Excercises Electric Machinery Fundamentals

Chapter 1 Problem 1-21



In [1]:

    
%pylab inline
%precision 1









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.1f'

Description

A linear machine has a magnetic flux density of 0.5 T directed into the page, a resistance of $0.25\,\Omega$, a bar length $l = 1.0\,m$, and a battery voltage of $100\,V$.



In [2]:

    
B = 0.5  #[T]
R = 0.25 #[Ohm]
l = 1.0  #[m]
Vb = 100 #[V]

(a)

What is the initial force on the bar at starting? What is the initial current flow?

(b)

What is the no-load steady-state speed of the bar?

(c)

If the bar is loaded with a force of 25 N opposite to the direction of motion, what is the new steady-state speed? What is the efficiency of the machine under these circumstances?

SOLUTION

(a)

The current in the bar at starting is $i = \frac{V_B}{R}$:



In [3]:

    
i = Vb/R
i









    Out[3]:





400.0

Amperes. Therefore, the force on the bar at starting is

$$ \vec{F} = i(\vec{l}\times\vec{B}) = ilB$$



In [4]:

    
F = i*l*B
F









    Out[4]:





200.0

Newton, to the right.

(b)

The no-load steady-state speed of this bar can be found from the equation

$$ V_B = e_\text{ind} = (\vec{v}\times \vec{B}) \cdot \vec{l} = vBl $$$$ \rightarrow v = \frac{V_B}{Bl} $$



In [5]:

    
v = Vb / (B*l)
v









    Out[5]:





200.0

metres/second.

(c)

With a load of 25 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 [6]:

    
Fapp = 25  #[N]
i = Fapp / (B*l)
i









    Out[6]:





50.0

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]:





87.5

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]:





175.0

m/s. The input power to the linear machine under these conditions is $P_\text{in} = V_B i$



In [9]:

    
Pin = Vb*i
Pin









    Out[9]:





5000.0

Watts. The output power from the linear machine under these conditions is $P_\text{out} = V_B i$



In [10]:

    
Pout = e_ind * i
Pout









    Out[10]:





4375.0

Watts. Therefore, the efficiency of the machine under these conditions is

$$ \eta = \frac{P_\text{out}}{P_\text{in}} $$



In [11]:

    
eta = Pout/Pin * 100  # [%]
eta









    Out[11]:





87.5

percent.