Fluid Mechanics Questions and Answers – Gradually Varied Flow(GVF) – 4

This set of Fluid Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Gradually Varied Flow(GVF) – 4”.

1. What is the expression for head loss in case of a GVF?
a) h_f = ^L⁄₂ S_f
b) h_f = LS_f
c) h_f = 2LS_f
d) h_f = 3LS_f
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2. What is the expression for the length of the backwater curve?
a) L = \(\frac{E_2-E_1}{S_f-S_0}\)
b) L = \(\frac{E_2-E_1}{S_f}\)
c) L = \(\frac{E_2-E_1}{S_0-S_f}\)
d) L = \(\frac{E_2-E_1}{S_0}\)
View Answer

Answer: c
Explanation:

3. Calculate the head loss if the length of the back water curve is 25000m and S_f=0.00006.
a) 1m
b) 1.5m
c) 2.0m
d) 2.5m
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4. Estimate the slope of energy line in a GVF having length of the back water curve 30000m and head loss of 1m.
a) 1.33×10^-5
b) 2.33×10^-5
c) 3.33×10^-5
d) 4.33×10^-5
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5. Determine the length of the back water curve if E₁=2.8m and E₂=5.6m. Given:S₀=0.00009 S_f= 0.00004.
a) 26000m
b) 36000m
c) 46000m
d) 56000m
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Answer: d
Explanation:

6. If the difference between specific energies is 2m calculate the rate of change of specific energies if the length of the back water curve is 26314m.
a) 6.6×10^-5m
b) 7.6×10^-5m
c) 8.6×10^-5m
d) 9.6×10^-5m
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Answer: b
Explanation:

7. Calculate the bed slope of the channel if the slope of the energy line 0.00024 and the length of the back water curve is 104166.67m. Given:E₁-E₂=3m.
a) 2.28×10^-5
b) 3.28×10^-5
c) 4.28×10^-5
d) 5.28×10^-5
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Answer: d
Explanation:

8. If the depths in a channel are 2m and 4m and the velocities are 0.5 m/s and 0.3m/s, calculate the difference between specific energies.
a) 2m
b) 3m
c) 4m
d) 5m
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Answer: a
Explanation:

9. Calculate the slope of the energy line if the bed slope of the channel is 4.81×10^-5 if the depths of the channel are 2.7m and 4.7m and velocities are 1 m/s and 0.5m/s respectively.
a) 0.00002
b) 0.00003
c) 0.00004
d) 0.00005
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Answer: d
Explanation:

10. The dimensions of a rectangular channel is 3m×2m and the bed slope of the channel is 1 in 1000, calculate the rate of change of depth if the rate of change of specific energy is 2×10^-5m. Given: n = 0.010
a) 1.43×10^-5m
b) 2.43×10^-5m
c) 3.43×10^-5m
d) 4.43×10^-5m
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Answer: c
Explanation:

Sanfoundry Global Education & Learning Series – Fluid Mechanics.

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