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1-January/1-13_Notes.md

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+Good sense of humor, or so it appears. Nice and talkative.
+
+Adjunct professor, so won’t always be here
+
+Fits like Cliver fairly well, likes to get on a little bit of a tangent.
+
+Knows a bit about legal-ese
+
+“Bring you a flavor of reality. What you’re really gonna run across with transformers, motors, and generators. I will use the book as a basis. I will cover some things not in the book, but they will be used extensively in the industry.”
+
+We’ll talk about some things that are skimmed in the book, but the professor believes to be important.
+
+You should know enough jargon to understand whats going on, and you’ll know what questions you need to ask to understand.
+
+Should be an active classroom, he doesn’t want yes-men and bobble-heads.
+
+If you can understand the homework, you can understand the class.
+
+The exams can have a 1-page cheat sheet, both sides. Don’t memorize, understand principles. Those who are naught but function repeaters will not do well in this class.
+
+Lecture is not all-encompassing. They are mainly talking points. If you can’t find them… he’s not extremely technologically literate.
+
+“There is no risk of getting the wrong version, because the author is dead.” - Professor Garrett
+
+
+
+2 exams: 60%
+
+Homework: 10% completion grade
+
+final: 30%
+
+***If you have a 90% exam average, you’re exempt from the final***
+
+Fun Fact: Licensing for Professional Engineer is $450 per state every 3 years
+
+Most in class examples are from his wastewater processing. Will also talk about receiving from utilities.
+
+
+
+There’s crossover from AC circuits.
+
+Most motors are induction motors. Most generators are synchronous generators.
+
+DC motors are making a comeback, because computers.
+
+Gonna skim over power systems, sprinkle it out through the semester.
+
+
+
+# Actual Notes
+
+We will use conventional current flow. Please.
+
+Current goes from + (source) to + (load). Thats great for DC, but what about AC?
+
+We use + and - to talk about conventional current at a 0^o^ for + and - for 180^o^.
+
+Remember phasor diagrams? God I hope so.
+
+
+
+We are not going to talk about a combined AC/DC sources. We can do that, but we won’t deal with it because that just makes things needlessly complex. Everything is symmetric around zero.
+
+
+
+e/E and v/V both mean voltage. Don’t worry about it.
+
+e/v/i is instantaneous. E~m~/V~m~/I~m~ is peak
+
+$cos(2\pi ft+ \theta)$ is our standard for both current and voltage. $\theta$ is in radians.
+
+Frequency in our walls is 60Hz. Thus, most of our problems will be in 60Hz.

1-January/1-15_Notes.md

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+# Review
+
+Lets assume we found a house on an asland and we checked the voltage in an outlet and found the following.
+$$
+E_m=2V
+\\
+I_m=1.5A
+\\
+f=60Hz
+\\
+\theta = {\pi \over 3}rad
+$$
+What is the period?
+
+$1 \over 60$ seconds, or $0.1 \overline{66}$ seconds.
+
+
+
+Conversion from cos to sin
+
+cos -> sin 
+
++90^o^
+
+sin -> cos
+
+-90^o^
+
+
+
+## Effective Value (RMS)
+
+$V_{eff}=V_{RMS}=E_m* {\sqrt{2}\over 2}= E_m*0.707$
+
+
+
+We will be working with RMS values in this class. If you ever need to work with max values, you will need to calculate it. But we’ll probably just work with RMS in general.
+
+
+
+## Phasor shit
+
+Usually used for voltage (and occasionally power)
+
+$120<20 ^{\circ}$
+
+### Leading vs lagging
+
+Everything is a voltage source. The current is a result. Thus, if it is leading, the current is ahead of the voltage. If it is lagging, the current is behind the voltage.
+
+
+
+3 phase means 3 voltage sources $120^{\circ}$ apart. This is a fixed constant. This means that all the loads should be the same values.
+
+
+
+If the current relative to the voltage is a capacitive current, it leads by $90^{\circ}$. Inductive lags by the same
+
+***Inductive circuits lag. Capacitive circuits lead.***
+
+## Harmonics
+
+Fundamental harmonic is the frequency of the system.
+
+each harmonic is x times the fundamental frequency.
+
+## Impedance
+
+$Z=R-j(2\pi fL<90^{\circ}-{1\over 2\pi fC}<(-90^{\circ}))$ or in phasor form
+
+Cap leads by $90^{\circ}$
+
+inductor lags by $90^{\circ}$
+
+## Right hand rule
+
+$$
+E=N{\Delta \phi \over \Delta t}
+\\
+where\\
+E=Voltage\\
+N=turns\\
+\Delta\phi=\Delta flux (webers)\\
+\Delta t=\Delta time
+$$
+
+## Examples
+
+### Example 1
+
+$V(t)=12cos(60t+45^{\circ})$
+
+phasor $\ne 12<45^{\circ}$
+
+phasor = $6\sqrt{2}<45^{\circ}$
+
+
+
+# Chapter 7
+
+Power and wattage are the same. Positive watts is power in, negative watts is power out. 
+
+True power is a little more difficult.
+
+$P=IV(cos(\theta_V-\theta_I))$
+
+Circulating power, or reactive power is measured in voltamps reactive (VAR)
+
+$Q=VI(sin(\theta_V-\theta_I))$
+
+Complex power is the total power, measured in voltamps (VA)
+
+$S=|V||I|$
+
+Ratio of active power to apparent power is the power factor.
+
+$pf={P\over S}=cos(\theta_V-\theta_I)$
+
+$745.7W =1hp$
+

1-January/1-27_Notes.md

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+# Three phase generator
+
+Voltage induced in a 3 phase generator is
+$$
+E=BLV\\
+E = V_{instant} (V)\\
+B= flux_{constant}\\
+L= length_{conductor}\\
+V= speed_{poles}
+$$
+length and speed are **GENERALLY **in SI.
+
+## Phasing
+
+Phases are 1,2,3. Coloring is either Blue, Orange, Yellow, or Black, Blue, White
+
+If you hook up the wrong phases, things will generally spin in the wrong direction. To fix it, swap the leads..
+
+But why are we using 3-phase. 
+
+If we use less than 3, we get pulsing power. If we use more than 3, its way less efficient.
+
+
+
+The circuits have 3 parts:
+
+- Sources
+- Transmission lines
+- loads
+
+There’s 4 possible 3-phase system configurations.
+
+| Source   | Load     | System Description |
+| -------- | -------- | ------------------ |
+| $Y$      | $Y$      | $Y-Y$              |
+| $Y$      | $\Delta$ | $Y-\Delta$         |
+| $\Delta$ | $\Delta$ | $\Delta-\Delta$    |
+| $\Delta$ | $Y$      | $\Delta-Y$         |
+
+All the power sources in lab are $Y$ sources. You can tell, because there’s a neutral.
+
+### Requirements for a balanced circuit
+
+- All sources have the same
+    - amplitude
+    - frequency
+- line imedances are equal on all phases
+- loads are equal
+- sources are $120^\circ$
+
+
+
+# Y config
+$$
+V_{ac}=V_{ab}=V_{bc}=V_{line}\\
+V_{an}=V_{bn}=V_{cn}=V_{phase}\\
+V_{line}=V_{phase}\times \sqrt{3}\\
+I_{ab}=I_{bc}=I_{ac}=I_{line}\\
+I_{an}=I_{bn}=I_{cn}=I_{phase}\\
+I_{line}=I_{phase}
+$$
+
+# $\Delta$ config
+
+$$
+V_{ac}=V_{ab}=V_{bc}=V_{line}\\
+V_{an}=V_{bn}=V_{cn}=V_{phase}\\
+V_{line}=V_{phase}\\
+I_{ab}=I_{bc}=I_{ac}=I_{line}\\
+I_{an}=I_{bn}=I_{cn}=I_{phase}\\
+I_{line}=I_{phase}\times\sqrt{3}
+$$
+
+# Power
+$$
+S_{phase}=V_{phase}\times I_{phase}={V_{line}\times I_{line}\over \sqrt{3}}\\
+S_{line}=V_{line}\times I_{line}={V_{phase}\times I_{phase}\over \sqrt{3}}\\
+S_{total}=3\times S_{phase}=\sqrt{3}\times V_{line}\times I_{line}\\
+$$
+

2-February/2-10_Notes.md

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+![](Untitled.assets/realTransformerEquivalentCircuit.png)
+
+# Transformers
+
+$R_m$ is the heat losses in the transformer.
+
+$X_m$ accounts for inperfections in power transfer.
+$$
+R_m>>R_f\\
+X_m>>X_f\\
+\therefore\ ignore\ voltage\ across\ R_f\ and\ X_f\\
+R_m={E_p^2\over P_m}\\
+S_m=E_p\times I_m\\
+X_m={E_p^2\over Q_m}\\
+Q_m=\sqrt{S_m^2-P_m^2}
+$$
+$R_p$ total transformer resistance referred to primary side
+
+$X_p$ total transformer reactance referred to on the primary side
+
+Short the secondary, then:
+$$
+R_p=R_{f_1}+a^2R_{f_2}\\
+X_p=X_{f_1}+a^2X_{f_2}\\
+Z_p=\sqrt{R_p^2+X_p^2}\\[16pt]
+Find\ I_p\ and\ P_p\ to\ find\ values\ of\ R_p\ and X_p\\[16pt]
+if\ X_p>>5R_p\\ neglect\ R_p
+$$
+
+
+
+Subtract S from the rating to get the actual usable power

2-February/2-12_Notes.md

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+<1000V is low voltage, because normal insulation will still work
+
+1kV-35kV is mid voltage, usually going down the road on poles. (on poles, the higher, the more voltage)
+
+35kV-100kV is high voltage
+
+100kV+ is specific/very high voltage. At this point, everything is specially built

2-February/2-17_Notes.md

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+# Equivalent Transformer Example
+
+## Example 1
+
+### Open Circuit
+
+Measure from secondary because you will be more likely to have that source handy
+$$
+500kVA;\ 69kV-4.160kV;\ 60Hz\\[16pt]
+Measured\ Values\ (from\ secondary):\\
+V_s=4160V;\ I_s=2A;\ P_s=5kW\\[12pt]
+R_s={V^2\over P}={4160V^2\over 5kW}=3461\Omega\\
+S_s=VI=4160V\times2A=8320VA\\
+Q_s=\sqrt{S^2-P^2}=\sqrt{(8320VA)^2-(500W)^2}\\Q=6650VAR\\
+X_s={V^2\over Q_s}={4160V^2\over6650VAR}=2602\Omega\\
+a={V_p\over V_s}={69000V\over4160V}=16.586\\a^2=275\\
+R_{p_m}=275\times3461\Omega=952k\Omega\\
+X_{p_m}=275\times2602\Omega=715k\Omega
+$$
+
+### Short Circuit
+
+$R_m$ and $X_m$ can be ignored
+
+Measure from primary, because you want to keep the voltage as low as possible.
+$$
+I=4A;\ P=2.4kW\\
+Z_f={V\over I}={2600V\over4A}=650\Omega\\
+R_f={P\over I^2}={2400W\over (4A)^2}=150\Omega\\
+X_f=\sqrt{Z^2-R^2}=\sqrt{(650\Omega)^2-(150\Omega)^2}\\X_f=632\Omega
+$$
+
+
+# Name Plate Data
+
+Remember, this is all the data you’re given from the nameplate.
+
+$S_n$=nominal apparent power rating (generally KVA)
+
+$E_{n_p}$=Nominal primary voltage rating (this is the higher of the two)
+
+$E_{n_s}$=nominal secondary voltage rating (this is the lower of the two)
+
+
+
+Using these, you can calculate $I_{n_p}={S_n\over E_{n_p}}$ and $I_{n_s}={S_n\over E_{n_s}}$
+
+## Per Unit Methodology
+
+Always choose a base:
+
+- Base Power $S_n$
+- Base voltages
+    - $E_{n_p}$
+    - $E_{n_s}$
+
+$$
+Nominal:\\
+Z_{n_p}={E_{n_p}\over I_{n_p}}={E_{n_p}^2\over S_n}\\
+Z_{n_s}={E_{n_s}\over I_{n_s}}={E_{n_s}^2\over S_n}\\
+Z_p(per\ unit)={Z_p\over Z_{n_p}}\\
+Z_s(per\ unit)={Z_s\over Z_{n_s}}\\
+$$
+
+Per unit impedance = % impedance
+
+==Note:==
+
+==For $S<100kVA\implies {X\over R}\approx1$ so we can then assume $Z=\sqrt2R=\sqrt2X$==
+
+==For $S>100kVA\implies {X\over R}>4$ so we can then assume $Z\approx X$==
+
+
+
+### Example 1
+
+$$
+250kVA;\ 4160/480V;\ 60Hz;\ 5.1\%\ Impedance\\[16pt]
+Nominal:\\
+Z_{n_p}={E_{n_p}^2\over S_n}={4160V^2\over 250kVA}=69\Omega\\
+Z_{n_s}={E_{n_s}^2\over S_n}={480V^2\over 250kVA}=0.92\Omega\\[16pt]
+Actual:\\
+Z_p=Z_p(pu)\times Z_{n_p}\\
+Z_p=5.1\%\times69\Omega=3.52\Omega\\
+X_p=3.52\Omega;\ R_p\approx0\Omega
+$$
+
+### Example 2
+
+Find $I_L$ given the following:
+$$
+100kVA;\ 480V-120V;\ 5\%\ impedance;\ {X\over R}>4\\
+X\approx Z;\ R_L=100\Omega\\[20pt]
+Z_{n_p}={E_{n_p}^2\over S_n}={480V^2\over 100kVA}=2.3\Omega\\
+Z=Z_{n_p}\times Z_p\\
+Z=2.3\Omega\times0.05=0.1152\Omega\\
+move\ R_L\ across\ transformer\\
+simple\ current\ calculations
+$$
+

2-February/2-19_Notes.md

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+# Voltage Regulation
+$$
+VR(\%) ={E_{NL}-E_{FL}\over E_{FL}}\\where\\
+E_{NL}=V_s(no\ load)\\
+E_{FL}=V_s(Full\ rated\ load)\\
+Z_p=\sqrt{R_p^2+X_p^2}\\
+E_{FL}=I_{n_s}\times Z_{FL}
+$$
+
+
+
+Unrelated: Brown-outs are because there’s less power through an incandescent bulb, so the color changes to brown.
+
+
+
+# Construction of a Power Transformer
+
+Core is something metal, probably iron. This keeps $I_m$ low. 
+
+Windings are kept close together, to keep $X_f$ low, which keeps voltage regulation high.
+
+## Transformer Taps
+
+This changes the connection point to get the right voltage. 
+
+Some are motorized, these are utility values. Others are automatic, and based on the secondary voltage. They “tap” the voltage in one direction or the other.
+
+# Cooling methods
+
+- Dry
+    - non-vented (“AA”)
+    - vented (“AFA”)
+- Wet
+    - Oil convection (“OA”)
+        - think water cooling, but no pump
+    - Fans connected to the “radiator” (“FA”)
+
+### Hot spot temperatures
+
+Given $40^\circ C$ ambient, the following classes exist
+
+- $55^\circ C$ (wet)
+- $65^\circ C$ (wet)
+- $105^\circ C$ 
+- $115^\circ C$ (dry)
+- $130^\circ C$
+- $150^\circ C$ (dry)
+- $155^\circ C$
+- $180^\circ C$
+- special cases
+
+Some places run them “cold”, which lets them run longer. Utilities run them “hot”, which replaces them faster, but its cheaper.

2-February/2-24_Notes.md

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+
+
+# Chapter 11
+
+Withstand current = $I_{sc}$
+
+$Z_{thev} = {V_{nominal} \over I_{sc}}$
+
+## Dual Voltage Transformers
+
+usually single phase
+
+usually residential
+
+Medium voltage primary
+
+Secondary is (2) 120V secondaries. There’s 3 wires. One thats bare wire (neutral), and 2 wires for power, both at 120V. Just like the transformer that we used in lab.
+
+The transmission lines generally go to multiple households. This makes it easier to balance the load.
+
+
+
+Large buildings normally have $208V_{line}$, or $120V$ in a 3-phase, wye power. 
+
+![](2-24_Notes.assets/redundancyConnection.png)
+
+![](2-24_Notes.assets/highCurrentTransformer.png)
+
+## Autotransformers
+
+![](2-24_Notes.assets/autoTransformer.png)
+
+Primary turns $N_p$
+
+Secondary turns ($N_s$)
+
+$V_s=\frac{N_s}{N_p}\times V_p$
+
+Turns ratio $a=\frac{N_s}{N_p}$
+
+$I_pN_p=I_sN_s$
+
+$I_p$ and $I_s$ are both down
+
+
+
+There’s a problem: Theres no isolation between the primary and the secondary. Its cheaper, but noise problems are still there.
+
+
+
+They’re normally labeled, but our professor is an asshole, so we don’t know in lab.
+
+Don’t use a nail as a fuse.
+
+
+
+# Chapter 12: 3-Phase
+
+It… its just 3 transformers
+
+## Delta-Wye configuration
+
+![](2-24_Notes.assets/3phaseTransformerDeltaWye.png)
+
+Left is $\Delta$, right is $Y$
+
+Name-plate:
+
+$480-208$ = $\Delta-\Delta$
+
+$480/277-208/120$=$Y-Y$
+
+These can be in any combination
+

2-February/2-26_Notes.md

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+# Special Transformers
+
+## Instrumentation transformers
+
+6900V line. Its hard to make a meter for that. So, we make special transformers so we can measure it.
+
+### Potential Transformer
+
+book calls em “Voltage transformers”
+
+denoted as PTs, POTs, or VTs
+
+Used for monitoring voltage and power.
+
+This is what your power meter is on the side of your house
+
+Made to have a high impedance. This means that you don’t have much power on the low voltage side, so its more accurate.
+
+### Current Transformers
+
+used for monitoring current. 
+
+denoted as CTs
+
+Usually 1 turn around the incoming cable. 
+
+Used in contactless current measurement
+
+Used for exceedingly high currents (>400A)
+
+
+
+If you ever deal with a current transformer, don’t touch it. The voltage on the meter is GIANT. And WHATEVER YOU DO. ==DON’T BREAK THE CIRCUIT.==
+
+Always ground the secondary.
+
+
+
+# Large Design Problem
+
+Given voltage and short-circuit/widthstand current
+
+![](2-26_Notes.assets/powerProblem.png)
+$$
+V_{util}=U
+$$
+You may be given $X\over R$
+
+Awesome. Now we know what the utility looks like.
+
+This is all what we’ve done in class so far.
+
+
+
+Now, we’re gonna add a transformer.
+
+![](2-26_Notes.assets/powerProblemExpanded.png)
+$$
+Z_{base}=\frac{E_p^2}{S}=\frac{4160V^2}{500kVA}\\
+Z_{base}=34.6\Omega\\
+Z_f=Z_{base}\times5\%\\
+Z_f=1.73\Omega
+$$
+Now we’re gonna add a load.
+
+![](2-26_Notes.assets/powerProblemComplete.png)
+
+Alright. Given this, use the calculations above, and the following:
+$$
+Z_{util}={4160V\over10kA}=0.416\Omega
+$$
+You can now convert to a simple circuit like this:
+
+![](2-26_Notes.assets/powerProblemShrunk.png)[^1]
+
+This can help you find $I_p$. You can then calculate $I_s$, calculate power, etc. etc. etc.
+
+Or, you can find $V_p$ for the transformer. This can also be represented at $V_{FL}$
+
+Now we can find regulation:
+$$
+regulation={V_{NL}-V_{FL}\over V_{FL}}
+$$
+We can also find efficiency:
+$$
+Efficiency ={P_{out}\over P_{in}}={V_{load}\times I_{load}\over V_{util}\times I_p}
+$$
+
+## Simple Circuit Review
+
+Given a simple circuit, with just $Z_{in}$ and $V_{in}$
+
+All given numbers are in RMS
+
+If you are asked to graph the waveform, tack a $\sqrt2$ on the end.
+
+Inductors lag, capacitors lead
+
+$V_L=\frac{V_\phi}{\sqrt3}$
+
+Review delta vs wye
+
+
+
+[1]:a2 actually means $a^2$, but qucs doesn’t understand powers in component names

2-February/2-3_Notes.md

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+# Transformers math
+$$
+{E_1\over E_2}=a={N_1\over N_2}={I_2\over I_1}\\
+$$
+
+| Parameter     | Secondary $\implies$ Primary | Primary $\implies$ Secondary |
+| ------------- | ---------------------------- | ---------------------------- |
+| Impedance (Z) | $a^2Z$                       | $Z/(a^2)$                    |
+| Voltage (E)   | $aE$                         | $E/a$                        |
+| Current (I)   | $I/a$                        | $aI$                         |
+
+$$
+Eff={P_2\over P_1}
+$$
+
+Ideal transformers have no losses, and all the power in goes into the out

2-February/2-5_Notes.md

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+# Transformers
+
+We will always assume an iron core. It will be stated if you don’t have one. Two lines between inductors indicates transformer
+
+
+
+Riser diagrams go from bottom to top. This shows the physical locations of the circuitry. Traditional diagrams only show power, and are generally left to right, or vice versa
+
+
+
+![idealTransformer](2-5_Notes.assets/idealTransistor.png)
+
+
+$$
+I_p={I_S\over a}\\
+V_p=V_sa
+$$
+Given these two, your remembered calculations will find the power and the perceived impedance on either side
+
+
+
+You can find your source impedance for something you own. But for utilities, they give you the widthstand current or available short circuit current. Given that and the voltage, you can find the impedance. 
+
+---
+
+RIT recently announced the construction of a new building on campus. A $120V,\ 60Hz$ generator located 100 miles from campus will supply power to the building. The internal resistance of the generator is $0.01\Omega$.
+
+The proposed Transmission line has the resistance of $0.25\Omega \ per\ 1000ft$. The steady state load of the building is $1.0\Omega$. 
+
+- Draw a circuit based on this representation of the project.
+- calculate the resistance of the transmission line $R_{trans}$
+- Calculate the voltage at the building $V_B$
+- Calculate the power delivered to the building $P_B$
+
+
+The load of the transmission lines comes out to $132k\Omega$
+
+This transmission load is difficult to deal with without a transformer. So, lets add one. Actually, no. Lets add 2.

3-March/3-4_Notes.md

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+# Post test notes
+
+"The coronavirus is probably gonna give us the next ice age..." - unknown
+
+Reviewing what you SHOULD have done on the exam. Grades are not complete yet.
+
+"I've seen everything from good to... bad... so far."
+
+Everything on the exam was discussed in class.
+
+
+
+## Question 1
+
+### Part a
+
+$480V<0\circ$ source at 60Hz. Graph was fairly simple. He wanted time for the graph, rather than a radians or degrees.
+
+### Part b
+
+given a short circuit current of 1000A or 1kAIC, and a pf=1. This means we only have to find the resistive component of $Z_{in}$.
+
+This far could have been done without a calculator. $Z_{in}=0.48\Omega$
+
+### Part c
+
+Given a coil, with $X=0.5\Omega$.
+
+There were two answers accepted the $Z_T$:
+
+A:
+$$
+Z=\sqrt{R^2+X^2}\\
+where\ R=0.5\Omega;\ X=0.5\Omega\\
+or\\
+where\ R=0.48\Omega;\ X=0.48\Omega\\
+Z\approx 0.69\Omega<46^\circ\ or\ Z\approx 0.707\Omega<45^\circ\\
+I={V\over Z}={480V\over0.69}=695A
+$$
+
+Then, just graph it on top of the voltage graph
+
+## Question 2
+
+added simple transformer, just like in the notes. 480-240V, 10kVA, Z=10%. Magnetizing impedance is ignored. Z is all resistive.
+
+Find $Z_f$.
+
+$$
+a=2;\ a^2=4\\
+Primary\\
+Z_{base}={V^2\over S}={480V^2\over10000VA}=23\Omega\\
+Z_f=Z_{base}\times 10\%=2.3\Omega\\
+Secondary\\
+Z_{base}={V^2\over Z}={240V^2\over10000VA}=5.7\Omega\\
+Z_{f_{sec}}=5.7\times 10\%=0.57\Omega
+$$
+
+==***NOTE: These impedances are either one, or the other. It wasn't completely established, so no points lost, but it will be important.***==
+
+## Question 3
+
+Connect em all together, but no load yet.
+
+Don't bank on the transformer rating and the power input to be the same.
+
+$I_p=0A;\ \therefore\ V_s=240V$
+
+## Question 4
+
+Connect a $1\Omega$ resistor to the secondary.
+
+Convert that to a $4\Omega$ resistor on the opposite side of the transformer.
+
+total impedance of the circuit should be about $6.78\Omega$.
+
+$$
+I_p={480\over6.78}=70.8A\\
+I_s=I_pa=141.6A\\
+V_s=141.6V\\
+V_p=283.2V\\
+P_s=IV=(141.6A)(141.6V)\approx20kW\\
+P_{util}=34kVA\\
+\%VR={240-141.6\over240}=41\%
+$$
+
+Unasked questions:
+
+Whats the efficiency?
+$$
+eff={P_o\over P_i}\approx58\%
+$$
+
+## Question 5
+
+3 phase transformer. 4160/2400 volt generator in wye configuration. This is the 3 generators tied together in... series, ish?
+
+4160-480/277V transformer.
+
+277V across each coil, and 480 from phase to phase.
+
+Generator phase voltage is 2400V.
+
+The line voltage of the generator is 4160V.
+
+$V_p=V_l=4160V$ for the transformer primary. $V_p=277V;\ V_l=480V$ for the transformer secondary.
+
+To be continued after break...

4-April/exam2.md

@@ -0,0 +1,130 @@
+1.  Clearly label each voltage.
+
+![image-20200407153954860](exam2.assets/image-20200407153954860.png)
+
+2.  What is the synchronous speed of this motor?
+    $$
+    n_s=120*{f\over p}=120*{60Hz\over 6 poles}\\
+    \overline{\underline{|n_s=1200rpm|}}
+    $$
+    
+
+3.  How many poles does this motor have?
+    $$
+    poles = phases * 2 = 3 * 2\\
+    \overline{\underline{|6\ poles|}}
+    $$
+    
+
+4.  What is the slip of this motor?
+    $$
+    s={n_s-n\over n_s}={1200rpm-1076rpm\over1200rpm}\\
+    \overline{\underline{|s=0.10\overline3|}}
+    $$
+    
+
+5.  What is the mechanical output in BHP (brake horsepower) of this motor at full speed?
+    $$
+    rated\ hp * efficiency=10hp * 0.8\\
+    \overline{\underline{|8BHP|}}
+    $$
+
+6.  What is the total impedance $Z_{motor}$ of this motor at full load?
+    $$
+    Z={V\over FLA}={208V\over30A}=6.9\overline3\Omega\\
+    \theta=cos^{-1}(pf)=cos^{-1}(0.7071)\approx cos^{-1}({\sqrt2\over2})=45^\circ\\
+    \overline{\underline{|Z=6.9\overline3\Omega<45^\circ|}}
+    $$
+
+7.  What is the resistive impedance $R_{motor}$ of this motor at full load?
+    $$
+    |Z|*cos(\theta)={\sqrt2\over2}\\
+    R=4.9\Omega
+    $$
+
+8.  What is the reactive impedance $X_{motor}$ of this motor at full load?
+    $$
+    |Z|*sin(\theta)=6.9\overline3*{\sqrt2\over2}\\
+    X=4.9\Omega
+    $$
+    
+
+9.  What is the $X\over R$ ratio of the motor?
+
+10.  What is the full load apparent power drawn by this motor?
+     $$
+     S=FLA * V * \sqrt3=208V*30A*\sqrt3\\
+     S=10.8kVA
+     $$
+
+11.  What is the Full Load Active power drawn by this motor?
+     $$
+     P=S*cos(\theta)=10.8kVA*{\sqrt2\over2}\\\overline{\underline{|P=7.63668kW|}}
+     $$
+     
+
+12.  What is the full load reactive power drawn by this motor?
+     $$
+     Q=S*sin(\theta)=10.8kVA*{\sqrt2\over2}\\\overline{\underline{|Q=7.63668kW|}}
+     $$
+
+13.  What is the locked rotor apparent power drawn by the motor?
+     $$
+     S=V*LRA*\sqrt3=208V*180A*\sqrt3\\
+     \underline{\overline{|S=64.8kVA|}}
+     $$
+
+14.  Determine $I_{secondary_{line}}$.
+     $$
+     I={S\over V}={10.8kVA\over 208V}\\
+     \overline{\underline{|I=51.96A|}}
+     $$
+     
+
+15.  Determine $I_{secondary_{\phi}}$.
+     $$
+     I_{line}=I_\phi=51.96A
+     $$
+     
+
+16.  Determine $I_{primary_{\phi}}$.
+     $$
+     I_{primary}={I_{secondary}\over a}={51.96A \over 4}\\\overline{\underline{|I_{primary_\phi}=13A|}}
+     $$
+
+17.  Determine $I_{primary_{line}}$.
+     $$
+     I_{line}=I_\phi*\sqrt3=13A*\sqrt3\\
+     \underline{\overline{|I_{line}=22.5A|}}
+     $$
+
+18.  Determine $I_{utility_{line}}$.
+     $$
+     I_{utility}=I_{primary}=22.5A
+     $$
+
+19.  Determine $P_{secondary_\phi}$.
+     $$
+     P=VI=120V*51.96A=6.235kW
+     $$
+
+20.  Determine $P_{secondary_{3\phi}}$
+     $$
+     P_{3\phi}=3*P=3*6.235kW=18.7kW
+     $$
+
+21.  Determine $P_{primary_\phi}$.
+     $$
+     P=VI=480V*13A=6.24kW
+     $$
+
+22.  Determine $P_{primary_{3\phi}}$
+     $$
+     P_{3\phi}=3*P=3*6.24kW=18.7kW
+     $$
+
+23.  Determine $P_{utility}$.
+     $$
+     P_{utility}=P_{primary_{3\phi}}=18.7kW
+     $$
+     

4-April/exam3.md

@@ -0,0 +1,92 @@
+# Skyler MacDougall
+
+## Exam 3
+
+1.  ![https://cdn.discordapp.com/attachments/693545477475663886/699352158344052776/unknown.png](exam3.assets/unknown.png)
+
+    The current will go up, because the current requirements will still be the same for the motor. I can barely find anything saying the mathematics of this. Quora is the only place I could find any math remotely similar to what we were looking for, which follows below.
+
+    $2.4*20A=48A$
+
+    
+
+2.  Define the following terms:
+
+    -   Slip of a Synchronous Generator
+        The difference in speed (in percent) between the synchronous speed (the frequency of the output signal) and the rotational speed (the speed of the rotor).
+    -   Armature of a generator
+        The rotating portion of the generator. When in a motor, this is the rotor.
+    -   Rotating AC field generator vs Stationary AC Field Generator
+        Rotating AC field generators have stationary magnetic fields, while stationary AC field generators have rotating magnetic fields.
+    -   Permanent Magnet generator (explain benefits and whether its a stationary or rotating AC field generator)
+        Permanent Magnet generators are stationary AC fields. Permanent Magnet generators are cheaper to build.
+    -   Infinite bus
+        The bus who's voltage and frequency remains constant even after variation in the load. 
+
+3.  For a single Phase Capacitor Start Motor running at 240V<60Hz at a speed of 1700rpm, running a water pump. Calculate the following:
+
+    1.  What is the synchronous speed of the motor?
+        $$
+        n_s=120{f\over p}=\cancel{120}40*{\cancel{60}20Hz\over1\cancel{9}}\\
+        \overline{\underline{|n_s=1800rpm|}}
+        $$
+
+    2.  What is the slip of the motor?
+        $$
+        s={n_s-n\over n_s}={1800rpm-1700rpm\over1800rpm}\\
+        \overline{\underline{|s\approx5.\overline5|}}
+        $$
+
+    3.  How many poles does this motor have?
+        $$
+        9
+        $$
+
+    4.  Draw a circuit that shows this motor and how it works. Make sure you show both windings.
+        ![image-20200423142207040](exam3.assets/image-20200423142207040.png)
+
+    5.  Explain how the motor starts?
+        Close the switch, then rotate the rotor.
+
+    6.  After installing this motor, we find that the rotational direction of the motor is backwards. How do we change this so the pump will always run in the correct rotational direction?
+        If the rotor is in motion, stop the rotor. Then, reverse the polarity of the windings.
+
+4.  For a regular DC motor, define the following:
+
+    1.  Is the motor a stationary or rotational field motor? Explain.
+        The motor is like a rotational field motor because it uses permanent magnets.
+    2.  What is a Split Ring? What does it do?
+        A split ring is a cylindrical shell split in 2 halves, the power transfers from the split rings through the brushes and into the rotor. The split ring keeps the DC motor turning in one direction.
+    3.  What is a commutator?
+        A commutator switches the direction of the current at certain intervals which allows the motor to rotate in one direction continuously.
+    4.  Is the speed of a DC motor proportional to the applied voltage?
+        The speed of a DC motor is directly proportional to the applied voltage. 
+
+5.  Explain/define
+
+    1.  What does ECM mean?
+        Electronically commutated motor.
+    2.  What is an ECM motor?
+        A motor which uses electronics to control its speed.
+    3.  How does an ECM motor work?
+        An ECM motor works like any other motor except that its speed is dictated by a microprocessor housed in the motor's shell. 
+    4.  How is this motor different than a regular DC motor?
+        A regular motor requires a change in applied voltage to change its speed. An ECM motor can regulate its speed without a change in applied voltage.
+    5.  What is the armature of an ECM motor?
+        The armature of an ECM motor is its rotor.
+
+6.  a 150KVA 480VAC Permanent Magnet Generator (powered by a diesel engine) has an impedance Z of $5\Omega$ with a normal reactance X of $4\Omega$.
+
+    1.  The generator has a sub-transient reactance of $0.5\Omega$ (occurs for the first 3 cycles where there is a short circuit). What is the short circuit current available from the generator for the short circuit immediately after the short circuit occurs?
+        $$
+        I_b={S\over\sqrt3E_b}={150K\cancel{V}A\over\sqrt3*480\cancel{V}}=180.4A\\
+        I"={1\over X"}*X_s*I_b={1\over 0.5\Omega}*4\Omega*180.4A\\
+        \overline{\underline{|I"=1443A|}}
+        $$
+
+    2.  The generator has a transient reactance of $1\Omega$ (occurs after the short circuit has occurred and is maintained for a long period of time). What is the current from the generator if the short circuit is maintained?
+        $$
+        I_b=180.4A\\
+        I'={1\over X'}*X_s*I_b={1\over 1\cancel {\Omega}} *4 \cancel{\Omega} *180.4A\\
+        \overline{\underline{|I'=721A|}}
+        $$

_Homework/hw1.md

@@ -0,0 +1,104 @@
+## Skyler MacDougall
+
+### Homework 1: Due 1/22/2020
+
+10.  
+
+     1.  Draw a waveshape of a sinusoidal voltage having a peak value of 200V and a frequency of 5Hz.
+         ![homework1question10](hw1q10.PNG)
+
+     2.  If the voltage is zero at $t=0$, what is the voltage at $t=5ms$? $75ms$?$150ms$?
+
+         | Time (t, ms) | Voltage (V)   |
+         | ------------ | ------------- |
+         | 5            | $\approx31$   |
+         | 75           | $\approx 141$ |
+         | 150          | $-200$        |
+
+11.  A sinusoidal current has an effective value of 50A. Calculate the peak value of current.
+     $$
+50A*{\sqrt{2}\over2}\approx35A
+     $$
+     
+12. A sinusoidal voltage of 120V is applied to a $10\Omega$ resistor. Calculate
+
+    1. the effective current through the resistor.
+
+        $$
+        {120V\over 10\Omega}=12A
+        $$
+    
+2. the peak voltage across the resistor.
+   
+    $$
+        120V*\sqrt{2}\approx170V_{RMS}
+    $$
+    
+    3. the power dissipated by the resistor.
+        $$
+        120V*12A=1440W
+        $$
+    
+    4. The peak power dissipated by the resistor.
+        $$
+        170V_{RMS}*(12A*\sqrt{2})=2880W
+        $$
+    
+    
+    
+13. A distorted voltage contains an 11^th^ harmonic of 20V, 253Hz. Calcultate the frequency of the fundamental.
+
+    $$
+    235Hz*11^{th}Harmonic\approx2.6kHz
+    $$
+
+14. The current in a 60Hz single phase motor lags 36 degrees behind the voltage. Calculate the time interval between positive peaks of voltage and current.
+
+    $$
+    {1\over \cancel{60Hz}}*{\cancel{360^\circ}6^\circ\over 36^\circ}={1\over 6}seconds=0.1\overline{66}seconds
+    $$
+
+15. Determine the phase angle between the following phasors and, in each case, indicate which phasor is lagging.
+
+    ![homework1question13](hw1q13.PNG)
+
+    1. $I_1$ and $I_3$
+
+       
+        $$
+        \angle I_1-\angle I_3=-60^\circ\\
+        \therefore\\
+        I_1 lags I_3by60^\circ
+        $$
+        
+2. $I_2$ and $I_3$
+    
+    $$
+        \angle I_2-\angle I_3=-90^\circ\\
+        \therefore\\
+        I_2 lags I_3 by 90^\circ
+        $$
+    
+3.  $E$ and $I_1$
+    
+    $$
+        \angle E - \angle I_1=-150^\circ\\
+        \therefore\\
+        ElagsI_1by150^\circ
+        $$
+    
+16.  The voltage applied to an AC magnet is given by the expression $E=160sin\phi$, and the current is $I=20sin(\phi-60^\circ)$, all angles being expressed in degrees.
+
+     1.  Draw the phasor diagram for $E$ and $I$, using effective values.
+
+         ![homework1q16_2](hw1q16_2.png)
+
+     2.  Draw the waveshape of $E$ and $I$ as a function of $\phi$.
+
+         ![homework1q15](hw1q16.png)
+
+     3.  Calculate the peak positive power and the peak negative power in the circuit.
+
+         
+         
+         

_Homework/hw10.md

@@ -0,0 +1,101 @@
+# Skyler MacDougall
+
+## Homework 10: due 3/30/2020
+
+2. A wye-connected squirrel-cage motor having a synchronous speed of 900r/min has a stator resistance of $0.7\Omega$ and an equivalent rotor resistance of $0.5\Omega$. If the total leakage reactance is $5\Omega$ and the line-to-neutral voltage is 346V, calculate the following:
+
+    1. The value of $Z_1$ and the angle $\alpha$
+    $$
+        Z_1=\sqrt{R_1^2 +x^2}=\sqrt{0.7^2 +5^2}\\
+    Z_1=5.048\Omega\\
+        \alpha=tan^{-1}({X\over R_1})=tan^{-1}({5\Omega\over 0.7\Omega})\\
+    \alpha=82.03^\circ
+        $$
+    
+    
+    2. The speed when the breakdown torque is reached
+        $$
+        n_b=n_s(1-({R_2\over Z_1}))=900(1-({0.5\Omega\over 5.048\Omega}))\\
+        \underline{\overline{|n_b=810.85|}}
+        $$
+        
+    
+    3. the current $I_1$ at the breakdown torque (see below)
+    
+        ![](hw10.assets/hw10q2.png)
+        $$
+        I_1={E_L\over2 Z_1 cos({\alpha\over 2})}={346V \over 2 * 5.048 * cos( {82.03^\circ \over 2})}\\
+        \overline{\underline{|I_1=45.419A|}}
+        $$
+        
+    
+    4. The value of the breakdown torque [N*m]
+        $$
+        T_b={9.55(I_1^2Z_1)\over n_s}={9.55(45.419A^2* 5.048\Omega)\over 900r/min}\\
+        \overline{\underline{|T_b=110.498N*m|}}
+        $$
+        
+
+
+
+4. A 550V, 1780r/min, 3-phase, 60Hz, squirrel cage induction motor running at no-load draws a current of 12A and a total power of 1500W. Calculate the value of $X_m$ and $R_m$ per phase.
+
+    ![](hw10.assets/hw10q4.png)
+
+$$
+S=VA=550V*12A=6.6kVA\\
+Q=\sqrt{6.6kVA^2-1.5kW^2}=6.4kVAR\\
+R={V\over P}={550V\over 1500W}\\
+X={V\over Q}={550V\over 6.4kVAR}\\
+\underline{\overline{|R=0.3\overline6\Omega;X=0.09\Omega|}}
+$$
+
+
+
+9.  Consider the 5hp motor whose equivalent circuit shown below.
+    ![image-20200404160615051](hw10.assets/image-20200404160615051.png)
+
+    1.  Calculate the values of the inductances (in mH) of the leakage and magnetizing reactances.
+        $$
+        L_m={X\over 2\pi f}={110\Omega\over 2\pi(60Hz)}\\
+        L_x={X\over 2\pi f}={6\Omega\over 2\pi(60Hz)}\\
+        \overline{\underline{|L_x=15.9mH;\ L_m=291.8mH|}}
+        $$
+
+    2.  Determine the values of the leackage reactance and magnetizing reactance at a frequency of 50Hz.
+        $$
+        X_m=L_m*2\pi f=15.9mH*2\pi *50Hz\\
+        X_m=L_m*2\pi f=291.8mH*2\pi *50Hz\\
+        \overline{\underline{|X=5\Omega;\ X_m=91.6\Omega|}}
+        $$
+
+    3.  Calculate the 50Hz line to neutral voltage to obtain the same magnetizing current and compare it with the voltage at 60Hz.
+        $$
+        V'=V*{f_2\over f_1}={440\over\sqrt3}V*{50Hz\over60Hz}\\
+        V'\approx212V
+        $$
+
+10.  The 5hp motor represented by the equivalent circuit of the image above is connected to a 503V(line-to-line), 3-phase, 80Hz source. The stator and rotor resistances are assumed to remain the same.
+
+     1.  Determine the equivalent circuit when the motor runs at 2340r/min.
+
+         ![image-20200404172423179](hw10.assets/image-20200404172423179.png)
+
+     2.  Calculate the value of the torque [N*m] and the power [hp] developed by the motor. 
+         $$
+         P_r=({V\over\sqrt{X^2+R_1^2}})^2*R_2/s*phases\\
+         P_r=({290V\over\sqrt{8\Omega^2+(1.5\Omega+48\Omega)^2}})^2*R_2/s*3\ phases\\
+         P_r=({290V\over50.14\Omega})^2*{1.2\over 0.025}*3\ phases\\
+         P_r=(5.78A)^2*48\Omega*3\ phases\\
+         P_r=3819.39(A^2)*144\Omega\\
+         T=9.55{4817W\over 2400rpm}\\
+         P={nT\over 9.55}={2340rpm*19.2Nm\over 9.55}=4696W\\
+         \overline{\underline{|T=19.2Nm;P_r=6.29hp|}}
+         $$
+         
+
+$$
+s={n_s-n\over n_s}\\
+s*n_s=n_s-n
+$$
+

_Homework/hw11.md

@@ -0,0 +1,69 @@
+# Skyler MacDougall
+
+## Homework 11: due 4/1/2020
+
+
+
+9. A 3-phase, 225r/min synchronous motor connected to a 4kV, 60Hz line draws a current of 320A and absorbs 2000kW. Calculate:
+
+    1. The apparent power supplied to the motor.
+    2. The power factor.
+    3. The reactive power absorbed.
+    4. The number of poles on the rotor.
+
+10. A synchronous motor draws 150A from a 3-phase line. If the exciting current is raised, the current drops to 140A. Was the motor over- or under-excited before the excitation was changed?
+
+11.  
+
+    1. Calculate the approximate full-load current of the 3000hp motor in the image below, if it has an efficiency of 97%.
+
+        ![](hw11.assets/hw11q11.png)
+
+    2. What is the value of the field resistance?
+
+12. Referring to the image below, at what speed must the rotor turn to generate the indicated frequencies?
+    ![](hw11.assets/hw11q12.png)
+
+
+
+14. A synchronous motor has the following parameters, per phase
+    ![](hw11.assets/hw11q14.png)
+    $$
+    E=2.4kV;\ E_o=3kV\\
+    X_s=2\Omega\\
+    I=900A
+    $$
+    Draw the phasor diagram.
+
+    ![]()
+
+
+    Then, determine:
+
+    1. Torque angle $\delta$
+    2. Active power, per phase
+    3. Power factor of the motor
+    4. Reactive power absorbed/delivered per phase
+
+15.  
+
+    1. In problem 14, calculate the line current and the new torque angle $\delta$ if the mechanical load is suddenly removed.
+    2. Calculate the new reactive power absorbed/delivered by the motor, per phase.
+
+16. A 500hp synchronous motor drives a compressor and its excitation is adjusted so that the power factor is unity. If the excitation is increased without making any other change, what is the effect on the following:
+
+    1. The active power absorbed by the motor.
+    2. The line current.
+    3. The reactive power absorbed/delivered by the motor.
+    4. The torque angle.
+
+17. The 4000hp, 6.9kV motor shown below posesses a synchronous reactance of $10\Omega$ per phase. 
+
+    ![](hw11.assets/hw11q17.png)
+    The stator is connected in wye, and the motor operates at full-load (4000hp) with a leading power factor of 0.89. If the efficiency is 97%, calculate the following:
+
+    1. The apparent power
+    2. The line current
+    3. The value of $E_o$ per phase
+    4. The mechanical displacement of the poles from their no-load position
+    5. The total reactive power supplied to the electrical system.

_Homework/hw12.md

@@ -0,0 +1,58 @@
+# Skyler MacDougall
+
+## Homework 12: due 4/8/2020
+
+3. In analyzing a hydropower site, it is found that the turbines should turn at close to 350r/min. If the directly coupled generators must generate a frequency of 60Hz, calculate the following:
+
+    1. The number of poles on the rotor
+    2. The exact turbine speed
+
+4. An isolated 3-phase generator produces a no-load line voltage of 13.2kV. If a load having a lagging power factor of 0.8 is connected to the machine, must the excitation be increased or decreased in order to maintain the same line voltage?
+
+5. What conditions must be met before a generator can be connected to a 3-phase system?
+
+6. Calculate the number of poles on the generator in figure 12 below using the information given.
+
+    ![](hw12.assets/hw12q6.png)
+
+7. Calculate the number of poles on the aircraft generator shown in the figure 11 below. 
+
+    ![](hw12.assets/hw12q7.png)
+
+
+
+16. The 3-phase generator shown in the figure 16 below has the following characteristics:
+
+    ![](hw12.assets/hw12q16.png)
+    $$
+    E_o=2440V\\
+    X_s=144\Omega\\
+    R=17\Omega\\
+    Z=175\Omega (resistive)
+    $$
+    Calculate
+
+    1. The synchronous impedance $Z_s$ per phase
+    2. The total resistance of the circuit per phase
+    3. The total reactance of the circuit per phase
+    4. the line current
+    5. the line-to-neutral voltage across the load
+    6. the line voltage across the load
+    7. the power of the turbine driving the alternator
+    8. the phase angle between $E_o$ and the voltage across the load
+
+
+
+22. Referring to the figure 17 below, the following information is given about the generator:
+
+    ![](hw12.assets/hw12q22.png)
+    $$
+    E_o=12kV\\
+    E=14kV\\
+    X_s=2\Omega\\
+    E_o\ leads\ E\ by\ 30^\circ
+    $$
+
+    1. calculate the total active power output of the generator
+    2. draw the phasor diagram for one phase
+    3. calculate the power factor of the load

_Homework/hw13.md

@@ -0,0 +1,44 @@
+# Skyler MacDougall
+
+## Homework 13: due 4/15/2020
+
+8. Which of the motors discussed in this chapter is best suited to drive the following loads:
+
+    1. a small portable drill
+    2. a 3/4hp air compressor
+    3. a vacuum cleaner
+    4. a 1/100hp blower
+    5. a 1/3hp centrifugal pump
+    6. a 1/4hp fan for use in a hospital ward
+    7. an electric timer
+    8. a hi-fi turntable
+
+9. Referring to the figure 11 below, the effective impedance of the main and auxiliary windings under locked-rotor conditions were given as follows:
+    ![](hw13.assets/hw13q9.png)
+
+    |              | Effective Resistance | Effective Reactance |
+    | ------------ | -------------------- | ------------------- |
+    | Main winding | $4\Omega$            | $7.5\Omega$         |
+    | Aux winding  | $7.5\Omega$          | $4\Omega$           |
+
+    If the line voltage is 119V, calculate the following
+
+    1. the magnitude of $I_a$ and $I_s$
+    2. the phase angle between $I_a$ and $I_s$
+    3. the line current $I_L$
+    4. The power factor under locked rotor conditions
+
+
+
+12. A single phase motor vibrates at a frequency of 100Hz. What is the frequency of the power line?
+
+
+
+17. The motor described in the table has an LR power factor of 0.9 lagging. 
+
+    ![](hw13.assets/hw13q17.png)
+    It is installed in a workshop situated 600ft from a home, where the main service entrance is located. The line is composed of 2-conductor cable made of No. 12 gauge copper. The ambient temperature is $25^\circ C$ and the service entrance voltage is 122V. Calculate:
+
+    1. The resistance of the transmission line
+    2. The starting current and the voltage at the motor terminals
+    3. the starting torque [N*m]

_Homework/hw14.md

@@ -0,0 +1,17 @@
+# Skyler MacDougall
+
+## Homework 14: due 4/20/2020
+
+9. A separate excited DC generator turning at 1400r/min produces an induced voltage of 127V. The armature resistance is $2\Omega$ and the machine delivers a current of 12A. Calculate the terminal voltage.
+
+
+
+10. A separately excited DC generator produces a no-load voltage of 115V. What happens if:
+    1. the speed is increased by 20%?
+    2. the direction of rotation is reversed?
+    3. the exciting current is increased by 10%?
+    4. the polarity of the field is reversed?
+
+
+
+22. A 200W, 120V, 1800r/min DC generator has 75 communtator bars. The brush width is such as to cover 3 commutator segments. Show the duration of the commutation process is equal to 1.33ms.

_Homework/hw15.md

@@ -0,0 +1,28 @@
+# Skyler MacDougall
+
+## Homework 15: due 4/22/2020
+
+9. A 230V shunt motor has a nominal armature current of 60A. If the armature resistance is $0.15\Omega$, calculate the following:
+    1. The counter-emf [V]
+    2. the power supplied to the armature [W]
+    3. the mechanical power developed by the motor, [kW] and [hp]
+10.  
+    1. In the problem above, calculate the initial starting current if the motor is directly connected across the 230V line.
+    2. Calculate the value of the starting resistor needed to limit the initial current to 115A.
+
+
+
+12. A separately excited DC motor turns at 1200r/min when the armature is connected to a 115V source. Calculate the armature voltage required so that the motor runs at 1500r/min, and at 100r/min.
+
+13. The following details are known about a 250 hp, 230V, 435 r/min DC shunt motor:
+
+    - nominal full-load current = 862A
+    - insulation class = H
+    - weight = 3400kg
+    - external diameter of the frame = 915mm
+    - length of frame = 1260mm
+
+    1. Calculate the total losses and efficiency at full-load.
+    2. Calculate the approximate shunt field exciting current if the shunt field causes 20% of the total losses.
+    3. Calculate the value of the armature resistance as well as the counter-emf, knowing that 50% of the total losses at full-load are due to the armature resistance.
+    4. If we wish to attain a speed of 1100r/min, what should the approximate exciting current be?

_Homework/hw2.md

@@ -0,0 +1,118 @@
+## Skyler MacDougall
+
+### Homework 2: Due 1/22/2020
+
+7. A large motor absorbs 600kW at a power factor of 90%. Calculate the apparent power and reactive power absorbed by the machine.
+
+    $$
+    P=600kW; pf={P\over S};pf=0.9\\
+    {P\over pf}=S={600kW\over0.9}\\
+    666.\overline6kVA=S
+    $$
+    
+13. A single-phase motor draws a current of 16A from 240V, 60Hz. A wattmeter connected to the line gives a reading of 2765W. Calculate the power factor and the reactive power.
+    $$
+    P=2765W; pf={P\over S}\\
+    16A*240V=3840VA\\
+    pf={2765W\over 3840VA}\approx0.77\\
+    cos^{-1}(77)=\theta_{diff}\approx39^\circ\\
+    Q=VI(sin(\theta_{diff}))=16*240*sin(39^\circ)\\
+    Q\approx3775VAR\\
+    answers:\\
+    Q\approx3775VAR;pf\approx0.77or77\%
+    $$
+    
+14. if a capacitor having a reactance of $30\Omega$ is connected in parallel with the motor in question 13, calculate
+    1. The active power reading of the wattmeter.
+
+        Since capacitors are solely reactive, the real power used does not change. Therefore, it is still 2765W.
+
+    2. The total reactive power absorbed by the capacitor and the motor.
+        $$
+        {240V^2\over 30\Omega}=1920VAR\\
+        Q=3775VAR-1920VAR=1855VAR\\
+        $$
+
+    3. The apparent power of the AC line.
+        $$
+        S=\sqrt{2765W^2+1855VAR^2}\\
+        S\approx3329VA;
+        $$
+    
+4. The line current.
+        $$
+        {S\over V}=I\\
+        {3329VA\over 240V}\approx13.9A
+        $$
+        
+    5. The power factor of the motor/capacitor combination.
+    $$
+        {2765W\over 3329VA}=0.83
+    $$
+    
+
+
+16. An induction motor absorbs an apparent power of $400kVA$ at a power factor of 80%. Calculate
+    1. The active power absorbed by the motor.
+        $$
+        S=400kVA; pf=0.8;pf={P\over S}\\
+        pf*S=P=0.8*400kVA\\
+        P=320kW
+        $$
+    
+2. The reactive power absorbed by the motor.
+        $$
+        \theta_{diff}\approx37^\circ\\
+        S(sin(\theta_{diff})=Q\\
+        S(sin(37^\circ)=Q\\
+        Q\approx241kVAR
+        $$
+        
+    3. What purpose does the reactive power serve?
+
+        The reactive power is the power generated by non-resistive components, and is needed to produce a magnetic field.
+
+
+19. A motor having a power factor of 0.8 absorbs an active power of 1200W. Calculate the reactive power drawn from the line.
+    $$
+    pf={P\over S}\\
+    S={P\over pf}={1200W\over 0.8}\\
+    S=1500VAR
+    $$
+    
+22. The power factor at the terminals of a 120V source is 0.6 lagging, with a $2\Omega$ resistor in series with a $3\Omega$ inductor and an unknown. 
+
+    ![homework2question22](hw2q22.png)
+
+    Without phasor diagrams, calculate
+
+    1. The value of $E$
+        $$
+        pf=0.6;lagging=inductive; pf={P\over S}\\
+        600VA=S\\
+        P=pf*S=0.6*600VA=360V\\
+        cos^{-1}(0.6)\approx53^\circ\\
+        600*sin(53^\circ)=480VAR\\
+        {480VAR\over 120V}=4A_{Reactive}\\
+        {120V\over 4A_{Reactive}}=30\Omega\\
+        30\Omega-3\Omega=27\Omega=Z\\
+        4A_{Reactive}*27\Omega=108V\\
+        108V=E
+        $$
+    
+2. The impedance of the load $Z$
+        $$
+        Z=27\Omega (shown ^. above)
+        $$
+        
+    
+24. A single phase capacitor has a rating of 30kVAR, 480V, 60Hz. Calculate the capacitance in microfarads.
+    $$
+    X_C={1\over2\pi fC};Q=VI;X={V\over I}\\
+    I={Q\over V}={30kVAR\over 480V}=62.5A=I\\
+    X_C={V\over I}={480V\over 62.5A}=7.68\Omega\\
+    7.68\Omega={1\over 2\pi (60Hz)(C)}\\
+    C={1\over 2\pi (60Hz)(7.68\Omega)}\\
+    C=345\mu F
+    $$
+    

_Homework/hw3.md

@@ -0,0 +1,112 @@
+# Skyler MacDougall
+
+## Homework 3: due 2/3/2020
+
+12. A single-phase motor draws a current of $12A$ at a power factor of $60\%$. Calculate the in phase and quadrature components of current $I_p$ and $I_q$ with respect to the line voltage. 
+    $$
+    I_S=12A;\ pf=0.6\\
+    I_p=I_S(pf);\ I_q=I_S(sin(cos^{-1}(pf)))\\
+I_p=12A\times0.6\\|\overline{\underline{I_p=7.2A}}|\\
+    I_q=12A\times(sin(cos^{-1}(0.6)))=12A\times0.8\\
+    |\overline{\underline{I_q=9.6A}}|
+    $$
+
+
+
+15. Using only power triangle concepts and without drawing any phasor diagrams, calculate the impedance of the circuits below.
+
+    ![hw3Circuits](hw3.assets/hw3Circuits.png)
+
+    1. $$
+        Q_L={V^2\over X_L}={(40V)^2 \over 4\Omega}=400VAR\\
+        P={V^2\over R}={(40V)^2\over 10\Omega}=160W\\
+        S=\sqrt{P^2+Q^2}=\sqrt{(160W)^2+(400VAR)^2}\\
+        S\approx431VA\\
+        Z={V^2\over S}\angle(cos^{-1} ({P\over S}))={(40V)^2 \over 431VA} \angle (cos^{-1}({160W\over 431VA}))\\
+        |\overline{\underline {Z \approx 3.71\Omega \angle 68^\circ}}|
+        $$
+
+    2. $$
+        Q_L=I^2X_C=(5A)^22\Omega=-50VAR\\
+        P=I^2R=(5A)^26\Omega=150W\\
+        S=\sqrt{P^2+Q^2}=\sqrt{(150W)^2+(-50VAR)^2}\\
+        S\approx158VA\\
+        Z={S\over I^2}\angle(cos^{-1} ({P\over S}))={158VA \over (5A)^2} \angle (cos^{-1}({150W\over 158VA}))\\
+        |\overline{\underline {Z \approx 6.3\Omega \angle 18^\circ}}|
+        $$
+
+    3. $$
+        Q_L=I^2X_C=(5A)^22\Omega=-50VAR\\
+        P=I^2R=(5A)^26\Omega=150W\\
+        S=\sqrt{P^2+Q^2}=\sqrt{(150W)^2+(-50VAR)^2}\\
+        S\approx158VA\\
+        Z={S\over I^2}\angle(cos^{-1} ({P\over S}))={158VA \over (5A)^2} \angle (cos^{-1}({150W\over 158VA}))\\
+        |\overline{\underline {Z \approx 6.3\Omega \angle 18^\circ}}|
+        $$
+
+
+
+20. If a $500VAR$ is put in parallel with a $3840VA; pf=0.72$ motor, calculate
+
+    1. Active power.
+        The active power is unchanged at $2765W$, because the only thing we added was a capacitor, which is purely reactive power. 
+
+    2. The apparent power of the system.
+        $$
+        S=\sqrt{P^2+Q^2};\ P=2765W;\\ Q_C=-500VAR;\ Q_L \approx 2664VAR\\
+        S=\sqrt{(2765W)^2+(2664-500VAR)^2}\\sqrt((2765^2)+(answer(18)−500)^2)
+        |\underline{\overline{S\approx3511VA}}|
+        $$
+
+    3. The power factor of the system.
+        $$
+        pf={P\over S};\ P=2765W;\ S=3511VA\\
+        pf={2765W\over3511VA}\\
+        |\overline{\underline{pf\approx0.79}}|
+        $$
+
+
+
+21. A coil having a reactance of $10\Omega$ and a resistance of $2\Omega$ is connected in parallel with a capacitive reactance of $10\Omega$. If the supply voltage is $200V$, calculate
+
+    1. The reactive power generated by the coil
+        $$
+        Q={V^2\over X_L}={(200V)^2\over 10\Omega}\\
+        |\overline{\underline{Q=4kVAR}}|
+$$
+        
+    2. The reactive power generated by the capacitor
+$$
+        Q={V^2\over X_C}={(200V)^2\over 10\Omega}\\
+    |\overline{\underline{Q=-4kVAR}}|
+    $$
+        
+    3. The active power dissipated by the coil
+        $$
+    P={V^2\over R}={(200V)^2\over 2\Omega}\\
+        |\overline{\underline{P=20kW}}|
+    $$
+        
+    4. The apparent power of the circuit
+        $$
+        S=\sqrt{P^2+(\sum Q)^2}\\
+        S=\sqrt{(20kW)^2+(4kVAR-4kVAR)^2}\\
+        |\overline{\underline{S=20kVA}}|
+        $$
+
+
+
+26. A capacitor bank has a resistor connected to it, to dissipate voltage after its been disconnected. It should have $\le50V$ after 1 minute of disconnection. Calculate the discharge resistance required for a $30kVAR,\ 480V$ capacitor. Then, calculate the wattage rating for the resistor.
+
+$$
+30kVAR;\ 480V;\ C\approx345\mu F\\
+V(0)=480V;\ V_0=480V\\
+V_0e^{-t\over RC}=V(t)\\
+R={60\over \approx345\mu F (ln({5\over48}))}\\
+R\approx76k\Omega\\
+P={V^2\over R}={480^2\over 76k\Omega}\\
+|\overline{\underline{R\approx76k\Omega;\ P\approx3W}}|
+$$
+
+
+

_Homework/hw4.md

@@ -0,0 +1,66 @@
+# Skyler MacDougall
+
+## Homework 4: due 2/10/2020
+
+4. The ideal transformer below has 500 turns on the primary and 300 turns on the secondary. 
+    ![hw4q4](hw4.assets/hw4q4.png)
+    The source produces a voltage $E_g=600V$ and the load is $Z=12\Omega$. Calculate
+
+    1. The voltage $E_2$
+        $$
+        {N_1\over N_2}=a={E_1\over E_2}\\
+        {500\over 300}={600V\over E_2}\\
+        E_2={600V\times3\over5}\\
+        \overline{\underline{|E_2=360V|}}
+        $$
+        The current $I_2$
+        $$
+        I_2=({360V\over 12\Omega})\\
+        \underline{\overline{|I_2=30A|}}
+        $$
+        
+        The current $I_1$
+        $$
+        {I_2\over I_1}=a\\
+        {3\over5}={30A\over I_1}\\
+    I_1={5\times30A\over 3}\\
+        \overline{\underline{|I_1=50A|}}
+        $$
+        
+    2. The power delivered to the primary [$W$]
+    $$
+    P=IV\\
+        P=50A\times600V\\
+    \overline{\underline{|P=30kW|}}
+    $$
+    
+    3. The power output from the secondary [$W$]
+        $$
+        P=IV\\
+        P=60A\times 360V\\
+        \overline{\underline{|P=21.6kW|}}
+        $$
+
+5.  In problem 4, what is the impedance seen by the source $E_g$?
+    $$
+    R_2={a^2R_1}\\
+    R_2={({5\over3})^2\times12\Omega}\\
+    \overline{\underline{|R=33\Omega|}}
+    $$
+    
+6. In the circuit below, calculate the voltage across the capacitor and the current flowing through it.
+    ![hw4q6](hw4.assets/hw4q6.png)
+
+$$
+{R_{real}\times a^2}=R_{observed}\\
+45k\Omega\angle26^\circ\times ({1\over100})^2=R_{observed}\\
+R_{observed}=4.5\Omega\angle26^\circ\\[16pt]
+R_s=4\Omega;\ X_C=2\Omega;\ X_L=5\Omega;\ Z_s=5\Omega\angle53^\circ\\
+I={V\over R}\\
+I={10V\over 5\Omega\angle53^\circ}\\
+I=4A\angle53^\circ\\[16pt]
+{I_2\over I_1}={V_1\over V_2}=a\\
+{I_2\over 4A}={10V\over V_2}={1\over 100}\\
+\overline{\underline{|V_2=1kV;\ I_2=40mA|}}
+$$
+

_Homework/hw5.md

@@ -0,0 +1,94 @@
+# Skyler MacDougall
+
+## Homework 5: due 2/19/2020
+
+14. The primary of a transformer has twice as many turns as the secondary. ($a=2$) The primary voltage is $220V$ and a $5\Omega$ load is connected across the secondary. Calculate the power delivered by the transformer, as well as the primary and secondary current.
+    $$
+    V_p=220V;\ Z_s=5\Omega\\
+    V_s={V_p\over a}={220V\over2}=110V\\
+    P={V_s^2\over Z_s}={110V^2\over5\Omega}\\
+    P=2420W\\[16pt]
+    I_s={V_s\over Z_s}={110V\over 5\Omega}\\
+    I_s=22A\\[16pt]
+    I_p={I_s\over a}={22A\over2}\\
+    I_p=11A\\[20pt]
+    \overline{\underline{|P=2420W;\ I_s=22A;\ I_p=11A|}}
+    $$
+    
+
+
+
+21. Explain why the secondary voltage of a practical transformer decreases with increasing resistive loads. 
+
+The voltage decreases, because there is more voltage going to the internal losses.
+
+
+
+25. A 66.7MVA transformer has an efficiency of 99.3% when it delivers full power to a load having a power factor of 100%.
+    1. Calculate the losses in the transformer under these conditions.
+        $$
+        efficiency={P_s\over P_p}\\
+        P_s=66.7MW*0.993=66.2331MW\\
+        losses=P_p-P_s=66.7MW-66.2331MW\\
+        \overline{\underline{|losses=466.9kW|}}
+        $$
+        
+    2. Calculate the losses and efficiency when the transformer delivers 66.7MVA to a load having a power factor of 80%.
+
+$$
+losses\ are\ consistent\\
+\therefore\\
+losses=466.9kW\\[16pt]
+P_s=66.2MW;\ S={P\over pf}={66.2MW\over0.8}=82.75MVA\\
+S_p={66.7MW\over0.8}=83.375MVA\\
+efficiency = {S_s\over S_p}={83.375MVA\over82.75MVA}\\
+\overline{\underline{|losses = 466.9kW;\ efficiency=98.7\%|}}
+$$
+
+
+
+30. During a short-circuit test on a 10MVA, 66kV-7.6kV transformer, the following results were obtained. 
+    $$
+    E_g=2640V\\
+    I_{sc}=72A\\
+    P_{sc}=9.85kW
+    $$
+    Calculate:
+
+    1. The total resistance and total leakage reactance referred to the 66kV primary side.
+        $$
+        10MVA>> 100kVA\\ \therefore \\ Z\approx X\\[16pt]
+        Z={V\over I}={2640V\over 72A}\approx37\Omega\\
+        \overline{\underline{|R\approx0\Omega;\ X\approx37\Omega|}}
+        $$
+    
+    2. The nominal impedance of the transformer referred to the primary side.
+        $$
+        Z_n={E^2\over S_n}={(66kV)^2\over10MVA}\\
+        \overline{\underline{|Z_n=435.6\Omega|}}
+        $$
+    
+    3. The percent impedance of the transformer.
+        $$
+        Z_p (pu)={Z_p\over Z_{n_p}}={37\Omega\over435.6\Omega}\\
+        \overline{\underline{|Z_p(pu)=8.42\%|}}
+        $$
+
+
+
+31. In the above problem, if the iron losses at rated voltage are 35kW, calculate the full-load efficiency of the transformer if the power factor of the load is 85%.
+
+$$
+35kW@pf=0.85;\ S={P\over pf}={35kW\over0.85}=41.2kVA\\
+efficiency={S_s\over S_p}={10MVA-41.2kVA\over10MVA}\\
+\underline{\overline{|efficiency=99.5\%|}}
+$$
+
+33. If a transformer were actually built according to the below diagram, it would have very poor voltage regulation. Explain why, and propose a method for improving it.
+    ![](hw5.assets/hw5q33.png)
+
+
+
+When the load is reflected across the transformer, it becomes significantly smaller, where the internal losses are significant.
+
+One can change this in as simple a step as changing the sides that the power and the load are connected to, although if current or voltage requirements exist, it may be more difficult.

_Homework/hw6.md

@@ -0,0 +1,32 @@
+# Skyler MacDougall
+
+## Homework 6: due 2/26/2020
+
+7. A single-phase transformer has a rating of 100kVA, 7200V-600V, 60Hz. If it is reconnected as an autotransformer having a ratio of 7800V-7200V, calculate the load it can carry.
+
+$$
+100kVA= S;\ 7200V = V_1;\ 600V=V_2\\
+100kW=P_1
+$$
+
+8. In the above problem, How should the transformer terminals ($H_1$, $H_2$, $X_1$, $X_2$) be connected?
+
+$$
+-
+$$
+
+14. Many airports use series lighting systems in which the primary windings of a large number of current transformers are connected in series across a constant current, 60Hz source. In one installation, the primary current is kept at 20A. The secondary windings are individually connected to a 100W, 6.6A incandescent lamp. 
+    1. Calculate the voltage across each lamp.
+        $$
+        
+        $$
+    
+    2. The resistance of the secondary winding is $0.07\Omega$ while that of the primary is $0.008\Omega$. Knowing that the magetizing current and the leakage reactance are both negligible, calculate the voltage across the primary winding of each transformer.
+        $$
+        
+        $$
+    
+    3. If 140 lamps, spaced at 50m intervals, are connected in series using No. 14 wire, calculate the minimum voltage of the power source. Assume the wire operates at a temperature of $105^\circ C$.
+        $$
+        
+        $$

_Homework/hw7.md

@@ -0,0 +1,33 @@
+# Skyler MacDougall
+
+## Homework 7: due 3/2/2020
+
+7. In order to meet an emergency, three single-phase transformers rated at 100kVA, 13.2kV-2.4kV are connected in wye-delta on a 3-phase, 18kV line.
+    1. What is the maximum load that can be connected to the transformer bank?
+        $$
+        100kW=P_\phi\\
+        P_{3\phi}\approx173kW
+        $$
+    
+    2. What is the outgoing line voltage?
+        $$
+        a={N_p\over N_s}={13.2kV\over2.4kV}=5.5\\
+        V_s={V_p\over a}={18kV\over5.5}\\
+        \overline{\underline{|V_s=3.\overline{27}kV|}}
+        $$
+8. Two transformers rated at 250kVA, 2.4kV-600V are connected in open-delta to supply a load of 400kVA.
+    1. Are the transformers overloaded?
+        $$
+        I_S={S\over I}={250kVA\over 600V}=416.\overline6A\\
+        S=\sqrt3EI=\sqrt3\times416.\overline6A\times600V\\
+        S=433.01kVA\\
+        \therefore\\
+        The\ Transformer\ is\ not\ overloaded
+        $$
+    
+    2. What is the maximum load the bank can carry on a continuous basis?
+        $$
+        Calculated\ above\ to\ be\\
+        S=433.01kVA
+        $$
+        

_Homework/hw8.md

@@ -0,0 +1,126 @@
+# Skyler MacDougall
+
+## Homework 8: due 3/30/2020
+
+1.  Name the principal components of an induction motor.
+    Motors have two parts; a stator and a rotor.
+    
+2.  Explain how a revolving field is set up in a 3-phase induction motor.
+    Current is pushed through the stator.
+    
+3.   If we double the number of poles on the stator of an induction motor, will its synchronous speed also double?
+    No, it will be halved.
+    
+4.   The rotor of an induction motor should never be locked while full voltage is being applied to the stator. Explain.
+    The windings induct current in the rotor, and if left for long enough, it could melt or become otherwise damaged.
+    
+5.   Why does the rotor of an induction motor turn slower than the revolving field?
+    The rotating fields would align, stopping the induced voltage and current, slowing the rotor back down.
+    
+6.   What happens to the rotor speed and the rotor current when the mechanical load on an induction motor increases?
+
+The rotor will slow down, and the current will increase.
+
+10.  A 3-phase, 20-pole induction motor is connected to a 600V, 60Hz source.
+     1.  What is the synchronous speed?
+         $$
+         n_s={120f\over p}\\
+         n_s={120(60Hz)\over 20\ poles}\\
+         n_s=360r/min
+         $$
+         
+     2.  If the voltage is reduced to 300V, will the synchronous speed change? 
+         Because the voltage is not in the function shown above, it will not change the synchronous speed.
+
+
+
+12.  Calculate the approximate values of the starting current, full-load current, and the no-load current of a 150 horsepower, 575V, 3-phase induction motor.
+     $$
+     I_{full}={600P_h\over E}={600(150hp)\over575V}\\
+     \underline{\overline{|I_{full}\approx156A|}}\\
+     I_{nl}=0.3I_{full}=0.3(156A)\\
+     \underline{\overline{|I_{nl}\approx47A|}}\\
+     I_{start}=I_{LR}=6I=6(156A)\\
+     \underline{\overline{|I_{start}=939A|}}
+     $$
+     
+
+
+
+14. How can we change the direction of rotation of a 3-phase induction motor?
+    Swap the second and third phases on the stator.
+
+
+
+15. 
+    1. Calculate the synchronous speed of a 3-phase, 12-pole induction motor that is excited by a 60Hz source.
+        $$
+        n_s={120f\over p}\\
+        n_s={120(60Hz)\over12\ poles}\\
+        n_s=600r/min
+        $$
+        
+    2. What is the nominal speed if the slip at full load is 6%?
+        $$
+        s={n_s-n\over n_s}\\
+        0.06={600-n\over600}\\
+        600-n=36\\
+        n_{nominal}=564r/min
+        $$
+        
+
+
+
+18. A 3-phase 75hp, 440V induction motor has a full-load efficiency of 91% and a power factor of 83%. Calculate the nominal current per phase.
+    $$
+    75hp*0.742kW/hp=55.65kW\\
+    {55.65kW\over0.91\%\ efficiency}=61.15kW\\
+    {61.15kW\over0.83\ pf}=73.68kVA\\
+    P=3V_{phase}I_{phase}\\
+    73.68kVA=3({440V\over\sqrt3})I_{phase}\\
+    I_{phase}={73.68kVA*\sqrt3\over3*440V}\\
+    I_{phase}=96.68A
+    $$
+
+
+
+23. A large, 3-phase, 4kV, 60Hz, squirrel-cage induction motor draws a current of 385A and a total active power of 2344kW when operating at full load. The corresponding speed is accurately measured at 709.2r/min. The stator is connected in wye and the resistance between the two stator terminals is $0.10\Omega$. The total iron losses are 23.4kW and the windage and friction losses are 12kW. Calculate the following:
+    1. the power factor at full load
+        $$
+        
+        $$
+    
+    2. the active power supplied to the rotor
+        $$
+        P_{js}=3I^2R=3*(385A)^2*(0.10\Omega/2)\\
+        P_{js}\approx22.2kW\\
+    P_e=2344kW\\
+        P_f=23.4kW\\
+        P_r=P_e-P_f-P_{js}=2344kW-23.4kW-22.2kW\\
+        \overline{\underline{|P_r=2298kW|}}
+        $$
+        
+    3. the total $I^2R$ losses in the rotor
+        $$
+        s=0.004\\
+        P_{jr}=sP_r=0.004*2298kW\\
+    \overline{\underline{|S_{jr}=3.3kW|}}
+        $$
+        
+    4. The load mechanical power [kW], torque [$kN*m$], and efficiency [%].
+        $$
+        P_m=P_r-P_{jr}=2298kW-3.3kW\\
+        \overline{\underline{|P_m=2295kW|}}\\
+        T=9.55P_r/n_s={9.55*2298kW\over 720r/min}\\
+        \overline{\underline{|T=30480n*m|}}
+        $$
+
+
+
+24.  If we slightly increase the rotor resistance of an induction motor, what effect does this have (increase or decrease) upon
+     1.  Starting torque
+     2.  Starting current
+     3.  Full-load speed
+     4.  Efficiency
+     5.  Power factor
+     6.  Temperature rise of the motor at its rated power output.

_Homework/hw9.md

@@ -0,0 +1,77 @@
+# Skyler MacDougall
+
+## Homework 9: due 3/25/2020
+
+1.  What is the difference between a drip-proof motor and an explosion-proof motor?
+    Drip-proof motors are protected from dripping liquids up to 15 degrees from the vertical. An explosion-proof motor, by contrast, are fully sealed, and the frames are designed to withstand the enormous pressure.
+
+2.  What is the approximate life expectancy for a motor?
+    It depends on the size, speed, horsepower, what voltage and current its run at, and how long it is run at one time.
+
+3.  Explain what a NEMA Design D motor is unsatisfactory for driving a pump.
+    A pump should be a consistent speed, and shouldn't take long to get to full speed. A NEMA Design D motor achieves neither of these.
+
+4.  Identify the motor components shown in Fig. 3
+    ![](hw9.assets/hw9q4.png)
+
+5.  Show the flow of active power in a 3-phase induction motor when it operates
+
+    1.  As a motor.
+        Power is being put into the system in the same direction as rotation.
+    2.  As a brake.
+        Power is being put in in the opposite direction of rotation. As such, a lot of power is lost as heat.
+
+    ![image-20200401140723691](hw9.assets/image-20200401140723691.png)
+
+6.  Will a 3--phase motor continue to rotate if one of the lines becomes open? Will the motor be able to start on such a line?
+    The motor needs all 3 phases to start. It will be able to run with only two phases, but there will be a lot of excess heat, and the motor will wear out faster. 
+
+14. A 300hp, 2.3kV, 3-phase, 60Hz, squirrel-cage induction motor turns at a full-load speed of 590r/min. Calculate the approximate value of rotor $I^2R$ losses. 
+    $$
+    P_{jr}=({120{f\over p}-n\over 120{f\over p}})*(3P)\\
+    P_{jr}=({120{60Hz\over 6\ poles}-590rpm\over 120{60Hz\over 6\ poles}})*(3(300hp*746W/hp))\\
+    P_{jr}=({1200rpm-590rpm\over 1200rpm})*(3(223.8kW))\\
+    P_{jr}=(0.508\overline3)*(3(223.8kW))\\
+    \underline{\overline{|P_{jr}=341.295kW|}}
+    $$
+    If the line voltage then drops to 1944V, calculate the following:
+    
+    1. The new speed, knowing that the load torque remains the same.
+        $$
+        n=1200(1-s({V_1\over V_2})^2)=1200(1-0.508\overline3({2300V\over 1944V})^2)\\
+        n=346rpm
+        $$
+    
+    2. The new power output.
+        $$
+        P'=P*{n'\over n}=300hp*{346rpm\over 590rpm}\\
+        P'=176.34hp\approx131.6kW
+        $$
+        
+    3. The new $I^2R$ losses in the rotor.
+        $$
+        P_{jr}=({120{f\over p}-n\over 120{f\over p}})*(3P)\\
+        P_{jr}=({120{60Hz\over 6\ poles}-346rpm\over 120{60Hz\over 6\ poles}})*(3(131.6kW))\\
+        P_{jr}=0.711\overline6*3(131.6kW)\\
+        P_{jr}=280.6kW
+        $$
+        
+
+
+
+27.  The bearings in a motor have to be greased regularly, but not too often. The following schedule applies to two motors:
+     Motor A: 75hp, 3550r/min; lubricate every 2200 hours of running time.
+     Motor B: 75hp,   900r/min; lubricate every 10000 hours of running time.
+     Motor A runs continually, 24 hours per day. Motor B drives a compressor and operates about 6 hours per day. How often should the bearings of each motor be greased per year?
+     Motor A should be greased about once every quarter, or 4 times in a year.
+     Motor B should be greased about once every 5 years.
+
+28.  A 40hp 1780r/min, 460V, 3-phase, 60Hz, drip proof Baldor Super E premium energy induction motor has a power factor of 86% and an efficiency of 93.6%. The motor, priced at $2243, runs at full-load 12 hours a day, 5 days a week. Calculate the driving cost of the motor during a 3-year period, knowing that the cost of energy is \$0.06/kWh.
+     $$
+     12\ hours*5\ days/week\ *52\ weeks/year*3\ years=9360\ hours\\
+     P_{in}={P_{out}\over eff}={40hp*746W/hp\over0.936}=31.9kW\\
+     Cost=price/kWh*kW*h=\$0.06*9360\ hours*31.9kW\\
+     Running\ cost=\$17,904.00\\
+     Cost\ w/\ motor=\$20,147.00
+     $$
+