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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