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Engr 103

by: rebecca goodrich
rebecca goodrich

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1 semester worth of notes
Engineering Foundations
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This 20 page Bundle was uploaded by rebecca goodrich on Wednesday January 20, 2016. The Bundle belongs to Engr 103-008 at University of Alabama - Tuscaloosa taught by in Fall 2016. Since its upload, it has received 107 views. For similar materials see Engineering Foundations in Engineering and Tech at University of Alabama - Tuscaloosa.

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Date Created: 01/20/16
ENGR 103 Lecture 19 Resistance- the ability of an object to restrict or resist the flow of electrons in an electrical circuit - measured in ohms (Ω) Resistor- the circuit element responsible for restricting the flow of electrons - absorb energy from the electrons and dissipate it as heat - since there is a transfer of energy through a resistor, this means a difference in energy levels occur across the resistor constituting a voltage Ohms’s Law - Math: V = IR= 1 volt is equal to 1 ohm multiplied by the current (I) o Also one ohm= 1 volt divided by 1 Amp (I) - Passive sign convention must always be applied ; there for resistance is always positive - Circuit elements can be connected together in either a series or parallel way Series: elements in a series form a connection from end to end (like a train) - N number of Resistors in series can be combined to create a single equivalent resistance o Req Series= R1+R2+R3+…+RN - The current through each resistor is the Same o Iseries= I1=I2=I3=..,=IN - The total voltage drop across the equivalent resistance is equal to the sum of all individual resistor voltage drops o V series= V1+V2+V3…+VN Parallel Resistance= elements connect across each other like rungs of a ladder. - N number of resistors in parallel can be combined to create a single equivalent resistance o 1/ Req par. = 1/R1+1/R2+…+1/RN - Voltage across each resistor is the same o VPar= V1=V2=V3=…=VN - The total current passint through the equivalent resistance = sum of individual resistor currents o Itotal= I1+I2+I3+…+IN o ENGR 103 Notes 9/21 Heat engines Heat flows from hot to cold 1 law of thermodynamics Q in +out out - W outs useful work - Qoutis wasted energy so why do we have it ? o IT keeps the flow from high energy to low energy; without Qout there is no work, and the system isn’t functional because there is no temperature difference 2nd law of thermodynamics - Energy conversions naturally occur in one direction from higher quantity to lower quantity - 2nd law of thermodynamics Heat engines - a device that converts heat to work Thermal energy reservoir - a body with a very large thermal capacity; it can supply or receive large amount s of thermal energy without experiencing any change in temperature - Common thermal energy reservoirs o large bodies of water Efficiency- an engineering quantity that is used to measure the performance of a system (usually as a percent) n= desiredoutput requiredinput For a heat engine= nth= Wout/Qin Qin = Qout+Wout arrow Wout= Qin-Qout Qin−Qout Qout Nth= = 1- Qin Qin Qout TL For an ideal heat engine: Qin = TH Karno efficiency= maximum efficiency= 1-TL/TH - when using equation we need to have the Temperature converted to K or R - (Absolute temperature systems not relative) (TL= the “sink” the low temperature of the system) (TH= the “source” the high temperature of the system) Why can’t a heat engine be 100% efficient- because the low can never be 0, and the high can never be infinity. Power Power = Energy per unit time = P= E/t - SI unit is 1 Watt = 1 Joule per second - 1W=1J/s Problem 1. 1) Find gravitational Work= Wg = m g z a. M= 75, z =15 ft= 4.572, g= 9.81 m/s^2 b. Wg = 3,363.8 J 2) Find Power output Required: P= E/t a. E=Wg=3,363.8J and t= 30.1s b. Pout= 111.76W 3) Find power input required= Pout=n*Pinput Problem 2 1) n= desired out/ required in= PWout/PQin= 4MW/9MW=0.4444= 44.4% Qin=Qout+Wout t Problem 3 1. Given a. Qin= 25MW b. Source= 400 C c. Qout= 15 MW d. Sink= 30 C 2. Power out ? 3. Solution a. 400C= K+273 i. K= 127 b. 30C=K+273 i. K= -147 c. Pwout=pqin-pqout= 25-15=10…. Engr 103 Day 2 Solve like an engineer - when covering an area (triangle) with 5 “ by 5” tiles when the short sides are 10 in long - consider factors o grout lines- need less tile o splintering/ breaking a tile (order extra) o how to be efficient  put one square down, cut the second in half Documentation Standards for engineering work - identify the problem (or the assignment) at the top of the page - list any assumptions required in the analysis - utilize sketches as appropriate - make sure the problem is worked in a logical sequence that another person can easily follow - draw a box around (or underline) the final answer - Make conclusions and recommendations as appropriate - Double check all calculations - Staple all papers together in the correct order - Fold in half lengthwise from left to right (so the holes are on top) and label with: o First and last name o Date Submitted o Course and Section number o Homework assignment number ENGR 103- Lecture 21 KCL KVL Kirchhoff’s Voltage Law (KVL) - A loop is any closed path in an electrical circuit - The algebraic sum of all voltage drops in any and all loops are zero Resistors in series can be combined into a single equivalent resistance Voltage in series are added together to get total Current in series is constant Kirchhoff’s Current Law - the algebraic sum of all currents entering the node is zero Resistors in parallel can be combined into a single equivalent resistance Voltage in parallel is constant Individual Currents are added together to get total Simple circuits- most of the time- little or no extra info is given ENGR 103­004 Day 27 Dot Product Position vector= (xi+yj+zk) Unit vector= if mag of v = m then (x/m i+y/m j+ z/m k) Force vecor= f(x/m i+y/m j+ z/m k) Dot product= A . B = |A| |B| cos θ = AxBx +AyBy+AzBz 0< θ< 180 0< θ< 90 then A . B = positive  θ = 90 then A . B = 0  90< θ< 180 then A . B = negative  A . B= B . A S (A.B)= (sA).B = A. (sB) (A+B).C= (A.C)+(B.C) - ENGR 103- Lesson 15 Graphical Integration and differentiation Integral- the area under the curve Derivative- slope of a curve Taking an integral is taking the anti derivative of a function; therefore if you derive and then integrate a function you will get what you started. Common relationships: Acceleration is the derivative of velocity, which is the derivative of distance - Therefore distance is the integral of velocity, which is the integral of acceleration Force is the derivative of work; work is the integral of force Power is the derivative of energy, energy is the integral of power Current is the derivative of charge, charge is the derivative of current Integral formulas Integral is area under a curve, so for a square= b*h Integral is area under a curve, so for a triangle=1/2 b*h Integral is area under a curve, so for a trapezoid=1/2 (b1+b2)*h Integral of negative area (meaning the curve is below the x axis - subtract the absolute value of the area of the negative part from the positive part Useful diagrams when it comes to slope: Derivatives Derivatives = slope so use rise over run, x/y For a square, there is a slope of 0 so the derivative =0 For a triangle= slope= rise/run= (y2-y1)/(x2-x1) ENGR 103-004 Lecture 23 Vectors Scalar- a quantity characterized by one number (magnitude) - Ex. Length, mass, time, temp, speed, density, volume, energy, work, resistance Vector- a quantity with both magnitude (size) and direction - Ex. Force, pressure, stress, velocity, acceleration, momentum, impulse - Position vector- location relative to coordinate plane - Displacement- change in location relative to coordinate system - Velocity- speed in a defined direction (speed + direction) - Weight (force of gravity)- mass in a gravitational field - Magnetic Force= strength of field and direction - Vector nomenclature - - - Write the unit vector in the direction of OA = - 2 ways of expressing a vector o Cartesian Form  Vector is described as the sum of its vector components (x,y,z)= <i,j,k>  o Polar Form  Vector is described by is magnitude and angle formed with the x- axis   Vector components   Trig= SohCahToa  Sin θ= opposite side/hypotenuse  Cos θ= adjacent side/hypotenuse  Tan θ= opposite side/hypotenuse  Magnitude of the vector = side (x or y) * trig angle between  X- component= R(magnitude)*cos θ  Y- component= R(magnitude)*sin θ  Conversions between nomenclature - Unit vector= dimensionless vector that has a unit magnitude (vector=1) ENGR 103-004 Lecture 25 Vector Addition and Subtraction Properties - if there are two angles and one is facing the opposite direction, then choose one to be positive and the other two be negative, then add the two together. -Vector subtraction is vector addition, just with one number being negative - vector addition using the cartesion components 1. choose your coordinate plane 2. separate vectors into x and y components 3. add all the x components 4. add all the y- direction components For the parallelogram method (R) the resulting vector= F1 +F2 Engr 103- CH 3 - Glass box unfolded= 6 principal views - Only 3 views are necessary to define an object (usually) - 3 primary (regular) views= top, right, and front views - A sketch or drawing should contain only the views needed to clearly and completely describe the object - “general rule” only use as many views as are necessary- sometimes 3, 2, and even 1 1 angle and first angle systems - 1 angle- look up in book rdo changes how they are oriented - 3 angle projection o you rotate the object to find the view o look up in book Principal Views Placement on a page - front view = bottom left - top view = upper left - right view = bottom right Techniques of lines - Focus on center, visible, hidden, leader, and dimensional lines Precedence of lines - visible line will always take precedence (will always be shown if stacked) - hidden lines will be shown not center (if stacked) - Engr 103 Freebody diagram Newton’s first law - an object in motion stays in motion, an object at rest stays at rest - sum of F in x direction+ sum of F in y direction = 0 - Sum of Fx= 0 - Sum of Fy=0 - Newton’s second law F= ma Newton’s third law - for every reaction there is an equal and opposite reaction - statics- a ball rests on the ground: the ball’s Force of gravity is down at the same magnitude of Force normal up. - Dynamics Free body diagram- - isolate a body/ combination of bodies - identify all external forces acting on the body o no external forces o direction is important - replace physical contacts with equivalent forces ENGR 103 Vectors- Cross product Day 28 Vector Cross product is used to find a) the direction perpendicular to a plane with 2 vectors (direction normal) and b) the moment produced by a force (torque). Cross product between Vectors A and B = A x B= (|A||B|sinθ)u Θ= the angle between lines of vectors A and B. (0 < θ < 180) u= unit vector normal (perpendicular) to the plane containing A and B the cross product results in the creation of a new vector the cross product is also called the vector product A x B= -B x A S(Axb) = (sA)xB=Ax(sb) (A+B)xC=(AxC)+(BxC) Determine the direction perpendicular to a plane (Normal Direction) (Direction C) with C = AxB The unit vector u = C/ |C| ENGR 103- 004 Lecture 24 3D Vectors Position 2D vectors= r= (x -X )iH(yT-y )j=Hr T TH Expressing 3D vectors Cartesian= v= v i+v j+v k x y z Always use right hand coordinate system Polar = v=|v|(cos θxi + cos θyj+ cos θzk) Unit vector= v= (v i+v j+v k)/ 2 2 2 x y z √ vxi +vy j +vzk Direction cosines satisfy cos^2θx+ cos^2θy +cos^2θz=1


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