Download Exam 2 practice

ELEC 316 Practice Exam #2 2015
Q1 Consider the simplified DC machine shown below.
Given; R = 2.0 Ω, VB = 25V,
K = 2, Ф = 0.5, l =0.5m,
a) What is the machines maximum starting current (Io)?
b) What is the no load armature current (I1)?
c) What is the full load current (I2) at an applied torque of 5 Nm?
a) Io = _12.5 A_ b) I1 = __0 A____ c) I2= _5 A__
Q2. A 120V, dc motor has shunt field windings of 500 turns and field resistance of RF = 100 Ω and
armature resistance of RA = 0.10 Ω. The full load current is 40 A, and it produces an armature
reaction of 100 A-Turns.
Magnetization Curve at 2400 rpm
140
2400 rpm
Armature Voltage (V)
120
100
80
60
40
20
a) Equivalent circuit goes in this box.
0
(a)
(b)
(c)
0
0.5
1
Field Current (A)
1.5
2
Sketch the DC motor equivalent circuit in the box above. b) NFL = 2534 rpm
Show all components and label currents and voltages.
What is the full load speed (NFL) of this motor in rpm?
c) T = __17 Nm
What is the full load torque (T) available at the shaft?
Q3. A DC generator is rated at 180 kW, 430 V, 400A and 1800 rpm. The armature resistance is
0.04 Ω. The separately excited field circuit has 1000 turns per pole, a resistance of 90 Ω and is
supplied 450 V. The prime mover turns the generator at a constant 1600 rpm. The magnetization
curve (for N = 1800 rpm) is shown below.
(a) Find the no-load terminal voltage.
(b) Find the % voltage regulation assuming a 400A load, and no armature reaction?
For parts c and d) assume an armature reaction of 500 A turns
(c) Find the terminal voltage for a 400 A load?
(d) At this load, find the field current required to boost the terminal voltage to 380 V?
a) V = __377.7 V___
b) %VR =__4.42%___
c) V = ___339.6 V____
d) IF = ____6.5 A____