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The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 K and 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglect potential energy. If the pressure and temperature of the surroundings are 1 bar and 300 K, respectively, the available energy in kJ/kg of the air stream is
- a)170
- b)187
- c)191
- d)213
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The pressure, temperature and velocity of air flowing in a pipe are 5 ...
Given data:
Pressure (P) = 5 bar
Temperature (T) = 500 K
Velocity (V) = 50 m/s
Specific heat at constant pressure (Cp) = 1.005 kJ/kgK
Specific heat at constant volume (Cv) = 0.718 kJ/kgK
Surrounding pressure = 1 bar
Surrounding temperature = 300 K
Calculation:
1. Enthalpy of the air stream:
The enthalpy of the air stream can be calculated using the following formula:
h = Cp * T
h = 1.005 * 500
h = 502.5 kJ/kg
2. Internal energy of the air stream:
The internal energy of the air stream can be calculated using the following formula:
u = Cv * T
u = 0.718 * 500
u = 359 kJ/kg
3. Available energy of the air stream:
The available energy of the air stream can be calculated using the following formula:
Available energy = h - (P-P0) / ρ + gZ - u - To (S - So)
where P0 = surrounding pressure = 1 bar
ρ = density of air = P / (R * T)
R = gas constant = 0.287 kJ/kgK
g = acceleration due to gravity = 9.81 m/s^2
Z = height above the reference plane = 0 (neglecting potential energy)
To = surrounding temperature = 300 K
S = entropy of the air stream
So = entropy of the surroundings
Now, let's calculate each term of the equation:
(P-P0) / ρ = (5-1) / (0.287 * 500) = 0.0278 kJ/kg
S - So = Cp * ln(T/T0) - R * ln(P/P0) = 1.005 * ln(500/300) - 0.287 * ln(5/1) = 1.193 kJ/kg
So, available energy = 502.5 - 0.0278 + 0 - 359 - 300 * 1.193 = 186.96 kJ/kg
Therefore, the available energy in kJ/kg of the air stream is 187 kJ/kg (approximately).
Hence, option B is the correct answer.
Pressure (P) = 5 bar
Temperature (T) = 500 K
Velocity (V) = 50 m/s
Specific heat at constant pressure (Cp) = 1.005 kJ/kgK
Specific heat at constant volume (Cv) = 0.718 kJ/kgK
Surrounding pressure = 1 bar
Surrounding temperature = 300 K
Calculation:
1. Enthalpy of the air stream:
The enthalpy of the air stream can be calculated using the following formula:
h = Cp * T
h = 1.005 * 500
h = 502.5 kJ/kg
2. Internal energy of the air stream:
The internal energy of the air stream can be calculated using the following formula:
u = Cv * T
u = 0.718 * 500
u = 359 kJ/kg
3. Available energy of the air stream:
The available energy of the air stream can be calculated using the following formula:
Available energy = h - (P-P0) / ρ + gZ - u - To (S - So)
where P0 = surrounding pressure = 1 bar
ρ = density of air = P / (R * T)
R = gas constant = 0.287 kJ/kgK
g = acceleration due to gravity = 9.81 m/s^2
Z = height above the reference plane = 0 (neglecting potential energy)
To = surrounding temperature = 300 K
S = entropy of the air stream
So = entropy of the surroundings
Now, let's calculate each term of the equation:
(P-P0) / ρ = (5-1) / (0.287 * 500) = 0.0278 kJ/kg
S - So = Cp * ln(T/T0) - R * ln(P/P0) = 1.005 * ln(500/300) - 0.287 * ln(5/1) = 1.193 kJ/kg
So, available energy = 502.5 - 0.0278 + 0 - 359 - 300 * 1.193 = 186.96 kJ/kg
Therefore, the available energy in kJ/kg of the air stream is 187 kJ/kg (approximately).
Hence, option B is the correct answer.
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Community Answer
The pressure, temperature and velocity of air flowing in a pipe are 5 ...
¥=(h-h0)-T0(S-S0)+v^2/2
entropy is given S=CplnT-Rlnp
since h=CpT
¥=Cp(T-T0)-T0{Cpln(T/T0)-Ron(P/P0)}+v^2/2
submit values above
¥=1.005(500-300)-300{1.005*ln(500/300)-0.287*ln(5/1)}+50^2/2*1000
¥=187 kj/kg.
entropy is given S=CplnT-Rlnp
since h=CpT
¥=Cp(T-T0)-T0{Cpln(T/T0)-Ron(P/P0)}+v^2/2
submit values above
¥=1.005(500-300)-300{1.005*ln(500/300)-0.287*ln(5/1)}+50^2/2*1000
¥=187 kj/kg.
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The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 Kand 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglectpotential energy. If the pressure and temperature of the surroundings are 1 barand 300 K, respectively, the available energy in kJ/kg of the air stream isa)170b)187c)191d)213Correct answer is option 'B'. Can you explain this answer?
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The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 Kand 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglectpotential energy. If the pressure and temperature of the surroundings are 1 barand 300 K, respectively, the available energy in kJ/kg of the air stream isa)170b)187c)191d)213Correct answer is option 'B'. 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 The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 Kand 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglectpotential energy. If the pressure and temperature of the surroundings are 1 barand 300 K, respectively, the available energy in kJ/kg of the air stream isa)170b)187c)191d)213Correct answer is option 'B'. 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 The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 Kand 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglectpotential energy. If the pressure and temperature of the surroundings are 1 barand 300 K, respectively, the available energy in kJ/kg of the air stream isa)170b)187c)191d)213Correct answer is option 'B'. Can you explain this answer?.
The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 Kand 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglectpotential energy. If the pressure and temperature of the surroundings are 1 barand 300 K, respectively, the available energy in kJ/kg of the air stream isa)170b)187c)191d)213Correct answer is option 'B'. 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 The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 Kand 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglectpotential energy. If the pressure and temperature of the surroundings are 1 barand 300 K, respectively, the available energy in kJ/kg of the air stream isa)170b)187c)191d)213Correct answer is option 'B'. 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 The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 Kand 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglectpotential energy. If the pressure and temperature of the surroundings are 1 barand 300 K, respectively, the available energy in kJ/kg of the air stream isa)170b)187c)191d)213Correct answer is option 'B'. Can you explain this answer?.
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