Introduction to Fluid Mechanics - Su Momentum equation The effect of moving reference frames.

Transcription

1 Introduction to Fluid Mechanics - Su Momentum equation The effect of moving reference frames. Reading: Text, 4.4, 4.5. In using the momentum equation, we want to keep in mind two things: The derivation of the Reynolds transport theorem, which was our starting point for arriving at both the equation for mass conservation and the momentum equation, is totally valid for moving control volumes as long as all velocities are measured relative to the moving control volume. As long as you use relative velocities that way, the equations basically can t tell that the control volume is moving. Newton s second law, that the rate of change of the linear momentum of a system equals the net forces on that system, is only valid when the linear momentum is measured in an inertial (non-accelerating) reference frame. Thus, even though the Reynolds transport theorem is fine for any moving control volumes, the momentum equation as we ve seen it only works for non-accelerating control volumes. To consider accelerating control volumes, we will need to make further modifications. Example 1. Moving, inertial reference frames. (This example is drawn from J.A. Roberson and C.T. Crowe, Engineering Fluid Mechanics, 3rded.) y x HO 2 =1000 kg/m 3 d V J V S Figure 1: Schematic for example 1. Figure 1 shows a tank of water resting on a wheeled sled. A jet of water, with diameter d, issues from the orifice on the side of the tank. Assume that top of the tank is filled with pressurized air, and that the water jet velocity relative to the tank, V J, is fixed. The sled moves to the left at a constant speed, V S. The ground imposes a horizontal force, F gr, on the sled through the wheels (the exact nature of this force is not important). What is the magnitude and direction of this force? We ll specify the control volume to be a rectangle containing the entire tank and sled, which cuts perpendicularly through the water jet, and with the bottom of the control surface being adjacent to the ground. The control volume moves at the same speed as the tank and sled; the speed is constant, so the control volume represents an inertial reference frame. The general form of the momentum equation is I {}}{ F = II { }}{ ρ V dv + III {}}{ ρ V ( V da). (1) 1

2 All of the velocities in (1) are measured relative to the moving control volume, as discussed above. Equation 1 is exactly the same momentum equation we ve previously used for stationary control volumes; we are able to apply it here also because physics tells us that an observer in an moving reference frame can t tell that the reference frame is moving as long as the motion is inertial. This is also why we need to use velocities measured in the frame of reference of the moving control volume. We further observe that only the x-component of this vector equation will be important, since the sled only moves in the horizontal plane. Let s consider each of these terms separately. Term I. The forces in the x-direction acting on the control volume are the force imposed by the ground, F gr, and the pressure acting on the left and right sides of the control volume. The sides of the control volume are bordered by the atmosphere, except for where the water jet cuts through it. However, we know that free jets in the atmosphere have the same pressure as the atmosphere, so the pressure forces balance to zero. We therefore have for the x-component of force F,x = F gr. (2) Term II. This term describes the momentum contained in the control volume. In the control volume, the sled, tank, and the water in the tank move with zero velocity relative to the speed of the control volume, so their contribution to term II is zero. The portion of the jet located within the control volume, however, has non-zero V relative to the control volume, so we need to look at its contribution more closely. The integral works out to be equal to ρv J (πd 2 /4)L J,whereL J is the length of the jet that s inside the. All of those terms are constant, so the time derivative of the integral is zero. So the contribution of the jet to term II is zero too. We are left with ρ V dv =0. (3) Term III. This describes the flux of momentum through the control surface, which occurs only through the jet of water, and has the form of an outflow (thus having positive sign). The area of the jet is πd 2 /4, and its velocity relative to the moving control volume is V J,so term III becomes ρ V ( V da) = π 4 ρv J 2 d 2. (4) Solution. Collecting (2), (3) and (4), we have F gr = π 4 ρv 2 J d 2. (5) Two observations: the force imposed by the ground is in the positive x-direction, which makes sense, since otherwise the jet would cause the sled to accelerate continuously to the left; and deriving (5) is pretty trivial, which is the point, since the moving, inertial control volume we chose is the logical one for this problem. 2

3 Non-inertial control volumes. While inertial control volumes are pretty widely applicable, what happens when the velocity of the control volume varies with time? For some problems, it just makes more sense to define an accelerating control volume, and to measure velocities relative to that control volume. What we want is a momentum equation that s valid for such a non-inertial control volume. The main difficulty is that Newton s second law requires inertial reference frames, so we ll need to modify the momentum equation. Figure 2 depicts an accelerating control volume. y y z x V () t z (inertial frame) x Figure 2: Schematic of control volume moving at a non-constant velocity. Define the unprimed, x y z coordinate system as being stationary in space, and the x y z system as moving with the control volume; also, let unprimed velocity terms be measured in the x y z system, and primed velocities be measured in the x y z system, i.e. V = velocity measured relative to the stationary coordinate system V = velocity measured relative to the moving control volume. (6) Using this convention, we also have V (t) = velocity of the control volume relative to the stationary coordinates. (7) By (6) and (7), the primed and unprimed velocities relate to each other as V = V + V. (8) We know that we will need to apply Newton s second law, since we re dealing with momentum and force. Consider a fluid system with mass m, and assume it s so small that the entire system has the same velocity. We can define the linear momentum of the system in two ways; in terms of its velocity in the x y z reference frame, V, or its velocity in the x y z frame, V.Thatis P = m V = momentum in x y z frame P = m V = momentum in x y z frame. (9) Newton s law can only be applied to P, since the law requires that the reference frame be inertial ( d ) P F sys = = m d V = m a, (10) sys where a is the acceleration of the system in the x y z frame. Equation (10) is, of course, fine, but we would prefer to have it in terms of velocities measured relative to the control volume. We can write d V/= d V / + d V / using (8), and plugging this into (10), we get F sys = m d V + m d V (11) 3

4 which is totally logical (11) says that the for a fixed-mass system in a non-inertial reference frame, any forces applied first have to account for the acceleration of the reference frame, then the remaining forces can accelerate the system relative to that frame. Rearranging (11), d P = F sys m a (12) where a = d V / is the acceleration of the control volume, and we have applied the definition of the linear momentum in the control volume reference frame, P. Equation 12 can be thought of as Newton s law for non-inertial reference frames; it tells us that the rate of change of a system s momentum, as measured relative to a non-inertial frame, is equal to the force on the system minus the force m a associated with the acceleration of the reference frame. We can use this knowledge in applying the Reynolds transport theorem, which we re familiar with as ( ) dn = ηρ dv + ηρ( V d A ), (13) sys where we ve written the flux term using V to emphasize that the proper velocity to use is the velocity relative to the control volume. For our problem, define the extensive and intensive properties of interest as Then, (13) gives N = P, extensive property η = V, intensive property. ( d ) P = F sys sys IV { }}{ a ρdv = V ρdv + V ρ( V d A ), (14) where the term IV arises from integrating the acceleration a over the entire system, and we recall that the system and the control volume coincide at the particular time of interest. Equation (14) is the appropriate momentum equation when the control volume of interest is accelerating. (Technically, a could be taken out of the integral, since the acceleration of the x y z frame is uniform in space, but (14) is consistent with the text.) Example 2. The rocket problem. The rocket is the ubiquitous example for illustrating accelerating control volumes. We will assume the following: Initial mass of the rocket and its fuel is m 0 Friction drag from the air is negligible Fuel consumption rate ṁ (units mass per unit time), is constant Exhaust velocity relative to the rocket, V e, is constant The pressure in the exhaust is atmospheric Flow pattern of the fuel within the rocket is steady (why is this necessary?) 4

5 y x y V e x (stationary frame) Figure 3: Upward acceleration of a rocket. Rocket is initially at rest. We want to derive the equation that describes the vertical velocity of the rocket as a function of time in the x y reference frame, V (t). We are only interested in the y-component of the momentum equation (14), which can be written A {}}{ F,y B { }}{{ }}{ a,y ρdv = V y ρdv + C D { }}{ ρv y ( V d A ). (15) The appropriate choice of the control volume encloses the rocket and cuts horizontally through the exhaust. We ll work out the terms in (15) individually. Term A. The possible forces are gravity and pressure. The pressure forces all cancel because the pressure surrounding the control volume is atmospheric (including in the exhaust). That leaves gravity, so F,y = m g where m is the mass in the control volume. Rather than write the full mass conservation equation, just observe that the initial mass is m 0, and the rate at which fuel is consumed and shot out the end of the rocket is constant at ṁ; thus,wecanwrite F,y = (m 0 ṁt)g. (16) Term B. The vertical acceleration of the control volume, a,y = dv (t)/, is uniform in space, so we can take it out of the integral, leaving dv (t) a,y ρdv = ρdv. But the remaining integral just gives the mass in the control volume, so we get Term B = dv (t) dv (t) m = (m 0 ṁt). (17) 5

6 Term C. For this term, we separate the contents of the control volume into the rocket s structure, and the fuel it contains. We argue first that the structure of the rocket moves with the speed of the control volume, so its relative velocity V y is zero, and thus its contribution to term C is zero. For the fuel contained in the rocket, we make use of the assumption that the flow pattern of fuel inside the rocket is steady, so the time derivative of its part of the integral in term C is zero. Thus Term C = 0. (18) Term D. The flux term is nonzero only on the bottom, horizontal surface of the control volume, where the exhaust cuts the control surface. On that surface, the relative velocity of the exhaust is V e (since it s in the negative y-direction). We can write ρv y( V d A )= V e ρv e da, (bottom) where the last integral is positive, because the flux is out of the control volume. We can also easily see from the units of the integral that it represents the mass flow rate in the exhaust, which has to be equal to the rate of fuel consumption in the rocket, namely ρv e da = ṁ, so we get (bottom) Solution. Collecting (16), (17), (18) and (19), we get Term D = V eṁ. (19) dv (t) (m 0 ṁt)g (m 0 ṁt) = V eṁ, which we can rewrite as dv = V eṁ m 0 ṁt g, and, separating the variables (hopefully a familiar trick from differential equations), dv = V eṁ m 0 ṁt g. To get V (t), we need to integrate this from t =0,whereV = 0 (since the rocket is initially at rest), to some arbitrary t: which gives V (t) 0 t dv = V eṁ 0 m 0 ṁt g t ( V (t) = V e ln 1 ṁt ) gt, (20) m 0 0 which is commonly used to approximate rocket velocities. 6