HW2 - Kelompok 2
HW2 - Kelompok 2
HW2 - Kelompok 2
1. Hydropower plant consists of 4 turbines each with 150 MW capacity, and runs with a mean
head of approximately H = 600 m. Assuming that the efficiency of the turbines are 95%, find
the maximum flow rate through the turbines!
Answer:
P = ρ*g*H*Q*ε
Where : P = Power (W)
g = Gravity acceleration (m/s2)
ρ = density (kg/m3)
H = head (m)
d = flow rate (m3/s)
ε = efficiency (%)
Turbine Output = 150 MW * 4 = 600 MW = 600 x 106 W
631.600.000 W = ρ*g*H*d* ε
631.600.000 W = 9,8 m/s2 * (95%) * 600 m * d * 0,95
d= 600.000.000 W
9,8 m/s2 * 1000 kg/m3 * 600 m * 0,95
d = 107,42 m3/s (maximum flow rate)
2. The above hydropower plant has two of the turbines that can be reversed and pump the water
from the low to the high water reservoir. It is used for seasonal storage, and the high reservoir
has a capacity of 3×109 m3 of water. Use the numbers given in (1) and find how many days of
stored energy this corresponds to when the electricity consumption is 120 TWh per year.
Answer:
Total power production = 3x109 m3 / 107,42 m3/s
Period = 3x109 m3 / 107,42 m3 x 60 x 60
= 7757,71 hour