) Figure 4: PWM generation technique for CHB.Control Strategy:Upon successful development of the design and modulation of all power converters of SST, it isimportant to develop a control model to provide a reliable, resilient, and efficient SST for thegrid application. The high and low voltage balancing operation across the DC bus capacitor hasbeen very challenging for SST configuration especially for parallel-connected DAB and multi-port application [15]. In this paper, a d-q vector control-based DC voltage and load voltagebalancing technique for both rectifier and inverter stages are presented. A simple classic d-qvector controller is applied in this system.Rectifier StageThe high voltage grid is connected to the front end of the SST that is
, and their ability tocontrol, confine, and enhance light-matter interactions at the nanoscale. Results & DiscussionNanostructures, such as dolmens and oligomers of nanoparticles as well as several plasmonicmetasurface structures, have shown to exhibit Fano resonances in the optical domain. An example ofsuch nanostructure is the ultrathin Babinet-inverted metasurface made up of asymmetric split-ringapertures fabricated in a metal plate, which produces high-quality-factor (high-Q) Fano resonances. TheFano resonances originate from the interaction of bright modes and dark modes that give rise toasymmetric linewidth profiles in the scattering parameters, such as absorption or reflection spectra. All
b0 ? 1 for b 0 (9) b0 ? 0 for b>0 (10)Referring to the beam ab in Fig. 1, we may, for illustrative purposes, employ the rudiments ofsingularity functions and observe the defined sign conventions for beams to write the loadingfunction q, the shear force V, and the bending moment M for of this beam as follows:6-8 q ? Va > x @ /1 - M a > x @ /2 / P > x / xP @ /1 - K > x / xK @ /2 w1 / w0 > x / xw @ 0 / > x / xw @1 (11) L / xw
Tpolycarb (x) = 13320 C m x + 122 o C o LStudents will obtain a different temperature profile for each material they study and canthen use Fourier’s Law to determine the conduction heat transfer rate. dT kq''x = 1k = (Tw,s 1 Tp,s ) dx LBy providing the thermal conductivity, k, or the heat flux, q''x , it is possible to calculatethe other parameter. Alternatively, students could determine heat flux from the steadystate heat generation experiment outlined below. 0.21W mKq''x = (122 o C 1 88.8 o C) = 697 W 2 0.01m mOne Dimensional Conduction Through Composite Systems:Steady state heat conduction
; .:4 } 1996 ASEE Annual Conference Proceedings ‘Q,.,pllll’;? .— . Figure 1. Graph of F. Since formula (2) requires multiple uses of [t], which in turn involves (l), approximations would beuseful to avoid cumbersome calculations. As shown in Figure 1, the function F has a parabola-like graph.The fact that $’ mimics a parabola so much inspired us to find quadratic functions to approximately match it.We replace [z] with z in (2) to obtain the quadratic function ql(~) . *so that ql (x) agrees with F(z) for integral x. We note that ql is a lower bound of F. In order to find anupper bound, we shift the
and free combined)dof(fixdof)=1;free = find(dof==0);We reduce the structure force vector F and the structure stiffness matrix K to form the correspon-ding quantities Ffree and KFree , solve the set of linear equations for the vector q free of structure dis-placements and finally add the prescribed zero displacements to the solution vector using thefollowing statements.%initialize displacement vectorq = zeros(dim,1);%reduce stiffness matrix (eliminate rows and columns representing fixed dofs)Kfree = K(free,free);%reduce force vectorFfree = F(free);%solve equationsqfree = Kfree \ Ffree;%include fixed degrees of freedom in displacement vectorq(free) = qfree
from the previous semester completed the survey. While the assignment has been run foryears, the data was from the last year that the assignment was performed. The following showsthe survey questions and the students’ responses: 1. Q: The project was interesting? Student Response: Likert Scale 6.3/7.0 2. Q: The level of complexity of the assignment was adequate for this course? Student Response: Likert Scale 6.1/7.0 3. Q: You feel that this assignment should be included in this course for future students? Student Response: Likert Scale 6.3/7.0 4. Q: What changes would you make to the assignment? Summary of student responses: Most often stated was that there should not be any changes. Other suggestions
AnalysisFluid flow characterization in most hydraulic systems can be derived from the Navier Stokesequation ( V ) 0 …. (1)tWhere, = density, V = flow velocity vector divergence operator of a general flow field.Considering energy input and energy loss in a pump, equation (1) for one dimensional fluid flowsimplifies as Bernoulli’s equationP1 V12 P V2 z1 hi hL 2 z2 2 …. (2) 2g 2gwhere hi and hL are input energy head and head loss between inlet and exit of a pump.Utilizing this at pump inlet and exit, the efficiency of the pump can be expressed in terms ofoutput pressure P, flow rate Q, torque and angular velocity as PQe F1 ( P, Q
. Page 11.395.4The other equation that are used in this VI are equations to calculate the RMS values of voltageand current, the maximum value of current from voltage and impedance information, and the realpower (P), reactive power (Q), and total power (S). Vm Im V m ∠θ vV = ...... I = ......I m = ............(6) 2 2 Z∠θ zP = V I cos θ .........Q = V I sin θ .............S = P + jQ................(7)θ = θ v − θ i ....................................................................(8)The front panel of this VI consists of (a) the user inputs (controls) such as maximum voltage,angle of the voltage, impedance, angle of the
Images to Strengthen Learning 2ndedition, Corwin Press, CA[4] LaFosse, Michael (2009), Money Origami Kit, Tuttle Publishing Page 24.126.20AppendixDemographics Figure A.1 Race Distribution Figure A.2 Age Distribution Page 24.126.21 Figure A.3 Gender DistributionQuestionnaire Q: I feel developing intuition is important Figure A.4 Student Feedback on developing intuition Q: I feel visualizing algorithms is important for my learning
x2 x2/2 2. The integral of (1/x) dx is: x 2x ln x x2 3. The integral of 6x2 dx is: 2x3 6x3 6x 3x2 4. The derivative of 2x2 is: 2x dx 4x dx 4x3 dx x3 dx 5. The derivative of 2 ln x is: 2x dx (2/x) dx (1/x) dx 2 ln x dx 6. The integral of x2 dx from x=1 to x=3 is: 26/3 -26/3 26 none of these 7. Which of the following is a statement of the first law of thermodynamics? Q = mc(ΔT) W = ∫ p dV Q - W
of the abovemethod, we can say that it did appear to pass this test as few students complained that theevaluations were unfair.IV. An Inventory Policy ProblemThe second problem was an inventory policy problem where the students first saw the problemstory and action form shown in Figure 2. The story described a continuous review inventorysituation that included ordering costs, holding costs and shortage costs with stochastic demand Page 6.1120.3and lead times. The problem required students to determine a reorder quantity and reorder point(Q, R) policy. Demand was required to be estimated from samples acquired at a cost. Proceedings of the
528 mV Figure 8 – Waveform of vout3 C. The Butterworth Second Order Active High Pass FilterTo further prevent interference from other light sources, a Second Order Active High Pass filterwith a Q value of 6 is used. Since this filter has a high Q, around the center frequency it behaves Page 5.355.8very similarly to a band pass filter. The normalized frequency response of this filter is in Figure 9below. Magnitude 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0
. Page 2.127.3 Frequency Resolution: The FFT finds coefficients of the harmonic signal at anincremental frequency,∆f, which is determined by the data sampling rate divided by the numberof points acquired. This ∆f can be interpreted as the width of a frequency bin that is centered onfcent. The smaller the width of the bin, the higher the resolution of frequency. It is important tonote that changing the sample size or sampling rate will change the resolution, and that changingsampling rate alone modifies both the Nyquist cut-off frequency and the center frequency. F R E Q U E N C Y R E S O L U TIO N (H Z) 10.00 1.0 0
. a2 w2 50 D A dimensions in mm z 100 C 100 w1 q y
students’reports.q “It was relatively simple to understand the algorithm in the Maple code; however, it was a bit troublesome to convert the code line by line. We had to reevaluate our thinking process and convert the code block by block. The conversion of an idea was much easier to implement than converting a line of code. Once we understood the methodology behind a chunk of code, it was relatively straightforward to express that idea in Mathematica’s syntax…. The reason why this process was straightforward was due to our knowledge of Mathematica on an intermediate level.”q “The Maple code was not very helpful as a template. Basically, the Maple code gave a general idea of what each objective required. Some commands in Maple have
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2 ρ 2where Pi is the pressure, Vi is the velocity, αi is the kinetic energy correction factor and Zi is theelevation at point i, ρ is the density, g is acceleration due to gravity, WP is the pump work, hfp isthe frictional losses associated with the pump, and hf are the frictional losses due to skin friction,sudden contractions and expansions, and pipe fittings. Students don’t readily connect equation(1) with the energy balance they see in thermodynamics. They are most familiar with the energybalance for a single input, single output system operating at steady state in the form [Smith et al.,2005] ∆H + ∆KE + ∆PE = Q + W (2)where H is enthalpy, KE
multiple times to students that valves areabout the only thing that can be adjusted in a chemical process, and that adjustments in Page 26.233.2temperature, pressure, and composition, for example, all occur by turning a valve.Heat Exchangers Zoned Analysis Required. In many organic chemical processes, a subcooled liquid streammust be vaporized and superheated for a vapor-phase, catalytic reaction. A typical heat source issteam condensing at constant temperature from saturated vapor to saturated liquid. Anapproximate T-Q diagram is shown in Figure 1. The solid lines represent the actual situation.The dashed line represents the situation often
) qe Q QbCeAgain, the students can use Excel to plot equation (4) and determine the parameters. Theconstant Q represents the maximum adsorbate that can be adsorbed onto the surface, andb is the isotherm constant. If b is large, and the quantity Q b is much larger than one, the € The implications can be discussed in class.isotherm is favorable.Modeling the adsorption kinetics is more complicated, and requires a differentialequation. This is probably not too difficult for first year engineering students, but may befor students with less mathematical background. To assist nonengineering studentsunderstand a first order process, it is helpful to first show them some examples: flow offluid from a tank, or the braking of an
respective crystallization Page 9.683.3temperature, Tpeak, crystallization increases from 503.4 ºC to 556.3 ºC.Proceedings of the 2004 American Society for Engineering Education Annual Conference & Exposition Copyright © 2004, American Society for Engineering EducationThe Kissinger model describes crystallization kinetics during continuous heating and is given by: Ã h Ô /Q lnÄÄ 2 Õ? ÕÕ RT - constant ÄT
Engineering Education Annual Conference & Exposition Copyright © 2003, American Society for Engineering EducationOutcome links represent an estimate of the Course Outcome’s importance to that ProgramOutcome. In a similar manner, Program Outcome – Program Objective links are an estimate ofthe Program Outcome’s importance to a Program Objective. The curriculum emphasis for eachProgram Outcome can be estimated using the expression in Figure 2. for (p = 0; p < M; p++) // each program outcome {programOutcome [p] = 0; for (q = 0; q < L; q++) // each course {for (r = 0; r < K; r++) // each course outcome {ProgramOutcome [p
the following is a statement of the first law of thermodynamics? Q = mc(∆T) W = ∫ p dV Q - W = ∆U all of these 8. Which of the following is a definition of enthalpy? Q/T Q–W mcp 7 U + pV 9. The typical electric power plant relies on which energy conversion cycle? Brayton Otto Rankine Diesel 10. Which of the following expresses the second law of thermodynamics?: (a) Work cannot be completely converted into heat. (b) Heat cannot be completely converted into work. (c) Energy can be neither created nor destroyed in a system. (d) The rate of mass flowing into a system equals the mass flow rate
biomedical imagingapplications is Alzheimer's disease classification. The proposed paper talks about the feature extraction of the MRImages in Alzheimer’s disease. Here we introduce the concept of Non-Separable Wavelet Transform which can beused as an image segmentation technique. We use the Q-shift 10 length filter bank combination which reveals moreinformation in the low frequency signals, thus segmenting the image to highlight the portion of concern. Then weuse the area technique to classify the image. Finally, a GUI is used to show the results based on the proposedmethod.Keywords: - MR Images, DWT, 2DT-CWT, Alzheimer’s disease, Segmented Image.Introduction:Diagnosing patients affected by Alzheimer’s disease at an early stage plays a crucial role
), and self-identity (shyness and confidence toengage in class). In addition, foreign students entering a Mechanics of Materials course havepoor foundational knowledge in subjects like Physics and/or Statics and encounter difficultieslearning in a virtual course delivery model (82% asynchronous and 18% live Q&A sessions).The combination of these challenges exacts a tremendous obstacle to student learning, studentretention, and student persistence.The developed instructional approach uses reduced question sets for homework assignments,which aim to improve the lower-level questioning found in Bloom’s Taxonomy and Costa’sLevels of Thinking and reducing the higher-level questioning. This model helps to decrease thecognitive load that is placed
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
: function MEAN EXAM SCORE BY QUINTILE(exam, quintile) 2: points = 0 3: max points = 0 4: for q in get questions(exam) do 5: mean = question score by quintile(q, quintile) 6: points = points + mean 7: max points = max points + get max points(q) 8: end for 9: return points/max points10: end functionWe then define the unfairness of a collection of exams for a given quintile as the standarddeviation of the expected scores for that quintile across all of the exams.To be clear, a collection of exams is not necessarily unfair if there is high variance in the studentscores when students are given different exams from this collection. We expect such a variance inscore resulting from a variance in student abilities. We
for a fixed number of stages in the column.In addition, this paper provides the VBA code to find real roots of any cubic equation: such afunction can also be useful in other Excel applications.The inputs to the spreadsheet are the x-y equilibrium data, the feed composition and “q-value”(usually, the liquid mole fraction of the feed: formally defined as the heat required to vaporizeone mole of feed at the entering conditions divided by the molar latent heat of vaporization of thefeed8 (equation 11.4-12, page 710)), the desired tops and bottoms purity, the reflux ratio, and theMurphree efficiency. The outputs are the location of the azeotrope (if present), the intersectionpoint of the feed line with the equilibrium curve, the required number of
included theselabs and design project. Each lab was then granted a score (0-3 or 0-4) in each evaluativecategory depending on the lab’s level of adoption of that category. The two researchers thendiscussed and reconciled their results into one final result set, which is what is presented in theresults that follows.Results and DiscussionAfter the final agreement was met on the scores, a summary of the overall scores wasgenerated, as seen in Table 1. Table 1: Summary Statistics from Final Data SetIt can be seen that every lab failed to attain a majority of total points with the exception ofthe Software Design Project (SDP), which ranked first in most categories. The Quality andProductivity (Q&P) lab was also much higher
-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