· 9 years ago · Jan 24, 2017, 03:48 PM
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10\hyphenation{op-tical net-works semi-conduc-tor}
11
12\begin{document}
13 \title{PLACEHOLDER TITLE}
14
15 \author{\IEEEauthorblockN{F.J.J. Hoens\\} % put here you name (initials and last name)
16 \IEEEauthorblockA{f.j.j.hoens@student.tue.nl, ID 0964741} % put here you email address and your ID number
17 \thanks{Lab 1: setup 3, J. Roberts% put here the number of the laboratory setup an the names of your partner during the experiments
18
19 Lab 2: setup 3, J. Roberts
20
21 Lab 3: setup 1, J. Roberts
22 }
23 }
24\maketitle
25
26\begin{abstract}
27 content...
28\end{abstract}
29\section{Introduction}
30\IEEEPARstart{T}{he} world is becoming increasingly more electric. Mechanical processes are constantly replaced by their electric and digital counterparts. One of the areas which is in the middle of this transition is the field of the motor. Compared to other electrical transitions, the motor one seems to be slower. The transition can be sped up however if high quality models are created of electro motors, this allows engineers to model more efficient electric motors and then create an actual motor out of that design.
31
32In this paper three electrical machines will be compared to their respective model. The accuracy of the model will be discussed and recommendations on how to improve the model will be included.
33\section{Methods}
34\subsection{Measuring and Cutting the data}
35To measure the required data, a torque step sequencer is set up with a step size of 20. The torque can go from -5 Nm to 5 Nm, this means each step is equal to .5 Nm. The torque step sequencer has to run a couple of time with different armature voltages, in order to compare the efficiency of the machines at different armature voltages.
36
37Before the data can be analysed, it has to be processed. First off, it has to be cut from 128000 values to the step size 20. One second of measurement data records 1600 values, the used step time was 4 seconds. It takes a bit of time for the machines to settle in, that is why the first 400 values are cut off. The other 1200 values are used to calculate a mean. This is done for every torque and for every armature voltage.
38\begin{equation}
39Cut_{start} = 400+1600T_s (i-1)+1+20T_s 1600 (k-1)
40\end{equation}
41\begin{equation}
42Cut_{end} = 20T_s 1600 (k-1) + 1600T_s
43\end{equation}
44Where $T_s$ is step time, i is step number and k is armature voltage step (first armature voltage, second...). These cut values can now be used to calculate the mean of the measured data. This can be done in MATLAB like this\\
45\texttt{$I_{a,cut}$ = mean($I_{a,measured}$($Cut_{start}$:$Cut_{end}$));\\
46 $n_{a,cut}$ = mean($n_{measured}$($Cut_{start}$:$Cut_{end}$));\\
47 $V_{a,cut}$ = mean($V_{a,measured}$($Cut_{start}$:$Cut_{end}$));.
48}
49Where $I_{a,measured}$ is the measured armature current, $n_{measured}$ is the measured rotation speed and $V_{a,measured}$ is the armature voltage.
50\subsection{DC Machine Measurements}
51
52\begin{table}
53 \label{tab:DCM}
54 \centering
55 \caption{Parameters of the DC machine}
56 \begin{tabular}{ l |l| l| l }
57 Symbol&Variable&Value&Unit\\
58 \hline\hline
59 K$\phi_p$&Machine and flux constant&1.05&$\frac{Vs}{rad}$\\
60 $R_f$&Field resistance&259&Ohm\\
61 $V_f$&Field voltage&207&Volt\\
62 $I_f$&Field current&0.8&Ampere\\
63 $R_a$&Armature resistance&0.98&Ohm\\
64 $f_c$&Coulomb constant&0.8445&Nm\\
65 $f_v$&Viscous constant&0.005792&$\frac{Nms}{rad}$\\
66 \hline
67 \end{tabular}
68\end{table}
69
70After all the data has been cut, the efficiency can be calculated. This means the friction has to be found, the friction is given by
71\begin{equation}
72 T_{FW} = f_c n_{cut} + f_v
73\end{equation}
74where $f_c$ is the first friction constant and $f_v$ is the second friction constant. The developed torque is obtained by
75\begin{equation}
76 T_d = K \phi_p * I_a
77\end{equation}
78where K is the machine constant and $\phi$ the flux. The mechanical speed is defined by
79\begin{equation}
80\omega = 2\pi n_{cut} .
81\label{eq:omega}
82\end{equation}
83The developed power is
84\begin{equation}
85P_{developed} = K \phi_p \omega I_{a, cut}.
86\end{equation}
87The shaft torque is then given by
88\begin{equation}
89T_s = T_{dc} - 0.8T_{FW},
90\label{eq:Ts}
91\end{equation}
92where $T_{FW}$ is multiplied by 0.8 due to the fact that 2 machines are coupled, an estimation has been made that 80\% of the measured friction is due to the DC machine, and the other 20\% due to the PMSM. To find the efficiency of the model, the electric and mechanical power has to be calculated. The equation for electric power is given by
93\begin{equation}
94P_{electrical}=I_f V_f + I_{a, cut} V_{a, cut}, \label{Pelectrical}
95\end{equation}
96where $I_f$ is the field current and $V_f$ is the field voltage.
97The equation for mechanical power is given by
98\begin{equation}
99P_{mechanical}=2\pi n T_s. \label{Pmechanical}
100\end{equation}
101The efficiency is then given by
102\begin{equation}
103\eta = \frac{P_{mechanical}}{P_{electrical}}. \label{efficiency}
104\end{equation}
105It is important to note that Equations \ref{Pelectrical}, \ref{Pmechanical} and \ref{efficiency} will be different if the torque is negative because the DC machine would not act as a motor but as a generator. This changes the equations slightly.$P_{electrical}$ and $P_{mechanical}$ become negative, and the efficiency is now given by
106\begin{equation}
107\eta = \frac{P_{elecitrical}}{I_fV_f+P_{mechanical}}.
108\end{equation}
109There will be a point where the efficiency will be 0. This occurs when the the shaft torque $T_s$ is smaller than 0 and the negative friction torque $T_{FW}$ is smaller than the shaft torque $T_s$.
110
111\subsection{DC Machine Model}
112
113Now that the measured data has been processed, a model has to be created to compare the data to. A simulation of the torque step sequencer can be made by creating a variable starting at -5 and constantly incrementing by the step size until it reaches 5. Combining this variable which will be called $T_{ref}$ together with the friction $T_{FW}$ into one equation, the $T_{dc}$ can be obtained
114\begin{equation}
115T_{dc} = T_{FW} - T_{ref}.
116\end{equation}
117As soon as $T_{dc}$ is known, it becomes possible to calculate the shaft torque using equation \ref{eq:Ts}. The armature voltage is then defined by
118\begin{equation}
119I_a = \frac{T_{dc}}{K\phi_p}
120\end{equation}
121where K$\phi_p$ is the machine and flux constant given in table \ref{tab:DCM}. The excitation voltage E can be deducted from
122\begin{equation}
123E=V_a-R_aI_a.
124\end{equation}
125Where $R_a$ is the armature resistance and $V_a$ is the armature voltage. The frequency of the DC machine can be given by
126\begin{equation}
127n = \frac{E}{2\pi K\phi_p}.
128\label{eq:nErelation}
129\end{equation}
130Now that all these values have been modelled and calculated, the efficiency can be determined through the same method as used in subsection B.
131\subsection{PMSM Measurements}
132\begin{table}
133 \label{tab:PMSM}
134 \centering
135 \caption{Parameters of the PMSM}
136 \begin{tabular}{ l |l| l| l }
137 Symbol&Variable&Value&Unit\\
138 \hline\hline
139 $R_s$&Stator resistance&0.75&Ohm\\
140 $L_s$&Synchronous inductance&3.15&mH\\
141 p&Poles&8&None\\
142 $\lambda$&Flux linkage&0.134&Wb-t\\
143 $f_c$&Coulomb constant&0.8445&Nm\\
144 $f_v$&Viscous constant&0.005792&$\frac{Nms}{rad}$\\
145 \hline
146 \end{tabular}
147\end{table}
148
149\begin{figure}
150 \centering
151 \includegraphics{PMSMcircuit.eps}
152 \caption{Equivalent circuit of a permanent magnet synchronous machine}
153 \label{fig:PMSMcircuit}
154\end{figure}
155
156
157The test machine used is the SEM HJ130G8-130S PMSM, it's parameters are shown in table \ref{tab:PMSM}.
158
159An important first step is to note that the armature current $I_a$ has to be negative when the PMSM is in generator mode, therefore $I_a$ should be multiplied by -1 as long as the torque is less than 0. The phase current can be found by substituting,
160\begin{equation}
161\theta = -\arccos(PF)
162\end{equation}
163into
164\begin{equation}
165I_{a, phase} = |I_a \cos(\theta) + 1j \sin(\theta)|
166\end{equation}
167where PF is the measured power factor. The electrical rotation be obtained by
168\begin{equation}
169\omega_{elec} = \frac{\omega_{mech} p}{2}
170\end{equation}
171where p is the number of poles and $\omega_{mech}$ is defined by equation \ref{eq:omega}. Using the equivalent circuit of a PMSM as seen in figure \ref{fig:PMSMcircuit} the definition of $E_f$ is defined by
172\begin{equation}
173E_{f} = |V_{an} - I_{a, phase}(R_s + 1j L_s \omega_{elec})|
174\end{equation}
175where $V_{an}$ is the measured phase voltage, $R_s$ is the stator resistance and $L_s$ is the synchronous inductance. Then the developed torque can be obtained through
176\begin{equation}
177P_d = \frac{E_f I_a \cos(\delta - \theta)}{\omega_{mech}}
178\label{eq:PMSMPd}
179\end{equation}
180where $\delta$ is the phase angle of $E_{f, phase}$, and the shaft torque as
181\begin{equation}
182T_s = 3 T_d - 0.2 (f_c n_{cut} + f_v)
183\end{equation}
184where $f_c$ and $f_v$ are the friction constants and $n_{cut}$ is the cut rotation speed. A multiplication by 0.2 is done because the PMSM accounts for 20\% of the total friction. Just like in the DC machine, there has to be accounted for the motor and generator mode. If the shaft torque is bigger or equal to zero, the efficiency is given by
185\begin{equation}
186\eta = \frac{T_s \omega_{mech} }{3 PF V_{an} I_a}.
187\label{eq:Efficiency1}
188\end{equation}
189However, if the shaft torque is smaller than the negative friction torque ($f_c n_{cut} + f_v$) the efficiency is obtained by
190\begin{equation}
191\eta = \frac{- 3 PF V_{an} I_a}{- T_s \omega_{mech} }.
192\label{eq:Efficiency2}
193\end{equation}
194The final case is when the shaft torque is smaller than zero, but bigger than the negative friction torque. In this case the efficiency should be zero.
195\subsection{PMSM Model}
196In the permanent magnet synchronous machine model, the armature current $I_a$ has to increment constantly by the step size of the model. Moreover multiple values of the speed n should be chosen, these should be the same values of n as used in the actual measured data. Knowing these values, $\omega_{mech}$, $\omega_{elec}$ and $X_s$ can be calculated using the equations given in subsection D. Using the flux linkage given in table \ref{tab:PMSM}, the excitation voltage is defined by
197\begin{equation}
198E_f = \lambda \omega_{elec}
199\end{equation}
200where $\lambda$ is the flux linkage. As B is equal to 0 in the model, the excitation voltage is in phase with the armature current. The phase voltage is therefore obtained by
201\begin{equation}
202V_{an} = |E_f + I_a R_s + j I_a X_s|
203\end{equation}
204where $I_a$ is the armature voltage, $R_s$ is the stator resistance and $X_s$ is the stator reluctance. As the phase angle is zero in the model, the equation for the developed power is equal to equation \ref{eq:PMSMPd} with $\cos(0)$. Now all the same steps as in subsection D can be followed, however the power factor PF is not know yet. The power factor is defined by
205\begin{equation}
206PF = \cos(\theta)
207\label{eq:PF}
208\end{equation}
209where $\theta$ is the angle of the phase voltage $V_{an}$. And finally the efficiency $\eta$ can be calculated exactly as done before.
210\subsection{Induction Machine Measurements}
211\begin{table}
212 \label{tab:IM}
213 \centering
214 \caption{Parameters of the Induction machine}
215 \begin{tabular}{ l |l| l| l }
216 Symbol&Variable&Value&Unit\\
217 \hline\hline
218 $R_s$&Stator resistance&0.75&Ohm\\
219 $L_s$&Synchronous inductance&3.15&mH\\
220 p&Poles&8&None\\
221 $\lambda$&Flux linkage&0.134&Wb-t\\
222 $f_c$&Coulomb constant&0.8445&Nm\\
223 $f_v$&Viscous constant&0.005792&$\frac{Nms}{rad}$\\
224 \hline
225 \end{tabular}
226\end{table}
227The induction machine that was used in this paper, is the 3.0 kW SEW DRE100L2/F1/EV7C.
228\subsection{Induction Machine Model}
229For this model, the frequency is a constant and the slip will be an incrementing variable which starts at -1 and stops at 1. The RMS phase voltage is defined by
230\begin{equation}
231V_{an} = a n_s + b
232\end{equation}
233where a and b are controller constants and $n_s$ is the stator frequency. The slip frequency can be determined using equation \ref{eq:omega}, then the model frequency can be determined by
234\begin{equation}
235n_{model} = \frac{\omega_{mech}}{\omega_{elec}} = \frac{\omega_s (1-s)}{0.5\omega_s p}.
236\end{equation}
237\begin{figure}
238 \centering
239 \includegraphics{IMcircuit.eps}
240 \caption{Equivalent circuit of a single phase of the induction machine}
241 \label{fig:IMcircuit}
242\end{figure}
243We have to model the impedances of the three inductors as seen in figure \ref{fig:IMcircuit}, each of the three impedances can be defined as
244\begin{equation}
245Z = \omega_{elec} L + R
246\end{equation}
247where L is the specific inductance of that particular inductor and R is equal to the specific resistance of the resistor, as seen in the equivalent circuit in figure \ref{fig:IMcircuit}. Now that all impedances are known, it is possible to find out the total impedance $Z_{tot}$ and the currents $I_1$ and $I_2$.
248\begin{equation}
2493T_d = \frac{3 I_2^2 R_2}{s \omega_s}
250\end{equation}
251The power factor can be derived from equation \ref{eq:PF} where the equation has to be absolute due to the fact that the PF becomes negative as soon as $I_1$ is bigger than 90 degrees. The shaft torque is defined by
252\begin{equation}
253T_s = 3T_d - 0.2 T_{FW}
254\end{equation}
255where $T_{FW}$ is friction torque measured. $T_{FW}W$ is multiplied by 0.2 because 80\% of the measured friction torque comes from the attached DC machine. The efficiency is now defined by equation \ref{eq:Efficiency1} in case the slip is between 0 and 1 while the efficiency is defined by equation \ref{eq:Efficiency2} in case the slip is less than 0.
256
257
258\section{Results}
259\subsection{Error measurement}
260To find out how big of an error exists between the model and the actual measurements. The average discrepancy is calculated, this is defined as
261\begin{equation}
262\delta =mean(~\frac{|Modelled - Measured|}{Measured}~).
263\end{equation}
264where Modelled contains the modelled value, and Measured contains the measured value. All these values are recorded in table \ref{tab:Errors}. The average discrepancy $\delta$ is a mean value and does not include far outliers, which helps to find a stable and accurate error measurement. From these values it is very simple to find the inaccuracy percentage, this is done by multiplying all the values by 100.
265\begin{table}
266 \centering
267 \caption{Relative errors for all machines}
268 \begin{tabular}{ l |l| l}
269 Symbol&Variable&Value\\
270 \hline\hline
271 $\delta n_{dc}$&Average speed discrepancy of DCM&0.023\\
272 $\delta \eta_{dc}$&Average efficiency discrepancy of DCM&0.081\\
273 $\delta n_{sm}$&Average speed discrepancy for PMSM&$1.09 \cdot 10^-5$\\
274 $\delta \eta_{sm}$&Average efficiency discrepancy for PMSM&0.0887\\
275 $\delta P_{f_{sm}}$&Average power factor discrepancy for PMSM&0.0091\\
276 \hline
277 \end{tabular}
278 \label{tab:Errors}
279\end{table}
280\subsection{DC Machine}
281\begin{figure}
282 \centering
283 \includegraphics{DCMfigure.eps}
284 \caption{Plots of the Efficiency and Frequency compared with the Torque, measured and model values.}
285 \label{fig:DCMGraph}
286\end{figure}
287When comparing the efficiency of the measurement with that of the model. A quick glance at figure \ref{fig:DCMGraph} shows that the model is a bit inaccurate. Most of this is caused by the estimation of the friction torque $T_{FW}$. Moreover the inaccuracy is not equally distributed through the three different armature voltages. This can be explained by the fact that the inaccurate $T_{FW}$ will affect a higher $V_a $ more due to the relationship with the speed $n_s$ as seen in equation \ref{eq:nErelation}. Interestingly the efficiency of the created model is lower than the measured efficiency, while the rotational speed n of the model is higher than the measured speed. This can be explained by equations \ref{Pmechanical} and \ref{efficiency}. $P_{mechanical}$ rises due to the higher value of n, and this decreases the value of $\eta$ because $P_{mechanical}$ is in the denominator.
288
289In the end, the inaccuracy of the speed is 2\% and the inaccuracy for the efficiency is 8\%. 8\% is pretty significant but is totally understandable due to the estimated values used in the model. The model can become more accurate by approximating the friction more accurately.
290\subsection{Permanent Magnet Synchronous Machine}
291\begin{figure}
292 \centering
293 \includegraphics{PMSMfigure.eps}
294 \caption{Plots of the Efficiency, Frequency and Power factor compared to the Torque, measured and model values.}
295 \label{fig:DCMGraph}
296\end{figure}
297A trend starts to occur when
298
299
300The most glaring error in the model is immediately visible, the power factor does not drop to 0.95 but to 0.97. Which is a significant outlier if that value is compared to the average discrepancy $\delta P_{f_{sm}}$. This is however still an acceptable inaccuracy of 2.1\%.
301\subsection{Induction Machine}
302\begin{figure}
303 \centering
304 \includegraphics{IMfigure.eps}
305 \caption{Plots of the Efficiency, Torque and Power factor compared to the Speed, measured and model values.}
306 \label{fig:IMGraph}
307\end{figure}
308The Induction machine has no data in table \ref{tab:Errors}, this is because an error has been made in the model. Looking at \ref{fig:IMGraph} shows that the model does not accurately represent what has been measured. This means that the induction machine model created in this paper can not be used to make accurate predictions of the behaviour of an induction machine. The error is mostly likely due to the usage of slip as a variable. The model is not entirely flawed and can easily be improved to get rid of the current error.
309\section{Conclusion}
310\end{document}