Test: Energy Equation & Its Applications - 2 - Civil Engineering (CE) MCQ

# Test: Energy Equation & Its Applications - 2 - Civil Engineering (CE) MCQ

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## 9 Questions MCQ Test - Test: Energy Equation & Its Applications - 2

Test: Energy Equation & Its Applications - 2 for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Test: Energy Equation & Its Applications - 2 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Energy Equation & Its Applications - 2 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Energy Equation & Its Applications - 2 below.
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Test: Energy Equation & Its Applications - 2 - Question 1

### If a1 and a2 are the cross-sectional areas of a tank and orifice, h the height of water level intank above the centre of the orifice, the velocity of approach is given by

Test: Energy Equation & Its Applications - 2 - Question 2

### If the coefficient of velocity of flow through a orifice is cv and coefficient of resistance is cr, then

Detailed Solution for Test: Energy Equation & Its Applications - 2 - Question 2

Test: Energy Equation & Its Applications - 2 - Question 3

### If Cv, Cc, Dd and Cr are the hydraulic coefficients of an orifice, then

Test: Energy Equation & Its Applications - 2 - Question 4

The momentum correction factor for triangular velocity distribution for flow between parallel plates as shown in the figure is

Detailed Solution for Test: Energy Equation & Its Applications - 2 - Question 4

Triangular velocity distribution is achieved for Couette flow.
Average velocity,Vav = V/2
Momentum correction factor

Energy correction factor,

Test: Energy Equation & Its Applications - 2 - Question 5

If the velocity is zero over half of the cross- sectionai area and is uniform over the remaining half, then the momentum correction factor is

Detailed Solution for Test: Energy Equation & Its Applications - 2 - Question 5

Momentum correction factor,

Test: Energy Equation & Its Applications - 2 - Question 6

Match List-I with List-I and select the correct answer using the codes given below the lists:
List-I
A. Moment of momentum equation
B. Bernoulli’s equation
C. Euler’s equation
D. Hagen-Poiseuille equation

List-II
1. Equation to find energy loss in a pipeline
2. Equation of motion for one-dimensional steady flow of ideal and incompressible fluid
3. Equation based on conservation of momentum principle applicable to circulatory flows
4. Three-dimensional equation ofmotion based on Principle of conservation of momentum for ideal and incompressible fluid flow

Detailed Solution for Test: Energy Equation & Its Applications - 2 - Question 6

Bernoulli's equation is derived from equation of motion for one-dimensional steady flow of ideal and incompressible fluid. \

Test: Energy Equation & Its Applications - 2 - Question 7

If velocity is zero over 1/3rd of a cross-section and is uniform over remaining 2/3rd of the cross-section, then the correction factor for kinetic energy is

Detailed Solution for Test: Energy Equation & Its Applications - 2 - Question 7

∴ v = 3/2 v
∴ Kinetic energy correction factor

Test: Energy Equation & Its Applications - 2 - Question 8

A jet of water issue from a 5cm diameter nozzle, held vertically upwards, at a velocity of 20 m/sec. If air resistance consumes 10% of the initial energy of the jet, then it would reach a height, above the nozzle, of

Detailed Solution for Test: Energy Equation & Its Applications - 2 - Question 8

Using Bernoulli’s equation,

Test: Energy Equation & Its Applications - 2 - Question 9

The kinetic energy correction factor is

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