ECE 3614
QUIZ # 2
Name ____________________________________
Grade:_______
“As a Mississippi State University student I will conduct myself with honor and integrity at all times. I will not
lie, cheat, or steal, nor will I accept the actions of those who do.” – Mississippi State University Honor Code
http://students.msstate.edu/honorcode
1) [10 pts] In a few words define the following terms:
a. Commutation
b. Voltage buildup
c. Flux weakening
d. Stray losses
e. Armature Reaction Voltage Drop
2) [30 pts] A DC generator is connected in a short-shunt, cumulative compound configuration,
with a 3% reduction in MMF due to armature reaction.
a. Draw the full equivalent circuit, showing the windings (with polarity marking), all
current paths, and voltages
b. Describe the physical meaning of each component in the circuit, and include the
effective MMF equation for this configuration.
S
Rs
It
+
Ra
Is
Ia
+
Rf
Vt
If
Ea
F
Vt,It represent voltage and current at the terminal
S represents the series field coil, negative polarity (opposing MMF)
Rs is the resistance of the series field coil
Is represents the current through the series field coil
F represents the shunt field coil, positive polarity
Rf is the resistance of the shunt field coil
If represents the current through the shunt field coil
Ea represents the armature voltage, which includes losses such as brush voltage drop
Ra represents the resistance in the armature windings, usually a small resistance
NfIf eff =0.97 ∙ (NfIf - NsIs)
3) [30 pts] A 15 hp, 180 V DC machine, has the parameters listed below. When the machine is
connected as a shunt motor, it will draw 10 A under no-load condition. When a mechanical
load is attached, the constant voltage source delivers 15 kW to the motor. Consider the
mechanical load is something different from the machine’s rated loading capacity.
Armature Winding Resistance = 0.125 Ω
Series Field Winding Resistance = 0.08 Ω
Shunt Field Winding Resistance = 36 Ω
a.
b.
c.
d.
e.
No Load Speed = 1200 rpm
Series Field Winding = 3 turns
Shunt Field Winding = 1600 turns
Calculate the internally developed voltage, when the motor is loaded.
Calculate the % speed regulation of the motor for this load.
Calculate the torque developed by the motor.
Calculate the efficiency of this mechanically loaded motor.
If an additional starting resistance is selected as 2.5Ω, calculate the armature
current when the motor is started (no-load).
V
a) If = Rt =
f
180
36
=5𝐴
Ia,FL = It,FL − If =
15,000
− 5 = 78.33 A
180
Ea,FL = Vt − R a Ia = 180 − (0.125)(78.33) = 170.21 𝑉
b)
Ea,FL
=
Ea,NL
2πnFL
)
60
2πnNL
𝐾𝑎 Φ𝑝 (
)
60
𝐾𝑎 Φ𝑝 (
=
nFL
nNL
Ia,NL = It,NL − If = 10 − 5 = 5 A
Ea,NL = Vt − R a Ia = 180 − (0.125)(5) = 179.375 V
nFL = (
Ea,FL
170.21
) (1200 rpm) = 1139 rpm
) nNL = (
Ea,NL
179.375
1200 − 1139
∙ 100% ⟹ 𝑆𝑅 = 5.4%
1139
𝑆𝑅 =
c) Pdev = Ea Ia = (170.21)(78.33) = 13,333 𝑊
ωm =
Te =
2πn
2π(1139)
=
= 119.2 rad⁄s
60
60
Pdev
13,333
=
⟹ 𝑇𝑒 = 111.8 𝑁 ∙ 𝑚
ωm
119.2
d) Prot = Ea,NL Ia,NL = (179.375)(5) = 897 𝑊
2
Pcu = If2 R f + Ia,FL
R a = (5)2 (36) + (78.33)2 (0.125) = 1,667 W
η=
Pout
15,000 − 897 − 1667
∙ 100% =
∙ 100% ⟹ 𝜂 = 82.9%
Pin
15000
e) Ia,start =
Vt −Ea,start
Ra +Rext
180−0
= 0.125+2.5 ⟹ 𝐼𝑎,𝑠𝑡𝑎𝑟𝑡 = 68.57 𝐴
4) [30 pts] A 3 kW, 225 VDC generator has the following characteristics provided below. This
machine requires 3100 A-turns to deliver power to a rated load. For flat voltage regulation,
it is connected as a long-shunt cumulative compound generator. Armature reaction for the
generator is 150 A-turns.
Armature Winding Resistance = 0.16 Ω
Shunt Field Winding Resistance = 100 Ω
No Load Test Speed = 1800 rpm
Shunt Field Winding = 1200 turns
a. Calculate the number of turns required for a series coil.
b. Calculate the electromagnetic power and torque developed for a rated load.
c. If rotational loss is 50 W and series winding resistance is 0.12 Ω, find the efficiency
of the generator for a rated load.
225 𝑉
a) 𝐼𝑓 = 100 Ω = 2.25 𝐴 ; 𝐼𝑠 = 𝐼𝑎 = 𝐼𝑡 + 𝐼𝑓 =
3000 𝑊
225 𝑉
− 2.25 = 15.6 𝐴
ℱ𝑛𝑒𝑡 = (𝑁𝑓 ∙ 𝐼𝑓 ) + (𝑁𝑠 ∙ 𝐼𝑠 ) − ℱ𝐴𝑅
𝑁𝑠 =
ℱ𝑛𝑒𝑡 − (𝑁𝑓 ∙ 𝐼𝑓 ) + ℱ𝐴𝑅
≈ 35 𝑡𝑢𝑟𝑛𝑠
𝐼𝑠
b) Ea = Vt + R a Ia = 225 + (0.16 + 0.12)(15.6) = 228.74 V
Pdev = Ea Ia = (228.74)(15.6) = 3,564 𝑊
ωm =
Te =
2πn
60
=
2π(1800)
60
= 188.5 rad⁄s
Pdev
3564
=
⟹ 𝑇𝑒 = 18.91 𝑁 ∙ 𝑚
ωm
188.5
c) Prot = 50 𝑊
Pcu = If2 R f + Ia2 (R a + R s ) = (2.25)2 (100) + (15.6)2 (0.12 + 0.16) = 564 W
η=
Pout
3,000
∙ 100% =
∙ 100% ⟹ 𝜂 = 79.5%
Pin
3000 + 50 + 536