Thermodynamics: SFEE Homework Solution

In summary, the question asks to determine the mass flow rate and enthalpy of air entering and leaving a jet engine diffuser. Using the given values of temperature, pressure, velocity and inlet area, the mass flow rate is calculated to be 78.76kg/s. The enthalpy of the air leaving the diffuser is not given, but it can be assumed to be very small compared to the inlet velocity, which is confirmed by assuming an exit velocity of 0 or less than 3m/s. This results in an enthalpy of 303.14kJ/kg, which is rounded to 2 decimal places as per the question's guidance.
  • #1
JamesB93
7
1

Homework Statement



This is a question from an exam that my university determined to be so harsh that they allowed resits with uncapped grades. I'm going to be taking such a resit thus I'm trying to figure out the paper:

Air at 10oC and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200m/s and a specific enthalpy of 283.14kJ/kg. The inlet area of the diffuser is 0.4m2. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity.

Determine:

a) The mass flow rate of the air (5 marks)

b) The enthalpy of the air leaving the diffuser (5 marks)

Homework Equations



pv=mRT
modified SFEE: h1 + c12/2 = h2 + c22/2

volumetric flow = velocity x inlet area

R = 287J/kg/K

The Attempt at a Solution


[/B]
a) VF = 200m/s * 0.4m2 = 80m3/s
mass flow = (pressure x volumetric flow rate)/(gas constant of air x temperature)
mass flow = (80000 x 80)/(287 x 283.15) = 78.76kg/s

b) for this section I have no idea, all I can say is that the enthalpy is very large in comparison with the inlet. I have been given no values as far as I can tell.
 
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  • #2
JamesB93 said:

Homework Statement



This is a question from an exam that my university determined to be so harsh that they allowed resits with uncapped grades. I'm going to be taking such a resit thus I'm trying to figure out the paper:

Air at 10oC and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200m/s and a specific enthalpy of 283.14kJ/kg. The inlet area of the diffuser is 0.4m2. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity.

Determine:

a) The mass flow rate of the air (5 marks)

b) The enthalpy of the air leaving the diffuser (5 marks)

Homework Equations



pv=mRT
modified SFEE: h1 + c12/2 = h2 + c22/2

volumetric flow = velocity x inlet area

R = 287J/kg/K

The Attempt at a Solution


[/B]
a) VF = 200m/s * 0.4m2 = 80m3/s
mass flow = (pressure x volumetric flow rate)/(gas constant of air x temperature)
mass flow = (80000 x 80)/(287 x 283.15) = 78.76kg/s

b) for this section I have no idea, all I can say is that the enthalpy is very large in comparison with the inlet. I have been given no values as far as I can tell.

EDIT:

I had an idea I could just assume the exit velocity roughly to see how much difference it made. As the general guidance from the question is to give answers in kJ/kg and rounded to 2dp, a velocity of less than 3m/s makes absolutely no difference to the outcome. With the emphasis on "very small" in the question, would it be safe to assume that it's less than 3m/s (1.5% of the original)?

eg. @ 3m/s

h1 + c12/2 - c22/2 =

283140 + (2002 - 32)/2 = h2
283140 + 19995.5 = 303135.5J/kg = 303.14kJ/kg
 
  • #3
Why didn't you just take the exit velocity as zero?
 
  • #4
Chestermiller said:
Why didn't you just take the exit velocity as zero?

I was thinking that, wasn't sure if it was acceptable to do so. My lecturers haven't been too clear through this year. It gives the same result so I'm guessing that's what I'm supposed to do.

Thanks :)
 
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FAQ: Thermodynamics: SFEE Homework Solution

What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, but can only be transferred or converted from one form to another. This means that in any thermodynamic process, the total energy of the system and its surroundings remains constant.

What is the second law of thermodynamics?

The second law of thermodynamics states that in any natural process, the total entropy of a closed system will always increase over time. Entropy can be thought of as a measure of disorder or randomness within a system. This law also implies that energy will always flow from a higher temperature to a lower temperature, and cannot spontaneously flow in the opposite direction.

How is the efficiency of a thermodynamic process calculated?

The efficiency of a thermodynamic process is calculated by dividing the work output by the energy input. This can be expressed as a percentage, with a higher percentage indicating a more efficient process. The efficiency of a heat engine, for example, is given by the equation: efficiency = (work output / energy input) x 100%.

What is the significance of the Carnot cycle in thermodynamics?

The Carnot cycle is a theoretical thermodynamic cycle that is used as a benchmark for the maximum possible efficiency of a heat engine. It consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. The Carnot cycle helps to demonstrate the importance of maximum efficiency in thermodynamic processes and serves as a model for designing more efficient engines.

How is thermodynamics related to other branches of science?

Thermodynamics is a fundamental part of physics, and it has important applications in chemistry, biology, and engineering. It is closely related to other branches of science such as heat transfer, fluid mechanics, and statistical mechanics. Thermodynamics also plays a significant role in understanding natural phenomena, such as weather patterns and the behavior of stars and planets.

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