In [1]:
%pylab inline
A 480-V, 250-kVA, 0.8-PF-lagging, two-pole, three-phase, 60-Hz synchronous generator’s prime mover has a no-load speed of 3650 r/min and a full-load speed of 3570 r/min.
It is operating in parallel with a 480-V, 250-kVA, 0.85-PF-lagging, four-pole 60-Hz synchronous generator whose prime mover has a no-load speed of 1800 r/min and a full-load speed of 1780 r/min. The loads supplied by the two generators consist of 300 kW at 0.8 PF lagging.
In [2]:
n_nl1 = 3650.0 # [r/min]
n_fl1 = 3570.0 # [r/min]
p_1 = 2
n_nl2 = 1800.0 # [r/min]
n_fl2 = 1780.0 # [r/min]
p_2 = 4
S_1 = 250e3 # [VA]
S_2 = 250e3 # [VA]
PF_1 = 0.8
PF_2 = 0.85
Pload = 300e3 # [W]
The no-load frequency of generator 1 corresponds to a frequency of:
$$f_\text{nl1} = \frac{p_1 \cdot n_m}{120}$$
In [3]:
fnl_1 = n_nl1*p_1 / 120
print('''
fnl_1 = {:.3f} Hz
================='''.format(fnl_1))
The full-load frequency of generator 1 corresponds to a frequency of:
$$f_\text{fl1} = \frac{p_1 \cdot n_m}{120}$$
In [4]:
ffl_1 = n_fl1*p_1 / 120
print('''
ffl_1 = {:.3f} Hz
================'''.format(ffl_1))
The no-load frequency of generator 2 corresponds to a frequency of:
$$f_\text{nl2} = \frac{p_2 \cdot n_m}{120}$$
In [5]:
fnl_2 = n_nl2*p_2 / 120
print('''
fnl_2 = {:.3f} Hz
================='''.format(fnl_2))
The full-load frequency of generator 1 corresponds to a frequency of:
$$f_\text{fl2} = \frac{p_2 \cdot n_m}{120}$$
In [6]:
ffl_2 = n_fl2*p_2 / 120
print('''
ffl_2 = {:.3f} Hz
================='''.format(ffl_2))
In [7]:
SD_1 = (n_nl1 - n_fl1) / n_fl1
print('''
SD_1 = {:.2f}%
============'''.format(SD_1*100))
and
In [8]:
SD_2 = (n_nl2 - n_fl2) / n_fl2
print('''
SD_2 = {:.2f}%
============'''.format(SD_2*100))
In [9]:
P1_fl = S_1 * PF_1
print('P1_fl = {:.0f} kW'.format(P1_fl/1e3))
The full load power supplied by generator 2 is:
$$P_\text{2,full load} = S\cos{\theta_2}$$
In [10]:
P2_fl = S_2 * PF_2
print('P2_fl = {:.1f} kW'.format(P2_fl/1e3))
The power supplied by generator 1 as a function of frequency is given by:
$$P_1 = s_\text{P1}(f_\text{nl1} - f_\text{sys})$$and the power supplied by generator 2 as a function of frequency is given by:
$$P_2 = s_\text{P2}(f_\text{nl2} - f_\text{sys})$$The power curve’s slope for generator 1 is:
$$s_\text{P1} = \frac{P}{f_\text{nl} - f_\text{fl}}$$
In [11]:
sp_1 = P1_fl / (fnl_1 - ffl_1)
print('sp_1 = {:.3f} MW/Hz'.format(sp_1/1e6))
The power curve’s slope for generator 2 is:
$$s_\text{P2} = \frac{P}{f_\text{nl} - f_\text{fl}}$$
In [12]:
sp_2 = P2_fl / (fnl_2 - ffl_2)
print('sp_2 = {:.3f} MW/Hz'.format(sp_2/1e6))
The total power that they must supply is 300 kW, so the system frequency can be found from the equations:
\begin{align*} P_\text{LOAD} &= P_1 + P_2\\ P_\text{LOAD} &= s_\text{P1} (f_\text{nl1} - f_\text{sys}) + s_\text{P2} (f_\text{nl2} - f_\text{sys})\\ \leadsto f_\text{sys} & = \frac{s_\text{P1} f_\text{nl1} + s_\text{P2} f_\text{nl2} - P_\text{LOAD}}{s_\text{P1} + s_\text{P2}} \end{align*}
In [13]:
f_sys = (sp_1*fnl_1 + sp_2*fnl_2 - Pload) / (sp_1 + sp_2)
print('''
f_sys = {:.3f} Hz
================='''.format(f_sys))
In [14]:
P1 = sp_1 * (fnl_1 - f_sys)
print('''
P1 = {:.0f} kW
==========='''.format(P1/1e3))
The power supplied by generator 1 is:
$$P_2 = s_\text{P2}(f_\text{nl2} - f_\text{sys})$$
In [15]:
P2 = sp_2 * (fnl_2 - f_sys)
print('''
P2 = {:.0f} kW
==========='''.format(P2/1e3))