TMA 4195 Matematisk modellering Exam Tuesday December 16, :00 13:00 Problems and solution with additional comments

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1 Problem F U L W D g m 3 2 s kg Table : Dimension matrix TMA 495 Matematisk moellering Exam Tuesay December 6, :00 3:00 Problems an solution with aitional comments The necessary force (F ) to keep a ship at a constant spee (U) epens on its shape; primarily the length (L), with (W ), an its epth into the water (D). In aition, the water ensity,, the water viscosity,, an the acceleration of gravity, g, play a part. Use imensional analysis to n an expression for the force which inclues the two most famous imensionless numbers in ship esign: Froue number: F r = U= p Lg; Reynols number: Re = LU=: Ieally, a scale moel ) of the ship shoul be teste experimentally in water by keeping the imensionless numbers for the moel equal to those of the original ship. Is this really possible? Hints: [F ] = kgm=s 2, [] = kg/m 3, [] = m 2 =s, [g] = m=s 2. ) A scale moel is a moel of the ship with the same geometric shape, but with a smaller size (Say, L =m for the moel, compare to 200m for the original ship). Solution: Using the information provie in the problem, we assume F = f (U; L; W; D; ; ; g) : () The imension matrix follows immeiately an is shown in Table. The rank is 3, an there are several possibilities for core variables (avoiing F ): (U; L; ), (g; D; ), (; ; W ),. However, if one aims for the Froue an Reynols numbers, the choice (U; L; ) looks reasonable. With 8 variables, there are 8 3 = 5 imensionless combinations. Since Re involves an Fr involves g, it is easy to arrive at the formula F = U 2 L 2 Re; F r; W L ; D : (2) L

2 The scale moel keeps W=L an D=L unchange, so if we forget those, we nee to map the function = F = (Re; F r) (3) U 2 L2 for ranges of Re an Fr typical for the original ship. Assume that length of the moel, L m, is equal to rl, where r is about 0 2. If we aim to keep F r, we have to run the moel with U m = p ru, which looks feasible. However, for the same, the Reynols number woul then be a factor r 3=2 o. The only way to compensate this woul actually be n a ui with a corresponingly small viscosity, but this oes not exist. If we start by keeping the Reynol number (rather unrealistic!) we run into similar problems. Scale testing of ships (e.g. at the Tyholt wave basins), is carrie out by keeping the Froue number (relate to wave inuce resistance) an neglecting the variation in the Reyleigh number. The rst analysis of this situation ates back to one of the founers of Flui Mechanics, Lor Kelvin. Rea more about ship resistance on the Internet. NB! This problem has many i erent solutions, all equally goo as long as the set of variables is the same, an the core variables are chosen properly. 2 Problem A chemical reactor consists of a tank lle with a soli catalyst. A ui containing a issolve chemical with concentration c I ows into the tank. The ui ow Q is measure in volume per secon an the available volume for the ui insie the reactor is V. Some of the chemical is converte into a prouct by the catalyst an this prouct leaves the tank with the ui stream. The concentration of non-converte chemical in the tank is c (t ), an this is also the concentration in the ui leaving the tank. No volume changes are involve in these reactions, an the reactor is well-mixe with no signi cant concentration graients insie. The catalyst s e ciency is measure in terms of a quantity a, an the amount of chemical converte per time an volume unit is a c. Moreover, the change in a per time unit is k c a. At the start of reaction, t = 0, we have a (0) = a I an c (0) = c I. (a) Formulate the equations for the concentration of the chemical an the catalyst s e ciency, an erive the time scales What are these scales expressing? In the following it is known that T T 2. (b) T = a I ; (4) T 2 = k c I ; (5) 2

3 Use one of the time scales, write t = T, an scale the equations so that Explain the meaning of " an. c = ca + ( c) ; a = "ca; (6) c (0) = ; a (0) = ; 0: Determine the functions C 0 an A 0 in the perturbation expansions c () = C 0 ()+"C ()+ an a () = A 0 () + "A () +. (NB! Here an in the following two points we assume for simplicity that = ). (c) Use the other time scale an show that this leas to a singular perturbation system with Eqn. 6 as the inner system. Determine the outer solution to leaing orer, i.e., n a 0 an c 0 in the perturbation expansions c (t) = c 0 (t) + "c (t) + an a (t) = a 0 (t) + "a (t) +. (Hint: Derive an solve a separable i erential equation for a 0, which, however, leas to an implicit expression for the actual solution). () Carry out a matching of the inner an outer solutions above an state leaing orer uniform solutions. Solution: (a) The conservation law for the chemical may be expresse in i erential form as (V c ) t = V a c + Qc I Qc : (7) (All terms express change in mass per time unit). For the e ciency we have similarly a t = k c a : (8) The time scale T is erive from the ecay of the chemical in a close volume at the beginning of the reaction, c t = a Ic ; + c = C exp ( t = (=a I )) : (9) The time scale T 2 relates to the ecay in the e ciency of the catalyst, a t = k c I a : (0) 3

4 There is also a thir time scale, T 3 = V=Q. This is simply the time it takes to ll a volume V with the ow Q. The most elegant erivation of the time scales is to follow Lin&Segel s recipe: T = T 2 = (b) We consier the following scales This leas to max jc j max jc =t j = max ja j max ja =t j = It now obvious that T shoul be equal to T, an Moreover, jc Ij jc I a I j = a I ; ja Ij jk c I a I j = : k c I c = c I c; a = a I a; t = T : () c a I T = ac + Q ( c) ; a I V a = ca: (2) T k c I " = =a I = (k c I ) = T T 2 : (3) = T T 3 : (4) Since T T 2, " is a small imensionless parameter, an Eqns. 6 represents a regular perturbation problem. Now, =, an the system is Inserting the power series in ", we obtain to the leaing orer c = ca + ( c) ; a = "ca: (5) C 0 = C 0A 0 + ( C 0 ) ; A 0 = 0: (6) In aition, the initial conitions, which may be satis e exactly, are C 0 (0) = ; A 0 (0) = : (7) This leas to A 0 () =, an C 0 + 2C 0 = ; (8) 4

5 that is, C 0 () = 2 + e 2 : (9) (c) By changing the time scale to T = T 2 in Eqns. 2, an using t = T 2 t, we obtain " c = ac + ( c) ; t a = ca: (20) t which is a typical singular perturbation system with a small parameter in front of the highest erivative. We also observe that = t=", in accorance with Eqns. 6 being the inner system of Eqn. 20. Inserting c (t) = c 0 (t) + "c (t) + an a (t) = a 0 (t) + "a (t) + gives to leaing orer From the rst equation we obtain an hence, 0 = a 0 c 0 + ( c 0 ) ; a 0 t = c 0a 0 : (2) c 0 = + a 0 ; (22) a 0 t = a 0 + a 0 : (23) This is a separable equation which may be written + a 0 = a 0 t; (24) an integrate to ln a 0 + a 0 = t + t 0 ; (25) where t 0 is an unknown constant. It is not possible to express a 0 as an elementary function of t. () We have now obtaine the inner solution A 0 () = ; which also ful l the initial conitions, an the outer solution, with one unknown constant t 0. C 0 () = 2 + e 2 ; (26) c 0 (t) = + a 0 (t) ; ln a 0 (t) + a 0 (t) = t + t 0 (27) 5

6 The matching principle in its simplest form requires that lim C 0 () = lim c 0 (t) ;! t!0 lim A 0 () = lim a 0 (t) : (28)! t!0 Since A 0 =, we therefore nee that lim t!0 a 0 (t) =. This is ful lle only if t 0 =. In aition, we also have to check that 2 = lim! C 0 () = lim t!0 c 0 (t) = lim t!0 + a 0 (t) ; (29) which is true since lim t!0 a 0 (t) =. This nally leas to the uniform solutions t a u 0 (t) = A 0 + a 0 (t) lim A 0 () = a 0 (t) ; "! t c u 0 (t) = C 0 " + c 0 (t) lim! c 0 () = 2 + e 2t=" + + a 0 (t) 2 = e 2t=" a 0 (t) : (30) 3 Problem Explain that in orer to etermine the stability of an equilibrium point u 0 for the moel u=t = f (u), it is usually su cient to consier the sign of f (u) aroun u 0. Apply this for making a sketch of the bifurcation iagram inicating stable an unstable equilibrium points in the (; u)-plane for the moel u t = + u2 2u ( + u) : (3) (It is not necessary to state the exact expressions for the equilibrium points or consier points requiring higher orer analysis). Solution: We observe that u (t) increases when f (u) > 0 an ecreases when f (u) < 0. If we start from a point u near u 0, we will rift towars u 0 if f (u ) < 0 an u > u 0, or f (u ) > 0 when u < u 0. The opposite gives a rift away from u 0 (Nees to be more careful if f 0 (u 0 ) = 0). Relating this to the erivative test requires Taylor s Formula: f (u) (not neee for the exam). Here the equilibrium points occur for f (u 0 ) = f (u) = f u (u 0) (u u 0 ) + o (ju u 0 j) : (32) + u 2 2u ( + u) = 0; (33) 6

7 u µ 2 Figure : Bifurcation iagram for Problem 3. Re is unstable an green is stable. that is, or u 0 = ; (34) u 2 0 2u 0 = : (35) The rst equation represents a line an the secon a parabola in the (; u)-plane, as sketche on Fig.. It is convenient rst to raw the curves an etermine the (unique!) sign of f (u) in each region (shown with blue symbols on the graph). Then stability/instability is etermine by moving vertically aroun the equilibrium points. 4 Problem A part of a water cleaning system is moelle as a tube (along the x -axis) of length L where pollute water ows with constant velocity V. The tube also contains absorbers that remove the pollution. The concentration of pollutant in the water is c, measure as amount per length unit of pipe. Similarly, the amount of absorbe pollutant per length unit of pipe is enote. Some of the absorbe pollutant will over time re-enter the water stream. The absorption an re-entering is moelle by the = k (A ) c k 2 : (36) State the integral conservation law for the pollutant an show that it leas to the i (c + (V c ) = 0: (37) 7

8 Base on the integral law, establish that a iscontinuity in the concentrations, moving with velocity U, has to ful l U c 2 c = V; (38) (c 2 + 2) (c + ) where (c ; ) an (c 2; 2) are the concentrations on the respective sies of the iscontinuity. (b) Introuce suitable scales an show that the equations may (c + ) + = = ( ) c : Explain the meaning of " an (Hint: Use the same scale for an c ). Assume that the tube is in nitely long in both irections an consier analytic solutions of Eqns. 39 an 40 in the form of fronts, travelling with velocity U: c (x; t) = C (x Ut) ; (4) (x; t) = R (x Ut) : (42) We limit ourselves to the special case where C () an R () ( = x Ut) satisfy lim C () = ; (43)! lim C () = 0;! (44) lim! + ; (45) lim! R () = 0: (46) (c) Insert 4 an 42 into Eqn. 39, integrate once, an use the behaviour at an to etermine U an a simple relation between C an R. Use this information an Eqn. 40 to etermine C () an R (). How is the behaviour of the solution when "! 0? (Hint: The equation has a solution for 0 < y < M). () y = y + y M y () = M + exp (47) (48) 8

9 Assume " = 0 in Eqn. 40 so that the system (39 an 40) simpli es to Consier the initial conition = c c + ; Q (c) = Q = 0; (49) c ; < x < ; t 0: (50) c + c (x; 0) = ; x < 0; 0; x > 0: Show that the corresponing solution of Eqn. 49 evelops a shock. Determine the shock velocity from the expression in point (a) an compare to the result in (c). Solution: (a) Consier a control volume between x an x 2 incluing both the ui an the absorbers. Since the ux is V c, we obtain t Z x 2 x (5) (c + ) x + V c (x 2; t ) V c (x ; t ) = 0: (52) The stanar argument, interchanging erivative an integral an the integral mean value theorem (c + (V c ) = 0: (53) In orer to erive U, we put the control volume aroun the iscontinuity. The conservation of pollution requires ( refers to the left of the shock, an 2 to the right): Thus, t Z x 2 x = lim t!0 (c + ) x + V c (x 2; t ) V c (x ; t ) [(c 2 + 2) U = (c + )] Ut t + (V c 2) (V c ) = 0: c 2 c (c 2 + 2) (c + ) V: (b) The equations an the hint suggest the following scaling: x = Lx; t = L V t Thus = A; c = (c + ) + = = ( ) c ; 9

10 where " = = (k A) L=V ; (57) = = (k A) =k 2 : (58) Similarly to Problem 2, both " an are quotients between time scales. The time scale L=V is the transverse time for the ui, = (k A) is the scale for absorption, an =k 2 the typical time for re-entrance of pollution into the stream. (c) We introuce the trial solution in Eqn. 55 an let = x Thus, U C U R + C Ut: = 0: (59) U [C () + R ()] + C () = const. (60) Letting! shows that the constant is equal to 0, whereas! gives U + + = 0; (6) + or Since C ( U) UR = 0, we have C = U = an the ratio between C an R is thus constant for all -s! Inserting this into the secon equation, leas to + : (62) + U R = ( + ) R; (63) U "U R = ( R) ( + ) R R = R ( + ) R 2 : (64) From the equation in the hint, which is a simple moi cation of the logistic equation, we obtain (not bothering about an arbitrary shift of the origin): R (x; t) = + + exp x Ut "U C (r; t) = ( + ) R (x; t) = U = + : + 0 ; + exp x Ut "U ; (65)

11 When "! 0, the solution approaches a shock occurring at x = Ut. () Even if it is simple to work irectly with Eqn. 49, the equation may be transforme to our familiar form by rst observing that Q q (c) = Q (c) c = @x = + is always strictly positive, we may just as well consier = 0: (66) (c + ) = 0: (68) The kinematic velocity, =q (c), is strictly increasing with c, an hence characteristic lines starting from x < 0 will overtake the lines starting at x > 0, thus leaing to a shock situation (the characteristics move into the shock). The spee of the shock was consiere in point (a), an the scale version for the shock velocity is U = c 2 c (c ) (c + ) = 0 = : (69) + This is equal to the velocity in (c), an shows that for this case, the continuous front solutions of the full system ecay nicely into the shock solution of the egenerate system (" = 0).