Exercise 4 - Hydraulic Systems

Transcription

1 Exercise 4 - Hyraulic Systems 4.1 Hyraulic Systems Hyraulic systems are, in general, escribe by the Navier-Stokes equations as you might have learne in flui ynamics courses. In orer to simplify the moeling of such systems, it is convenient to use simpler formulations. Typical elements which constitute hyraulic systems are: ucts, compressible noes, valves (treate within the next few weeks) Hyraulic Ducts A sketch for a water uct is epicte in Figure 1. The quantities that we nee to moel a uct are the height ifference h, the top an bottom pressures p 1 (t), p 2 (t), the length of the uct l, the velocity of the water flowing into the uct v(t), the ensity of the flowing flui ρ, an the cross-sectional area of the uct A. In general, we are intereste e y p 1 (t) A l h v(x, y, t) = v(t) α p 2 (t) e x Figure 1: Sketch of a water uct. in moeling the velocity of the flui, i.e. fining an equation of the form t v(t) = f(p 1(t), p 2 (t), v(t), h, ρ, A, l). (4.1) A free boy iagram of the flui is shown in Figure 2. This material is base on the HS17 teaching assistance taught by Nicolas Lanzetti an Gioele Zarini. This ocument can be ownloae at

2 p 1 (t) A F fric F gravity p 2 (t) Figure 2: Free boy iagram of the flui. Applying Newton s law to the chunk of water insie the uct gives m v t = F pressure + F gravity,x F fric. (4.2) Uner the assumption of a constant pressure istribution over the cross-sectional area A we may write F pressure = (p 1 (t) A p 2 (t) A) The mass of the flui in the uct is given by ρ V = ρ A l m = ρ A x. To compute the contribution of gravity, we integrate over the whole uct F gravity = g m uct l [ ] sin(α) = g ρ A x 0 cos(α) [ ] sin(α) = ρ g A l cos(α) Using the facts that the elevation angle α satisfies sin(α) = h l, we may write F gravity,x = h ρ g A. This ocument can be ownloae at 2

3 The friction force epens on the shape factor l an reas F fric = 1 2 ρ v2 sign(v) λ Al (4.3) where λ is a constant coefficient an is the iameter of the pipe. Plugging the foun results in Equation (4.2) one gets the conservation law along the longituinal axis of the uct m t v(t) = ρ l A t v(t) = A (p 1(t) p 2 (t)) + A ρ g h F fric (t). (4.4) Compressibility of Ducts The compressibility is the property of ucts (an possibly other flui containers) to eform uner the effect of an applie pressure. Mathematically, it is efine as σ 0 = 1 V 0 V p, (4.5) where V [m 3 ] is the volume, p [Pa] is the pressure, an σ 0 [Pa 1 ] the compressibility. V (t) k = 1 σ 0 V 0 V in (t) V = 0 p(t) V out (t) Figure 3: Sketch of the hyraulic compressibility In general, since the flui entering the element is not equal to the one exiting, we have a time-varying volume t V (t) = V in (t) V out (t). (4.6) Changes in volume cause a irect increase in the acting pressure accoring to p(t) = 1 V (t) +p 0, σ 0 V }{{ 0 } V (t) = V (t) V 0, (4.7) p compress where p 0 an V 0 enote the static (unloae) pressure an volume of the elastic element respectively. This ocument can be ownloae at 3

4 4.1.3 Compressibility of Fluis Analogously, we efine compressibility of matter (fluis in particular) as σ 0 = 1 V 0 V p, (4.8) where V [m 3 ] is the volume, p [Pa] is the pressure, an σ 0 [Pa 1 ] the compressibility. Note that the only ifference with the above is a minus sign in the efinition, that is p(t) = 1 σ 0 V (t) V 0 + p 0 V (t) = V (t) V 0. (4.9) This ocument can be ownloae at 4

5 4.2 Tips Formulate the mass balance for the water tank as t m W = ρ A W t h W = ṁ in ṁ out with a fictitious height h W. Then, you may fin the true height h W with ( ) ( ) 2 h W = h W + λ W sign t h W t h W. Moreover, the mass flow of a incompressible flui through a valve with opening area A can be moele as ṁ(t) = c A 2ρ (p before p after ), where c is the so-calle ischarge coefficient. This ocument can be ownloae at 5

6 4.3 Example Your SpaghETH is growing every week more an although no particular prouction issues occur you are concerne about ecology. Since each tank of pasta you cook nees water an a correct salt seasoning for it to taste that elicious, you nee a lot of salt an water, which are often waste. For this reason, you open a research branch in your startup which ecies to esign a uct-hyraulic system to counteract the waste of water an salt. The tank where the pasta cooks is connecte to a uct (iameter T ), the Tunnel. Before the water enters a secon uct (iameter T ), the Seasoner, a pump increases its pressure by p. Note that only compressibility effects of the Tunnel shoul be taken into account. The pressure at the water s surface p is assume to be known. Assume a circular tunnel, whose area reas A T = π 2 T /4. The area of the water tank is A W (with A W A T ). A sketch of the system with the relative parameters is shown in Figure 4. Salt l D l T λ AP, ρ, v AP (t) T h T Pump h V h(t) λ BP, ρ, v BP (t), σ Figure 4: Sketch of the system. 1. List all the reservoirs an the relative state variables. 2. Fin the pressure p 1 (t) at the beginning of the Tunnel as a function of the velocity in the Tunnel v BP (t). 3. Formulate the ifferential equation for v BP (t) as function of the pressure right before the pump p 2 (t). 4. Exploiting the compressibility of the Tunnel, fin the pressure p 2 (t) explicitly. 5. Formulate the ifferential equation for v AP (t). 6. Formulate the ifferential equation for h(t). This ocument can be ownloae at 6

7 Solution. 1. The reservoirs of the system an their relative state variables are: The Tunnel kinetic energy. The state variable of this reservoir is the velocity of the water in the Tunnel v BP (t). The Seasoner kinetic energy. The state variable of this reservoir is the velocity of the water in the Tunnel v AP (t). The water mass in the tank. The state variable of this reservoir is the water s height h(t). The compressibility of the Tunnel. The state variable of this reservoir is the water volume V (t). 2. Bernoulli: First, we nee to use Bernoulli s law from the water s surface to the beginning of the Tunnel, in orer to fin the local pressure. This reas p ρ + g h(t) = v BP(t) 2 + p 1(t) + g h V 2 ρ p 1 (t) = p + g ρ (h(t) h V ) ρ vbp(t) 2. 2 (4.10) Here, we are making the assumption that the area of the tank is much larger than that of the tunnel, thus ḣ(t) Tunnel: The impulse equation for the Tunnel reas ρ l T A T t v BP(t) = A T (p 1 (t) p 2 (t)) F fric (t), (4.11) where F fric (t) is the friction force, which reas F fric (t) = A T λ BP lt ρ 2 T sign(v BP (t)) v BP (t) 2. (4.12) The pressure p 2 (t) is obtaine from the compressibility of the Tunnel. 4. Compressibility: As learne in the lecture, the pressure before the valve p 2 (t) is compute as where p 2 (t) = 1 σ The volume balance reas V (t) V 0 + p stat = 1 σ V (t) V 0 V 0 + ρ g (h(t) h V ) + p, V (t) t (4.13) V 0 = l T π 2 T. (4.14) 4 = V in V out = A T (v BP (t) v AP (t)). (4.15) This ocument can be ownloae at 7

8 5. Seasoner: The impulse equation for the Seasoner reas ρ (l D +h T h V ) A T t v AP(t) = A T (p 2 (t) + p p )+A T ρ g (h V h T ) F fric (t), where F fric (t) is the friction force, which reas (4.16) F fric (t) = A T λ AH (l D + h T h V ) ρ 2 T sign(v AP (t)) v AP (t) 2. (4.17) 6. Water Tank: The water tank stores water s mass. Its state variable is the height of the water tank h(t). The mass balance reas t m(t) = ρ A W h(t) t = m in m out = ρ A T (v AP (t) v BP (t)). (4.18) This leas to a relation for the change of the water height t h(t) = A T A W (v AP (t) v BP (t)). (4.19) This ocument can be ownloae at 8