Answer: An electric power system is a network of electrical components deployed to supply, transfer, and use electric power.
Explanation:
Known :
Q = 300 L/s = 0.3 m³/s
D1 = 350 mm = 0.35 m
D2 = 700 mm = 0.7 m
g = 9.81 m/s²
Solution :
A1 = πD1² / 4 = π(0.35²) / 4 = 0.096 m²
A2 = πD2² / 4 = π(0.7²) / 4 = 0.385 m²
hL = (kL / 2g) • (U1² - U2²)
hL = (kL / 2g) • Q² (1/A1² - 1/A2²)
hL = (1 / 2(9.81)) • (0.3²) • (1/(0.096²) - 1/(0.385²))
hL = 0.467 m
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