9.19 Generate Bode magnitude and phase plots (straight-line approximations) for the following voltage transfer functions. (a) H(
ω) = 4×104(60+j6ω) (4+j2ω)(100+j2ω)(400+j4ω) (b) H(ω) = (1+j0.2ω)2(100+j2ω)2 (jω)3(500+jω) (c) H(ω) = 8×10−2(10+j10ω) jω(16−ω2 +j4ω) (d) H(ω) = 4×104ω2(100−ω2 +j50ω) (5+j5ω)(200+j2ω)3
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Answer:
a) The velocity is 42.0833 ft/s
b) The flow rate is 3366.664 ft³/s
c) The Froude number is 0.2345
d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft
e) The critical depth is 3.8030 ft
Explanation:
Given data:
80 ft wide channel, L
1 ft and 10 ft water depths, d₁ and d₂
Questions: a) Velocity of the faster moving flow, v = ?
b) The flow rate (discharge), q = ?
c) The Froude number, F = ?
d) The flow energy dissipated, E = ?
e) The critical depth, dc = ?
a) For the velocity:
Solving for F:
F = 7.4162
Here, g = gravity = 32.2 ft/s²
b) The flow rate:
c) The Froude number:
d) The flow energy dissipated:
e) The critical depth: