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Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  Fluid Mechanics for Civil Engineering  >  Test: Hydraulic Gradient & Total Energy Line - Civil Engineering (CE) MCQ

Test: Hydraulic Gradient & Total Energy Line - Civil Engineering (CE) MCQ


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10 Questions MCQ Test Fluid Mechanics for Civil Engineering - Test: Hydraulic Gradient & Total Energy Line

Test: Hydraulic Gradient & Total Energy Line for Civil Engineering (CE) 2024 is part of Fluid Mechanics for Civil Engineering preparation. The Test: Hydraulic Gradient & Total Energy Line questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Hydraulic Gradient & Total Energy Line MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Hydraulic Gradient & Total Energy Line below.
Solutions of Test: Hydraulic Gradient & Total Energy Line questions in English are available as part of our Fluid Mechanics for Civil Engineering for Civil Engineering (CE) & Test: Hydraulic Gradient & Total Energy Line solutions in Hindi for Fluid Mechanics for Civil Engineering course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt Test: Hydraulic Gradient & Total Energy Line | 10 questions in 20 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study Fluid Mechanics for Civil Engineering for Civil Engineering (CE) Exam | Download free PDF with solutions
Test: Hydraulic Gradient & Total Energy Line - Question 1
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Energy gradient line takes into consideration

Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 1

Explanation: EGL is obtained by plotting total head at various points along the axis of the pipe.

where H is the total head, P / γ is the pressure head, z is the potential head and v2 / 2g is the velocity head. Hence, EGL is also called Total Energy Line (TEL).

Test: Hydraulic Gradient & Total Energy Line - Question 2
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 Hydraulic gradient line takes into consideration

Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 2

Explanation: HGL is obtained by plotting piezometric head at various points along the axis of the pipe.
Hp = P ⁄ γ + z
where Hp is the piezometric head, P ⁄ γ is the pressure head and z is the potential head.

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Test: Hydraulic Gradient & Total Energy Line - Question 3
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Which of the following is true?

Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 3

Explanation: EGL is obtained by plotting total head at various points along the axis of the pipe. Since the total head decreases in the direction of flow, EGL will always drop in that direction.

Test: Hydraulic Gradient & Total Energy Line - Question 4
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Which of the following is true?
Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 4

Explanation: HGL is obtained by plotting piezometric head at various points along the axis of the pipe. Since pressure may either rise or fall in the direction of flow, HGL may or may not change in that direction.

Test: Hydraulic Gradient & Total Energy Line - Question 5
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Which of the following is true?
Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 5

Explanation: EGL is obtained by plotting total head and HGL is obtained by plotting piezometric head at various points along the axis of the pipe.

Hp = P ⁄ γ + z
where H is the total head, P ⁄ γ is the pressure head, z is the potential head, Hp is the piezometric head, and v2 / 2g is the velocity head.
H = Hp + v2 / 2g Since Hp < H, HGL can never be above EGL.

Test: Hydraulic Gradient & Total Energy Line - Question 6
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The vertical intercept between EGL and HGL is equal to
Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 6

Explanation: EGL is obtained by plotting total head and HGL is obtained by plotting piezometric head at various points along the axis of the pipe.

Hp = P ⁄ γ + z
where H is the total head, P ⁄ γ is the pressure head, z is the potential head, Hp is the piezometric head, and v2 / 2g is the velocity head.
H – Hp = v2 / 2g, the vertical intercept between EGL and HGL is equal to the kinetic head.

Test: Hydraulic Gradient & Total Energy Line - Question 7
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 The slope of HGL will be
Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 7

Explanation: The vertical intercept between EGL and HGL is equal to the kinetic head. For a pipe of uniform cross-section, there will be no change in the velocity of flow across the pipe. Since the kinetic head remian constant, the slope of HGL will be equal than that of EGL.

Test: Hydraulic Gradient & Total Energy Line - Question 8
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For a nozzle, the vertical intercept between EGL and HGL
Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 8

Explanation: The vertical intercept between EGL and HGL is equal to the kinetic head. For a nozzle, the cross-sectional area decreases in the direction of flow leading to an increase in the velocity of flow across the pipe. Since the kinetic head increases, the vertical intercept between EGL and HGL will increase.

Test: Hydraulic Gradient & Total Energy Line - Question 9
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For a diffuser, the vertical intercept between EGL and HGL
Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 9

Explanation: The vertical intercept between EGL and HGL is equal to the kinetic head. For a diffuser, the cross-sectional area increases in the direction of flow leading to a decrease in the velocity of flow across the pipe. Since the kinetic head decreases, the vertical intercept between EGL and HGL will decrease.

Test: Hydraulic Gradient & Total Energy Line - Question 10
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Which of the following is true?
Detailed Solution for Test: Hydraulic Gradient & Total Energy Line - Question 10

Explanation: EGL is obtained by plotting total head at various points along the axis of the pipe.

where H is the total head, P ⁄ γ is the pressure head, z is the potential head, , and v2 / 2g is the velocity head.
Hence, there is no relation whatsoever between the slope of EGL and that of the axis of the pipe.

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