Engineering Hydrology Questions and Answers – Streamflow – Area-Velocity Method – Set 2

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

1. If ‘y’ is the average depth of a section, ‘v’ is the resultant velocity of current meter and ‘t’ is the time taken by a boat to cross the section, the discharge through the section by moving-boat method is given as:
Q=v2.y.t.sin⁡ θ cos⁡θ
What is denoted by ‘ϴ’ in the above equation?
a) Angle between boat and streamflow
b) Angle between boat and resultant
c) Angle between streamflow and resultant
d) Angle between either streamflow or boat and the resultant
View Answer

Answer: d
Explanation: The angle between the boat and the stream is kept as close to perpendicular as possible. Generally, the angle between the resultant and boat is measured. But for the sake of the final equation, the angle measured may be with respect to either the boat or stream as sin and cos are complementary.

2. A moving boat (perpendicular to stream) is studied between a section AB. The start and end depths of section AB are yA and yB. A current meter attached to the boat makes an angle of α with the boat and records a velocity vR. If tAB is the time taken by the boat to cross AB, what is the correct equation to estimate the discharge through section AB?
a) QAB=2(yA+yB).\(v_R^2\).sin⁡ 2α.tAB
b) QAB=\(\frac{1}{2}\) (yA+yB).\(v_R^2\).sin⁡ 2α.tAB
c) QAB=\(\frac{1}{4}\) (yA+yB).\(v_R^2\).sin 2α.tAB
d) QAB=\(\frac{1}{8}\) (yA+yB).\(v_R^2\).sin⁡ 2α.tAB
View Answer

Answer: c
Explanation: Let vb and vs be the velocity of boat and stream, respectively. Let WAB be the width of section AB. The discharge through AB is given as,
QAB = (Average depth of AB)*WAB*vs
⇒ QAB = \(\left(\frac{y_A+y_B}{2}\right)\).(vb*tAB).(vR sin⁡ α) = \(\left(\frac{y_A+y_B}{2}\right)\).(vR cos⁡α*tAB).(vR sin⁡ α)
⇒ QAB = \(\left(\frac{y_A+y_B}{2}\right).v_R^2.\left(\frac{2 sin⁡α cos⁡α}{2}\right).\)tAB = \(\frac{1}{2} \left(\frac{y_A+y_B}{2}\right).v_R^2.\)(2 sin⁡α cos⁡α ).tAB
∴ QAB = \(\frac{1}{4}\) (yA+yB).\(v_R^2\).sin⁡ 2α.tAB

3. A boat moves with a speed of 12 km/h perpendicular to the flow of a stream. It crosses a particular section whose end depths are 2.8 m and 2.95 m in 11 seconds. A current meter attached to the boat makes of an angle of 10° with the boat direction. Calculate the discharge through the given section during the time the boat crosses it.
a) 62 m3
b) 620 L
c) 680 m3
d) 6800 L
View Answer

Answer: c
Explanation: The average depth of the section is, y=\(\frac{2.8+2.95}{2}\) = 2.875 m
The width of the section is, W=speed of boat*time taken=12kmph*17 sec
= \(\left(\frac{12}{3.6}\right)\) m/s *17sec=36.67 m
The velocity of stream, Vf = VR sin⁡ θ = \(\left(\frac{V_b}{cos⁡ θ}\right)\)*sin⁡θ = Vb tan⁡θ=\(\left(\frac{12}{3.6}\right)\)*tan⁡10=0.59 m/s
Therefore, discharge through the section =Depth*Width*stream velocity
= 2.875*36.67*0.59=62.2 m3/s
Discharge from the section in 11 seconds = 62.2 * 11 = 684.2 m3 ≅ 680 m3

4. Find the average width of any intermediate section of a stream divided into N sections (N ≥ 3). The width of each section is 2.5m and the depth varies linearly from the edges to 6 m at the centre.
a) 1.25 m
b) 2.5 m
c) 2.8125 m
d) 5 m
View Answer

Answer: b
Explanation: For intermediate sections, average width is given as,
\(\overline{W}_i = \frac{w_i+w_{i+1}}{2} = \frac{2.5+2.5}{2} = \frac{5}{2}\) = 2.5 m
Since the stream is divided into sections of equal widths, the average width of each intermediate section will be 2.5 m, but the end sections will have different average widths.

5. A stream is divided into N (N ≥ 4) sections, with end sections having widths ‘x’ m and intermediate sections having width ‎’y’ m. If the first average width and the second average width are same and y = 3 m, what is the relationship between ‘x’ and ‘y’?
a) x = 2 y
b) y = 2x
c) x = y2
d) y = x2
View Answer

Answer: b
Explanation: Considering the first and second average width,
\(\overline{W}_1 = \overline{W}_2 ⇒ \frac{\left(x+\frac{y}{2}\right)^2}{2x}=\frac{y}{2}+\frac{y}{2} \)
⇒ \(\left(x^2+\frac{y^2}{4}+xy\right)\) = 2xy ⇒ x2+\(\frac{y^2}{4}\) = xy⋯⋯⋯(1)
Substituting y = 3 in (1),
⇒x2+\(\frac{9}{4}\) = 3x ⇒ 4x2-12x+9=0
Solving the quadratic equation, we get x = 1.5 m.
∴ x=\(\frac{y}{2}\) or y=2x

6. A stream divided into N sections of equal width will have N-1 average widths of equal width.
a) True
b) False
View Answer

Answer: b
Explanation: Let us take the width of each section to be 1m.
The average width of end sections = \(\frac{\left(1+\frac{1}{2}\right)^2}{2*1}\) = 1.125 m
The average width of intermediate sections = \(\frac{1}{2}+\frac{1}{2}\)=1 m
Therefore, the average width of end sections will be different to that of the intermediate sections.

7. The area-velocity method is carried out on a stream by dividing it into 6 sections of widths (from left bank to right bank) – 3.5m, 3.2m, 3.8m, 4m, 3.9m and 4.2m. Find the difference (value only) between the average width of 1st section and average width of 5th section. (Consider left bank as starting point).
a) 0
b) 0.7
c) 0.787
d) 1.574
View Answer

Answer: c
Explanation: Since the stream has 6 initial sections, the 5th average width is the last average width.
\(\overline{W}_1 = \frac{\left(w_1+\frac{w_2}{2}\right)^2}{2w_1} = \frac{\left(3.5+\frac{3.2}{2}\right)^2}{2*3.5}\) = 3.7157 m
and, \(\overline{W}_5 = \frac{\left(w_6+\frac{w_5}{2}\right)^2}{2w_6} = \frac{\left(4.2+\frac{3.9}{2}\right)^2}{2*4.2}\) = 4.5027 m
∴ |\(\overline{W}_1-\overline{W}_5\)|=|3.7157-4.5027|=0.787

8. A river is divided into an odd number of sections. The first section and each alternating section have a width of 4.5 m and rest of the sections have a width of 4.2 m. What will be the average width of the last section?
a) 4.35 m
b) 4.84 m
c) 4.95 m
d) 5.19 m
View Answer

Answer: b
Explanation: Since there are odd number of sections, the last section will have a width of 4.5 m and second to last section will have a width of 4.2 m. So the average width of the last section is given as,
\(\overline{W}_{last} = \frac{\left(w_{last}+\frac{w_{second \, last}}{2}\right)^2}{2*w_{last}} =\frac{\left(4.5+\frac{4.2}{2}\right)^2}{2*4.5}\) = 4.84 m
Therefore, average width of the last section is 4.84 m.

Sanfoundry Global Education & Learning Series – Engineering Hydrology.


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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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