Advanced Electrical Machines and Drives Assignment - ENEE20002

This assignment solution addresses several key concepts in electrical engineering. Question 3 analyzes the feasibility of an 800 MW HV link, comparing costs for overhead and subsea routes using both AC and LCC HVDC options, calculating relative costs and breakeven distances. Question 5 delves into synchronous generators, determining the magnitude of the synchronous internal voltage, field current, and flux linkages based on given inductance parameters and operating conditions. The assignment also references concepts related to motor analysis and control systems. This document is designed to aid students in understanding and solving complex electrical engineering problems by providing detailed solutions and calculations.

ASSIGNMENT
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Question 3
You are working for a small consultancy company hired to assess the feasibility of connecting
800 MW HV link. Three routes are possible:
(a) Overhead direct 220 km (overhead line)
(b) Overhead direct 350 km (overhead line)
(c) Subsea coastal 400 km (cable)
Both AC and LCC HVDC options are possible:
AC sub-station cost: $20,000/MW
DC substation cost: $240,000/MW
AC overhead line cost: $1400/MW-km
DC overhead line cost: $900/MW-km
AC cable cost: $16,000 /MW-km (cable+laying)
DC cable cost: $10,000 /MW-km (cable+laying)
i. Calculate the relative costs of each option?
ii. What is the breakeven distance in each choice?
Solution
Part I: Relative costs of each option
Relative cost of each station=Cost of each substation+cost of cable laying
Applyingthe equation of relative cost ¿ each station
a) Overhead direct 220 km
HV AC cost= ( AC substationcost∗HV link ) +( AC overhead line cost∗HVlink∗overhead line)
¿ ( $ 20,000∗800 )+ ( $ 1,400∗800∗220 )
¿ $ 262,400,000
HV DC cost= ( DC substation cost∗HV link )+(DC overhead line cost∗HVlink∗overhead line)
¿ ( $ 240,000∗800 )+ ( $ 900∗800∗220 )
¿ $ 350,400,000
b) Overhead direct 350 km(overhead line)
HV AC cost= ( AC substationcost∗HV link ) +( AC overhead line cost∗HVlink∗overhead line)
¿ ( $ 20,000∗800 )+ ( $ 1,400∗800∗350 )
¿ $ 408,000,000

HV DC cost= ( DC substation cost∗HV link )+(DC overhead line cost∗HVlink∗overhead line)
¿ ( $ 240,000∗800 )+ ( $ 900∗800∗350 )
¿ $ 44,000,000
c) Subsea coastal 400 km(cable)
HV AC cost= ( AC substationcost∗HV link ) +( AC cable cost∗HVlink∗overhead line)
¿ ( $ 20,000∗800 )+ ( $ 16,000∗800∗400 )
¿ $ 5,136,000,000
HV DC cost= ( DC substation cost∗HV link )+(DC cable cost∗HVlink∗overhead line)
¿ ( $ 240,000∗800 )+ ( $ 10,000∗800∗400 )
¿ $ 3,392,000,000
Part II: Breakeven distance in each choice.
Beakdeven distance ( x ) ,is the distance at which the HV AC cost at x distance=HV DC cost at x distance
a) Overhead direct
( 20,000∗800 ) + ( 1,400∗800∗x )= ( 240,000∗800 )+ ( 900∗800∗x )
Solving for the distance x ∈km
¿ 16,000,000+1,120,000 x=192,000,000+720,000 x
¿ ( 1,120,000−720,000 ) x= ( 192,000,000−16,000,000 )
¿ 400,000 x=176,000,000
x= 176,000,000
400,000 =440 km
b) Subsea coastal
¿(20,000∗800)+( 16,000∗800∗x)=(240,000∗800)+(10,000∗800∗x)
Solving for the distance x ∈km
¿ 16,000,000+12,800,000 x=192,000,000+8,000,000 x
¿ ( 12,800,000−8,000,000 ) x= ( 192,000,000−16,000,000 )
¿ 4,800,000 x=176,000,000
x= 176,000,000
4,800,000 =36.667 km

Question 5
A 50 Hz three-phase synchronous generator with small armature resistance has the following
inductance parameters:
𝐿𝑠 = 2.7656 𝑚H; 𝑀𝑓 = 31.6950 𝑚H
𝑀𝑠 = 1.3828 𝑚H; 𝐿𝑓 = 433.68 𝑚H
The machine is rated at 645 MVA, 0.85 power factor lagging, 3800 rpm and 22 kV. When
operating under rated voltage conditions, the line to neutral terminal voltage and line current of
phase a may be written:
𝑉𝑎 = 19596 cos(𝜔t) 𝑉; 𝑖𝑎 = 21603 cos (𝜔t− 31.380) 𝐴
Solution
i. Determine the magnitude of the synchronous internal voltage.
Solving for V a max∧I a max
V a max = 22000∗√2
√ 3 =17962.9248 V
I a max =645∗106∗ √2
√ 3∗22∗103 =23938.1952 A
Calculating the internal voltage by V a∧I a ¿ the equation
ea
' = √2 |Ei|cos ( ωt + σ )=V a+(lM + Ms ) d ia
dt
Substituting the values
ea
' =19596∗cos ( ωt )+( ( 2.7656+1.3828 )∗10−3 ) d
dt ¿
taking the derivative of cos=sin
ea
' =19596∗cos ( ωt ) − ( 4.1484 )∗10−3∗21603 sin ( ω t−31.38 0 )
θ=cos−1 (0.85)=31.7883 ω=2 πf =100 π rad
s
solving the equation withthe calculated values
ea
' =19596∗cos ( ωt ) − ( 4.1484 )∗10−3∗w∗21603 sin ( ω t−31.78 83 )
ea
' =19596∗cos ( ωt ) −(4.1484∗10−3∗100 π∗21603)sin ( ω t−31.78 83 )
ea
' =19596∗cos ( ωt ) −28154.2890(sin wt cos 31.7883+cos wt sin 31.7883)
Expanding the Equationabove gives
ea
' =19596∗cos ( ωt ) +14831.1792 coswt −23931.1536 sinwt
ea
' =34427 .1792 coswt −23931.1536 sinwt
ea
' =34423. 1792 cos ( ωt ) −23931.1536 sinwt

√ 2| Ei|cos ( ωt +σ ) =41925.29 cos (ωt +34.8)
Equating the variables ¿ the equation above
√2|Ei|=41925.29
Ei =41925.29 V
σ =34.8
ii. The field current and the flux linkages with the field winding.
Field current
I f = √2|Ei|
ω∗mf
= 41925.29
100 π∗31.6950∗10−3 =4210.5172 A
Flux linkages
λf =lff∗¿ If − 3 mf
√ 2 | Ia| sinθa ¿
θa=θ+δ=31.38+34.8=66.18
|Ia|sin θa= 21603
√2 sin(66.18)=13976. 4307 A
∴ λf =433.68∗10−3∗4210.5172−3∗31.695∗13976.4307∗10−3
√2 sin(66.18)
¿ 966.35 wb . turns
References
Fuchs, E. & Masoum, M. A. S., 2015. Power Quality in Power Systems and Electrical Machines. Second
Edition ed. s.l.:Academic Press/Elsevier.
Vukosavic., S. N., 2013. Electrical machines. s.l.:Springer.