Engineering Hydrology Questions and Answers – Streamflow – Velocity Measurement – Set 2

This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Streamflow – Velocity Measurement – Set 2”.

1. What is the facility used for the calibration of a current meter?
a) Towing tank
b) Wind tunnel
c) Pelton wheel turbine
d) Vortex generator
View Answer

Answer: a
Explanation: A towing tank is a long channel filled with still water and having the provision to move objects along the water surface longitudinally. The current meter is set up on a carriage which is towed at a given speed and the revolutions of the current meter are noted. This is repeated for complete range of stream velocities.

2. What equipment is required for measuring stream velocity using current meters by wading?
a) Boat
b) Cable way
c) Surface floats
d) No equipment required
View Answer

Answer: d
Explanation: Wading is the act of walking through water and is most suitable for streams of very shallow depth. The current meter is held at a suitable depth by the observer and no additional equipment is required. It is a simple and quick process but is applicable only to small streams.

3. Which of the following is true with respect to using current meters from bridges?
a) Measurement should be done on upstream section
b) Measurement should be done on downstream section
c) Location of measurement is determined using theodolite
d) Total station is necessary to carry out the observations
View Answer

Answer: b
Explanation: Bridges may not be the most suitable location for using current meters but have the advantage of easy accessibility. It should be taken care that measurement be done on a downstream section of the river, so as to prevent the damage of the current meter due to waves knocking it into the bridge piers.
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4. For a very shallow stream, what is the average velocity in a vertical section taken as?
a) Velocity at 0.2 times the depth below the stream surface
b) Velocity at 0.2 times the depth above the stream bottom
c) Velocity at 0.6 times the depth below the stream surface
d) Velocity at 0.6 times the depth above the stream bottom
View Answer

Answer: c
Explanation: It is time consuming to determine the velocity at a vertical section accurately. Hence, for very shallow streams, the average velocity at a vertical section is simply taken as the velocity at that section at a depth of 0.6 times the depth below the water surface.

5. For a stream of average depth of 4.5m, a particular vertical section AA had a depth of 5m. What is the average velocity (in m/s) at section AA? The velocities at different depths at section AA are given below.

Depth of point below water surface (in m)

1 2 3 4 5
Velocity (in m/s) 0.6 0.55 0.48 0.25 0.05

a) 0.33
b) 0.40
c) 0.43
d) 0.48
View Answer

Answer: c
Explanation: The given stream is moderately deep and single point observation method cannot be applied as the depth is more than 3m. The two velocities at section AA to be considered are that at 0.2 times the depth and 0.8 times the depth below the water surface, i.e., v0.2 and v0.8 respectively. Since the depth of given section is 5m, the velocities at (0.2*5=) 1m and (0.8*5=) 4m depth are required.
Average velocity at section AA=\(\frac{v_{0.2}+v_{0.8}}{2}=\frac{0.6+0.25}{2}\)=0.425 m/s≅0.43 m/s

6. For flooding rivers, average velocity at a section is obtained by taking velocity measurements near the bed of the river.
a) True
b) False
View Answer

Answer: b
Explanation: The only measurement taken for flooding rivers is the surface velocity within depths of 50cm from the river surface. The final velocity is then obtained by multiplying it with a reduction factor.
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7. What is the minimum mass of a sounding weight (in kg) required for a current meter to measure an average stream velocity of 0.28 m/s and for a flow depth of 3.5 m?
a) 0.5
b) 5
c) 8.8
d) 9.8
View Answer

Answer: b
Explanation: The minimum weight of a sounding weight (in N) is given as,
W=50* v*d=50*0.28*3.5=49 N=\(\frac{49}{9.81}\) kg=4.995 kg≅5 kg

8. Which of the following is true with respect to rod floats used to measure stream velocity?
a) Weights are present on both ends of the rod float
b) The rod float should be immersed maximum upto half the depth of flow
c) The rod should be made to float in a horizontal position on the water surface
d) The rod should not touch the river bottom at any time
View Answer

Answer: d
Explanation: Rod floats are devices used to predict the stream velocity in a vertical section directly. They are made to float in a vertical position with a weighted end at the bottom. They are generally immersed at more than one-fourth the depth of water but can be immersed upto 0.9 times the depth of flow but should not touch the bottom.
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9. What disadvantage of the surface float does the canister float overcome?
a) Leakage problems
b) Inability to be used in flooding rivers
c) Disturbance by surface winds
d) Inability to measure the velocity in the vertical
View Answer

Answer: c
Explanation: Surface floats are simple floating objects immersed upto very shallow depths below the water surface. It can easily be affected by winds blowing over the water surface which affects its ability to give correct readings. A canister float overcomes this problem by having a weight attached to it at the bottom.

10. A current meter recorded 15 and 36 revolutions per minute corresponding to stream velocities of 0.32 m/s and 0.68 m/s, respectively. For a stream velocity of 1 m/s, how many revolutions would the current meter undergo in one minute?
a) 47
b) 55
c) 68
d) 83
View Answer

Answer: b
Explanation: From the given data, two current meter equations can be formed taking current meter constants as ‘a’ and ‘b’.
v1=aN1+b ⇒ 0.32=\(\frac{15}{60}\) a+b ⇒ 0.25a+b=0.32⋯⋯⋯(1)
v2=aN2+b ⇒ 0.68=\(\frac{36}{60}\) a+b ⇒ 0.6a+b=0.68⋯⋯⋯(2)
Solving (1) and (2) we get,
a=1.02857 and b=0.062857
So, the calibration equation for the given current meter is,
v=1.02857N+0.062857
Now velocity is given as 1 m/s, so the corresponding revolutions per second will be,
⇒1=1.02857N+0.062857 ⇒ N=\(\frac{1-0.062857}{1.02857}\) = 0.911 rps
∴ No.of revolutions per minute=0.911*60=54.66≅55

Sanfoundry Global Education & Learning Series – Engineering Hydrology.

To practice all areas of Engineering Hydrology, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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