Theoretical competition. Tuesday, 15 July 20141/1
Problem 1 (9 points)
This problem consists of three independent parts.
Part A (3 points)
A small puck of mass ?? is carefully placed onto the inner surface of the
thin hollow thin cylinder of mass ?? and of radius ??. Initially, the cylinder
rests on the horizontal plane and the puck is located at the height ?? above
the plane as shown in the figure on the left. Find the interaction force ??
between the puck and the cylinder at the moment when the puck passes
the lowest point of its trajectory. Assume that the friction between the
puck and the inner surface of the cylinder is absent, and the cylinder
moves on the plane without slipping. The free fall acceleration is ??.
Part B (3 points)
A bubble of radius ??=5.00 cm, containing a diatomic ideal gas, has the soap film of thickness ?=
N10.0 μm and is placed in vacuum. The soap film has the surface tension ??=4.00?10?2 m and the density
??=1.10 3. 1) Find formula for the molar heat capacity of the gas in the bubble for such a process when cmthe gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; 2) Find formula for the frequency ?? of the small radial oscillations of the bubble and evaluate it under the assumption that the heat capacity of the soap film is much greater than the heat capacity of the gas in the bubble. Assume that the thermal equilibrium inside the bubble is reached much faster than the period of oscillations.
Hint: Laplace showed that there is pressure difference between inside and outside of a curved
2??surface, caused by surface tension of the interface between liquid and gas, so that???= ?? g
Part C (3 points)
Initially, a switch ?? is unshorted in the circuit shown in the figure on the right, a
capacitor of capacitance 2?? carries the electric charge ??0, a capacitor of
capacitance ?? is uncharged, and there are no electric currents in both coils of
inductance ?? and 2??, respectively. The capacitor starts to discharge and at the
moment when the current in the coils reaches its maximum value, the switch ?? is
instantly shorted. Find the maximum current ??max through the switch ?? thereafter.
Theoretical competition. Tuesday, 15 July 2014 1/4
Problem 1
Solution
Part A
Consider the forces acting on the puck and the cylinder and
depicted in the figure on the right. The puck is subject to the
gravity force ???? and the reaction force from the cylinder ??. The
cylinder is subject to the gravity force ????, the reaction force from
the plane ??1, the friction force ?????? and the pressure force from the
puck ??′=???. The idea is to write the horizontal projections of
the equations of motion. It is written for the puck as follows
??????=??sin??, (A.1)
where ???? is the horizontal projection of the puck acceleration.
For the cylinder the equation of motion with the
acceleration ?? is found as
????=??sin?????????. (A.2)
Since the cylinder moves along the plane without sliding its
angular acceleration is obtained as
??=??/?? (A.3) Then the equation of rotational motion around the center of mass of the cylinder takes the form ????=????????, (A.4) where the inertia moment of the hollow cylinder is given by
??=????2. (A.5) Solving (A.2)-(A.5) yields
2????=??sin??. (A.6) From equations (A.1) and (A.6) it is easily concluded that
??????=2????. (A.7)
Since the initial velocities of the puck and of the cylinder are both equal to zero, then, it follows from
(A.7) after integrating that
????=2????. (A.8) It is obvious that the conservation law for the system is written as
??????=2+2+2,(A.9) where the angular velocity of the cylinder is found to be ?? ??=??, (A.10) since it does not slide over the plane.
Solving (A.8)-(A.10) results in velocities at the lowest point of the puck trajectory written as
??=2 (2??+??)??=?? (2??+??)??????????????????2????2????2(A.12) (A.13) In the reference frame sliding progressively along with the cylinder axis, the puck moves in a circle of radius ?? and, at the lowest point of its trajectory, have the velocity
????????=??+?? (A.14) and the acceleration
??rel=?? (A.15) At the lowest point of the puck trajectory the acceleration of the cylinder axis is equal to zero, therefore, the puck acceleration in the laboratory reference frame is also given by (A.15).
???????=??
then the interaction force between the puck and the cylinder is finally found as
?? ??=3???? 1+3?? .2??????????2??
(A.16) (A.17)
Theoretical competition. Tuesday, 15 July 2014 2/4
Part B
1) According to the first law of thermodynamics, the amount of heat transmitted ???? to the gas in the bubble is found as
????=??????????+??????, (B.1) where the molar heat capacity at arbitrary process is as follows
1?????????? ??=??????=????+??????(B.2)
Here ???? stands for the molar heat capacity of the gas at constant volume, ?? designates its pressure, ?? is the total amount of moles of gas in the bubble, ?? and ?? denote the volume and temperature of the gas, respectively.
Evaluate the derivative standing on the right hand side of (B.2). According to the Laplace formula, the gas pressure inside the bubble is defined by 4?? ??=??, (B.3) thus, the equation of any equilibrium process with the gas in the bubble is a polytrope of the form
??3??=const. (B.4) The equation of state of an ideal gas has the form
????=??????, (B.5) and hence equation (B.4) can be rewritten as
??3???2=const. (B.6) Differentiating (B.6) the derivative with respect to temperature sought is found as
????3?? = (B.7) ????2?? Taking into account that the molar heat capacity of a diatomic gas at constant volume is
5 ????=2??, (B.8)
and using (B.5) it is finally obtained that
3J ??=????+2??=4??=33.2 mole?K. (B.9)
2) Since the heat capacity of the gas is much smaller than the heat capacity of the soap film, and there is heat exchange between them, the gas can be considered as isothermal since the soap film plays the role of thermostat. Consider the fragment of soap film, limited by the angle ?? as shown in the figure. It's area is found as
??=??(????)2. (B.10) and the corresponding mass is obtained as
??=??????. (B.11) Let ?? be an increase in the radius of the bubble, then the
Newton second law for the fragment of the soap film mentioned
above takes the form
???? =??′??′???????????, (B.12)
where ?????????? denotes the projection of the resultant surface tension
force acting in the radial direction, ??′ stands for the gas pressure
beneath the surface of the soap film and
????′=?? 1+2?? .
?????????? is easily found as
??????????=????????=???2?2??[ ??+?? ??]???. (B.13)
Since the gaseous process can be considered isothermal, it is
written that
??′??′=????. (B.14)
Assuming that the volume increase is quite small, (B.14) yields
113?? ??′=????≈??≈?? 1??? . (B.15) 1+ ?? 1+??Thus, from (B.10) - (B.16) and (B.3) the equation of small oscillations of the soap film is derived as
8?? ?????? =???2??
(B.16)
Theoretical competition. Tuesday, 15 July 2014 3/4
with the frequency
??= ??????2=108 s?1. 8?? (B.17) Part C The problem can be solved in different ways. Herein several possible solutions are considered. At the moment when the current in the coils is a maximum, the total voltage across the coils is equal to zero, so the capacitor voltages must be equal in magnitude and opposite in polarity. Let ?? be a voltage on the capacitors at the time moment just mentioned and ??0 be that maximum current. According to the law of charge conservation
??0=2????+????, (C1.1) thus, ??0 ??=3??. (C1.2)
Then, from the energy conservation law
=2+2?2??the maximum current is found as ??0=2??02????022????02+????22+2????22 (C1.3) (C1.4) After the key ?? is shortened there will be independent oscillations in both circuits with the frequency ??= (C1.5) and their amplitudes are obtained from the corresponding energy conservation laws written as
2????22????2+2????0 +2=2(C1.7) 2Hence, the corresponding amplitudes are found as
??1= ??0, (C1.8)
??2= 0. (C1.9)
Choose the positive directions of the currents in the circuits as shown in the
figure on the right. Then, the current flowing through the key is written as follows
??=??1???2. (C1.10)
The currents depend on time as
??1 ?? =??cos????+??sin????, (C1.11)
??2 ?? =??cos????+??sin????, (C1.12)
The constants ??,??,??,?? can be determined from the initial values of the currents and their amplitudes by putting down the following set of equations
??1 0 =??=??0, (C1.13) 2 ??2+??2=??1, (C1.14) ??2 0 =??=??0, (C1.15) 2 ??2+??2=??2. (C1.16)
Solving (C1.13)-(C1.16) it is found that
??=2??0, (C1.17) ??=???0, (C1.18) The sign in ?? is chosen negative, since at the time moment of the key shortening the current in the coil 2?? decreases.
Thus, the dependence of the currents on time takes the following form
??1 ?? =??0(cos????+2sin????), (C1.19) ??2 ?? =??0(cos?????sin????).(C1.20)
In accordance with (C.10), the current in the key is dependent on time according to
?? ?? =??1 ?? ???2 ?? =3??0sin????. (C1.21)
Hence, the amplitude of the current in the key is obtained as ??max=3??0=????0=
(C1.22) 222????0=2????1222????2(C1.6)
Theoretical competition. Tuesday, 15 July 2014 4/4 Instead of determining the coefficients ??,??,??,?? the vector diagram shown
in the figure on the right can be used. The segment ???? represents the current sought
and its projection on the current axis is zero at the time of the key shortening. The
current ??1 in the coil of inductance ?? grows at the same time moment because the
capacitor 2?? continues to discharge, thus, this current is depicted in the figure by
the segment ????. The current ??2 in the coil of inductance 2?? decreases at the time of
the key shortening since it continues to charge the capacitor 2??, that is why this
current is depicted in the figure by the segment ????.
It is known for above that ????=??0,????= ??0,????= 0. Hence, it is
found from the Pythagorean theorem that
????= ?????????=2??0, (C2.1)
????==??0, (C2.2)
Thus, the current sought is found as ??max=????=????+????=3??0=????0= (C2.3)
It is clear that the current through the key performs harmonic oscillations with the frequency
??= (C3.1) and it is equal to zero at the time of the key shortening, i.e.
?? ?? =??maxsin????. (C3.2)
Since the current is equal to zero at the time of the key shortening, then the current amplitude is equal to the current derivative at this time moment divided by the oscillation frequency. Let us find that current derivative. Let the capacitor of capacitance 2?? have the charge ??1. Then the charge on the capacitor of capacitance ?? is found from the charge conservation law as
??2=??0???1. (C3.3)
After shortening the key the rate of current change in the coil of inductance ?? is obtained as =??1 ??1 (C3.4) 2????whereas in the coil of inductance 2?? it is equal to
=???0???1 ??2(C3.5) 2????Since the voltage polarity on the capacitors are opposite, then the current derivative with respect to
time finally takes the form
???2=??0=??2??0. ??=??1 (C3.6) 2????Note that this derivative is independent of the time of the key shortening!
Hence, the maximum current is found as
??max=??=????0= and it is independent of the time of the key shortening!
??
(C3.7)
Theoretical competition. Tuesday, 15 July 20141/3
Problem 2. Van der Waals equation of state (11 points)
In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered.
Taking into account the finite size of the molecules, the gaseous equation of state takes the form
?? ????? =????, (1)
where ??,??,?? stands for the gas pressure, its volume per mole and temperature, respectively, ?? denotes the universal gas constant, and ?? is a specific constant extracting some volume.
With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter
?? ??+ ????? =????. (2) ??where ?? is another specific constant.
At temperatures ?? below a certain critical value ???? the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment АВ drawn at some constant pressure ??????. This straight segment is located between the volumes ???? and ????, and corresponds to the equilibrium of the liquid phase (indicated by ??) and the gaseous phase (referred to by ??). From the second law of thermodynamics J. Maxwell showed that the pressure ?????? must be chosen such that the areas I and II shown in Figure 1 must be equal. Part А. Non-ideal gas equation of state (2 points)
Figure 1. Van der Waals isotherm of gas/liquid Figure 2. Several isotherms for van der Waals
(curve 1) and the isotherm of an ideal gas (curve 2). equation of state.
With increasing temperature the straight segment ???? on the isotherm shrinks to a single point when the temperature and the pressure reaches some values ???? and ??????=????, respectively. The parameters ???? and ???? are called critical and can be measured experimentally with high degree of accuracy.
Theoretical competition. Tuesday, 15 July 20142/3
Part B. Properties of gas and liquid (6 points)
This part of the problem deals with the properties of water in the gaseous and liquid states at temperature ??=100 °C. The saturated vapor pressure at this temperature is known to be ??????=??0=1.0?
kg105 Pa, and the molar mass of water is ??=1.8?10?2 mole. It is reasonable to assume that the inequality ??????? is valid for the description of water properties in a gaseous state.
Almost the same volume ????0 can be approximately evaluated using the ideal gas law.
If the system volume is reduced below ????, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value ????min.
0. The condition of mechanical stability of supercooled gas at constant temperature is written as:
????<????For the van der Waals’ description of water in a liquid state it is reasonable to assume that the following inequality holds ???
??/??2.
Assuming that ?????????, find the following characteristics of water. Do not be surprised if some of the data evaluated do not coincide with the well-known tabulated values!
Part С. Liquid-gas system (3 points)
From Maxwell’s rule (equalities of areas, by applying trivial integration) and the van der Waals’ equation of state together with the approximations made in Part B, it can be shown that the saturated vapor pressure ?????? depends on temperature ?? as follows
??ln??????=??+??,(3)
where ?? and ?? are some constants, that can be expressed in terms of ?? and ?? as ??=ln ??2 ?1; ??=?????????
Theoretical competition. Tuesday, 15 July 20143/3
W. Thomson showed that the pressure of saturated vapor depends on the
curvature of the liquid surface. Consider a liquid that does not wet the material
of a capillary (contact angle 180°). When the capillary is immersed into the
liquid, the liquid in the capillary drops to a certain level because of the surface
tension (see Figure 3).
Figure 3. Capillary Metastable states, considered in part B3, are widely used in real
experimental setups, such as the cloud chamber designed for registration of immersed in a liquid that elementary
particles. They also occur in natural
phenomena such as the does not wet its material formation of morning dew. Supercooled vapor is subject to condensation by forming liquid droplets. Very small droplets evaporate quickly but large enough ones can still grow.
Theoretical competition. Tuesday, 15 July 2014 1/3
Problem 2.Van der Waals equation of state
Solution
Part А. Non-ideal gas equation of state
A1.If ??=??is substituted into the equation of state, then the gas pressure turns infinite. It is obvious that this is the moment when all the molecules are tightly packed. Therefore, the parameter ?? is approximately equal to the volume of all molecules, i.e. ??=??????3 (A1.1) A2.In the most general case thevan der Waals equation of state can be rewritten as
??????3? ??????+?????? ??2+?????????=0 (A2.1).
Since at the critical values of the gas parameters the straight line disappears, then, the solution of (A2.1) must have one real triple root, i.e. it can be rewritten as follows
????(???????)3=0(A2.2).
Comparing the coefficients of expression (A2.1) and (A2.2), the following set of equations is obtained
3????????=??????+??????
3??????2 (A2.3). ??=??
??????3??=????
Solution to the set (A2.3) is the following formulas for the van der Waals coefficients
??=
27??2????264??????????
??
(A2.4), (A2.5).
??=8??
The critical parameters are achieved in the presence of an inflection point in the isotherm, at which the first and second derivatives are both zero. Therefore, they are defined by thefollowingconditions
????
=0 (A2.6), ????and
??2??
????????
=0 (A2.7).
Thus, the following set of equations is obtained
??????2???+=0 ????? 2
??
which has the same solution (A2.4) and (A2.5).
A3.Numericalcalculationsforwaterproduce the following result
????=0.56
m6?Pa
??+?? ????? =????
???? ??????
2??????
??????? 3
?
????6??????
=0
(A2.8),
????=
A4.From equations (A1.4) and (A3.2) it is found that
3
mole?5m3
3.1?10mole ??
(A3.1). (A3.2). (A4.1).
????= ??=3.7?10?10m≈4?10?10m
??
Part B. Properties of gas and liquid
B1.Usingtheinequality???????, the van der Waals equation of state can be written as
??0+
??????????
????=????
4????
(B1.1), (B1.2).
which has the following solutions
????=2?? 1± 1?
????
Theoretical competition. Tuesday, 15 July 2014 2/3
Smaller root in (B1.2) gives the volume in an unstable state on the rising branch of thevan der Waals isotherm. The volume of gas is given by the larger root, since at ??=0an expression for the volume of an ideal gasshould be obtained, i.e.
????
????=2?? 1+ 1?
????
????4????0
????
(B1.3).
?3
For given values of the parameters the value ????=5.8?10. It can therefore be assumed ????0????
1? ??0??????????0
??????
? ??0????
that ?????1, then (B1.3)takes the form 2
????≈
B2. For an ideal gas
hence,
???
??0
=
4????0
≈
????0
(B1.4). (B2.1), (B2.2) (B3.1)
????0=
1
???? =
????0?????
????0
=2 1? 1?
????????=0.58%.
B3.Mechanical stability of a thermodynamic system is inpower provided that
????
<0. ????
??
The minimum volume, in which the mattercan still exist in the gaseous state, corresponds to a point in which
????
??????????→ ???? =0(B3.2).
Using the van der Waals equation of state (B3.2) is written as
????????2??
????=? +=0
??
(?????)
??
??
(B3.3). (B3.4). (B3.5). (B4.1) (B4.2).
From (B3.2) and (B3.3), and with the help of?????????????, it is found that
2??
??????????=???? Thus,
??????????????
2????
=
B4. Usingtheinequality?????/??, the van der Waals equation of state is written as
??
??????? =????, whose solution is
????=2???? 1± 1?
??
4??????
??
??2??2
2????0
=86
In this case, the smaller root shouldbe taken, since at??→0the liquid volume????=?? must be obtained according to (B4.1), i.e.
????=2???? 1? 1?
??
4??????
??
≈?? 1+
??????
. ??
(B4.3). (B5.1). (B6.1).
B5. Since (B4.3) givesthevolumeoftheonemoleofwaterits mass density is easily found as
??????2kg
????=??=≈=5.8?10 ??m3
??
B6. Inaccordancewith (B4.3) the volume thermal expansion coefficient is derived as
1???????????
??=???????=??+??????≈??=4.6?10?4К?1
??
?? 1+??
B7.The heat, required to convert the liquid to gas, is used to overcome the intermolecular forces that create negative pressure ??/??2, therefore,
????11
??=????≈ ??????2????=?? ????? (B7.1),
??
and using?????????, (B7.1) yields
??= ??=????
??
????
??
???? 1+??
≈????=1.0?106kg??J
(B7.2).
Theoretical competition. Tuesday, 15 July 2014 3/3
B8.Consider some water of volume??. To make a monolayer of thickness ?? out of it, the following work must be done
??=2???? (B8.1).
Fabrication of the monomolecular layer may be interpreted as the evaporation of an equivalent volume of water which requires the following amount of heat
??=???? (B8.2), where the mass is given by
??=?????? (B8.3).
Using (A4.1a), (B5.1)and(B7.2), one finally gets ??N ??=2??2????=0.12?10?2
m (B8.4).
Part С. Liquid-gas systems
C1.At equilibrium, the pressure in the liquid and gas should be equalat all depths. The pressure??in the fluid at the depth ?is related to the pressure of saturated vapor above the flat surface by
??=??0+???????(C1.1).
The surface tension creates additional pressure defined by the Laplace formula as
2?? ?????=?? (C1.2).
The same pressure??inthefluidatthedepth? depends on the vapor pressure ??? over the curved liquid surface and its radiusofcurvature as
2?? ??=???+?? (C1.3).
Furthermore, the vapor pressure at different heights are related by
???=??0+???????(C1.4).
Solving (C1.1)-(C1.4), it is found that
2?? ?=(?????)???? (C1.5).
Hence,the pressure difference sought is obtained as
2??????2???? ?????=??????0=???????=???????≈??????. ?????????? (C1.6). Note that the vapor pressure over the convex surface of the liquid is larger than the pressure above
the flat surface.
C2.Let ????be vapor pressure at a temperature ????, and ??????????be vapor pressure at a temperature ??????????. In accordance with equation (3) from problem statement, whentheambient temperature falls by an amount of ?????the saturated vapor pressure changes by an amount ?? ?????=?? (C2.1). ?????????????
In accordance with the Thomson formula obtained in part C1, the pressure of saturated vapor above the droplet increases by the amountof ?????. While a droplet is small in size, the vapor above its surface remains unsaturated. Whena droplet hasgrownuptoacertainminimumsize, thevaporaboveitssurface turns saturated.
Since the pressure remains unchanged, the following condition must hold
??????????+?????=???? (C2.2).
Assuming the vapor is almost ideal gas, its density can be found as
???? ????=??????????? (C2.3).
From equations (C2.1)-(C2.3), (B5.1) and (C1.6) one finds
2??????????????? =???? ??
Thus, it is finally obtained that
??=2????2????????????????? ?? ???????????? (C2.4). (C2.5). =1.5?10?8m
Theoretical competition. Tuesday, 15 July 2014 1/2
Problem3. Simplest model of gas discharge (10 points)
An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning.
Part А. Non-self-sustained gas discharge (4.8points)
In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates ??ext pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume.
When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events??recthat occurs in the gas per unit volume and per unit time is given by
??rec=??????????,
where ?? is a constant called the recombination coefficient, and ????,????denotethe electron and ion number densities, respectively.
Suppose that at time ??=0the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density ????(??) depends on time??as follows:
???? ?? =??0+??tanh????,
where ??0,?? and ??are some constants, and tanh??stands for the hyperbolic tangent.
Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be ????1=12?1010cm?3. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of ????2=16?1010cm?3.
Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers.
Assume that the gas fills in the tube between the two parallel conductive plates of area ?? separated by the distance ??? ?? is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube.
Assumethat both the electrons (denoted by thesubscript??) and the ions (denoted by the subscript??) acquire the same ordered speed??due to the electric field strength?? found as
??=????,
where ??isa constant called charge mobility.
Theoretical competition. Tuesday, 15 July 2014 2/2
PartB. Self-sustained gas discharge (5.2 points)
In this part of the problemthe ignition of the self-sustained gas discharge is consideredto show how the electric current in the tube becomes self-maintaining.
Attention!In the sequel assume that the external ionizer continues to operatewith the same ??extrate, neglect the electric field due to the charge carriers such that the electric field is uniform along the tube, and the recombination can be completely ignored.
For the self-sustained gas discharge there are two
important processes not considered above. The first process
isasecondary electron emission, and the second one isa
formation of electron avalanche. The secondary electron
emission occurs when ions hit on the negative electrode,
called a cathode, and the electrons are knocked out of it to
move towards the positive electrode, called an anode. The
ratio of the number of the knocked electrons????per unit
timeto the number of ions???? hitting the cathode per unit
time is called the coefficient of the secondary electron
emission,??=????/????.The formation of the electron
avalanche isexplained as follows. The electric field accelerates free electrons which acquire enough kinetic energy to ionize the atoms in the gas by hitting them. As a result the number of free electrons moving towards the anodesignificantly increases. This process is described by the Townsend coefficient ??, which characterizes an increase in the number of electrons??????due to moving ???? electrons that have passed the distance????, i.e.
??????=??????. ????The total current?? in any cross section of the gas tube consists of the ion ????(??)and the electron????(??)
currents which, in the steady state, depend on the coordinate ??, shown in the figureabove. The electron current ????(??) varies along the ??-axis according to the formula
???? ?? =??1????1??+??2,
where ??1,??2,??1 are some constants.
The ion current ????(??) varies along the ??-axis according to the formula
???? ?? =??2+??1????2??,
where ??1,??2,??2 are some constants.
Let the Townsend coefficient ??be constant. When the length of the tube turns out greater than some critical value, i.e. ??>??cr, the external ionizer can be turned off and the discharge becomes self-sustained.
Theoretical competition. Tuesday, 15 July 2014 1/3
Problem 3. Simplest model of gas discharge
Solution
Part А. Non-self-sustained gas discharge
A1.Let us derive an equation describing the change of the electron number density with time. It is determined by the two processes; the generation of ion pairs by external ionizer and the recombination of electrons with ions. At ionization process electrons and ions are generated in pairs, and at recombination processthey disappear in pairs as well.Thus, their concentrations are alwaysequal at any given time, i.e.
?? ?? =???? ?? =????(??)(A1.1).
Then the equation describing the numberdensityevolution of electrons and ions in time can be written as
????(??)2
=???????(??)(A1.2). ??????????
It is easy to show that at??→0 the function tanh????→0, therefore, by virtue of the initial condition ?? 0 =0,one finds
??0=0 (A1.3).
Substituting ???? ?? =??tanh???? in (A1.2) and separating it in the independent functions (hyperbolic, or 1 and????), one gets
??=
??????????
(A1.4),
??= ?????????? (A1.5).
A2.According to equation (A1.4) the number density of electronsat steady-state is expressed in terms of the external ionizer activity as
????1= ????2= ????=
????????1??????????2????
(A2.1), (A2.2), (A2.3).
????????1+????????2
Thus,the following analogue of the Pythagorean theorem is obtained as
10?3
????= ???? (A2.4) 1+????2=20.0?10cm.
A3.In the steady state, the balance equations of electrons and ions in the tube volume take the form
??
????????????=??????????????+?? (A3.1),
????????????=??????????????+??(A3.2).
It follows from equations (A3.1) and (A3.2) that the ion and electron currents are equal, i.e. ????=???? (A3.3). At the same time the total current in each tube section is the sum of the electron and ion currents ??=????+???? (A3.4). By definition ofthe current density the following relations hold
??
????=2=??????????=????????????(A3.5),
??
??
????
????=2=??????????=????????????(A3.6).
Substituting (A3.5) and (A3.6) into (A3.1) and (A3.2), the following quadratic equation for the current is derived
????????????=?????? 2???????? +2??
??
2
??
(A3.7). (A3.8). (A3.9).
The electric field strength in the gas is equal to
??
??=?? and solution to the quadratic equation (A3.7) takes the form
??=
????2??2??????3
?1± 1+
4????????????4??2??2
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