Michael S. Czahor

WESEP 502X Homework 1

Problem 1

Declare Variables

R1=.294 # Stator Resistance
X1=.503
Rc=1000
Xm=13.25
X2=.209  #Rotor Area 
R2=.061
f=60     #Frequency measured in Hertz
p=3     #Pole Pairs
Ws=125.664 #Stator Speed rad/second
V1=220/sqrt(3) #Line-Neutral Voltage
ws=377
i=complex(imaginary=1)  #Create Complex Number

Circuit Analysis

Rpar=(Rc*i*Xm)/(Rc+(i*Xm)) #Par Cor Impedance

slip= matrix(nrow = 251, ncol = 1)
for (t in 0:251) {
  slip[t,1] = (Ws-t)/Ws
}

row.names(slip)=1:251
colnames(slip)=c("slip")

head(slip)

##     slip
## 1 0.9920
## 2 0.9841
## 3 0.9761
## 4 0.9682
## 5 0.9602
## 6 0.9523

Req=R2*(1-slip)/slip
colnames(Req)="Req"
head(Req)

##         Req
## 1 0.0004893
## 2 0.0009865
## 3 0.0014919
## 4 0.0020055
## 5 0.0025277
## 6 0.0030586

Rs2=R2+Req+i*X2 #Series R Imp

Rs1=R1+i*X1 #Series Stator Impedance

Current Analysis

I1=V1/(Rs1+((Rpar*Rs2)/(Rpar+Rs2)));
colnames(I1)="I1"
head(I1)

##            I1
## 1 71.6-143.5i
## 2 71.6-143.4i
## 3 71.7-143.3i
## 4 71.7-143.2i
## 5 71.8-143.1i
## 6 71.9-143.1i

I2=(Rpar/(Rpar+Rs2))*I1; 
colnames(I2)="I2"
head(I2)

##            I2
## 1 71.1-140.9i
## 2 71.1-140.8i
## 3 71.2-140.8i
## 4 71.2-140.7i
## 5 71.3-140.6i
## 6 71.4-140.5i

I2a=abs(I2);

Torque Computation

Torque=(3*3*I2a^2*R2)/(ws*slip);
colnames(Torque)="Torque"
head(Torque)

##   Torque
## 1  36.56
## 2  36.83
## 3  37.11
## 4  37.40
## 5  37.68
## 6  37.98

Plot

length(Torque)

## [1] 251

wm=(1:251)

options(repos=structure(c(CRAN="http://streaming.stat.iastate.edu/CRAN/")))
install.packages("ggplot2")

## 
## The downloaded binary packages are in
##  /var/folders/s7/b77ffsr9155dls85f90m83lc0000gn/T//RtmpIBz5et/downloaded_packages

library(ggplot2)

Finalplot=data.frame(wm,Torque)
qplot(wm,Torque,data=Finalplot,xlab="WM in Rad/Second",ylab="Torque (pu)")

Problem 2a

Declare Variables

Pmech=-1 #Measured in MW
Slip=.3 

Calculate Ps and Pr

Ps=Pmech/(1-Slip)
Ps   #MW

## [1] -1.429

Pr=-Slip*Ps
Pr   #MW

## [1] 0.4286

2a Quick Facts

This is subsynchronous

Ps < 0 Pointing towards grid delivers power via stator

Pm < 0 Pointing away from rotor machine receives mechanical power

Pr > 0 Pointing towards rotor, machine receives power via rotor

Motoring

Problem 2b

Declare Variables

Pmech=-1 #Measured in MW
Slip=-.3 

Calculate Ps and Pr

Ps=Pmech/(1-Slip)
Ps   #MW

## [1] -0.7692

Pr=-Slip*Ps
Pr   #MW

## [1] -0.2308

2b Quick Facts

This is supersynchronous

Ps < 0 Pointing towards grid delivers power via stator

Pm < 0 Pointing away from rotor machine receives mechanical power

Pr < 0 Pointing towards rotor, machine delivers power via rotor

Generating

Problem 3

Double Fed Induction Generator rated at 2 Megawatts

Delivers grid a maximum of 2 MW

Declare Variables

Pmech=2 #MW
Slip=seq(-.3,.3,by=.01)

Solve For Ps

Ps= matrix(nrow = 61, ncol = 1)
for (t in 0:61) {
  Ps[t,1] = (Pmech/(1-Slip[t]))
}

row.names(Ps)=1:61
colnames(Ps)="MW"
head(Ps)

##      MW
## 1 1.538
## 2 1.550
## 3 1.562
## 4 1.575
## 5 1.587
## 6 1.600

Maximize Stator Winding Capacity

max(Ps)

## [1] 2.857

nrow(Ps)

## [1] 61

Ps[61]

## [1] 2.857

Slip[61]

## [1] 0.3

Comments on solution

Maximizing the power that the stator windings needed to be able to handle yielded a solution of 2.857 MW. This occured when the slip was .3. This represents a subsynchronous system. In this type of system the machine will be receiving power via the rotor.