Lab 3A: Modeling and Experimentation: Two-Can System

Transcription

1 Lab 3A: Modeling and Experimentation: Two-Can System Prof. R.G. Longoria Department of Mechanical Engineering The University of Texas at Austin June 26, 2014

2 1 Introduction 2 One-Can 3 Analysis 4 Two-Can 5 Simulation 6 Summary 7 Pre-Lab/LE

3 Overview This lab deals with modeling and experimentation with a system formed by cascading two cans so that one empties into the other, which then empties to a reservoir. These slides describe the derivation of a mathematical model for the one and two-can systems. The models are used to guide physical experimentation and simulation studies.

4 Lab 3A Specific Objectives One of the main goals of this lab is for you to practice building a model of a system and assessing how accurately it can predict actual system behavior, in the process learning about what is needed to properly parameterize the model using theoretical and/or experimental means. For the two-can system, you will: 1 Use your model to determine the initial volume of water in the top can that will result in the level of fluid in the bottom can reaching a specified level at some point in time 2 Determine how well you can exactly maximize the level in the bottom can without any spillage.

5 Outline Introduction One-Can Analysis Two-Can Simulation Summary Pre-Lab/LE You see tank emptying problems beginning in physics classes

6 Outline Introduction One-Can Analysis Two-Can Simulation Summary Pre-Lab/LE Then you study these problems in more detail in fluid mechanics From: Introduction to Fluid Mechanics, Fox and McDonald, 4th edition In this example, control volume equations are used to derive an ODE in terms of the height, h, of fluid in this tank.

7 One-can system model In systems modeling, we learn to model by identifying distinct basic elements and use physical laws to understand how they are interconnected. Fluid potential energy storage in a can (or tank) is modeled as a C element quasi-static assumption constant-area tank pressure-volume (P V ) constitutive relation Orifice flow induces energy dissipation modeled by using an R element Steady-flow Bernoulli model simplifying assumptions pressure-flow (P Q) constitutive relation

8 The hydraulic can or tank element A can only stores potential hydraulic energy (if we can ignore inertial effects), and we show that this is a capacitive element by recalling that, P = ρgh = ρg A V = 1 C V where the can is assumed to have constant cross-sectional area, so C is a hydraulic capacitance, C = A/(ρg). If the can or tank does not have constant cross-sectional area, as is common in many applications as illustrated below, then the constitutive law would not be linear. It may also be the case that you cannot ignore inertial effects of the fluid. What would the tank look like in such a case?

9 An Orifice/Valve as an R Element You can show that orifice flowrate, Q, can be directly related to pressure drop across the orifice, P, by, Q = C c A o 2P/ρ where C c is a discharge coefficient, A o the orifice area. We identify the flowrate as a flow variable and pressure as an effort variable, the product being the dissipated power, P = P Q. Since the orifice involves dissipative processes, we model the orifice as a resistive (R) element.

10 Orifice flow relation The flowrate relation derived is used extensively for orifice flow predictions in a wide range of applications. For example, it is commonly applied when predicting hydraulic valve flows. If the pressure can reverse sign, then the relation may need to be generalized to account for sign by using, Q = C c A o sgn(p ) 2 P /ρ where the sgn() function provides sign information and the absolute value of pressure is taken when used in computation. Note: detailed discussion of the orifice flow rate model can be found in the appendix.

11 One can dynamic system model We assemble the elements together using the mass continuity equation where, where assuming incompressible flow, For the one can system, ṁ stored = ṁ in ṁ out ρ V = ρq in ρq out V = Q in Q out where V, the volume stored in the can, is the system state. Recognize that this equation cannot be solved until the terms on the right-hand side are expressed as functions of states or inputs.

12 One can dynamic system model state equation To derive the state equation for the one-can system, use the constitutive relations discussed earlier. First, where But, d V dt = Q in Q out Q out = C c A o 2P/ρ. P = V/C Now, d V dt = Q in C c A o 2 V/ρC which can be simplified to the state equation, d V dt = Q in K V where K is a constant for a given can. This equation is in the form of a 1st order state equation.

13 Steady-state versus dynamic emptying A steady-state condition occurs when Q in is equal to Q out so, d V dt = Q in Q out = 0. The system is said to be in equilibrium or steady-state and volume will be constant. A dynamic state occurs when Q in Q out, so the tank dynamically empties and/or fills over time, with volume a dynamic state determined by, d V dt = Q in K V

14 Steady-state versus dynamic emptying Steady-state: In general, setting dx/dt = 0 for all your state equations gives n algebraic equations that can be solved for the equilibrium states. In this case, d V dt = 0. gives Q in = Q out, which can be solved for V e, the steady-state or equilibrium volume. Q. Can you solve for V e? Dynamic state: The special case Q in = 0 with known V o is the can emptying problem, d V dt = K V which can be solved for V (t) in closed form, given we know K. Q. Can you find V (t)?

15 Finding K from experiments The emptying can model can be used to design an experiment to determine the unknown parameter, K. The equation, d V dt = K V is a nonlinear ordinary differential equation that can be integrated from the form, d V = Kdt V An emptying experiment begins with an initial volume, V o, that empties in time, T e, as expressed through the integration, d V Te = Kdt V o V 0 0

16 Model analysis Integration of the equation, yields, giving, 0 0 d V = 2 V 1/2 0 V V o V o d V Te = Kdt V o V 0 Te = Kdt = Kt Te V o = KT e. This relation suggests that measurements of initial volume and time-to-empty can be used to estimate the parameter K. We use this as an example of model-based experiment design.

17 Data collection and analysis The relation, 2 V o = KT e, suggests there is a linear relation between initial volume and time-to-empty. Q. Do you think it is correct to force the trend line in this relation to go through (0,0)?

18 Determining K from a steady-state experiment The other way to find a K value is to use a steady-state experiment. 1 Describe what measurements would be needed to find K. 2 Explain how the physical conditions are different from the those in a dynamic can emptying experiment. 3 Is there a preference in choosing one of the experimental approaches described over the other?

19 Two-can system model The state equations for the two can system are found by simply applying continuity on the second can. You should be able to show that the state equations for the volumes in cans 1 (top) and 2 (bottom) are, respectively, V 1 = K 1 V1 V 2 = +K 1 V1 K 2 V2 where the K values are distinct for each can. Note that there is no input flow to top can, but it can be readily added. These equations can not be integrated analytically, but can be solved using numerical simulation.

20 Two-can system dynamics Consider the case where the top can (1) is filled to some level and the bottom can (2) is initially empty. When released, flow from the top can will begin to fill the bottom can, which then empties (to a reservoir). The dynamic response of each can volume should follow trends similar to those shown below.

21 Script for two-can The two-can system ODEs are formatted in a script function file below: function file function f = two_can(t,v) global K1 K2 if V(1)<0 Q1 = 0; % if V<0, can is empty else Q1 = K1*sqrt(V(1)); end if V(2)<0 Q2 = 0; % if V<0, can is empty else Q2 = K2*sqrt(V(2)); end f = [-Q1;Q1-Q2]; Note how special consideration must be made to prevent volume values becoming negative, which is not possible in this case.

22 Two-can system block diagram A block diagram for the two-can system is shown below. One way to keep volume greater than or equal to zero is to use a saturation block with a lower limit set to zero and upper limit set to Inf.

23 Experimental Testing and Evaluation of Model Testing a one-can model. Given an experimentally determined, K, repeated tests can be run to determine if time-to-empty is accurately predicted. To fully assess accuracy of the model, the volume needs to be measured over time and compared to simulation predictions. Testing a two-can model. One quick test is to begin with top can at a given volume, bottom can empty, and estimate peak level reached in the bottom can and time-to-empty. This gives a first assessment of accuracy of the combined system model. As for the one-can system, measuring volume over time would give a more accurate assessment.

24 Summary The one and two-can systems rely on physical modeling of hydraulic system elements, and introduce the application of state space modeling. The system model is described by two nonlinear differential equations in state space form. To make predictions about the two volume states of this system, it is necessary to solve the equations using a numerical simulation program. In addition, each can requires its own K value which must be experimentally determined.

25 Summary of Pre-Lab 3A see the clog for details 1 Develop a two-can simulation model in LabVIEW ready for use in lab. Submit screen shots of the front panel and block diagram to illustrate your code. The program should be readable and neatly organized. The following slide provides a table with can dimensions. Use Can Set 1 to find ideal values for K 1 (bottom can) and K 2 (top can). You will need to refer to the appendix for details. Submit your calculations as part of this pre-lab. Submit results from a sample simulation of Can Set 1 given that Can 1 is initially empty and Can 2 is 3/4 full. 2 Submit a description of a laboratory procedure for collecting data from one-can experiments for determining K. The only measurements you will be able to make are of time-to-empty, T e, (using a stop watch) and volume. You should think through any problems that can arise when running these experiments.

26 Can dimensions for two-can systems

27 LE 3A see the clog for details 1 In-lab: Run a single experiment to validate whether an initial volume specified for Can 1 achieves the desired height (volume) in Can 2. 2 In-lab: Use your simulation model of the two-can system to determine the initial volume in Can 1 that will maximize the volume in Can 2 without spilling. 3 Written: Submit a written LE with the following: abstract describing lab and results Describe the results and any discrepancies in your experiments on the maximum-level filling of Can 2. Updated lab procedure, making updates and refinements, especially adding any guidance you think would have improved your results. Compare your values for K1 and K 2 as measured in the lab to those from the theoretical (ideal) calculations. Explain any differences and whether they make sense given the assumptions made in the ideal model.

28 Orifice flow model Orifice flow model Begin by using mass continuity and the steady Bernoulli equation to derive a relation for the velocity of the fluid exiting at control surface 2 (see right). In this case, 0 = t CV ρd V + ρv da CS P 1 ρ + V gh 1 = P 2 ρ + V gh 2 to show that: 2gh V 2 = [1 (A 2 /A 1 ) 2 ] What is A 2? Approximate by using the jet area. If there is an orifice where the jet area is not equal to the orifice area, we use a contraction coefficient.

29 Orifice flow model Orifice contraction coefficients The contraction coefficient is defined by, C c = A jet A o Some common experimentally determined values:

30 Orifice flow model Velocity coefficient Using the contraction coefficient, our model for the exiting mass flowrate is, ṁ = ρq = ρa jet V = ρc c A o C v V ideal where we ve used the velocity coefficient, C v = V actual V ideal = velocity coefficient which accounts for friction effects.

31 Orifice flow model Orifice flowrate model Use the ideal velocity from the Bernoulli equation, 2gh V ideal = V 2 = [1 (A 2 /A 1 ) 2 ] and gauge pressure, P, at the bottom of the can so the flowrate from the orifice is, Q = ṁ ρ = C c C v A o 2P [1 Cc 2 (A 2 /A 1 ) 2 ] ρ = K 2P oa o ρ where ρ is assumed constant and, K o = C c C v [1 C 2 c (A 2 /A 1 ) 2 ]

32 Orifice flow model Ideal orifice flow constant, K ideal In the derived expression for K o, assume ideal contraction and velocity coefficients, C c = 1 and C v = 1 so, K o,ideal = 1/ [ 1 (A orifice /A can ) 2] and the ideal case flowrate is, 2 V 2 Q ideal = K o,ideal A o ρ C = K ρg o,ideala o V = Kideal V ρ A can where the ideal can coefficient is defined by, K ideal = K o,ideal A o 2g/Acan, a value calculated from easily measured dimensions.

33 Orifice flow model Example K ideal calculations Example calculation of ideal flow coefficient calculations given in different units are shown below.