ELECTROTECHNOLOGY
ELTK1200
REVIEW #3
1.
A 40kW load, operating at 0.707 lagging power factor, is supplied from 500V, 50Hz
mains. Calculate:
(a)
the capacitor value required to raise the line power factor to unity (C = 509F).
(b)
the capacitance required to raise the power factor to 0.95 lagging.
(C = 342F)
2.
A single-phase motor running from a 230V, 50Hz supply takes a current of 11.6A
when giving an output of 1.5kW, the efficiency being 80 per cent. If the power factor
of the supply is raised to 0.95 (lagging), calculate:
(a)
the original power factor (pf = 0.703 lagging).
(b)
the capacitance for power factor correction (C = 77.1F).
(c)
the kVAR rating of the capacitor (QC = 1.282kVAR).
(d)
the new supply current (I = 8.58A).
(e)
sketch the power triangle after correction.
Note: If no efficiency, given power is what device needs
(PIN = POUT, no losses). But if efficiency is mentioned in
question, given is POUT and you must calculate PIN.
3.
A 440V, 60Hz, 20kW, single-phase induction motor has
a full-load efficiency of 91.15% and operates at a power
factor of 0.87 (lagging). Find the value of the capacitor to
be connected in parallel to improve the circuit power
factor to 0.95 (lagging) (C = 71.6F).
4.
A coil of 1006 resistance and 0.1H inductance is connected in series with a 0.1F
capacitor to a 230V variable frequency a.c. supply. Calculate:
(a)
the resonant frequency (fR = 1591.6Hz).
(b)
the potential difference across the capacitor at resonance (VC = 2300V).
5.
A coil of 0.84H inductance and 506 resistance is connected in series with a capacitor
of 14F capacitance. Find:
(a)
resonant frequency (fR = 46.41Hz).
(b)
the voltage drops across the capacitor, the coil and the source voltage, if 5A
of current flows at resonant frequency (VC = VL = 1225V, VS = 250V).
(c)
repeat step (b) if the same current flows at 60Hz.
(VC = 947, VL = 1583V, VS = 683V) [ lagging circuit, VR = 250V ]
ELTK1200
Marine Institute
Rev 2
137
6.
A single-phase power transformer supplies a load of 20kV#A at a power-factor of 0.81
lagging. The iron loss of the transformer is 200W and the copper loss at this load is
180W. Calculate:
(a)
the efficiency ( = 97.7%).
(b)
if the load is now changed to 30kV#A at a power-factor of 0.91 lagging,
calculate the new efficiency ( = 97.83%).
[ load is 150% of original load, so Cu loss = (1.5)2 * 180W = 405W ]
7.
A 75kW, 400V, 3-phase, delta-connected induction motor has a full-load efficiency
of 91% and operates at a power factor of 0.9 lagging. Calculate the line and phase
currents at full load (IPHASE = 76.3A, ILINE = 123.2A).
8.
A 3-phase, wye-connected alternator (generator) supplies a delta-connected
induction motor with a phase voltage of 600V. If the line current is 40A and the
power factor is 0.95 lagging, find:
(a)
the phase voltage of the alternator (VPHASE = 600V).
(b)
the line voltage (VLINE = 1039V).
(c)
the phase voltage of the motor (VPHASE = 1039V).
(d)
the line current (ILINE = 40A).
(e)
the current in each phase of the motor (IPHASE = 23.1A).
(f)
the current in each phase of the alternator (IPHASE = 40A).
(g)
total power supplied to the load (PT = 68.4kW).
9.
A 3-phase, 440V, 60Hz, delta-connected generator supplies a balanced
wye-connected load, which consists of a coil of resistance 356 and inductive
reactance 256. Determine:
(a)
sketch the circuit.
(b)
current in each coil (IPHASE = 3.41A).
(c)
current in each phase of the alternator (IPHASE = 5.91A).
(d)
power factor of the load (pf = 0.8137 lagging).
(e)
total power supplied to the load (PT = 3663W).
ELTK1200
Marine Institute
Rev 2
138