HET 228
Tutorial 1(SOLUTION)
Semester 2, 2014
SWINBURNE UNIVERSITY UNIVERSITY OF TECHNOLOGY TECHNOLOGY (SARAWA K CAMPUS) HET HET 228 228 Electrical Actu ators and Sensor Sensor s Semester 2, 2014 Tutor ial 1 (SOLUTION) (SOLUTION)
Q1
A linear machine shown in Figure Q1 has a magnetic flux density of 0.5 T directed into the page, a resistance of 0.25 battery voltage of 100 V.
Ω, a bar length l = 1.0 m, and a
a) b) c) d)
Determine the Determine the initial force on the bar at starting. Determine the Determine the initial current flow. Determine the Determine the no-load steady-state speed of the bar. If the bar is loaded with a force of 25 N opposite to the direction of of motion, determine the determine the new steady-state speed. e) Under these circumstances, determine the determine the efficiency of the machine. (Hint: Efficiency,
t
η =
=0s
Pin
i
R
VB
Pout
× 100% )
= 0.5 T (into the page)
B
= 0.25 Ω
= 100 V
Figure Q1
SUTS
Page 1 of 16
HET 228
Tutorial 1(SOLUTION)
Semester 2, 2014
Q1 (SOLUTION) Given:
B = 0.5 T (into the page) R = 0.25 Ω l = 1.0 m
VB = 100 V
a) Determine the initial force: F = ilB = 400 × 1 × 0.5 = 200 N (to the right) b) Determine the initial current flow: i=
V B R
=
100 0.25
= 400 A
c) Determine the no-load steady-state: V B = eind = vBl
∴ vss =
V B Bl
=
100 0.5 × 1
= 200 m/s
d) Determine the applied force: F app = F ind = ilB
Determine the current: i=
F app Bl
=
25 0.5 × 1
= 50 A
Take note of the following correction: Determine the induced emf: eind = V B − iR = 100 − (50 × 0.25) = 87.5 V
Hence, the new steady-state speed v ss =
V B Bl
=
87.5 0.5 × 1
= 175 m/s
Correction:
v ss =
eind Bl
=
87.5 0.5 × 1
= 175 m/s
e) Determine the efficiency of the machine: input power = 100×50 = 5000 W output power = 87.5×50 = 4375 W η =
SUTS
Pout Pin
× 100% =
4375 5000
× 100% = 87.5%
Page 2 of 16
HET 228
Q2
Tutorial 1(SOLUTION)
Semester 2, 2014
The linear DC machine of Figure Q1 has been modified to have the following characteristics: Flux density, B = 5 T out of page Resistance, R = 0.5 Ω Battery voltage, VB = 150 V Note: the bar length is maintained at 1 m. a) If the bar has a load of 10 N attached to it opposite to the direction of motion, determine the steady state speed of the bar. b) If the bar runs into a region where the flux density falls to 4 T, determine the final steady state speed of the bar. c) If the circuit resistance is increased to 1 Ω, with everything else running as in part (b), determine the final steady state speed of the bar. d) Suppose VB is further increased to 175 V, with everything else running as in part (b), determine the new steady state speed of the bar.
Take note of the following corrections:
Q2 (SOLUTION) a) F app = F ind = ilB i
=
F app
=
10
=
5 ×1
Bl
2A
eind = V B − iR = 150 − (2 × 0.5) = 149V
∴ vss =
V B Bl
=
149 5 ×1
Correction:
vss =
= 29.8 m/s
eind
=
Bl
149 5 ×1
= 29.8 m/s Conclusion:
b) F app = F ind = ilB i=
F app
=
Bl
4 ×1
= 2.5 A When VB
eind = V B − iR = 150 − (2.5 × 0.5) = 148.75V
∴ vss = c)
eind
V B Bl
=
148.75 4 ×1
= 37.19 m/s
↓, vss ↓
Correction:
vss =
= V − iR = 150 − (2.5 × 1) = 147.5V
∴ vss =
↑, vss ↓
When B
10
eind
=
Bl
148.75 4 ×1
= 37.19 m/s
B
V B Bl
=
147.5 4 ×1
Correction:
= 36.88 m/s
vss =
eind
=
Bl
147.5 4 ×1
= 36.88 m/s
d) eind = V B − iR = 150 − (2.5 × 0.5) = 173.75V
∴ vss =
V B Bl
=
173.75 4 ×1
Correction:
= 43.44 m/s
eind
= V B − iR = 170 −
(2.5 × 0.5) = 173.75V
Correction: SUTS
v ss =
eind Bl
=
147.5 4 ×1
= 36.88 m/s
Page 3 of 16
HET 228
Q3
Tutorial 1(SOLUTION)
Semester 2, 2014
A ferromagnetic core is shown in Figure Q3. The depth of the core (into the page) is 0.05 m. The other dimensions of the core are as shown in Figure Q3. The relative permeability of the core is 1000 and is assumed to be constant. a) Determine the value of the current that will produce a flux of 0.005 Wb. b) With the current from part (a), determine the flux density at the top of the core. c) With the current from part (a), determine the flux density at the right side of the core. 0.05 m 0.1 m
0.2 m
0.15 m
i
500 turns
0.15 m
0.15 m
Core depth = 0.05 m
Figure Q3
Q3 (SOLUTION) ½ × 0.1 m
0.2 m
Step 1: Determine the mean path lengths
1 2
½ × 0.05 m
1 × 0.15 = 0.3 m 2
l1 = × 0.15 + 0.15 +
l 1 =
0.275 m
½ × 0.15 m
1 1 × 0.1 + 0.2 + × 0.05 = 0.275 m 2 2
l2 = l3
= l = 0.3 m
l4
=
1
l2
=
m 3 . 0 =
m 3 . 0 =
l
l
1
0.15 m
3
0.275 m ½ × 0.15 m l 4 =
SUTS
0.275 m
Page 4 of 16
HET 228
Step
Tutorial 1(SOLUTION)
2: Determine the (A = wi dth depth)
0.1 × 0.05 = 0.005 m 2
cross-sectional
Semester 2, 2014
areas
for
each
core
cross-sectional area for left side of the core
A1
=
A2
= 0.15 × 0.05 = 0.0075 m
2
cross-sectional area for the top of the core
A3
= 0.05 × 0.05 = 0.0025 m
2
cross-sectional area for right side of the core
A4
= A = 0.0075 m
2
leg
cross-sectional area for the bottom of the core
2
Step 3: Determine the reluctances for different cross-sectional areas (or the cor e legs) R 1
=
R 2
=
R 3
=
l1 µ o µ r A1
=
l2 µ o µ r A2
l3 µ o µ r A3
R 4 = R 2 =
=
=
0.3
(4π × 10− )× 1000 × 0.005 7
= 47746.48
0.275
(4π × 10− )× 1000 × 0.0075 7
0.3
(4π × 10− )× 1000 × 0.0025 7
A.t/wb
= 29178.41
A.t/wb
= 95492.97
A.t/wb
29178.41 A.t/wb
Step 4: Draw the magnetic c irc uit R 2
R 1
R TOT R 3
mmf
NI
Φ
mmf
NI
Φ
R 4
SUTS
Page 5 of 16
HET 228
Tutorial 1(SOLUTION)
Semester 2, 2014
Step 5: Answer the question (or sub-questions) a) Determine the value of the current that will produce a f lux of 0.005 Wb. Refer to Step 4, from the magnetic circuit: NI = ΦR TOT
∴ I =
(
= Φ R 1 + R 2 + R 3 + R 4
Φ (R 1 + R + R + R ) 2
3
4
N
=
) (
)
0.005 × 47746.48 + 29178.41 + 95492.97 + 29178.41 500
= 2.016 A
b) With the current from part (a), determine the flux density at the top of the core. The top of the core has cross-sectional area of A2 Btop =
Φ A2
=
0.005 0.0075
= 0.67 T
c) With the current from part (a), determine the flux density at the right side of the core. The right side of the core has cross-sectional area of A3 Bright =
SUTS
Φ A3
=
0.005 0.0025
= 2T
Page 6 of 16
HET 228
Q4
Tutorial 1(SOLUTION)
Semester 2, 2014
A ferromagnetic core with a constant relative permeability of 1500 is shown in Figure Q4. The dimensions are as shown in Figure Q4, and the depth of the core (into the page) is 0.07 m. The air gaps on the left and right sides of the core are 0.0007 m and 0.0005 m respectively. Because of fringing effects, the effective area of the air gaps is 5 percent larger than their physical size. a) If there are 300 turns in the coil wrapped around the centre leg of the core and if the current in the coil is 1.0 A, determine the flux in each of the left, centre, and right legs of the core. b) Determine the flux density in each air gap. 0.07 m
0.07 m
0.07 m
0.3 m
0.3 m
l 2
0.07 m
l 4
i l 1
0.3 m
Air gap length, l g2
Air gap length, l g1
300 turns
0.0007 m
l 5
0.0005 m
Take note of l 6
l 9 l 3
0.07 m
l 8
the following corrections:
l 7
Figure Q4
Core depth = 0.05 m
Q4 (SOLUTION) Step 1: Determine the mean path lengths l1
1 1 = × 0.07 + × (0.3 − 0.0007 ) = 0.18465 m 2 2
Correction: Core depth = 0.07 m
1 1 × 0.07 + 0.3 + × 0.07 = 0.37 m 2 2
l2 =
l9 = l1 = 0.18465 m l8 = l3 = l 4 = l7 = l 2 = 0.37 m
1 1 × 0.07 + × (0.3 − 0.0005) = 0.18475 m 2 2
l5 = l6
= l5 =
SUTS
0.18475 m
Page 7 of 16
HET 228
Step
Tutorial 1(SOLUTION)
2: Determine the cross-sectional (A = wi dth depth) & the air-gaps
Semester 2, 2014
areas
for
each
3
2
leg
cross-sectional area (throughout)
A1 = A2 = A3 = A4 = A5 = A6 = A7 = A8 = A9 = 0.07 × 0.07 = 4.9 × 10 −3 m 2 Ag 1 = Ag 2 = A1 × 105% = 5.145 × 10 − m
core
due to fringing effect
Step 3: Determine the reluctances for different cross-sectional areas (or the cor e legs) & the air-gaps R 1
=
l1 µ o µ r A1
l2
R 2 =
0.18465
=
(4π × 10− ) × 1500 × (4.9 × 10− ) 7
3
0.37
=
(4π × 10 ) × 1500 × (4.9 × 10 ) −7
µ o µ r A2
−3
R 8
= R 3 = R 4 = R 7 = R 2 = 40059.41
R 5
=
l5 µ o µ r A5
R 6 = R 5 =
R g 1
=
R g 2
=
=
= 19991.81
=
A.t/wb
40059.41 A.t/wb
A.t/wb
0.18475
(4π ×10 )×1500 × (4.9 ×10 − ) −7
3
= 20002.64
A.t/wb
20002.64 A.t/wb
l g1
0.0007
=
µ o µ r Ag1
lg 2 µ o µ r Ag 2
(4π ×10 )×1× (5.145 ×10 − ) −7
=
3
= 108268.67
A.t/wb
= 77334.76
A.t/wb
0.0005
(4π ×10 − )×1× (5.145 ×10 − ) 7
3
Step 4: Draw the magnetic c irc uit R 2
R 1
R 4
Loop 1
R 5
R 3
Loop 2
ΦΑ R g1
Loop 1 R g2
R A
ΦΒ R B
mmf NI Φ
Φ R 6
R 9 R 8
R 7
Φ = ΦΑ + ΦΒ
SUTS
Loop 2
ΦΑ
ΦΒ
mmf NI
R 3
Φ = ΦΑ + ΦΒ
Page 8 of 16
HET 228
Tutorial 1(SOLUTION)
R A
= R 1 + R 2 + R g1 + R 8 + R 9 = 228371.11
R B
= R 4 + R5 2 + R g 2 + R 6 + R 7 = 197458 .86
Semester 2, 2014
A.t/wb
A.t/wb
Step 5: Answer the question (or sub-questions)
a) Determine the flux in each of the left, centre, and right legs of the core.
Take note:
Flux in the left leg of the core = Φ A. (this is also the flux in air gap with length l g1 )
Flux in the right leg of the core = ΦB. (this is also the flux in air gap with length lg 2 )
Flux in the centre leg of the core = Φ.
Refer to Step 4, from the magnetic circuit: For Loop 1:
− Φ − R Φ = 0 ⇒ NI − R (Φ + Φ ) − R Φ = 0 ⇒ NI − R Φ − R Φ − R Φ = 0 ⇒ NI − (R + R )Φ − R Φ = 0 ⇒ (300 ×1) − (268430 .52Φ ) − (40059.41Φ ) = 0 ⇒ 300 − (268430 .52Φ ) − (40059 .41Φ ) = 0 (1)
NI R 3
A
A
A
3
B
A
3
3
3
A
A
B
A
A
A
3
A
B
A
B
A
B
For Loop 2: NI − R 3Φ − R B Φ B
(
=0
)
⇒ NI − R 3 Φ A + Φ B − R B Φ B =
0
⇒ NI − R 3 Φ A − R 3 Φ B − R B Φ B =
(
)
⇒ NI − R 3 + R B Φ B − R 3 Φ A =
0
0
(300 ×1) − (40059.41Φ A ) − (40059.41Φ B ) − (197458.86Φ B ) = 0 ⇒ 300 − (40059.41Φ A ) − (237518.27 Φ B ) = 0 (2 ) ⇒
SUTS
Page 9 of 16
HET 228
Tutorial 1(SOLUTION)
Semester 2, 2014
Let (1) – (2):
∴ Φ A =
197458 .86Φ B
= 0.8646Φ B (3)
228371 .11
Sub (3) into (1):
∴Φ B =
300 272144 .44
= 0.0011 Wb (4 )
Finally, sub (4) into (3): Φ A =
0.8646Φ B
=
∴Φ = Φ A + Φ B =
0.8646 × 0.0011 = 9.53 ×10 -4 Wb
(
0.0011 + 9.53 × 10 -4
) = 0.002055 Wb
b) Determine the flux density in each air gap. Bg1 =
Bg 2
SUTS
=
Φ A Ag1 Φ B
Ag 2
= =
9.53 × 10
−4
5.145 × 10
−3
0.0011 5.145 ×10 −3
= 0.1852 T =
0.2138 T
Page 10 of 16
HET 228
Q5
Tutorial 1(SOLUTION)
Semester 2, 2014
A ferromagnetic core with three legs is shown in Figure Q5(a). Its depth (into the page) is 0.08 m, and there are 400 turns on the centre leg. The remaining dimensions are shown in Figure Q5(a). The core is composed of a steel having the magnetization curve shown in Figure Q5(b). The relative permeability is not constant. Answer the following questions about the curve shown in Figure Q5(b) : a) Determine the required current to produce a flux density of 0.5 T in the centre leg of the core. b) Determine the required current to produce a flux density of 1.0 T in the centre leg of the core. Is it twice the current in part (a)? c) Determine the reluctances of the centre and right legs of the core under the conditions in part (a). d) Determine the reluctances of the centre and right legs of the core under the conditions in part (b). e) Draw conclusions about reluctances in real magnetic cores.
2
3
0.08 m
1
i
l 2
N = 400 turns
l 3
0.16 m
l 1 0.08 m
0.08 m
0.16 m
0.08 m
0.16 m
0.08 m
Core depth = 0.08 m
Figur e Q5(a)
Take Note: 1 =
SUTS
2 +
3
Page 11 of 16
