from the combustion of fuel is given by Qr = mf * HHV
The actual energy released in the combustion chamber is given by Qa = η_comb * Qr
3. Turbine:
The combustion gases leave the combustion chamber and enter the turbine at state 3.
The isentropic efficiency of the turbine is given by η_turbine = (h3s - h2) / (h3 - h2)
where h2 is the enthalpy at state 2 and h3s is the isentropic enthalpy at state 3.
We can calculate h2 using the specific heat capacity at constant volume cv = R / (k - 1) and temperature T2:
h2 = cv * T2
4. Back Work Ratio:
The back work ratio (BWR) is given by BWR = Wc / Qa
where Wc is the work done by the compressor and Qa is the actual energy released in the combustion chamber.
Calculations:
1. Air Compressor:
Using the ideal gas law, we can calculate the density of air at state 1 and state 2:
ρ1 = P1 / (RT1) = (100 * 10^3) / (R * 303)
ρ2 = P2 / (RT2) = (700 * 10^3) / (R * 533)
2. Combustion Chamber:
The mass of air is given by ma = m / (A/F + 1) = 12.6 / (60 + 1)
The mass of fuel is given by mf = m - ma
The energy released from the combustion of fuel is given by Qr = mf * HHV = mf
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Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer?
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Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer?.
Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer?.
Solutions for Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer?, a detailed solution for Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer? has been provided alongside types of Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Consider a gas turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 kPa and 700 kPa. Air enters the compressor at 30°C with a rate of 12.6 kg/s and leaves at 260°C. Now, this air and diesel fuel with an air-fuel ratio of 60 are burned in the combustion chamber with a combustion efficiency of 97%. Diesel has the heating value of 42000 kJ/kg. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85%. Treat the combustion gases as air with cp = 1.093 kJ/kg−K, cv = 0.806 kJ/kg−K and k = 1.357.What will be the back work ratio? (Answer up to two decimal places)Correct answer is '0.58'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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