Paper ID #6095Development of a Senior Mechatronics Course for Mechanical EngineeringStudentProf. John G. Cherng P.E., University of Michigan-Dearborn Received M.S and Ph. D. degrees from Mehanical and Aerospace Engineering, University of Tennessee, Knoxviell, Tennessee in 1975 and 1978 repectively. Teaching undergraduate and graduate courses in heat transfer, instrumentation, mechatronics, turbomachinery, vibrations and acoustics. Research interests inculde turbomachinery, vibration and acoustics, and nano materials.Dr. Ben Q. Li, University of Michigan, DearbornDr. Narasimhamurthi Natarajan, University of Michigan, Dearborn
angular velocity and rotation about a fixedaxis, the military tank shows even more value illustrating rotating reference frames and theconcepts of relative motion. For planar motion, the military tank can be thought of as twodistinct reference frames – the chassis reference frame and the turret/barrel reference frame.Students can relate to these two distinct reference frames because they already know theindependent motion of the turret with respect to the chassis. If the chassis (reference frame N)remains fixed to the ground and does not rotate but the turret (reference frame T) is free to rotate,the relative motion of two points, P and Q on the turret can be discussed (Figure 4). By fixingthe chassis of the tank to the ground, we can establish a
the firstreservoir. A pump with a characteristic curve (increase in head versus the flow rate) W s= 403.33 − 0.127 ⋅ Q + 0.004362 ⋅ Q 2 − 0.00003911⋅ Q 3 for 0 < Q < 150 gpm (7)where Ws is in ft-lbf/lbm when Q is in gpm, is in the system. The system consists of 2000 ft ofschedule 40 nominal 3-inch commercial steel pipe. Minor losses total K = 1000 and C = 0. Findthe flow rate the pump will produce in the system.Solution:The unified approach solution is provided in Figure 3. Much of the contents of the figure arespecifying the system boundary conditions, the physical properties, the friction factorrepresentation, and the units. As with the other examples, the solution is accomplished in thesolve block. Prior to the solve
recognizethe context in which a variable is being used to identify what the variable represents. However, Page 22.917.5to minimize confusion, adjustments were made where possible. Thermodynamics uses thevariable, h, to represent specific enthalpy while fluid mechanics uses the same variable torepresent vertical height. For clarity, height is represented by the non-italicized, lower-case “h”while specific enthalpy is represented by the italicized, lower case “h.” Other uses of h in fluidmechanics such as pump head (hp), turbine head (ht), and head loss (hL) incorporate a subscript,thus distinguishing these variables from others.The variable, Q
S1 =Entropy out. Ta = Ambient Temperature Figure 1. Simple pipe model for second law consideration. Application of the energy accounting equation as described by Foley [3], gives : Q& 1 / W& − ( m& e ) in / ( m& e ) out − E& gen ? E& CV . (1) Which for a steady flow, no work situation simplifies to : Q&1 ? m& .C p (T1 / T2 ) . (2) This is usually as far as we need to go in most practical problems, and hence any issues of entropy do not arise. However there is much to be learned by considering the entropy
: Page 12.327.5 d2x dx k1 3 Q - 2 dy 0 - y 0 2 x - x ? sin(y t ) dt 2 dt m m (4)where the spring force is kx - k1 x , m is the mass, d is the damping ratio of the system, y 0 3 m ) is the undamped linear natural frequency and Q and y are the amplitude and(given by kfrequency, respectively, of the external harmonic excitation. Setting k1 ? f k and introducing the
solver windowOne Dimensional Heat Conduction in FinsHeat transfer analysis of heat conduction in straight fins of uniform cross-sectional area isincluded in heat transfer text books. The analysis results in formulas for temperaturedistribution, the rate of heat exchange with the surrounding environment, and the fin efficiency.Boundary conditions used in the analysis will influence the resulting equations. For example, foran infinitely long fin of a uniform cross-sectional area the temperature distribution and heattransfer are given by the following equations. T − T∞ θ= = e − mx To − T∞ (1) q = hPkA(T − To
temperature, and that all of theevaporated water will be condensed and collected. This model will produce the equation: m% ? Q% / h fg where: Q% is the rate of solar energy incident on the “window” of the system. ( Q% = IA where I is the solar energy intensity (kW/m2) and A is the “window” area of your system (m2). hfg is heat of vaporization at water temperature. m% is the production rate of distilled water.To determine how much water your system can produce in 6 hours using this model, you willneed to research hourly values for solar energy intensity, I, (incident on a tilted surface?) inNew London in April and estimate the temperature of
tolengthen the mask and a carriage return before and after to increase the height of the mask.Example 1 also used “explicit,ALL” for several calculations. In this case, the definingequation was used as a mask for the “explicit,ALL” words. Several problems can be seen inthe hf calculation. The numerical value for V was not substituted into the equation. Instead,the previous equation defining V in terms of Q and d was inserted and the numerical valuesfor Q and d displayed. In addition, the Darcy friction value previously calculated is notrounded in the “explicit,ALL” equation. Attempts to resolve these problems were notsuccessful.Solving Equations Analytically. Equations can be solved symbolically (analytically) using[cntl .]. (The two keystrokes in
their writingabilities and previous experiences in ME 342W. Survey questions # 1 and #2 were open-ended questions. For survey question # 1, senior-level ME 440 students were asked: “Of all the engineering classes that you have taken at thisuniversity, which do you feel was best at helping you with your technical writing skills? Brieflydescribe why you selected this course.” Survey question # 2 followed up by asking students tothen specify their choice of second best course which fit this criteria. Table 1 summarizes theanswers to Question # 1 (Q#1) and Question # 2 (Q#2), by tallying all of the courses mentioned,whether by course code, explicit course title or some combination of each.Table 1: Student responses to 5 point Likert-scale
. Page 14.65.10 gal ( Flowrate : Start the flow after 100 min ) Q( t) 2∧ if t ≅ 100min min 0 otherwise Qgen1( T) 6600∧ W if T > KfromF( 130) ∧ K ( Heater ON if the temp. is below 130F ) 0 otherwise Qgen2( T) 4400∧ W if T > KfromF( 130) ∧ K 0 otherwiseUsing an energy balance to solve for the time rate of change (Slopes) of the temperatures : 1 s τ∧ C ∧ V1∧ τ∧ Cp∧ ∗ Tin / Ta+ ∧ Q( t∧ s
enthalpy profile basedon inlet boundary conditions (inlet flow and temperature), a nonuniform axial powershape, and a specified operating pressure. The first law is defined as follows1: dEcv V2 V2 Q cv W cv m i hi i gzi m e he e gz e (1) dt i 2 e 2 The pressure drop is based on the equation resulting from a mechanical energy balancefor an incompressible fluid2: pi Vi 2 pe Ve2 zi h ze hTurbine hL (2
𝑇𝑇133 Consider any arbitrary process involving external heat transfer to or from any CV as shown in figure 1.134 Multiple heat sources and/or sinks might exist and external work may/may not be done on/by the CV. If135 q′′ External is the instantaneous heat flux at any point on the surface of the CV, then the net external heat136 transfer is given by:137138 Q External = ∫ ∫ q′′ External dA dt = ∫ ∫ dQ External (6) t A t A139140 The external heat flux is integrated over the surface area of the CV, denoted by A. All of the external heat141 transfer must occur across the boundary of the CV. Therefore
))); % Angle of AoA to X axis Page 12.242.11% CALCULATE POSITION OF C AT ALL STEPSfor q=1:(steps+1) theta2=theta2min+(q-1)*(range)/steps; % Current theta2% CALL FUNCTION TO GET ESTIMATES OF THETAS 3 AND 4 thetabars=thetas(theta1,theta2,theta3,theta4,r1,r2,r3,r4); theta3=thetabars(1); % Set current theta3 to Newton-Raphsonestimate theta4=thetabars(2); % Set current theta4 to Newton-Raphsonestimate thth(q)=theta4; Xc(q)=r2*cos(theta2)+r5*cos(theta3+psic); % Put current Xc inmatrix Yc(q)=r2*sin(theta2)+r5*sin(theta3+psic); % Put current iYc inmatrix Xb(q)=r1*cos(theta1)+r4*cos(theta4); % Put current Xb inmatrix Yb(q)=r1*sin(theta1)+r4
_________________________ ________________ e. Q _________________________ ________________ f. h _________________________ ________________ g. v mP _________________________ ________________ h. q _________________________ ________________ B. The letter vee (upper and lower case) is used to represent several quantities in this course. List these quantities and indicate which vee is used for each. Page 22.219.10 Figure 3. Quiz to assess students
principle. To do so let Brepresent entropy. The second law of thermodynamics states that entropy can be created withinthe control volume by dissipative processes such as friction. Entropy can cross the controlsurface by convection or without bulk motion as a consequence of heat. The terms of the generalbalance principle applied to entropy are the following. $ $ ? D $ ? / t s V © dA, $ ? / q © dA BCV ? Ð t s dV, B- Ð dV, Bc Ð r Bnc Ð (11) CV CV T CS CS T $ is the dissipation rate, T is the absolute temperature, and q isHere s is the entropy
-equilibrium equation of the link about its other end. E In the hypothetical sub-mechanism, Figure 3, this means that B j is determined from the moment-equilibrium equation of link i about H E E E E ÂM js h ? Ri · B j - Rgi · fi - qi ? Ri e jsi · B j e j - Rgi e jsi · fi e ji i - q i ? Ri B j sin*s j / s i + - Rgi f i sin*i i / s i + - q i ? 0 (15) Therefore Page 13.101.8
of the system and the surroundings. Based on this study, a change seems warranted forthe first and second laws analysis relating to the electrical work.BackgroundA typical thermodynamic problem is solved by the traditional methodology of analysis thatcombines the first and second laws of thermodynamics. The first law describes the principle ofconservation of energy and is defined as the net energy transfer into the system being equal tothe change of the total energy in the system. Although there may be minor differences indescribing the principle by different authors, the first law of thermodynamics is usually given ina general equation form as Q – W = ΔE or Qin - Qout + Win – Wout = ΔE for closed systems where
performance in a subsequent course (e.g., intermediate thermodynamics or heat transfer).ConstraintsAs an important and fundamental restriction to this project, students are prohibited fromapproaching the problem by analyzing the refrigeration system, i.e., the refrigerant flow. Instead,students must explore the meaning of the quantities involved, particularly heat and work, byexamining the refrigerator compartments. In other words, instead of investigating where theenergy is going to, students should question where it is coming from.It should be noted that the actual COP may also be defined in terms of the rate of energy transfer: Q˙ c COP actual
Experiences for All Students Rotterdam, The Netherlands: Sense Publishers2 Patton, M. Q.(1990). Qualitative evaluation and research methods. Newbury Park, CA: Sage Publications3 Strauss, A., & Corbin, J. (1998). Basics of qualitative research: Techniques and procedures for developing groundedtheory (2nd ed.). Thousand Oaks, CA: Sage Publications Page 22.1519.21Memorandum To: Thermodynamic Analysis Team From: Sharon Parker, Cal Poly Supermileage Team Date: June 30, 2010 Re: Engine Performance AnalysisEach year, Cal Poly’s Supermileage Vehicle Team competes in the Shell Eco-Marathon
democratization of manufacturingand programmable electronics. The design experience in the class provides students an outlet forexercising their creativity at the highest level of Bloom’s taxonomy. ∗A web-based, private beta version was provided to students in the Fall 2015 to assist them with their projects.Figure 5: A few pictures of the projects from the Final Robot Design Project Demo day, Fall 2015;see complete gallery of pictures at https://goo.gl/photos/WhYmy4kxFoStwNQz7References1 Shen, Q., Al-Smadi, Y. M., Martin, P. J., Russell, K., and Sodhi, R. S., 2009, “An extension of mechanism design optimization for motion generation”, Mechanism and machine theory, 44(9), pp. 1759–1767.2 The National Academy of Engineering, 2005, Educating the
. Using equation (1) the property B can then be replaced with energy in its variousguises and the accounting of energy through the control volume becomes :DU CV •U int ernal ? - Â Q% net _ in / Â W% net _ out - Â (m% e) net _ in …….(2) Dt •t By treating the surroundings (i.e. Everything else) as just another control volume onecan apply the same accounting exercise to the surroundings:DU SUR •U int ernal ? - Â Q% net _ in / Â W% net _ out - Â (m% e) net _ in …….(3) Dt •tFinally, by adding the two rates of change, equation (2) and (3), it was hypothesized byJoule and Clausius that, in the absence of Nuclear reactions, the sum should always equalzero. That is, the energy is conserved. i.e.DU SUR DU CV
? ? Cv.TE − p E .v E ? u E − p E .v E (10) mReturning now to our control volume A,DU A ? ∑ Q A / ∑ W A − m I .(u I − p I .v I ) / m E .(u E − p E .v E ) (11)Again it is interesting to note that the convected terms comprise properties that arecompletely defined outside of the control volume A which we are ultimately analyzing.At this point convention is that there is ‘convenience’ to be had by grouping the terms in thebrackets into one term. i.e. the property enthalpy (h) is justified.Hence,DU A ? ∑ Q A / ∑ W A − m I .(hI ) / m E .(hE ) (12) This convenience however comes at a price. First, internal energy is a property that wasclosely related to pressure and temperature and fundamental in our
the heat exchanger. Location 4 is the cold water leaving the heat exchanger. Page 15.57.4The water in the tank is at temperature T1. The hot water enters the heat exchanger at thistemperature.From the first law of thermodynamics, the rate of heat removed from the hot water, Q, isequal to the rate of decrease of the internal energy of the hot water. d T1 Q ? / τ cV (1 ) dtwhere: Q = the rate of heat transfer ρ = density of the water in the tank c = specific heat of the water in the tank V = volume of water in the tank
experimental design, data collection, and data analysis; (q) acompetence in the use of computational tools; (r) a knowledge of chemistry; and (s)knowledge of calculus-based physics.Outcome can be assessed at the “program level” [2] using standardized testing such asfundamental of engineering exam, or it can be pursued at the “course level”. To ensure theteaching quality of the outcome based mechanical engineering program, systematicassessment of engineering curriculum is needed. More importantly, faculty needs to improveteaching techniques so that the student learning of each course can be measured againstprogram outcomes. In the past eight years, the current authors have developed a technique to
language generationsystem, and the PyGame 2D graphics engine. Only data for the second problem archetype, theideal gas, piston-cylinder problem, is shown for brevity.In figure 5 the input file is shown that includes most of the necessary data that defines thearchetype. The parameter “P-2digpc” in the third line is a parameter that tells the system that thetype of problem to be generated is of the ideal gas, piston cylinder archetype and to includeseveral default values for parameter ranges. ## Input File: ## 2D ideal gas piston-cylinder archetype "P-2digpc" ## Parameter List P1,V1,T1,P2,V2,T2,m,rho1,rho2,W,Q,U1,U2,v1,v2, w,q,u1,u2,deltaU,deltau,deltaT,deltaP
duration of time devoted to the students’presentations of the four different product ideas, as well as the free-form question-and-answersessions that followed each presentation. The quantity and distribution of verbal participationfrom individuals during each Q&A discussion was also calculated. Although limited in scope,results of this first study suggest a correlation between the duration of Q&A sessions,distribution of communication responsibility among individual team members, and final productselection. Furthermore, a total of 23 out of 24 students (96%) on Team A and 20 out of 24students (83%) on Team B asked and/or answered questions during the discussions throughoutthe meeting, suggesting that the stress and emotion of the high
. closed systems, evaluation of properties,state principle, internal energy vs. enthalpy, transient vs. steady state, realizing entropy is athermodynamic property, reversibility, and correct application of process equations vs. rateequations. A few examples are discussed here with specific strategies to promote studentlearning.Students often struggle to distinguish between isothermal and adiabatic processes. Students findit counter-intuitive that a system can absorb energy by a heat transfer, Q without a change intemperature during a process. In many cases the temperature increases with heating, but if thesystem undergoes a phase change at constant pressure the temperature remains constant. Aclassic example is boiling water trapped in a piston
∀ ∀0 2u > x > ♣and represents a concentrated load of magnitude one 2 at x ? 0 . Macauley represented the Diracdelta function at x ? a as a bracket with a subscript of –1. x / a /1 ? φ ∗x / a+ ? 0 x ≈ a (2) Page 15.549.3The unit doublet, or concentrated couple or moment can be represented graphically as follows: q(x,u) 1 u2 0 u 2u x 1 / u2 Figure 2: Graphical display2 of the unit doublet function at x=0.The unit doublet function at x ? 0 is defined as: ∀0 / ♣ ∞ x ∞ 0
filtration through the layer of particulates isdG P − Pi =k a (3) dτ PaThe harmonic fluctuation of the cylinder is described by the following expressions = A Sinωτ (4)andd 2s 2 = − Aω 2 Sinωτ (5)dτIntroducing the following variables ( Pa − Pi ) ⋅ F Pa G RTq= (6) α= (7) v = = (8) Pa m hF Pa Page 12.1489.5one has instead of equations (1), (2), and (3) d 2h d 2s + α q = − −g (9) dτ 2 dτ 2 (1 - q