tank

� � � � � � � ��

� �� ! ��!�� "$#&% ��' �(�)��* + ��% ,-#&.0/

1325476 8:9fanno model

for relativly short tube

Isothermal modelfor relativly long tube

;3<5=?> @BADCFEDGIHfanno model


Isothermal modelfor relativly long tube

Volume forced models

Volume is a function of pressrue or rigit(the volume can be also a function of inertia and etc)

Semi rigit tank

External forces that control the tank volume

Fig. 11.1: The two different classifications ofmodels that explain the filling orevacuating of a single chamber

In some ways the next two Chapterscontain materials that is new to the tra-ditional compressible flow text books1.It was the undersigned experience, thatin traditional classes for with compress-ible flow (sometimes referred to as gasdynamics) don’t provide a demonstra-tion to applicability of the class mate-rial aside to aeronautical spectrum evensuch as turbomachinery. In this Chap-ter a discussion on application of com-pressible flow to other fields like manu-facturing is presented2.

There is a significant impor-tance to the “pure” models such Isother-mal flow and Fanno flow which have im-mediate applicability. However, in manyinstances, the situations in life are far

1After completion of these Chapters, the undersigned discover two text books which to include somematerial related to this topic. These books are Owczarek, J. A., Fundamentals of Gas Dynamics,International Textbook Co., Scranton, Pennsylvania, 1964. and “Compressible Fluid Flow,” 2nd Edition,by M. A. Saad, Prentice Hall, 1985. However, these books contained only limit discussions on theevacuation of chamber with attached nozzle.

2Even if the instructor feels that their students are convinced about the importance of the compress-ible, this example can further strength and enhance this conviction.

183

184 CHAPTER 11. EVACUATING AND FILLING A SEMI RIGID CHAMBERS

more complicate. Combination of gascompressibility in the chamber and flowout or through a tube post a special in-terest and these next two Chapters are dealing with these topics. In the first Chap-ter models where the chamber volume is controlled or a function of the pressureare discussed. In the second Chapter models where the chamber’s volume is afunction of external forces are presented (see Figure 11.1).

11.1 Governing Equations and AssumptionsThe process of filing or evacuating a semi flexible (semi rigid) chamber through atube is very common in engineering. For example, most car today equipped withan airbag. For instance, the models in this Chapter are suitable for study of thefilling the airbag or filling bicycle with air. The analysis is extended to include asemi rigid tank. The term semi rigid tank referred to a tank that the volume is eithercompletely rigid or is a function of the chamber’s pressure.

As it was shown in this book the most appropriate model for the flow inthe tube for a relatively fast situation is Fanno Flow. The Isothermal model is moreappropriate for cases where the tube is relatively long in–which a significant heattransfer occurs keeping the temperature almost constant. As it was shown in Chap-ter 9 the resistance, �

�� , should be larger than �� . Yet Isothermal flow model isused as to limiting case.

�� fanno model


Isothermal modelfor a relativly long tube

A schematic of a direct connection

�� !�"fanno model


Isothermal modelfor a relativly long tube

The connection is through a narrow passage

reducedconnection 1 2 1 2

Fig. 11.2: A schematic of two possible connections of the tube to a single chamber

#%$'&)( *,+ - .

Control volume for the evacuating case

/10 2 3

4%5'6)7 8,9 : ;

Control volume for the filling case

<1= >

Fig. 11.3: A schematic of the control vol-umes used in this model

The Rayleigh flow model requiresthat a constant heat transfer supplied eitherby chemical reactions or otherwise. Thisauthor isn’t familiar with situations in whichRayleigh flow model is applicable. Andtherefore, at this stage, no discussion is of-fered here.

Fanno flow model is the most ap-propriate in the case where the filling andevacuating is relatively fast. In case the theTo put the dimensionless

analysis to discuss what isfast enough. filling is relatively slow (long �

�� than theIsothermal flow is appropriate model. Yetas it was stated before, here Isothermal flow

11.1. GOVERNING EQUATIONS AND ASSUMPTIONS 185

and Fanno flow are used as limiting or bounding cases for the real flow. Addition-ally, the process in the chamber can be limited or bounded between two limits ofIsentropic process or Isothermal process.

In this analysis, in order to obtain the essence of the process, some simpli-fied assumptions are made. The assumptions can be relaxed or removed and themodel will be more general. Of course, the payment is by far more complex modelthat sometime clutter the physics. First, a model based on Fanno flow model isconstructed. Second, model is studied in which the flow in the tube is Isothermal.The flow in the tube in many cases is somewhere between the Fanno flow modelto Isothermal flow model. This reality is an additional reason for the constructionof two models in which they can be compared.

Effects such as chemical reactions (or condensation/evaporation) are ne-glected. There are two suggested itself possibilities to the connection between thetube to the tank (see the Figure 11.2): one) direct two) through a reduction. Thedirect connection is when the tube is connect straight to tank like in a case wherepipe is welded into the tank. The reduction is typical when a ball is filled troughan one–way valve (filling a baseball ball, also in manufacturing processes). Thesecond possibility leads itself to an additional parameter that is independent of theresistance. The first kind connection tied the resistance, �

�� , with the tube area.

The simplest model for gas inside the chamber as a first approximation isthe isotropic model. It is assumed that kinetic change in the chamber is negligible.Therefore, the pressure in the chamber is equal to the stagnation pressure, ��(see Figure 11.4). Thus, the stagnation pressure at the tube’s entrance is the sameas the pressure in the chamber.

��

1 2

Fig. 11.4: The pressure assumptions in the chamberand tube entrance

The mass in the cham-ber and mass flow out are ex-pressed in terms of the cham-ber variables (see Figure 11.3.The mass in the tank for perfectgas reads

�� "!$#&% � (11.1)

And for perfect gas the mass at any given time is

�'% �)( �+*+, ( �+*-/. ( �+* (11.2)

The mass flow out is a function of the resistance in tube, �� and the pressure

difference between the two sides of the tube ��0�+!$# ( � ��21 ��3546�87 * . The initial condi-tions in the chamber are

. ( � * , �)( � * and etc. If the mass occupied in the tube is


neglected (only for filling process) the most general equation ideal gas (11.1) reads

��

�� ,-/.��

� 3�� 3�� 3 ( � ��21 � 7��3 * % � (11.3)

When the plus sign is for filling process and the negative sign is for evacuatingprocess.

11.2 General Model and Non-dimensionedIt is convenient to non-dimensioned the properties in chamber by dividing them bytheir initial conditions. The dimensionless properties of chamber as�. % . ( ��% ��+*

. ( �&% � *�, % , ( � % ��+*, ( ��% � *�� % �)( � % ��+*�)( ��% � * (11.4a)��% �� (11.4b)

where� �

is the characteristic time of the system defined as followed

� � % , ( � *�� -/. ( � * * (11.5)

The physical meaning of characteristic time,� �

is the time that will take to evacuatethe chamber if the gas in the chamber was in its initial state, the flow rate was atits maximum (choking flow), and the gas was incompressible in the chamber.

Utilizing these definitions (11.4) and substituting into equation (11.3) yields

�)( � *", ( � *� � -/. ( � * �� ,�. �! "� �� 3- �. 3 �)( � *. ( � * �

� # #%$� �� & � - �. 3 . ( � * � � �� ( ��+*�% � (11.6)

where the following definition for the reduced Mach number is added as�� % � 3 ( �+*� � �� (11.7)

After some rearranging equation (11.6) obtains the form�� ,�. �' � � �� -/. ( � *, ( � *

��3 �� 3� �. 3 �� % � (11.8)

11.2. GENERAL MODEL AND NON-DIMENSIONED 187

and utilizing the definition of characteristic time, equation (11.5), and substitutinginto equation (11.8) yields �

� �� ,�. � �� 3 �� . 3 % � (11.9)

Note that equation (11.9) can be modified by introducing additional param-eter which referred to as external time,

� � �� 3. For cases, where the process timeis important parameter equation (11.9) transformed to�

� �� ,�. � � � �� &3 �� . 3 % � (11.10)

when�� 1 �, 1 �. 1 and

�� are all are function of��in this case. And where

��% � 4 � � �� .It is more convenient to deal with the stagnation pressure then the actual

pressure at the entrance to the tube. Utilizing the equations developed in Chapter4 between the stagnation condition, denoted without subscript, and condition in atube denoted with subscript 1. The ratio of

�� is substituted by�� 3� �. 3 %�� . �� 7�� (11.11)

It is convenient to denote �� % �!�� 7 �"��#�$� �� #�� (11.12)

Note that

�� is a function of the time. Utilizing the definitions (11.11) and sub-

stituting equation (11.12) into equation (11.9) to be transformed into�� ,�. � �� ( ��+*

�� . % � (11.13)

Equation (11.13) is a first order nonlinear differential equation that can be solvedfor different initial conditions. At this stage, the author isn’t aware that is a generalsolution for this equation4. Nevertheless, many numerical methods are available tosolve this equation.

11.2.1 Isentropic process

The relationship between the pressure and the temperature in the chamber can beapproximated as isotropic and therefore�. % . ( �+*. ( � *

% � �)( �+*�)( � *

� �� % �� (11.14)

3This notation is used in many industrial processes where time of process referred to sometime asthe maximum time.

4To those mathematically included, find the general solution for this equation.


The ratios can be expressed in term of the reduced pressure as followed:�� . % �� % �� (11.15)

and �� . % �� (11.16)

So equation (11.13) is simplified into three different forms:�� , �� $� �� ( ��+*

�� % �

� �� , � �� ,� �� $� �� ( ��+*

�� % ��, � �� ,

� �� ( ��+*��

� � % � (11.17)

Equation (11.17) is a general equation for evacuating or filling for isentropic pro-cess in the chamber. It should be point out that, in this stage, the model in the tubecould be either Fanno flow or Isothermal flow. The situations where the chamberundergoes isentropic process but the the flow in the tube is Isothermal are limited.Nevertheless, the application of this model provide some kind of a limit where toexpect when some heat transfer occurs. Note the temperature in the tube entrancecan be above or below the surrounding temperature. Simplified calculations of theentrance Mach number are described in the advance topics section.

11.2.2 Isothermal Process in the Chamber

11.2.3 A Note on the entrance Mach number

The value of Mach number, � 3 is a function of the resistance, �� and the ratio

of pressure in the tank to the back pressure, �� 46�&3 . The exit pressure, ��7 isdifferent from � � in some situations. As it was shown before, once the flow becamechoked the Mach number, ��3 is only a function of the resistance, �

�� . Thesestatements are correct for both Fanno flow and the Isothermal flow models. Themethod outlined in Chapters 8 and 9 is appropriate for solving for entrance Machnumber, ��3 .

Two equations must be solved for the Mach numbers at the duct entranceand exit when the flow is in a chockless condition. These equations are combina-tions of the momentum and energy equations in terms of the Mach numbers. Thecharacteristic equations for Fanno flow (9.50), are

�� %� � �� 3 � � � �� 7 (11.18)

11.3. RIGID TANK WITH NOZZLE 189

and

� 7� � ( �+*

% � �� 7 7 � �� 3� 7�� 37 � 7 7�� 37 � 3 7� ��

�� (11.19)

where �� is defined by equation (9.49).

The solution of equations (11.18) and (11.19) for given �� and

�� # #%$ yieldsthe entrance and exit Mach numbers. See advance topic about approximate solu-tion for large resistance, �

�� or small entrance Mach number, ��3 .11.3 Rigid Tank with NozzleThe most simplest possible combination is discussed here before going trough themore complex cases A chamber is filled or evacuated by a nozzle. The gas in thechamber assumed to go an isentropic processes and flow is bounded in nozzlebetween isentropic flow and isothermal flow5. Here, it also will be assumed thatthe flow in the nozzle is either adiabatic or isothermal.

11.3.1 Adiabatic Isentropic Nozzle Attached

The mass flow out is given by either by Fliegner’s equation (4.47) or simply use� � � �� and equation (11.17) becomes� �� ( ��+*

�� % � (11.20)

It was utilized that�, %

and�� definition is simplified as

�� % . It can be noticed

that the characteristic time defined in equation (11.5) reduced into:

� � % , ( � *� � � - . ( � *+* (11.21)

Also it can be noticed that equation (11.12) simplified into�� % � �� 7�� % � � � � ��#�$� �� #�� (11.22)

Equation (11.20) can be simplified as� � � � � ��

�� % � (11.23)

5This work is suggested by Donald Katze the point out that this issue appeared in Shapiro’s BookVol 1, Chapter 4, p. 111 as a question 4.31.


Equation (11.23) can be integrated as� ��3 �

� � �� #��% � (11.24)

The integration limits are obtained by simply using the definitions of reduced pres-sure, at �)( ��% � *�%

and �)( ��% ��+*�% �� . After the integration, equation (11.24) and

rearrangement becomes�� % � � � � ��

� � �� (11.25)

Example 11.1:A chamber is connected to a main line with pressure line with a diaphragm andnozzle. The initial pressure at the chamber is

�� and the volume is

��

�� .

Calculate time it requires that the pressure to reach 5[Bar] for two different noz-zles throat area of 0.001, and 0.1 [

� 7 ] when diaphragm is erupted. Assumed thestagnation temperature at the main line is the ambient of

� � � � � .SOLUTION

The characteristic time is

� � �� % ,� � � % ,� � � % ��

�� 1� % �

�� (11.26)

And for smaller area

� � �� % ��

�� 1�� % � �

� � � � �� % �)( �+*�)( � *

% �� % � � �

The time is

��%�� # $

(11.27)

Substituting values into equation (11.27) results

��% ��

� � � � � � � � �� % �� (11.28)

11.3.1.1 Filling/evacuating the chamber under upchucked condition

The flow in the nozzle can became upchucked and it can be analytically solved.Owczarek [1964] found that analytical solution which described here.

11.4. RAPID EVACUATING OF A RIGID TANK 191

11.3.2 Isothermal Nozzle Attached

In this case the process in nozzle is assumed to isothermal but the process in thechamber is isentropic. The temperature in the nozzle is changing because thetemperature in the chamber is changing. Yet, the differential temperature changein the chamber is slower than the temperature change in nozzle. For rigid volume,�, %

and for isothermal nozzle�. %

Thus, equation (11.13) is reduced into� �� %

�� % � (11.29)

Separating the variables and rearranging equation (11.30) converted into� ��3� ��

�� #� � �� % � (11.30)

Here,

�� is expressed by equation (11.22). After the integration, equation

(11.30) transformed into �� % � � � � ��

�� %��( �$� �� * �� #�� (11.31)

11.4 Rapid evacuating of a rigid tank11.4.1 With Fanno Flow

The relative Volume,�, ( �+*)% , is constant and equal one for a completely rigid

tank. In such case, the general equation (11.17) “shrinks” and doesn’t contain therelative volume term.

A reasonable model for the tank is isentropic (can be replaced polytropicrelationship) and Fanno flow are assumed for the flow in the tube. Thus, the specificgoverning equation is

� ��

� � �� % � (11.32)

For a choked flow the entrance Mach number to the tube is at its maximum, � � ��and therefore

�� % . The solution of equation (11.37) is obtained by noticing that�� is not a function of time and by variables separation results in� �#

�� % � ��

3� ��

� � �� %

� ��

� � � ��3 �� (11.33)


direct integration of equation (11.40) results in��% ( � � * ��

�� (11.34)

It has to be realized that this is “reversed” function i.e.��

is a function of Pand can be reversed for case. But for the chocked case it appears as�� % � �� ( � � * ��

�� (11.35)

The function is drawn as shown here in figure 11.6. The figure 11.6 shows

V(t) = P (t)

V(t) = P (0)

P(t)

t̄

0 0.2 0.4 0.6 0.8 1.00

0.2

0.4

0.6

0.8

1.0

Fig. 11.5: The reduce time as a function of the modified reduced pressure

that when the modified reduced pressure equal to one the reduced time is zero.The reduced time increases with decrease of the pressure in the tank.

At certain point the flow becomes chockless flow (unless the back pressureis complete vacuum). The transition point is denoted here as �� . . Thus, equationThe big struggle look for sug-

gestion for better notation.


(11.40) has to include the entrance Mach under the integration sign as

�� % � ��

� � �� (11.36)

For practical purposes if the flow is choked for more than 30% of the charecteristictime the choking equation can be used for the whole range, unless extra longtime or extra low pressure is calculated/needed. Further, when the flow becamechockless the entrance Mach number does not change much from the chokingcondition.

Again, for the special cases where the choked equation is not applicablethe integration has to be separated into zones: choked and chockless flow regions.And in the choke region the calculations can use the choking formula and numericalcalculations for the rest.

Example 11.2:A chamber with volume of 0.1[

� �] is filled with air at pressure of 10[Bar]. The

chamber is connected with a rubber tube with

�% �

�� ,

� % �� and length of� % � �

�

��

SOLUTION

The first parameter that calculated is ��

�� % �11.4.2 Filling process

The governing equation is

� ��

� � �� % � (11.37)

For a choked flow the entrance Mach number to the tube is at its maximum, � � ��and therefore

�� % . The solution of equation (11.37) is obtained by noticing that�� is not a function of time and by variable separation results in� �#

�� % � ��

3� ��

� � �� %

� ��

� � � ��3 �� (11.38)

direct integration of equation (11.40) results in��% ( � � * ��

�� (11.39)

It has to be realized that this is a reversed function. Nevertheless, withtoday computer this should not be a problem and easily can be drawn as shownhere in Figure 11.6. The Figure shows that when the modified reduced pressure


Fig. 11.6: The reduce time as a function of the modified reduced pressure

equal to one the reduced time is zero. The reduced time increases with decreaseof the pressure in the tank.

At some point the flow became chockless flow (unless the back pressure isa complete vacuum). The transition point is denoted here as �� . . Thus, equation(11.40) has to include the entrance Mach under the integration sign as�� % � ��

� ��

� � �� (11.40)

11.4.3 The Isothermal Process

For Isothermal process, the relative temperature,�. %

. The combination of theisentropic tank and Isothermal flow in the tube is different from Fanno flow in thatthe chocking condition occurs at

4 � � . This model is reasonably appropriated

when the chamber is insulated and not flat while the tube is relatively long and theprocess is relatively long.

It has to be remembered that the chamber can undergo isothermal pro-cess. For the double isothermal (chamber and tube) the equation (11.6) reducedinto

�)( � *+, ( � *�� -/. ( � * �� ,�� "� �� 3- �)( � *. ( � * �

� # � $� �� - . ( � * � � �� ( ��"*�% � (11.41)

11.4.4 Simple Semi Rigid Chamber

A simple relation of semi rigid chamber when the volume of the chamber is linearlyrelated to the pressure as

, ( �+* %��)( �+* (11.42)

where

�is a constant that represent the physics. This situation occurs at least

in small ranges for airbag balloon etc. The physical explanation when it occursbeyond the scope of this book. Nevertheless, a general solution is easily can beobtained similarly to rigid tank. Substituting equation (11.42) into yields

��

�� % � (11.43)


Carrying differentiation result in�� $� ��

� � % � (11.44)

Similarly as before, the variables are separated as� �#��% ��

3��

� � (11.45)

The equation (11.50) integration obtains the form��% � 7��

� � ( � � � * ( �� * � � �� (11.46)

The physical meaning that the pressure remains larger thorough evacuating pro-cess, as results in faster reduction of the gas from the chamber.

11.4.5 The “Simple” General Case

The relationship between the pressure and the volume from the physical point ofview must be monotonous. Further, the relation must be also positive, increase ofthe pressure results in increase of the volume (as results of Hook’s law. After all, inthe known situations to this author pressure increase results in volume decrease(at least for ideal gas.).

In this analysis and previous analysis the initial effect of the chamber con-tainer inertia is neglected. The analysis is based only on the mass conservationand if unsteady effects are required more terms (physical quantities) have takeninto account. Further, it is assumed the ideal gas applied to the gas and this as-sumption isn’t relaxed here.

Any continuous positive monotonic function can be expressed into a poly-nomial function. However, as first approximation and simplified approach can bedone by a single term with a different power as

, ( �+* %�� (11.47)

When � can be any positive value including zero, � . The physical meaning of �% �

is that the tank is rigid. In reality the value of � lays between zero to one. When� is approaching to zero the chamber is approaches to a rigid tank and vis versawhen the ��

the chamber is flexible like a balloon.

There isn’t a real critical value to � . Yet, it is convenient for engineers tofurther study the point where the relationship between the reduced time and thereduced pressure are linear6 Value of � above it will Convex and and below itconcave.

6Some suggested this border point as infinite evocation to infinite time for evacuation etc. Thisundersigned is not aware situation where this indeed play important role. Therefore, is waiting to findsuch conditions before calling it as critical condition.


�� $� ��

�� % � (11.48)

Notice that when �%

equation (11.49) reduced to equation (11.43).After carrying–out differentiation results��

� � � ��

� � % � (11.49)

Again, similarly as before, variables are separated and integrated as fol-lows � �#

��%

��

3��

� � (11.50)

Carrying–out the integration for the initial part if exit results in��% � 7��

� � ( � � � � � � * ( �� * � � �� (11.51)

The linear condition are obtain when

� � � � � � % � � � % � � � � (11.52)

That is just bellow 1 ( �% �

� � � � �� ) for � % �� .

11.5 Advance TopicsThe term �

�� is very large for small values of the entrance Mach number whichrequires keeping many digits in the calculation. For small values of the Mach num-bers, equation (??) can be approximated as

�� %

� �� # 7 � � �

�7� � �� # 7 � �

�7 (11.53)

and equation (??) as

� � �� #� � ( �+*

% � �� # � (11.54)

The solution of the last two equations yields

� ��%

�� # #%$ � 7� � ��

(11.55)

11.5. ADVANCE TOPICS 197

This solution should used only for � ��

� ��1� �� ; otherwise equations (??) and

(??) must be solved numerically.The solution of equation (??) and (??) is described in “Pressure die cast-

ing: a model of vacuum pumping” Bar-Meir, G; Eckert, E R G; Goldstein, R JJournal of Manufacturing Science and Engineering (USA). Vol. 118, no. 2, pp.259-265. May 1996.

� � � � � � � ��

� �� ! �� * + ��% ,-#&. / � �� #&.��$� #&. �� !� % # * � � � .�

This chapter is the second on the section dealing with filling and evacuating cham-bers. Here the model deals with the case where the volume is controlled by exter-nal forces. This kind of model is applicable to many manufacturing processes suchas die casting, extraction etc. In general the process of the displacing the gas (inmany cases air) with a liquid is a very common process. For example, in die cast-ing process liquid metal is injected to a cavity and after the cooling/solidificationperiod a part is obtained in near the final shape. One can also view the exhaustsystems of internal combustion engine in the same manner. In these processes,sometime is vital to obtain a proper evacuation of the gas (air) from the cavity.

12.1 ModelIn this analysis, in order to obtain the essence of the process, some simplifiedassumptions are made. It simplest model of such process is when a piston isdisplacing the gas though a long tube. It assumed that no chemical reaction (orcondensation/evaporation) occur in the piston or the tube 1. It is further assumedthat the process is relatively fast. The last assumption is a appropriate assumptionin process such as die casting.

Two extreme possibilities again suggest themselves: rapid and slow pro-cesses. The two different connections, direct and through reduced area are com- again to add the dimensional

analysis what is rapid andwhat is slow.bined in this analysis.

1such reaction are possible and expected to be part of process but the complicates the analysis anddo not contribute to understand to the compressibility effects.

199

200CHAPTER 12. EVACUATING/FILING CHAMBERS UNDER EXTERNAL VOLUME CONTROL

12.1.1 Rapid Process

Clearly under the assumption of rapid process the heat transfer can be neglectedand Fanno flow can be assumed for the tube. The first approximation isotropicprocess describe the process inside the cylinder (see figure 12.1.

��

Fanno model

isontropic process 1 2

Fig. 12.1: The control volume of the “Cylinder”

Before introducing the steps of the analysis, it is noteworthy to think aboutthe process in qualitative terms. The replacing incompressible liquid enter in thesame amount as replaced incompressible liquid. But in a compressible substancethe situation can be totally different, it is possible to obtain a situation where thatmost of the liquid entered the chamber and yet most of the replaced gas can bestill be in the chamber. Obtaining conditions where the volume of displacing liquidis equal to the displaced liquid are called the critical conditions. These critical con-ditions are very significant that they provide guidelines for the design of processes.

Obviously, the best ventilation is achieved with a large tube or area. Inmanufacture processes to minimize cost and the secondary machining such astrimming and other issues the exit area or tube has to be narrow as possible. In theexhaust system cost of large exhaust valve increase with the size and in additionreduces the strength with the size of valve2. For these reasons the optimum sizeis desired. The conflicting requirements suggest an optimum area, which is alsoindicated by experimental studies and utilized by practiced engineers.

The purpose of this analysis to yields a formula for critical/optimum ventarea in a simple form is one of the objectives of this section. The second objectiveis to provide a tool to “combine” the actual tube with the resistance in the tube,thus, eliminating the need for calculations of the gas flow in the tube to minimizethe numerical calculations.

A linear function is the simplest model that decibels changes the volume.In reality, in some situations like die casting this description is appropriate. Nev-ertheless, this model can be extended numerical in cases where more complexfunction is applied.

, ( �+* %�, ( � *� �

�� (12.1)

2After certain sizes, the possibility of crack increases.

12.1. MODEL 201

Equation (12.2) can be non–dimensionlassed as

�, ( ��+*�% � ��(12.2)

The governing equation (11.10) that was developed in the previous chapter(11) obtained the form as� ��

� �, ��

�� ,� ��

�( � *�� % � (12.3)

where�� % � 4 � � �� . Notice that in this case that there are two different charac-

teristic times: the “characteristic” time,� �

and the “maximum” time,� � �� . The

first characteristic time,� �

is associated with the ratio of the volume and the tubecharacteristics (see equation (11.5)). The second characteristic time,

� � �� is as-sociated with the imposed time on the system (in this case the elapsed time of thepiston stroke).

Equation (12.3) is an nonlinear first order differential equation and can berearranged as follows

� �� # � # � ��

� � �� %� �� )( � *�% �� (12.4)

Equation (12.4) is can be solved only when the flow is chocked In which case

��

isn’t function of the time.The solution of equation (12.4) can be obtained by transforming and by

introducing a new variable � % �� and therefore�� %�� . The reduced Pres-

sure derivative,� �� % 7 �� 3 � � � ( � �� * � 3 � � Utilizing this definition and there implication

reduce equation (12.4) � � � ( � �� * � 3 � �( � � * ( �� *�� %

� �� (12.5)

where

�% # �� # � ��

�� And equation (12.5) can be further simplified as � �

( � � * ( �� * � %

� �� (12.6)

Equation (12.6) can be integrated to obtain ( � � *

� ��

�� % �

��

(12.7)

or in a different form

��

�� % ��

(12.8)


Now substituting to the “preferred” variable

� � # �� # � ��

� � �� 3

�� % ��(12.9)

The analytical solution is applicable only in the case which the flow is chokedthorough all the process. The solution is applicable to indirect connection. Thishappen when vacuum is applied outside the tube (a technique used in die castingand injection molding to improve quality by reducing porosity.). In case when theflow chockless a numerical integration needed to be performed. In the literature, tocreate a direct function equation (12.4) is transformed into

� �� % � � � # � # � ��

� � �� (12.10)

with the initial condition of

�)( � *�%

(12.11)

The analytical solution also can be approximated by a simpler equation as�� %� � � � � ��

(12.12)

The results for numerical evaluation in the case when cylinder is initially at anatmospheric pressure and outside tube is also at atmospheric pressure are pre-sented in figure 12.2. In this case only some part of the flow is choked (the laterpart). The results of a choked case are presented in figure 12.3 in which outsidetube condition is in vacuum. These figures 12.2 and 12.3 demonstrate the impor-tance of the ratio of

# �� # � . When# �� # �

the pressure increases significantly and

verse versa.Thus, the question remains how the time ratio can be transfered to param-

eters that can the engineer can design in the system.Denoting the area that creates the ratio

# �� # � % as the critical area, � �

provides the needed tool. Thus the exit area, � can be expressed as

� % �� (12.13)

The actual times ratio# �� # � �� can be expressed as

� � �� % � � ��

3� �� (12.14)

12.1. MODEL 203

∗�

∇

�o

�

DimensionlessArea, A/Ac

0.00.20.51.01.22.05.0

4fLD

=

1.0

1.4

1.8

2.2

2.6

3.0

3.4

3.8

4.2

4.6

5.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

P (t�

)P (0)

100.0

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗

∗∗

∗∗

∗∗

∗∗

∗

∗

∗

∗

∗

∗

� � � � � � � � � � � � � � � � � � ��

��

��

��

�

�

�

�

�

∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇∇

∇∇

∇∇

∇∇

∇∇

∇

∇

∇

∇

�o� �o� �o� �o� �o� �

o��o�

�o�

�o�

�o�

�o�

�o�

�o�

�o�

�o�

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�o�

�o�

�o�

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�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

∗�

∇

�o

�

DimensionlessArea, A/Ac

0.00.20.51.01.22.05.0

4fLD

=

Dimensionless Time, t�,

or, Cylinder Volume Fraction

1.0

1.4

1.8

2.2

2.6

3.0

3.4

3.8

4.2

4.6

5.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

P (t�

)P (0)

5.0

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗

∗∗

∗∗

∗∗

∗∗

∗

∗

∗

∗

∗

∗

� � � � � � � � � � � � � � � � � � ��

��

��

��

�

�

�

�

�

∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇∇

∇∇

∇∇

∇∇

∇

∇

∇

∇

∇

∇

�o� �o� �o� �o� �o� �

o��o�

�o�

�o�

�o�

�o�

�o�

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�o�

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o�

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o�

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o�

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�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

�

o�

Figure b

Figure a

Fig. 12.2: The pressure ratio as a function of the dimensionless time for chockless condition


P(t)

P(0)

DIMENSIONLESS TIME, t�

, or, CYLINDER VOLUME FRACTION

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

. . . . . . . . . . .

∗

×

�

0.0

0.1

0.5

1.0

1.5

4.0

.......

........

........

.......

.......

.....................................................

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗

∗∗

∗∗

∗∗

∗∗

∗∗

∗

∗

× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ××

×

�

�

�

�

�

�

�

�

�

�

�

�

�

�

��

Ac

A� �� =

Fig. 12.3: The pressure ratio as a function of the dimensionless time for choked condition

12.1. MODEL 205

According to equation (11.5)� �

is inversely proportional to to area,� �

�

4 � . Thus,

equation (12.14) the� � �� is canceled and reduced into

� � �� % �� (12.15)

Parameters influencing the process are the area ratio,�� , and the friction

parameter, �� . From other detailed calculations [?] it was found that the influence

of the parameter �� on the pressure development in the cylinder is quite small.

The influence is small on the residual air mass in the cylinder but larger on theMach number, �� # . The effects of the area ratio,

�� , are studied here since it is

the dominant parameter.It is important to point out the significance of the

# � # � . This parameter rep-resents the ratio between the filling time and the evacuating time, the time whichwould be required to evacuate the cylinder for constant mass flow rate at the max-imum Mach number when the gas temperature and pressure remain in their initialvalues. This parameter also represents the dimensionless area,

�� , according to

the following equationFigure 12.4 describes the pressure as a function of the dimensionless time

for various values of�� . The line that represents

�� %

is almost straight. For

Fig. 12.4: The pressure ratio as a function of the dimensionless time

large values of�� the pressure increases the volume flow rate of the air until a

quasi steady state is reached. This quasi steady state is achieved when the volu-metric air flow rate out is equal to the volume pushed by the piston. The pressureand the mass flow rate are maintained constant after this state is reached. Thepressure in this quasi steady state is a function of

�� . For small values of

�� there

is no steady state stage. When�� is greater than one the pressure is concave up-

ward and when�� is less than one the pressure is concave downward as shown

in Figures 12.4, which was obtained by an integration of equation (12.9).

12.1.2 Examples

Example 12.1:Calculate the minimum required vent area for die casting process when the die vol-ume is �

��1� � �� and �

�� % � . The required solidification time,� � �� % �

�� .

SOLUTION


12.1.3 Direct Connection

In the above analysis is applicable to indirect connection. It should be noted thatcritical area, � �

, is not function of the time. The direct connection posts more math-ematical difficulty because the critical area is not constant and time dependent.

To continue

12.2 SummaryThe analysis indicates there is a critical vent area below which the ventilation ispoor and above which the resistance to air flow is minimal. This critical area de-pends on the geometry and the filling time. The critical area also provides a meanto “combine” the actual vent area with the vent resistance for numerical simulationsof the cavity filling, taking into account the compressibility of the gas flow.

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