This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Hydrologic Channel Routing – Set 4”.

1. What should be the range of time interval t chosen in Muskingum routing to achieve reliable results? K is the storage-time constant and x is the weighting factor?

a) x > t > 2Kx

b) x < t < 2Kx

c) K > t > 2Kx

d) K < t < 2Kx

View Answer

Explanation: The time interval t in Muskingum routing should be chosen such that it satisfies the condition K > t > 2Kx for best results. If t < 2Kx the first routing coefficient will have a negative value.

2. The value of two Muskingum routing coefficients are 0.035 and 0.375. What is the value of the third coefficient?

a) 0.41

b) 0.59

c) 0.34

d) 0.66

View Answer

Explanation: Given C

_{0}= 0.035 and C

_{1}= 0.375. Now,

C

_{0}+C

_{1}+C

_{2}=1

⇒C

_{2}=1-C

_{0}-C

_{1}=1-0.035-0.375=0.59

3. The first Muskingum coefficient is multiplied to outflow value whereas the second and third coefficients are multiplied to inflow values in Muskingum routing equation

a) True

b) False

View Answer

Explanation: The Muskingum routing equation is given as

Q

_{2}=C

_{0}I

_{2}+C

_{1}I

_{1}+C

_{2}Q

_{1}

It can be clearly seen that the first two coefficients are multiplied to inflow values and the last coefficient is multiplied to an outflow value.

4. What is the value of the smallest Muskingum routing coefficient for a storage-time constant of 10 hours, weighting factor of 0.3 and time interval of 6 hours?

a) 0

b) -0.4

c) 0.05

d) 0.4

View Answer

Explanation: The first coefficient is given as,

C

_{0}=\(\frac{-Kx+0.5t}{K-Kx+0.5t}=\frac{(-10*0.3)+(0.5*6)}{10-(10*0.3)+(0.5*6)}\)=0

Therefore to satisfy the condition C

_{0}+C

_{1}+C

_{2}=1, the other two coefficients should be greater than zero. Hence, the value of the smallest coefficient is 0.

5. During a particular time interval the discharge at the head and tail of a reach are given as 30 m^{3}/s and 22 m^{3}/s, respectively. In the successive time interval the flow at the head increased to 40 m^{3}/s. If the routing coefficients are C_{0} = 0.1, C_{1} = 0.4 and C_{2} = 0.5 what is the discharge at the tail of the reach during this time interval?

a) 27 m^{3}/s

b) 30 m^{3}/s

c) 32 m^{3}/s

d) 34 m^{3}/s

View Answer

Explanation: Given I

_{1}= 30 m

^{3}/s, Q

_{1}= 22 m

^{3}/s and I

_{2}= 44 m

^{3}/s.

Representing the time interval as 1 and 2 we have the required discharge as,

Q

_{2}=C

_{0}I

_{2}+C

_{1}I

_{1}+C

_{2}Q

_{2}=(0.1*40)+(0.4*30)+(0.5*22)

⇒Q

_{2}=4+12+11=27 m

^{3}/s

Therefore, the discharge at the tail of the reach in the required time interval is 27 m

^{3}/s.

6. The inflow values of a channel reach are given below.

Time interval | 1^{st} hour |
2^{nd} hour |
3^{rd} hour |
4^{th} hour |
5^{th} hour |
6^{th} hour |

Inflow (in m^{3}/s) |
23 | 30 | 37 | 31 | 25 | 19 |

Find the peak outflow discharge (in m^{3}/s) from the reach of the Muskingum routing coefficients are given as C_{0} = 0.03, C_{1} = 0.22 and C_{2} = 0.75. Take the initial inflow and outflow values as 16 m^{3}/s.

a) 21.3

b) 25.05

c) 26.36

d) 27.44

View Answer

Explanation: Calculating the discharge for each time interval.

1

^{st}hour – I0 = 16 m

^{3}/s, Q

_{0}= 16 m

^{3}/s and I

_{1}= 23 m

^{3}/s

⇒Q

_{1}=C

_{0}I

_{1}+C

_{1}I

_{0}+C_2 Q

_{0}=(0.03*23)+(0.22*16)+(0.75*16)=16.21 m

^{3}/s

2

^{nd}hour – I

_{1}= 23 m

^{3}/s, Q

_{1}= 16.21 m

^{3}/s and I

_{2}= 30 m

^{3}/s

⇒Q

_{2}=C

_{0}I

_{2}+C

_{1}I

_{1}+C

_{2}Q

_{1}=(0.03*30)+(0.22*23)+(0.75*16.21)=18.12 m

^{3}/s

3

^{rd}hour – I

_{2}= 30 m

^{3}/s, Q

_{2}= 18.12 m

^{3}/s and I

_{3}= 37 m

^{3}/s

⇒Q

_{3}=C

_{0}I

_{3}+C

_{1}I

_{2}+C

_{2}Q

_{2}=(0.03*37)+(0.22*30)+(0.75*18.12)=21.3 m

^{3}/s

4

^{th}hour – I

_{3}= 37 m

^{3}/s, Q

_{3}= 21.3 m

^{3}/s and I

_{4}= 31 m

^{3}/s

⇒Q

_{4}=C

_{0}I

_{4}+C

_{1}I

_{3}+C

_{2}Q

_{3}=(0.03*31)+(0.22*37)+(0.75*21.3)=25.05 m

^{3}/s

5

^{th}hour – I

_{4}= 31 m

^{3}/s, Q

_{4}= 25.05 m

^{3}/s and I

_{5}= 25 m

^{3}/s

⇒Q

_{5}=C

_{0}I

_{5}+C

_{1}I

_{4}+C

_{2}Q

_{4}=(0.03*25)+(0.22*31)+(0.75*25.05)=26.36 m

^{3}/s

6

^{th}hour – I

_{5}= 25 m

^{3}/s, Q

_{5}= 26.36 m

^{3}/s and I

_{6}= 19 m

^{3}/s

⇒Q

_{6}=C

_{0}I

_{6}+C

_{1}I

_{5}+C

_{2}Q

_{5}=(0.03*19)+(0.22*25)+(0.75*26.36)=25.84 m

^{3}/s

The outflow values increase upto the 5th hour and then start to fall. The peak value of the outflow is 26.36 m

^{3}/s.

7. The inflow values into a river section during 6 am to 6 pm were recorded as given below.

Time | 6 am | 8 am | 10 am | 12 pm | 2pm | 4 pm | 6 pm |

Inflow (in m^{3}/s) |
50 | 81 | 113 | 134 | 107 | 78 | 50 |

What is the most likely time during which the outflow peak will occur? The outflow from the section at 6 am was recorded as 45 m^{3}/s. Use Muskingum method with coefficients C_{0} = 0.04, C_{1} = 0.38 and C_{2} = 0.58.

a) 12 pm

b) 1:30 pm

c) 3 pm

d) 4 pm

View Answer

Explanation: Calculating the discharge for each time.

At 8 am – I

_{1}= 50 m

^{3}/s, Q

_{1}= 45 m

^{3}/s and I

_{2}= 81 m

^{3}/s

⇒Q

_{2}=C

_{0}I

_{2}+C

_{1}I

_{1}+C

_{2}Q

_{1}=(0.04*81)+(0.38*50)+(0.58*45)=48.34 m

^{3}/s

At 10 am – I

_{2}= 81 m

^{3}/s, Q

_{2}= 48.34 m

^{3}/s and I

_{3}= 113 m

^{3}/s

⇒Q

_{3}=C

_{0}I

_{3}+C

_{1}I

_{2}+C

_{2}Q

_{2}=(0.04*113)+(0.38*81)+(0.58*48.34)=63.34 m

^{3}/s

At 12 pm – I

_{3}= 113 m

^{3}/s, Q

_{3}= 63.34 m

^{3}/s and I

_{4}= 134 m

^{3}/s

⇒Q

_{4}=C

_{0}I

_{4}+C

_{1}I

_{3}+C

_{2}Q

_{3}=(0.04*134)+(0.38*113)+(0.58*63.34)=85.04 m

^{3}/s

At 2 pm – I

_{4}= 134 m

^{3}/s, Q

_{4}= 85.04 m

^{3}/s and I

_{5}= 107 m

^{3}/s

⇒Q

_{5}=C

_{0}I

_{5}+C

_{1}I

_{4}+C

_{2}Q

_{4}=(0.04*107)+(0.38*134)+(0.58*85.04)=104.52 m

^{3}/s

At 4 pm – I

_{5}= 107 m

^{3}/s, Q

_{5}= 104.52 m

^{3}/s and I

_{6}= 78 m

^{3}/s

⇒Q

_{6}=C

_{0}I

_{6}+C

_{1}I

_{5}+C

_{2}Q

_{5}=(0.04*78)+(0.38*107)+(0.58*104.52)=104.4 m

^{3}/s

At 6 pm – I

_{6}= 78 m

^{3}/s, Q

_{6}= 104.4 m

^{3}/s and I

_{7}= 50 m

^{3}/s

⇒Q

_{7}=C

_{0}I

_{7}+C

_{1}I

_{6}+C

_{2}Q

_{6}=(0.04*50)+(0.38*78)+(0.58*104.4)=92.19 m

^{3}/s

The outflow values increase to 104.52 m

^{3}/s at 2 pm and it remains almost same 2 hours later at 4 pm. This implies that the outflow increases after 2 pm to a peak after which it decreases to 104.4 m

^{3}/s at 4 pm and then further decreases. Therefore, the outflow peak occurs sometime between 2 pm and 4 pm.

8. The inflow values for a river reach increases uniformly at the rate of 18 m^{3}/s for 3 hours after which it starts decreasing at a rate of 14 m^{3}/s until it reaches the initial value of 18 m^{3}/s. Given that the initial outflow is 15 m^{3}/s, find the approximate attenuation of peak between inflow and outflow hydrograph. Use Muskingum method with C_{0} = 0.1, C_{1} = 0.4 and C_{2} = 0.5.

a) 29 m^{3}/s

b) 24 m^{3}/s

c) 19 m^{3}/s

d) 15 m^{3}/s

View Answer

Explanation: Calculating the discharge for each hour.

1

^{st}hour – I0 = 18 m

^{3}/s, Q0 = 15 m

^{3}/s and I

_{1}= 18 + (18*1) = 36 m

^{3}/s

⇒Q

_{1}=C

_{0}I

_{1}+C

_{1}I

_{0}+C

_{2}Q

_{0}=(0.1*36)+(0.4*18)+(0.5*15)=18.3 m

^{3}/s

2

^{nd}hour – I

_{1}= 36 m

^{3}/s, Q

_{1}= 18.3 m

^{3}/s and I

_{2}= 18 + (18*2) = 54 m

^{3}/s

⇒Q

_{2}=C

_{0}I

_{2}+C

_{1}I

_{1}+C

_{2}Q

_{1}=(0.1*54)+(0.4*36)+(0.5*18.3)=28.95 m

^{3}/s

3

^{rd}hour – I

_{2}= 54 m

^{3}/s, Q

_{2}= 28.95 m

^{3}/s and I

_{3}= 18 + (18*3) = 72 m

^{3}/s

⇒Q

_{3}=C

_{0}I

_{3}+C

_{1}I

_{2}+C

_{2}Q

_{2}=(0.1*72)+(0.4*54)+(0.5*28.95)=43.28 m

^{3}/s

4

^{th}hour – I

_{3}= 72 m

^{3}/s, Q

_{3}= 43.28 m

^{3}/s and I

_{4}= 72 – (14*1) = 58 m

^{3}/s

⇒Q

_{4}=C

_{0}I

_{4}+C

_{1}I

_{3}+C

_{2}Q

_{3}=(0.1*58)+(0.4*72)+(0.5*43.28)=56.24 m

^{3}/s

5

^{th}hour – I

_{4}= 58 m

^{3}/s, Q

_{4}= 56.24 m

^{3}/s and I

_{5}= 72 – (14*2) = 44 m

^{3}/s

⇒Q

_{5}=C

_{0}I

_{5}+C

_{1}I

_{4}+C

_{2}Q

_{4}=(0.1*44)+(0.4*58)+(0.5*56.24)=55.72 m

^{3}/s

6

^{th}hour – I

_{5}= 44 m

^{3}/s, Q

_{5}= 55.72 m

^{3}/s and I

_{6}= 72 – (14*3) = 30 m

^{3}/s

⇒Q

_{6}=C

_{0}I

_{6}+C

_{1}I

_{5}+C

_{2}Q

_{5}=(0.1*30)+(0.4*44)+(0.5*55.72)=48.46 m

^{3}/s

Therefore, peak attenuation ≈ 72 – 57 = 15 m^{3}/s

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