Answer:
a) expected flow rate is 1852 gpm
b) head is 41.7 ft
c) input power is 21.1 hP
Step-by-step explanation:
from flow coefficient equation C_Q = Q/(W×D³)
Q ∝ d³
Qd₁/Qd₂ = d₁³/d₂³
where Qd₁ and Qd₂ are different discharges with respect to different diameter, (Qd₁ = 3200 gpm) and d₁ is diameter one ( 12 in) while d₂ is diameter 2 ( 10 in),
now we substitute
Qd₂ = Qd₁(d₂³/d₁³)
Qd₂ = 3200( 10/12)³
Qd₂ = 1852 gpm
∴ expected flow rate is 1852 gpm
next we calculate the head
h ∝ d²
hd₁/hd₂ = d₁²/d₂²
where hd₁ and hd₂ are different heads with respect to different diameter, (hd₁ = 60 ft) and d₁ is diameter one ( 12 in) while d₂ is diameter 2 ( 10 in),
now we substitute
hd₂ = hd₁(d₂²/d₁²)
hd₂ = 60 ( 10/12 )²
hd₂ = 41.7 ft
∴ head is 41.7 ft
Now calculate the input power
W ∝ d⁵
Wd₁/Wd₂ = d₁⁵/d₂⁵
where Wd₁ and Wd₂ are different power with respect to different diameter, (Wd₁ = 60 hp) and d₁ is diameter one ( 12 in) while d₂ is diameter 2 ( 10 in),
now we substitute
Wd₂ = Wd₁(d₂⁵/d₁⁵)
Wd₂ = 60 ( 10/12 )⁵
Wd₂ = 21.1 hP
∴ input power is 21.1 hP