Page 157 - Industrial Power Engineering and Applications Handbook

Static controls and braking of motors 611 37
R Y E (iii) Power required when discharge is controlled through
static control, i.e. by speed variation;
Head Hd3 = 0.45 Hd,
discharge = 0.67 Q
q = 0.82
0.45 Hdl x 0.67 Q . d
Power
36 x 0.82
.. Ratio of power through speed variation
Rated power required
0.45 x 0.67 o,85
0.82
= 0.31
I i.e. the power reduction through speed control is 69% that of
I the rated power requirement.
Thus the energy saving in this particular case by employing
the method of speed control over that of throttle control will
0 be
= 69%-11% = 58%
Example 6.2
To determine saving of costs through speed control, consider
the following parameters, for the above case:
0 P = 100 kW
Venturimeter -To measure the velocity of fluid
@ Probe to sense the velocity of fluid Discharge = 67% of the rated flow
@ Flow meter or sensor - To convert the velocity of fluid Duration of operation at reduced capacity = say, for 25%
to the rate of flow of total working hours.
@ Variable voltage (fixed frequency) static control
Energy tariff = say, Rs 1.2 per kWh
Figure 6.40 Energy conservation through static control
:. Total energy consumed per year, considering 300 operating
days/year
= 100 x 24 x 300
where
P = shaft input in kW = 720 000 kWh
Hd = head in bar
Q = discharge in m3/hour Energy consumed while operating at the reduced capacity
d = specific gravity of the liquid in gm/cm3 for 25% duration = 0.25 x 720 000
q = efficiency of the pump
= 180 000 kWh
(i) Rated power required by the pump
:. Energy saving = 0.58 x 180 000
= 104 400 kWh
And total saving in terms of cost = 1.2 x 104 400
(ii) Power required when discharge is controlled through the
throttle, = Rs 125 280 per year
head Hd, = 1.138 Hdl
Example 6.3
discharge = 0.670 Energy saving in a slip recovery system
Consider an I.D. fan of 750 kW, running at 75% of N, for at
q = 0.73 least 20% of the day. Considering the load characteristic in
cubic ratio of speed,
1.138 H,, x 0.67 Q. d
:. Power =
36 x 0.73 :. P at 75% speed
Ratio of Dower throuah throttle = (0.75)3 x 750 kW
Rated power required = 316 kW
1.138 x 0.67 o,85 and slip power,
0.73
= 0.89 f, = (1 -0.75) x 316 kW
= 0.25 x 316
i.e. the power reduction with this method is around 11% of
the rated power requirement. = 79 kW