Excercises Electric Machinery Fundamentals

Chapter 1 Problem 1-22



In [1]:

    
%pylab inline
%precision 1









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.1f'

Description

A linear machine has the following characteristics:

$B = 0.5\,T$ into the page
$R = 0.25\,\Omega$
$l = 0.5\,m$
$V_B = 120\,V$



In [2]:

    
B = 0.5   # [T]
R = 0.25  # [Ohm]
l = 0.5   # [m]
Vb = 120  # [V]

(a)

If this bar has a load of 20 N attached to it opposite to the direction of motion, what is the steady-state speed of the bar?

(b)

If the bar runs off into a region where the flux density falls to 0.45 T, what happens to the bar?
What is its final steady-state speed?

(c)

Suppose V B is now decreased to 100 V with everything else remaining as in part (b). What is the new steady-state speed of the bar?

(d)

From the results for parts (b) and (c), what are two methods of controlling the speed of a linear machine (or a real dc motor)?

SOLUTION

(a)

With a load of 20 N opposite to the direction of motion, the steady-state current flow in the bar will be given by $$ F_\text{app} = F_\text{ind} = ilB $$ $$ \rightarrow i = \frac{F_\text{app}}{Bl} $$



In [3]:

    
Fapp = 20  # [N]
i = Fapp / (B*l)
i









    Out[3]:





80.0

Amperes. The induced voltage in the bar will be $$ e_\text{ind} = V_B - iR $$



In [4]:

    
e_ind = Vb - i*R
e_ind









    Out[4]:





100.0

Volts, and the velocity of the bar will be

$$v = \frac{e_\text{ind}}{Bl}$$



In [5]:

    
v = e_ind / (B*l)
v









    Out[5]:





400.0

metres/second.

(b)

If the flux density drops to 0.45 T while the load on the bar remains the same, there will be a speed transient until $F_\text{app} = F_\text{ind} = 20\,N$ again. The new steady state current will be

$$ F_\text{app} = F_\text{ind} = ilB $$$$ \rightarrow i = \frac{F_\text{app}}{Bl} $$



In [6]:

    
B = 0.45  # [T]
Fapp = 20 # [N]
i = Fapp / (B*l)
i









    Out[6]:





88.9

Amperes. The induced voltage in the bar will be

$$ e_\text{ind} = V_B - iR $$



In [7]:

    
e_ind = Vb - i*R
e_ind









    Out[7]:





97.8

Volts, and the velocity of the bar will be

$$v = \frac{e_\text{ind}}{Bl}$$



In [8]:

    
v = e_ind / (B*l)
v









    Out[8]:





434.6

(c)

If the battery voltage is decreased to 100 V while the load on the bar remains the same, there will be a speed transient until $F_\text{app} = F_\text{ind} = 20\,N$ again. The new steady state current will be $$ F_\text{app} = F_\text{ind} = ilB $$ $$ \rightarrow i = \frac{F_\text{app}}{Bl} $$



In [9]:

    
Vb = 100  # [V]
B = 0.45  # [T]
Fapp = 20 # [N]
i = Fapp / (B*l)
i









    Out[9]:





88.9

Amperes. The induced voltage in the bar will be

$$ e_\text{ind} = V_B - iR $$



In [10]:

    
e_ind = Vb - i*R
e_ind









    Out[10]:





77.8

Volts, and the velocity of the bar will be

$$v = \frac{e_\text{ind}}{Bl}$$



In [11]:

    
v = e_ind/ (B*l)
v









    Out[11]:





345.7

m/s.

(d)

From the results of the two previous parts, we can see that there are two ways to control the speed of a linear dc machine. Reducing the flux density $B$ of the machine increases the steady-state speed, and reducing the battery voltage $V_B$ decreases the stead-state speed of the machine. Both of these speed control methods work for real dc machines as well as for linear machines.