ONE-DIMENSIONAL ISENTROPIC FLOW
Oscar Eduardo Vázquez Moncada 1691276 José Luis Toscano Esquivel 1584067 Enrique Alonso Baeza Martínez 1691515 Erick Antonio González García 1691344
INTRODUCTION
Dimension is the same as direction, as we analize vectors. One-dimensional flow means that the flow is flowing to one-direction.
INTRODUCTION
Steady flow, when the properties of the flow stay
constant with the time. Isentropic flow, when the flow does not have a change on entropy.
INTRODUCTION
Flow can be modeled assuming them to be steady, one-dimensional and isentropic.
Although, in real life, any flow is entirely isentropic, but many of them can be assumed to be isentropic.
INTRODUCTION
There are many cases in internal ducts where the effects and viscosity and heat transfer are only important in the wall boundaries. Similarly in external flows, effects of viscosity and heat transfer are restricted to the boundary layers and shock waves, the rest is assumed to be isentropic.
GOVERNING EQUATIONS
By definition, entropy remains constant in an isentropic flow (as remarked before), that means that in such flow:
=
GOVERNING EQUATIONS
If any two points, as 1 and 2 shown in this figure, in an isentropic flow are considered, from the last equation we get:
=
GOVERNING EQUATIONS
The general equation of state gives: = In an isentropic flow: = − = Recalling that = , and considering the steady flow adiabatic energy equation applied between points 1 and 2, this gives: = = = =
GOVERNING EQUATIONS
In other words…
= + +
However, = 1 = 2 2 2
Since last equations, it follows that
= + − + −
or
= + − + −
This equation applies in adiabatic flow, if the flow is isentropic. So with the isentropic state relations.
GOVERNING EQUATIONS
The continuity equation gives = Which canbe arranged as
=
All equations above are sufficient to determine the characteristics of one-dimensional isentropic flow.
*Note that momentum equation wasn’t used in this analysis cause will always give the same result as the energy equation.
GOVERNING EQUATION
The isentropic equation of state gives
=
And, substituting on the energy equation − − Which can be arranged as: +(−) +(−)
1 =1
=
That is the same result that using the energy equation. But is simpler to apply than the momentum.
EXAMPLE A gas that has a molar mass of 39.9 and a specific heat ratio of 1.67 is discharged from a large chamber in which the pressure is 500 kPa and the temperature is 30°C through a nozzle. Assuming one-dimensional isentropic flow, find a. If the pressure at some section of the nozzle is 80 kPa, the Mach number, temperature, and velocity at this section. b. If the nozzle has a circular cross section and if its diameter is 12 mm at the section discussed in (a) above, the mass flow rate through the nozzle.
SOLUTION
The following is given: Is assumed to the velocity is effectively zero. Steady, one-dimensional, isentropic flow is assumed. a) Since the velocity in the chamber and therefore the Mach number in the chamber are assumed to be zero:
SOLUTION
Since the Mach number is zero, this is the resultant equation:
Using this value of temperature and the equation of speed sound velocity:
Hence, Therefore, Mach number is 1.8, T= 145.3 K and velocity is 404.7 m/s
SOLUTION b) The mass flow is given by and using the values of temperature and pressure established in a), density can be found as: Hence
Therefore, the mass flow through the nozzle is: 0.121 kg/s.
Activity 1
A gas has a molar mass of 44 and a specific heat ratio of 1.3. At a certain point in the flow, the static pressure and temperature are 80 kPa and 15°C, respectively and the velocity is 100 m/s. The gas is then isentropically expanded until its velocity is 300 m/s. Find the pressure, temperature, and Mach number that exist in the resulting flow.
Stagnation Conditions
Stagnation conditions are those that would exist if
the flow at any point in a fluid stream was isentropically brought to rest. In other words we can define it as a point in the streamline where we can find a Velocity that is equal to 0.
Stagnation Conditions
To define the stagnation temperature, it is actually only necessary to require that the flow be adiabatically brought to rest. To define the stagnation pressure and density, it is necessary, however, to require that the flow be brought to rest isentropically.
If the entire flow is essentially isentropic and if the velocity is essentially zero at some point in the flow, then the stagnation conditions existing at all points in the flow will be those existing at the zero velocity point as indicated in Figure 4.4.
What does Isentropic Means?
Isentropic conditions or flows are those in which there is not heat exchange in the system (Even if is in or out simply no heat exchange).
Even when the flow is nonisentropic, the concept of the stagnation conditions is still useful, the stagnation conditions at a point then being the conditions that would exist if the local flow were brought to rest isentropically as indicated in Figure 4.5.
Assumptions As Ideal Gases
Equation Of State
p = p(ρ,T);
=
Enthalpy Equation of State: h = h(p,T);
=
Entropic Equation of state: s = s(p,T);
=
1st Law of thermodynamics Adibatic system, no work done, no potential e.
= =
Figure 17-4 Real State and Stagnation State, diagram h-s. Yunus A. Cengel, “Thermodynamics. An Engineering Approach”, Mc Graw Hill, 7th edition, NY, USA, 2011, pp. 849.
Sonic Equation Speed of an infinitesimal disturbance within the fluid
=
=
,
Variables in any section, could be expressed in function of variables of state in stagnation point or Mach number
=
=
1
−
=
1
−
1
−
(4.15)
(4.16)
(4.17)
Example (Stagnation Conditions)
A pitot–static tube is placed in a subsonic airflow. The static pressure and temperature in the flow are 96 kPa and 27°C, respectively. The difference between the pitot and static pressures is measured and found to be 32 kPa. Find the air velocity (a) assuming an incompressible flow and (b) assuming compressible flow.
DATA
Difference of preassure
= 32
Real Preassure = 96 kpa Temperature = 27 Celsius or 300 kelvin Our constant R = 287 J/kg*K (Given by the air at 300k)
We will use the equation of state , then we will use the density
= velocity. = , We need to make this in order to obtain our =
= 1.1149 ∗
A) Considering Compressive flow
=2
−=2 = 239.6 .
B) Now assuming compressible flow We know that
= 32 = 96 32 = 1.333 = 1 28 , = = 1 − . .− . 1.33 = 1 = . ∗1.333 . 1 = 0.6541
To find velocity is necessary to find the sound speed on those conditions, we have all of them.
= 300 = = 0.654 1.4 287 ∗ = 227.06
The actual velocity is 225.7 m/s, whereas when compressibility effects are neglected, the velocity is found to be 239.6 m/s.
Activity 2
A high-speed subsonic Boeing 707 airliner is flying at a pressure altitude of 12 km. A Pitot tubeon the vertical . At which tail measures a pressure of Mach number is the airplane flying?
2.9610
Rayleigh Flow
Advanced Rayleigh Pressure Loss Model for HighSwirl Combustion in a Rotating Combustion Chamber
Andreas Penkner and Peter Jeschke, J. Eng. ‘Gas Turbines Power 138’ , 021502 (Sep 01, 2015) (12 pages)
What’s the Point?
This paper considers the effect of excessive total pressure losses for heat transfer problems in fluid flows with a high circumferential swirl component. At RWTH Aachen University, a novel gas generator concept is under research.
During the predesign of the rotating combustion
chamber using computational fluid dynamics (CFD) tools, unexpected high total pressure losses were detected. To analyze this unknown phenomenon, a gasdynamic model of the rotating combustion chamber has been developed to explain the unexpected high Rayleigh pressure losses.
Diagrams
Andreas Penkner and Peter Jeschke, J. Eng. ‘Gas Turbines Power 138’ , 021502 (Sep 01, 2015) (12 pages)
What is the relation with stagnation conditions?
We found that stagnation temperature is the maximum in the critic state of M=1.
= 1
Therefore, if the critical stagnation temperature is 1000 K, air can not be cooled below 490 K in Rayleigh flow. Physically, this means that the flow velocity reaches infinity when the temperature reaches 490 K-which is impossible in reality.
Critical Conditions
Critical Conditions
The critical conditions are those that would exist if the flow was isentropically accelerated or decelerated until the Mach number change to the unity (1).
→1
Critical Conditions
In other words, they are the conditions that would exist if the Mach number is isentropically changed from M to 1.
These critical conditions are usually denoted by an asterisk, by the symbols V*, p*, ρ*, T*, and A*.
Critical Conditions
∗ =
∗
−
•
+ +
=
− + +
∗
−
+ +
− + +
∗
•
=
=
Critical Conditions
There is a relation between the critical condition and the stagnation condition.
To find that relation, we need to set equations shown before.
= 0 in the
Critical Conditions
This gives us:
= +
= + ∗
•
= +
•
= +
∗
∗
∗
Critical Conditions
For the case of air flow ( give
= 0.833, ∗
= 1.4), these equations
= 0.528, ∗
= 0.634 ∗
Another aspect that is very important is the area relation, between a generic section of area A and the section of area that is in critical conditions A*.
Critical Conditions
The relation has it’s minimum at M = 1, i.e., if the process isentropic, section area should less than theiscritical area. no If there is aof smaller sectionbethan the critical, the transit to reach the critical section it is not isentropic.
∗
=
+ + −
Critical Conditions Area Ratio vs Mach number
= ..
In the case of a convergent-divergent nozzle, if the inlet flow is subsonic, to ensure a supersonic flow in the outlet, the throat section should be critical.
If the throat section is greater than the critical, at any section sonic conditions occur.
Critical Conditions
If the section of the throat is less than the critical, both upstream and downstream of the throat, there is the critical section; in the upstream section, it has sonic condition, but downstream the conditions can not be calculated with isentropic equations, as in the transition between both the flow is with irreversibilities (shockwaves).
Example. A gas is contained in a large vessel at a pressure of 300 kPa and a temperature of 50°C. The gas is expanded from this vessel through a nozzle until the Mach number reaches a value of 1. Find the pressure, temperature, and velocity at this point in the flow if the gas is (a) air and (b) helium.
= 1.4 ℎ = 1.667 = 300 = 3 23 =1
= + ∗
.
∗ = 300 . =158.5 . . . =146.1 ∗ = 300 ℎ . 30.8°
= + ∗
∗ = (323 ) . = 269.2 = 3.8° ∗ = (323 ) = 242.2 = ℎ .
To calculate the velocity, we use the equation.
= = = .∙ = (1.4) (269.2 ) = 328.9 / ℎ =
. .∙ 1.4 ×
× 242.2 = 916.1 /
Activity
1.- What are the critical conditions?
2.-How do you denote critical conditions? 3.-Write the equations for critical conditions and the ralation with stagnation condition equations.
4.- Find
, , for helium ( = 1.667). ∗
∗
∗
Maximum Discharge Velocity.
The “maximum discharge velocity” or “maximum escape velocity” is the velocity expanded that would generated if a gas was adiabatically until be its temperature had dropped to absolute zero. Using the adiabatic energy equation gives the maximum discharge velocity as.
= = 2 2
Maximum Discharge Velocity.
This can be rearranged to give. = 2 = 2 = 2 1 = 21
There is therefore a definite maximum velocity that can be generated in a gas having a given stagnation temperature. However, since the temperature is zero when this maximum velocity is reached, the Mach number will be infinite since, under these conditions, the speed of sound is 0.
Maximum Discharge Velocity.
It should be noted that the maximum discharge velocity given by the above equation could not be obtained in reality because at very low temperatures the assumptions used in deriving the above equations cease to apply, i.e., the properties of the gas change and once the temperature gets low enough the gas will liquefy.
Maximum Discharge Velocity.
It should also be noted that the maximum discharge velocity has nothing to do with the existence of a maximum velocity at which a body can move relative to a gas, no such limit existing according to the laws of conventional mechanics.
Isentropic Relations in Tabular and Graphical Form and from Software
The thermodynamic state of a fluid particle is defined by its properties (p, T,also u, ith,requires s); but from the the standpoint of mechanics knowing velocity of the particle and possibly its position in a gravitational field.
The thermodynamic properties are called state properties; They are the values measured with instruments are static relative to the fluid.
Isentropic Relations in Tabular and Graphical Form and from Software
However, isentropic flow calculations have been undertaken using sets of tables or graphs that give the variations of such quantities as p0/p, and T0/T with M in isentropic flow for a fixed value of the specific heat ratio γ. A typical set of tables would have the following headings.
Ƿ M ∗ Ƿ
Isentropic Relations in Tabular and Graphical Form and from Software
Such a table can be conveniently used in the calculation of the properties of a one-dimensional isentropic flow. For example, if the flow through a variable area channel is being considered and if the Mach number and pressure at one section are known, say M1 and p1, and if the pressure at some other section is known, say p2, then to find the Mach number at the second section the value of M1 is used with the table to find p1/p0.
Isentropic Relations in Tabular and Graphical Form and from Software
=
which allows p0/p2 to be found. Then using the tables, the value of M2 corresponding to this value of p0/p can be found.
Isentropic Relations in Tabular and Graphical Form and from Software
As mentioned above, today with the widespread availability of programmable calculators and computers there has been a considerable reduction in the use of isentropic flow tables and charts for calculating the properties of isentropic flows.
Isentropic Relations in Tabular and Graphical Form and from Software
Example Air flows from a large vessel in which the pressure is 300 kPa and the temperature is 40°C through a nozzle. If the pressure at some section of the discharge nozzle is measured as 200 kPa, find the temperature and velocity at this section. If the Mach number at some other section of the nozzle is 1.5, find the pressure, temperature, and velocity at this section. Assume that the flow is steady, isentropic, and one-dimensional.
Isentropic Relations in Tabular and Graphical Form and from Software
Isentropic Relations in Tabular and Graphical Form and from Software Solution
In this case, because the vessel is large and the velocity in it effectively zero, P0=300kPa, T0=313°K At the first section, here termed section 2, p0/p = 300/200 = 1.5; thus, isentropic relations or tables for γ = 1.4 or software gives M = 0.78 and T0/T = 1.12, and so at this section ° T= 313/1.12= 279K= 6 C
Isentropic Relations in Tabular and Graphical Form and from Software
Solution Hence, using V = Ma, it follows that
1.4
V= 0.78 x
. .
279.2 = 261.1 /
At the second section M = 1.5 so isentropic relations or tables for γ = 1.4 or software give T0/T = 1.45 and p0/p = 3.67, and so at this section T= 313/1.45= 216°K= -57°C
V= 1.5 x
P= 300/3.67= 81.7Kpa
1.4
. .
216 = 441.9 /
Isentropic Relations in Tabular and Graphical Form and from Software
Equations that are important and should take into account are the following. p0/p T0/T V= Ma
= =
/ / / /
Activity
The velocity, pressure, and temperature at a certain point in a steady air flow are 600 m/s, 70 kPa, and 5°C, respectively. If the pressure at some other point in the flow is 30 kPa, find the Mach number, temperature, and velocity that exist at this second point. Assume that the flow is isentropic and onedimensional.
