Thermodynamics-adiabatic diffuser

[yecko]

Homework Statement

(for higher resolution, the pdf contains the same information)

Homework Equations

(in the pic)

The Attempt at a Solution

Why can the V2 assumed to be 0 m/s in part a?
As from part two, it is not 0 m/s indeed.

Moreover, why do h1 is using data of ideal gas behavior at certain temperature without taking account of the pressure?

Thanks.

 

[Chestermiller]

The answer to your first question is that V2 shouldn't be assumed zero in part a.

The answer to your second question is that the enthalpy of an ideal gas is independent of pressure.

 

[DoItForYourself]

Hello,

V₂² can be considered negligible comparing to V₁².

180²=32,400 but 34.2²=1,169.6. You can try to insert the value of 34.2 in the equation of part a. You will see that your result will be almost the same. However, in the mass balance, you cannot consider that the outlet has zero velocity (and consequently zero mass), because you automatically set the one side of the equation equal to zero.

As for the enhalpy h₁, maybe table A-17 is for air at atmospheric pressure. Both pressures are very close to the atmospheric pressure (90 kPa and 100 kPa are 0.888 and 0.987 atm respectively.).

 

[Chestermiller]

Hello,

V₂² can be considered negligible comparing to V₁².

180²=32,400 but 34.2²=1,169.6. You can try to insert the value of 34.2 in the equation of part a. You will see that your result will be almost the same. However, in the mass balance, you cannot consider that the outlet has zero velocity (and consequently zero mass), because you automatically set the one side of the equation equal to zero.

As for the enhalpy h₁, maybe table A-17 is for air at atmospheric pressure. Both pressures are very close to the atmospheric pressure (90 kPa and 100 kPa are 0.888 and 0.987 atm respectively.).

Are you saying that the enthalpy of an ideal gas (required by the called-for assumptions) is not independent of pressure?

 

[yecko]

Hello,

V₂² can be considered negligible comparing to V₁².

180²=32,400 but 34.2²=1,169.6. You can try to insert the value of 34.2 in the equation of part a. You will see that your result will be almost the same. However, in the mass balance, you cannot consider that the outlet has zero velocity (and consequently zero mass), because you automatically set the one side of the equation equal to zero.

As for the enhalpy h₁, maybe table A-17 is for air at atmospheric pressure. Both pressures are very close to the atmospheric pressure (90 kPa and 100 kPa are 0.888 and 0.987 atm respectively.).

Why can the V2 assumed to be 0 m/s in part a?
As from part two, it is not 0 m/s indeed.

Moreover, why do h1 is using data of ideal gas behavior at certain temperature without taking account of the pressure

Do you mean both of my answer is because the difference is negligible?

V22 can be considered negligible comparing to V12"

How can I know that this before I calculate for part B?

Thanks

 

[DoItForYourself]

Are you saying that the enthalpy of an ideal gas (required by the called-for assumptions) is not independent of pressure?

It was my fault, the enthalpy is independent of pressure according to the assumptions of the problem. I forgot the assumption number 2.

 

[yecko]

the enthalpy is independent of pressure according to the assumptions of the problem

where did you get that? (i can't find this piece of information in assumption...)

The answer to your first question is that V2 shouldn't be assumed zero in part a.

so why do the solution put zero as v2?

thanks

 

[DoItForYourself]

Do you mean both of my answer is because the difference is negligible?How can I know that this before I calculate for part B?

Thanks

For your first question : As long as there is only a small change in pressure and the area becomes 4 times the inlet area, you can assume that the velocity in the outlet would be approximately 4 times smaller, because of the mass balance (for volumetric flow).

Volumetric flow=velocity x area through which the flow occurs. For approximately the same volumetric flow (small pressure increase), for 4 times bigger area, the velocity will be 4 times smaller (approximately). In order to be more accurate, the solution must have included the V₂², so we would have a 2x2 system. I think the reason why the solution does not inlude V₂², is because the problem would be more difficult (you must go to the table A-17 each time to determine the temperature from enthalpy), but the result would be almost the same.

For your second question : The enthalpy of an ideal gas is independent of pressure. Assumption 2 states that air is an ideal gas.

 

[Chestermiller]

where did you get that? (i can't find this piece of information in assumption...)

thanks

The solution tells you to assume that the gas is an ideal gas. In addition to PV=nRT, another axiomatic characteristic of an ideal gas that you should be aware of is that the enthalpy and internal energy are functions only of temperature (and not volume or pressure at a given temperature).

 

[yecko]

As long as there is only a small change in pressure and the area becomes 4 times the inlet area, you can assume that the velocity in the outlet would be approximately 4 times smaller, because of the mass balance (for volumetric flow).

so if there is ideal gas I can assume the v2 is zero?
how can I assume an increase of even a 4% increase or 4 times of inlet area is with or without change of velocity in the calculation?
thanks

 

[DoItForYourself]

You can assume v₂² is zero comparing to v₁², because you have a small increase in pressure and the area becomes 4 times bigger:

You could (initially) say that the velocity would be 4 times smaller, so v₂² would be roughly 1/16 v₁². Consequently, If you assume v₂ is zero in part a, the error would be small.

 

[Chestermiller]

so if there is ideal gas I can assume the v2 is zero?
how can I assume an increase of even a 4% increase or 4 times of inlet area is with or without change of velocity in the calculation?
thanks

You have two coupled algebraic equations in 2 unknowns, T2 and V2. One way of solving a set of equations like this is by " the method of successive substitutions." You assume a value for one of the unknowns, in this case V2 = 0, and then easily solve for the other variable T2. Then you take the value of T2 you obtained, substitute it into your other equation, and solve for a "corrected" value of V2. Then you take the corrected value of V2 you obtained, and solve for a new value of T2. Etc. If this scheme converges, you will have your answer after several iterations (when the calculated values of T2 and V2 stop changing significantly). Why don't you try this full scheme and see how it plays out?