Excercises Electric Machinery Fundamentals

Chapter 4 Problem 4-10



In [1]:

    
%pylab notebook
%precision 2









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.2f'

Description

Three physically identical synchronous generators are operating in parallel. They are all rated for a full load of 100 MW at $PF=0.8$ lagging. The no-load frequency of generator A is 61 Hz, and its speed droop is 3 percent. The no-load frequency of generator B is 61.5 Hz, and its speed droop is 3.4 percent. The no-load frequency of generator C is 60.5 Hz, and its speed droop is 2.6 percent.



In [2]:

    
Pn     = 100e6 # [W]
PF     =   0.8
f_nl_A =  61.0 # [Hz]
SD_A   =   3   # [%]
f_nl_B =  61.5 # [Hz]
SD_B   =   3.4 # [%]
f_nl_C =  60.5 # [Hz]
SD_C   =   2.6 # [%]

(a)

If a total load consisting of 230 MW is being supplied by this power system:

What will the system frequency be and how will the power be shared among the three generators?

(b)

Create a plot showing the power supplied by each generator as a function of the total power supplied to all loads (you may use Python to create this plot).
At what load does one of the generators exceed its ratings?
Which generator exceeds its ratings first?

(c)

Is this power sharing in (a) acceptable? Why or why not?

(d)

What actions could an operator take to improve the real power sharing among these generators?

SOLUTION

(a)

Speed droop is defined as:

$$SD = \frac{n_\text{nl}-n_\text{fl}}{n_\text{fl}} \cdot 100\% = \frac{f_\text{nl} - f_\text{fl}}{f_\text{fl}} \cdot 100\%$$

so,

$$f_n = \frac{f_\text{nl}}{\frac{SD}{100} + 1}$$

Thus, the full-load frequencies of generators A, B and C are:



In [3]:

    
f_fl_A = f_nl_A / (SD_A / 100.0 +1)
f_fl_B = f_nl_B / (SD_B / 100.0 +1)
f_fl_C = f_nl_C / (SD_C / 100.0 +1)
print ('f_fl_A = {:.3f} Hz'.format(f_fl_A))
print ('f_fl_B = {:.3f} Hz'.format(f_fl_B))
print ('f_fl_C = {:.3f} Hz'.format(f_fl_C))









    



f_fl_A = 59.223 Hz
f_fl_B = 59.478 Hz
f_fl_C = 58.967 Hz

and the slopes of the power-frequency curves are:

$$s_P = \frac{P}{f_\text{nl} - f_\text{fl}}$$



In [4]:

    
sp_A = Pn / (f_nl_A - f_fl_A)
sp_B = Pn / (f_nl_B - f_fl_B)
sp_C = Pn / (f_nl_C - f_fl_C)
print('''
sp_A = {:.2f} MW/Hz
sp_B = {:.2f} MW/Hz
sp_C = {:.2f} MW/Hz
'''.format(sp_A/1e6, sp_B/1e6, sp_C/1e6))









    



sp_A = 56.28 MW/Hz
sp_B = 49.45 MW/Hz
sp_C = 65.23 MW/Hz

The total load is 230 MW, so the system frequency can be optained form the load power as follows:

$$P_\text{load} = s_\text{PA}(f_\text{nl,A} - f_\text{sys}) + s_\text{PB}(f_{nl,B} - f_\text{sys}) + s_\text{PC}(f_\text{nl,C} - f_\text{sys})$$

$$\leadsto f_\text{sys} = \frac{s_\text{PA} f_\text{nl,A} + s_\text{PB} f_{nl,B} + s_\text{PC}f_\text{nl,C} - P_\text{load}}{s_\text{PA} + s_\text{PB} + s_\text{PC}}$$



In [5]:

    
Pload = 230e6 # [W]
f_sys = (sp_A*f_nl_A + sp_B*f_nl_B + sp_C*f_nl_C - Pload) / (sp_A + sp_B + sp_C)
print('''
f_sys = {:.2f} Hz
================'''.format(f_sys))









    



f_sys = 59.61 Hz
================

The power supplied by each generator will be:

$$P = s_{P_x} \cdot (f_{\text{nl}_x} - f_{\text{sys}_x})$$



In [6]:

    
Pa = sp_A * (f_nl_A - f_sys)
Pb = sp_B * (f_nl_B - f_sys)
Pc = sp_C * (f_nl_C - f_sys)
print('''
Pa = {:.1f} MW
Pb = {:.1f} MW
Pc = {:.1f} MW
============'''.format(Pa/1e6, Pb/1e6, Pc/1e6))









    



Pa = 78.3 MW
Pb = 93.5 MW
Pc = 58.1 MW
============

(b)

Generate a vector of different load power values



In [7]:

    
Pload_plot = arange(0,300.1,5) * 1e6   # [W]

Calculate the system frequency as function of $P_\text{load}$ using

$$f_\text{sys} = \frac{s_\text{PA} f_\text{nl,A} + s_\text{PB} f_{nl,B} + s_\text{PC}f_\text{nl,C} - P_\text{load}}{s_\text{PA} + s_\text{PB} + s_\text{PC}}$$

from part (a):



In [8]:

    
f_sys = (sp_A*f_nl_A + sp_B*f_nl_B + sp_C*f_nl_C - Pload_plot) / (sp_A + sp_B + sp_C)

Calculate the power of each generator



In [9]:

    
PA = sp_A * (f_nl_A - f_sys)
PB = sp_B * (f_nl_B - f_sys)
PC = sp_C * (f_nl_C - f_sys)

Plot the power sharing versus load:



In [10]:

    
title('Power Sharing Versus Total Load')
xlabel('Total Load [MW]')
ylabel('Generator Power [MW]')
plot(Pload_plot/1e6, PA/1e6, 'g--', linewidth = 2)
plot(Pload_plot/1e6, PB/1e6, 'b', linewidth = 2 )
plot(Pload_plot/1e6, PC/1e6, 'm.', linewidth = 2)
plot([0, 300], [Pn/1e6, Pn/1e6], 'r', linewidth = 2)
plot([0, 300], [0, 0], 'r:', linewidth = 2)
legend(('Generator A','Generator B','Generator C','upper power limit', 'lower power limit'), loc=4, framealpha=1);
grid()

This plot reveals that there are power sharing problems both for high loads and for low loads. Generator B is the first to exceed its ratings as load increases. Its rated power is reached at a total load of



In [11]:

    
interp(Pn, PB, Pload_plot)/1e6  # using the interpolate function to determine
                                # the exact crossover of PB and Pn @Pload









    Out[11]:





252.35

MW.

On the other hand, Generator C gets into trouble as the total load is reduced. When the total load drops below



In [12]:

    
interp(0, PC, Pload_plot)/1e6  # using the interpolate function to determine
                               # the exact crossover of PC and 0 @Pload









    Out[12]:





77.59

MW, the direction of power flow reverses in Generator C.

(c)

The power sharing in (a) is acceptable, because all generators are within their power limits.

(d)

To improve the power sharing among the three generators in (a) without affecting the operating frequency of the system, the operator should decrease the governor setpoints on Generator B while simultaneously increasing them in Generator C.