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A rectangular channel had a positive surge of velocity 10 m/s moving down the channel. If depth of flow and velocity after the passing of the surge are 4.0 m and 5.0 m/s respectively, the velocity before the passage of the surge is _____m/s.
- a)3.14
- b)4.14
- c)3.19
- d)4.19
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A rectangular channel had a positive surge of velocity 10 m/s moving d...
Let Vw = Velocity of surge wave
Superimpose a velocity Vw to the left to simulate steady flow as in figure
Given, y2 = 4.0 m, V2 = 5 m/s and Vw = 10 m/s
By the continuity equation,
⇒ y1 (Vw - V1 )= y2 (Vw - V2 )
⇒ y1 (10 - V1 )= 4(10 - 5)
Most Upvoted Answer
A rectangular channel had a positive surge of velocity 10 m/s moving d...
Given information:
- Depth of flow after surge passage (h2) = 4.0 m
- Velocity after surge passage (V2) = 5.0 m/s
- Surge velocity (Vs) = 10 m/s
To find:
- Velocity before surge passage (V1)
We can use the principle of conservation of energy to solve this problem. The total energy of the flow in the channel consists of the kinetic energy and the potential energy.
1. Calculate the specific energy before the surge passage (E1):
- E1 = h1 + (V1^2/2g)
- Here, h1 is the depth of flow before the surge passage and g is the acceleration due to gravity (9.81 m/s^2).
2. Calculate the specific energy after the surge passage (E2):
- E2 = h2 + (V2^2/2g)
3. Since the surge velocity is positive and moving down the channel, the specific energy after the surge passage must be greater than the specific energy before the surge passage (E2 > E1). Therefore, we can equate the two expressions for specific energy:
h2 + (V2^2/2g) = h1 + (V1^2/2g)
4. Rearranging the equation, we get:
(V2^2/2g) - (V1^2/2g) = h1 - h2
5. Substituting the given values:
(5^2/2g) - (V1^2/2g) = 4 - 0
6. Simplifying the equation:
25/2g - V1^2/2g = 4
7. Since g is a constant, we can remove it from both sides of the equation:
25 - V1^2 = 8g
8. Substituting the value of g:
25 - V1^2 = 8 * 9.81
9. Simplifying the equation:
25 - V1^2 = 78.48
10. Rearranging the equation, we get:
V1^2 = 25 - 78.48
V1^2 = -53.48
11. Since velocity cannot be negative, we discard the negative solution. Therefore:
V1 = √(25 - 78.48)
V1 = √(-53.48)
12. Taking the square root of a negative number is not possible in the real number system. Hence, there is no real solution for V1.
Therefore, the statement "Correct answer is option 'C'" is incorrect.
- Depth of flow after surge passage (h2) = 4.0 m
- Velocity after surge passage (V2) = 5.0 m/s
- Surge velocity (Vs) = 10 m/s
To find:
- Velocity before surge passage (V1)
We can use the principle of conservation of energy to solve this problem. The total energy of the flow in the channel consists of the kinetic energy and the potential energy.
1. Calculate the specific energy before the surge passage (E1):
- E1 = h1 + (V1^2/2g)
- Here, h1 is the depth of flow before the surge passage and g is the acceleration due to gravity (9.81 m/s^2).
2. Calculate the specific energy after the surge passage (E2):
- E2 = h2 + (V2^2/2g)
3. Since the surge velocity is positive and moving down the channel, the specific energy after the surge passage must be greater than the specific energy before the surge passage (E2 > E1). Therefore, we can equate the two expressions for specific energy:
h2 + (V2^2/2g) = h1 + (V1^2/2g)
4. Rearranging the equation, we get:
(V2^2/2g) - (V1^2/2g) = h1 - h2
5. Substituting the given values:
(5^2/2g) - (V1^2/2g) = 4 - 0
6. Simplifying the equation:
25/2g - V1^2/2g = 4
7. Since g is a constant, we can remove it from both sides of the equation:
25 - V1^2 = 8g
8. Substituting the value of g:
25 - V1^2 = 8 * 9.81
9. Simplifying the equation:
25 - V1^2 = 78.48
10. Rearranging the equation, we get:
V1^2 = 25 - 78.48
V1^2 = -53.48
11. Since velocity cannot be negative, we discard the negative solution. Therefore:
V1 = √(25 - 78.48)
V1 = √(-53.48)
12. Taking the square root of a negative number is not possible in the real number system. Hence, there is no real solution for V1.
Therefore, the statement "Correct answer is option 'C'" is incorrect.
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A rectangular channel had a positive surge of velocity 10 m/s moving down the channel. If depth of flow and velocity after the passing of the surge are 4.0 m and 5.0 m/s respectively, the velocity before the passage of the surge is _____m/s.a)3.14b)4.14c)3.19d)4.19Correct answer is option 'C'. Can you explain this answer?
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A rectangular channel had a positive surge of velocity 10 m/s moving down the channel. If depth of flow and velocity after the passing of the surge are 4.0 m and 5.0 m/s respectively, the velocity before the passage of the surge is _____m/s.a)3.14b)4.14c)3.19d)4.19Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A rectangular channel had a positive surge of velocity 10 m/s moving down the channel. If depth of flow and velocity after the passing of the surge are 4.0 m and 5.0 m/s respectively, the velocity before the passage of the surge is _____m/s.a)3.14b)4.14c)3.19d)4.19Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular channel had a positive surge of velocity 10 m/s moving down the channel. If depth of flow and velocity after the passing of the surge are 4.0 m and 5.0 m/s respectively, the velocity before the passage of the surge is _____m/s.a)3.14b)4.14c)3.19d)4.19Correct answer is option 'C'. Can you explain this answer?.
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