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needed.The conservation of mass applied to two nodes such as A and B provides two equations. Qa = Q1 +Q2 (1) Q1 = Qb + Q3 (2)The conservation of energy applied to the loop A, B, and C provides the third one. h1 + h3 = h2 (3)In this equation, h1, h2, and h3 are the head losses in each pipe.This system of equations cannot be easily solved since the expression for the head losses h asfunction of Q is very complicated. The head loss is given by L V2 h= f
in my energy systems class, a technical elective taken by mechanical engineering students.The course is basically an applied thermodynamics class, covering conventional andunconventional power and refrigeration systems. Over the course of two class periods, I derivedthe equation resulting in the “Hubbert curve,” and then had students use it to model theproduction rates of various fossil fuels, predicting the years of peak production rate anddepletion.The exponential modelThe exponential model assumes that the instantaneous rate of production is proportional to thecumulative production, Q: dQ ? aQ (1
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
. Each question is repeated for clarity and some general commentary concerningthe results of each is given. Page 11.22.3Q-2 How many years have you been attending Northeastern University? Q-2 TOTAL Average 3.84375 Median 3 Maximum 12 Minimum 0 EET Average 4.025 Median 3 Maximum 12 Minimum 0.25 MET Average 4.543478
+ Q = F (t ) (1) dt CHere t denotes time, Q denotes the charge on the capacitor at time t, R denotes the resistance ofthe resistor, C denotes the capacitance of the capacitor, and F(t) denotes the applied voltage. Oneway to view equation (1), which we shall call the mathematician’s point of view, is that thismodel equation is simply a first-order linear ordinary differential equation for which there is awell-established solution procedure. (See, for example, Edwards and Penney4 pp. 46 – 47.) Thispoint of view is satisfying to the mathematician because it demonstrates the utility ofmathematics. However, regarding the RC circuit as a “solved problem” does not
the unit vector nˆ3 .Bodies A and B are slender uniform rods with mass mA and mB , and length LA and LBrespectively. A torsional spring with the spring constant K A acts between body A and the ground.q and q are generalized coordinates. The basis vectors aˆ , bˆ , and nˆ (i = 1,2,3) are fixed on 1 2 i i ibody A, body B and the ground respectively. A force FQ is applied to point Q in the direction bˆ1 . Figure 1: A Simple Case Study – the Rigid Body Double PendulumThough the double pendulum case is simple, it contains basic features that are necessary to Page
"stdio.h"#include "math.h"// Declaration of variablesdouble AR;double Mach = 0.158;double RHO = 0.002378; //Slugs/ft^3double h_initial, h;double delta_elevator, alpha, alpha_initial;double a = 1116.45; //(ft/min);double k;double gust;double CL, CD, CM, CL_alpha, CD_alpha, CM_alpha, CMq, CM_alpha_dot;double CL0, CM0, CD0, CM0_wing, CM0_tail, CM0_fuselage ;double CL_delta_elevator, CM_delta_elevator;//double CDu, CLu;// Forces and Moments:double L, Lx, Lz, M, D, Dx, Dz, thrust;double W, m, g; // Weight of aircraftdouble Ix, Iy, Iz, Ixz; // Moments of Inertiadouble Q, S, b, c; // Dynamic rpessure, Planform Area, span and mean Chorddouble X_cg; // Location of Center of Gravity//velocities, angular rates and anglesdouble Vtrue_initial, Vtrue,ub
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
Figure 2. Asynchronous templateFigure 3 below shows a classic SR latch, the most fundamental memory circuit studied inintroductory digital circuit courses. Figure 4 shows exactly the same circuit, but drawndifferently to emphasize the single feedback path, which holds the one state variable in thecircuit. The circuit remembers which of the two input variables, S or R, was most recently a 1,by recording on the output variable, Q, a 1 if it was S or a 0 if it was R. By realizing that this SRlatch, the most fundamental memory circuit in any static memory device, is actually anasynchronous finite state machine, one realizes the fundamental nature of this topic. S S
Page 11.748.7 f0 1 L Selectivity Factor ? Q ? ? B R CThe quantity B is the 3-dB bandwidth.Assume that the circuit is initially relaxed and excited at t ? 0 at the resonant frequency by aninput sinusoid of the form vi (t ) ? Vi sin 2r f 0tBy either Laplace transform analysis or by solving the differential equation, the voltage vo (t )across the output resistor can be determined as Vi e /r Bt vo (t ) ? Vi sin 2r f 0t / sin 2r f 0 Ç 1 / (1/ 2Q) 2 t
vector of random numbers mean( v ) = 1.044 v := rnorm( n , µ , σ) the vector of random numbers mean( v ) = 1.00111 h := histogram ( intvls , v ) q := pnorm h ( 〈0〉 ) 〈0〉 ( + .05, µ , σ − pnorm h − .05, µ , σ ) h := histogram ( intvls , v) q := pnorm h
. Block diagram of the PXI communication system configured for wireless use.3 Educational Uses of the PXI SystemWhile inexpensive audio-band signal analyzer systems are useful at showing concepts of signalspectrum and I/Q modulation, the benefits of using a professional grade tool for displaying theproperties of real-world signals captured in real-time are difficult to understate. In the followingsections we provide a few examples of the many ways the PXI system can be used to enhanceclassroom teaching. We use the PXI system primarily as a classroom demonstrator as it’s cost(about $39K including educational discount) usually precludes purchasing one for every studentworkbench. However, the system can be used for some student lab exercises with
; L2 = 35 km; Transformer Information L3 = 40 km; T1: 100 MVA, 13.8–230 kV, Ä–Y, L4 = 15 km; X = 0.1 pu L5 = 50 km T2: 200 MVA, 15.0–230 kV, Ä–Y, X = 0.1 pu Power Flow Data Transformer neutrals are solidly grounded Bus 1: Slack bus Bus 2, 3, 4, 5, 6: Load buses System Base Quantities Bus 7: Generator (PV) bus; V = 15 kV; Sbase = 100 MVA (three-phase) P = 180 MW; -87 MVAR < Q < 87 MVAR Vbase = 13.8 kV (line-to-line) at Bus 1 Figure 2. Power Flow Results for Base Case of System Shown in Figure 1
-based demonstrations previously mentioned. This new board interconnects aTexas Instrument (TI) C6711 or C6713 DSP starter kit (DSK) to an Analog Devices (AD)quadrature modulator (AD9857). This modulator is capable of operating at up to 200 millionsamples per second (MS/s), with a resulting carrier or intermediate frequency of up to 80 MHz(i.e., 40% of the system’s sample frequency). An onboard 32-bit direct digital synthesizer (DDS)is used to generate the carrier waveform values. Baseband 14-bit in-phase and quadrature (I/Q)data are presented to the modulator, which can be programmed to interpolate the data at rates of4x to 252x. The AD9857 is interfaced to the DSK using an Altera Cyclone FPGA. The FPGAprovides queuing of the I/Q data, and the
transmitter section Page 11.41.3digitizes an analog input signal into 8-bit parallel values. For BPSK modulation, these values Fig. 2. Data Recovery Circuit Boardare multiplexed into a single output stream. For QPSK modulation, the values are multiplexedinto parallel I and Q output streams of even and odd bits.The receiver section performs the reverse operation. A phase locked loop extracts a clock fromthe incoming digital bit stream(s). For BPSK, a single incoming bit stream is converted to8-bit parallel values. For QPSK, two bit streams are first recombined into a single stream andthen parallelized. The parallel
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. 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
Method, a series of two equations must be used. First, the log mean temperature difference must be found using the equation below: ∆T1 − ∆T2 (1) ∆Tlm = ln(∆T1 / ∆T2 ) where ∆T1 is the temperature difference at the hot fluid inlet, and ∆T2 is the temperature difference at the hot fluid exit. Once this is found, the overall heat transfer coefficient, U, is found using the equation: Q&= U As ∆Tlm (2) where
finite line of charge of length and linear charge density of tL(C/m2) along the z-axis. For a differential element of charge dQ = tLdz, one can computethe electric force exerted on the charge q located at a distance t (cylindrical coordinatest,h,z are used here) from the line using Coulomb’s law asdFy = q dQ cos c/(4rg0R2) (g0 = 8.854 x 10-12 F/m is the free space permittivity) Page 11.523.5where c is the angle from the y-axis and R2 = t2 + z2. With cos c = t/R and integratingto get total force (also utilizing the symmetry of the problem)Fy = 2 Ð q tL t d z /(4rg0R3) where integration is from z = 0 to z = /2.Fy = q tL t I /(2rg0) with
Int Ext Int Ext Int Ext Int Ext Int Ext Int Ext California Polytechnic State University S √ University of California, Irvine Q √ University of California, Riverside Q √ √ University of Central Florida S √ √ University of Colorado at Boulder S √ Colorado State University S √ Columbia University S √ University of Delaware
methods to evaluate physics instruction and assessesundergraduate engineering students’ understanding of certain topics in thermal physics.PER documents students’ difficulties with the conceptions of heat and temperature5, 6, 7.Much of this research suggests that many students hold an intuitive belief about theconceptual relationship of heat (Q) and temperature (T) which might be represented bythis proportionality: Q∝TAs opposed to the established physics principle that heat transfer is proportional to thechange in temperature: Q ∝ ∆TMany research-based conceptual diagnostic surveys are openly available for assessinglearning in physics. The Heat and Temperature and
. Furnace W Win + z 0 Ww Top View Ti Q H Q To η Q purchased
, normal to the free stream flow. Performance characteristicsof airfoils are normally given in terms of the dimensionless lift coefficient and drag coefficient.For a specific angle of attack, α, the lift coefficient CL is a constant, even for different velocitiesV. Table 1 summarizes the relationships between the parameters that determine the lift forces of Page 11.141.7the airfoils. Table 1: Airfoil parameters No. Parameters Equations Notes 1 Lift coefficient C L = Lf (q 0 S) S: planform area of airfoil 2 Drag
Final Examination Question Number Avg. Max.Obj. Learning Objective Method Q-1 Q-2 Q-3 Q-4 Q-5 Q-6 10 Perform kinematic analysis of particle FinalO1 motion for rectilinear and curvilinear motion Exam. 6.79 7.23 7.01 Calculate the position, velocity andO2 acceleration at an instant given the expressions for the displacement 6.79 7.23 7.01 Draw motion diagrams and solve kinematicO3 problems
, Madison, WI. 2003[6] Timpson, W, Tang, R, Borrayo, E & Canetto, S. 147 Practical Tips for Teaching Diversity. Atwood Publishing,Madison, WI. 2003[7] Davis, Howard. The Culture of Building. Oxford University Press, Inc. 1999[8] http://www.seattle-chinese-garden.org/elements/[9] http://www.aviewoncities.com/rome/sanpietro.htm[10] http://www.glnckman.com/pei.htm[11] http://www.hcs.harvard.edu/~hapr/summer97_culture/roots.html[12] http://www.nps.gov/dsc/dsgncnstr/gpsd/ch4.htmlFigure 1: Vatican City and St. Peter’s in Rome Italyhttp://images.google.com/images?q=st.+peter%27s&hl=en&btnG=Search+ImagesFigure 2: Forbidden City in Beijing Chinahttp://images.google.com/images?q=forbidden+city&svnum=10&hl=en&lr=&start=20&sa
stm,3 + Qin = m air h av,4 + m ∀,5h ∀,5 + Q out Page 11.1220.3 2 Figure 1 Handout on Property Definitions Air/Water Vapor Mixtures Basic DefinitionsMole Fraction (yi): This is the mole percent of component i in the mixture. number of moles of component 1 in the mixture N1 y1 = = total number of moles in the mixture N totwith ∑ yi = 1 iPartial Pressure (Pi): This is the apparent pressure
Y m,n-1 X X Figure 1. An interior node at location (m,n) and its neighbors. Applying the heat balance equation around grid point (m,n) we get the followingapproximating algebraic equation, also known as the finite difference equation: ∂T q m−1,n + q m +1,n + q m,n −1 + q m,n +1 + S∆x∆y = ρc∆x∆y (1) ∂t Where
between work and heat is that work can always be converted completely to heat but that only a fraction of work can ever be transferred to heat. For that reason alone, we can argue that of the two forms, work is the “higher value” form of energy transfer. More on this with the second law introduction later. So taking B as Energy in our Reynolds Transport Equation we can write the First law of Thermodynamics so :- Q% / W% - (m% e) in / (m% e) out - E% gen ? E% CV First Law of Thermodynamics Net Energy accumulated in theDirect Energy
, temperature and electrical current and voltage in the PTAC experiments (left). Their results suggested a violation of the First Law of Thermodynamics (right). Page 11.963.6When the PTAC is considered as a control volume, the First Law of Thermodynamics can bewritten for the unit as P - Q% in / Q% out / KE% net ? 0 (2)where Q% is the rate that heat is taken into the PTAC from the air, Q% is the rate that heat is in outrejected from the PTAC, and KE% net represents the net change