# Engineering Hydrology Questions and Answers – Hydrologic Channel Routing – Set 4

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

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

Explanation: Given C0 = 0.035 and C1 = 0.375. Now,
C0+C1+C2=1
⇒C2=1-C0-C1=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

Explanation: The Muskingum routing equation is given as
Q2=C0 I2+C1 I1+C2 Q1
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

Explanation: The first coefficient is given as,
C0=$$\frac{-Kx+0.5t}{K-Kx+0.5t}=\frac{(-10*0.3)+(0.5*6)}{10-(10*0.3)+(0.5*6)}$$=0
Therefore to satisfy the condition C0+C1+C2=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 m3/s and 22 m3/s, respectively. In the successive time interval the flow at the head increased to 40 m3/s. If the routing coefficients are C0 = 0.1, C1 = 0.4 and C2 = 0.5 what is the discharge at the tail of the reach during this time interval?
a) 27 m3/s
b) 30 m3/s
c) 32 m3/s
d) 34 m3/s

Explanation: Given I1 = 30 m3/s, Q1 = 22 m3/s and I2 = 44 m3/s.
Representing the time interval as 1 and 2 we have the required discharge as,
Q2=C0 I2+C1 I1+C2 Q2=(0.1*40)+(0.4*30)+(0.5*22)
⇒Q2=4+12+11=27 m3/s
Therefore, the discharge at the tail of the reach in the required time interval is 27 m3/s.

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

 Time interval 1st hour 2nd hour 3rd hour 4th hour 5th hour 6th hour Inflow (in m3/s) 23 30 37 31 25 19

Find the peak outflow discharge (in m3/s) from the reach of the Muskingum routing coefficients are given as C0 = 0.03, C1 = 0.22 and C2 = 0.75. Take the initial inflow and outflow values as 16 m3/s.
a) 21.3
b) 25.05
c) 26.36
d) 27.44

Explanation: Calculating the discharge for each time interval.
1st hour – I0 = 16 m3/s, Q0 = 16 m3/s and I1 = 23 m3/s
⇒Q1=C0 I1+C1 I0+C_2 Q0=(0.03*23)+(0.22*16)+(0.75*16)=16.21 m3/s
2nd hour – I1 = 23 m3/s, Q1 = 16.21 m3/s and I2 = 30 m3/s
⇒Q2=C0 I2+C1 I1+C2 Q1=(0.03*30)+(0.22*23)+(0.75*16.21)=18.12 m3/s
3rd hour – I2 = 30 m3/s, Q2 = 18.12 m3/s and I3 = 37 m3/s
⇒Q3=C0 I3+C1 I2+C2 Q2=(0.03*37)+(0.22*30)+(0.75*18.12)=21.3 m3/s
4th hour – I3 = 37 m3/s, Q3 = 21.3 m3/s and I4 = 31 m3/s
⇒Q4=C0 I4+C1 I3+C2 Q3=(0.03*31)+(0.22*37)+(0.75*21.3)=25.05 m3/s
5th hour – I4 = 31 m3/s, Q4 = 25.05 m3/s and I5 = 25 m3/s
⇒Q5=C0 I5+C1 I4+C2 Q4=(0.03*25)+(0.22*31)+(0.75*25.05)=26.36 m3/s
6th hour – I5 = 25 m3/s, Q5 = 26.36 m3/s and I6 = 19 m3/s
⇒Q6=C0 I6+C1 I5+C2 Q5=(0.03*19)+(0.22*25)+(0.75*26.36)=25.84 m3/s
The outflow values increase upto the 5th hour and then start to fall. The peak value of the outflow is 26.36 m3/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 m3/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 m3/s. Use Muskingum method with coefficients C0 = 0.04, C1 = 0.38 and C2 = 0.58.
a) 12 pm
b) 1:30 pm
c) 3 pm
d) 4 pm

Explanation: Calculating the discharge for each time.
At 8 am – I1 = 50 m3/s, Q1 = 45 m3/s and I2 = 81 m3/s
⇒Q2=C0 I2+C1 I1+C2 Q1=(0.04*81)+(0.38*50)+(0.58*45)=48.34 m3/s
At 10 am – I2 = 81 m3/s, Q2 = 48.34 m3/s and I3 = 113 m3/s
⇒Q3=C0 I3+C1 I2+C2 Q2=(0.04*113)+(0.38*81)+(0.58*48.34)=63.34 m3/s
At 12 pm – I3 = 113 m3/s, Q3 = 63.34 m3/s and I4 = 134 m3/s
⇒Q4=C0 I4+C1 I3+C2 Q3=(0.04*134)+(0.38*113)+(0.58*63.34)=85.04 m3/s
At 2 pm – I4 = 134 m3/s, Q4 = 85.04 m3/s and I5 = 107 m3/s
⇒Q5=C0 I5+C1 I4+C2 Q4=(0.04*107)+(0.38*134)+(0.58*85.04)=104.52 m3/s
At 4 pm – I5 = 107 m3/s, Q5 = 104.52 m3/s and I6 = 78 m3/s
⇒Q6=C0 I6+C1 I5+C2 Q5=(0.04*78)+(0.38*107)+(0.58*104.52)=104.4 m3/s
At 6 pm – I6 = 78 m3/s, Q6 = 104.4 m3/s and I7 = 50 m3/s
⇒Q7=C0 I7+C1 I6+C2 Q6=(0.04*50)+(0.38*78)+(0.58*104.4)=92.19 m3/s
The outflow values increase to 104.52 m3/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 m3/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 m3/s for 3 hours after which it starts decreasing at a rate of 14 m3/s until it reaches the initial value of 18 m3/s. Given that the initial outflow is 15 m3/s, find the approximate attenuation of peak between inflow and outflow hydrograph. Use Muskingum method with C0 = 0.1, C1 = 0.4 and C2 = 0.5.
a) 29 m3/s
b) 24 m3/s
c) 19 m3/s
d) 15 m3/s

Explanation: Calculating the discharge for each hour.
1st hour – I0 = 18 m3/s, Q0 = 15 m3/s and I1 = 18 + (18*1) = 36 m3/s
⇒Q1=C0 I1+C1 I0+C2 Q0=(0.1*36)+(0.4*18)+(0.5*15)=18.3 m3/s
2nd hour – I1 = 36 m3/s, Q1 = 18.3 m3/s and I2 = 18 + (18*2) = 54 m3/s
⇒Q2=C0 I2+C1 I1+C2 Q1=(0.1*54)+(0.4*36)+(0.5*18.3)=28.95 m3/s
3rd hour – I2 = 54 m3/s, Q2 = 28.95 m3/s and I3 = 18 + (18*3) = 72 m3/s
⇒Q3=C0 I3+C1 I2+C2 Q2=(0.1*72)+(0.4*54)+(0.5*28.95)=43.28 m3/s
4th hour – I3 = 72 m3/s, Q3 = 43.28 m3/s and I4 = 72 – (14*1) = 58 m3/s
⇒Q4=C0 I4+C1 I3+C2 Q3=(0.1*58)+(0.4*72)+(0.5*43.28)=56.24 m3/s
5th hour – I4 = 58 m3/s, Q4 = 56.24 m3/s and I5 = 72 – (14*2) = 44 m3/s
⇒Q5=C0 I5+C1 I4+C2 Q4=(0.1*44)+(0.4*58)+(0.5*56.24)=55.72 m3/s
6th hour – I5 = 44 m3/s, Q5 = 55.72 m3/s and I6 = 72 – (14*3) = 30 m3/s
⇒Q6=C0 I6+C1 I5+C2 Q5=(0.1*30)+(0.4*44)+(0.5*55.72)=48.46 m3/s

Therefore, peak attenuation ≈ 72 – 57 = 15 m3/s

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