# Fluid Mechanics Questions and Answers – Gradually Varied Flow(GVF) – 3

This set of Fluid Mechanics Objective Questions & Answers focuses on “Gradually Varied Flow(GVF) – 3”.

1. Determine the value of the hydraulic depth of a channel having dydx = 4×10-4 and dEdx = 3.73×10-4. Given:V = 1m/s
a) 0.5m
b) 1.5m
c) 2.5m
d) 3.5m

Explanation:

2. A rectangular channel has depth of 1m and width of 2m, calculate the rate of change of depth if the rate of change of specific energy is 2×10-5m. Given: Fr=0.48.
a) 0.60×10-5m
b) 1.60×10-5m
c) 2.60×10-5m
d) 3.60×10-5m

Explanation:

3. The top width of a triangular channel section is 6m and the side slope of the channel is 1H:4V, calculate the rate of change of specific energy if dydx = 2×10-5m and V=2 m/s
a) 1.86×10-5m
b) 2.86×10-5m
c) 3.86×10-5m
d) 4.86×10-5m

Explanation:

4. If dEdx = 2.5×10-4 m and dydx = 3.5×10-4 m, calculate the value of Fr.
a) 0.23
b) 0.33
c) 0.43
d) 0.53

Explanation:

5. The rate of change of specific energy is given by x2/2 and x ranges from 0 to 3, calculate the value of specific energy.
a) 2.5m
b) 3.5m
c) 4.5m
d) 5.5m

Explanation:
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6. The rate of change of depth is given by 1/x2 and the rate of change of specific energy is given by 3x2 with x ranging from 0 to 0.2, calculate the value of Froude’s number.
a) 0.96
b) 0.97
c) 0.98
d) 0.99

Explanation:

7. The hydraulic depth of a channel is 0.94m and the velocity of flow is 2 m/s. Calculate the rate of change of depth if the rate of change of specific energy is 2.5×10-4m.
a) 2.33×10-4m
b) 3.33×10-4m
c) 4.33×10-4m
d) 5.33×10-4m

Explanation:

8. Calculate the value of Froude’s number for a rectangular channel having depth 1.5m and width 2.5m if the value of C = 30 and S0=1 in 1000.
a) 0.1
b) 0.2
c) 0.3
d) 0.4

Explanation:

9. The dynamic equation for the slope of water surface in a GVF is not valid for super critical flow.
a) True
b) False

Explanation:

10. If the ratio of dEdx and dydx is 0.2823, estimate the value of Froude’s number.
a) 0.65
b) 0.75
c) 0.85
d) 0.95

Explanation: Ratio = 1 – Fr2 = 0.2823; Fr=0.85.

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