Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Book Icon
Chapter 10, Problem 10.51P
<
To determine

(a)

If the flow of the unfinished concrete duct is critical.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

The flow is super critical.

Explanation of Solution

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s

Froude number is given as:

Fr=Vgy

(1)

Where,V=flow velocityg=specific densityy=hydraulic depth

Velocity of the flow

V=QA

(2)

A=areaQ=discharge

Calculation:

From equation (2)

V=QA

=QA=Qπ4(1.5)283.144(1.5)2=83.144(2.25)=4.527m/s Angle is calculated for the circular channel which is partially filled:

θd=900250650

θr=65*π1801.13rad

[units to radians]

Area of the flow is given as:

A=(D2)2(θrsinθdcosθd)

Substituting the values:

=(1.52)2(1.13sin65cos65)=(2.252)2(1.13sin65cos65)=1.2656(1.13sin65cos65)=0.420m2

Hydraulic depth is given as:

y=Awidth at top

=A2(D2)sinθ0.4202(1.52)sin65=0.3m

Substituting all equations (1):

Fr=Vgy

=Vgy=4.5279.81*0.32.685

The flow is determined to be super critical as the Froude’s number is more than 1.

Conclusion:

Thus, the flow of the unfinished concrete duct is determined.

<
To determine

(b)

The critical flow rates.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

Critical flow rate is Qc=3.07m3/sec

Explanation of Solution

Given Information:

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s

Concept Used:

Qc=αARb2/3*Sc1/2n

(1)

where,A=areaSc=Critical slopeR=hydraulic channelα=conversion factornn=mannings factor 

Calculation:

Substituting we have,

8=1.76×1.486×0.25×1.5×Sc1/20.014Sc1/2=8×0.0141.76×1.486×0.25×1.5Sc1/2=8×0.0140.980760.1120.98076

Sc=0.001

From equation (1) we have,

=1.486×1.76×(0.25×1.5)2/3×0.0011/20.014=2.61536×(0.375)2/3×0.0011/20.014=3.07m3/sec

Conclusion:

Thus, the critical flow rate is 3.07m3/sec.

<
To determine

(c)

The critical slope.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

Critical slope is Sc=0.0019

Explanation of Solution

Given Information:

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s

Concept Used:

Qc=αARb2/3*Sc1/2n

(1)

where,A=areaSc=Critical slopeR=hydraulic channelα=conversion factornn=mannings factor 

Calculation:

Substituting we have:

nQcαARb2/3=Sc1/2Sc1/2=3.07×0.0140.98076Sc=0.0019

Conclusion:

Thus, the critical slope is determined.

<
To determine

(d)

The Froude’s number.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

Froude’s number is Fr=1.027

Explanation of Solution

Given Information:

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s.

Concept Used:

Froude number is given as:

Fr=Vgy

(1)

Where,V=flow velocityg=specific densityy=hydraulic depth

Velocity of the flow

V=QA

(2)

A=areaQ=discharge

Calculation:

Substituting all equation (1)

Fr=Vgy

=VgyQAgy=3.071.767×19.81×0.31.0127

The flow is determined to be critical.

Conclusion:

Thus, the Froude number is determined.

<
To determine

(e)

The slope of the duct when the flow is uniform.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

Sc=0.2698

Explanation of Solution

Given Information:

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s .

Concept Used:

nQcαARb2/3=Sc1/2where:A=areaSc=Critical slopeR=hydraulic channelα=conversion factornn=mannings factor 

Calculation:

The slope is equal in the duct with the uniform flow. So the duct slope with the critical flow is given as:

Sc=0.2698

Conclusion:

Thus, the slope of the duct when the flow is uniform is determined.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
I submitted the below question and received the answer i copied into this question as well. Im unsure if it is correct, so looking for a checkover. i am stuck on the part tan-1 (0.05) = 0.04996 radians. Just unsure where the value for the radians came from. Just need to know how they got that answer and how it is correct before moving on to the next part. If any of the below information is wrong, please feel free to give me a new answer or an entire new explanation.   An Inclining experiment done on a ship thats 6500 t, a mass of 30t was moved 6.0 m transvesly causing a 30 cm deflection in a 6m pendulum, calculate the transverse meta centre height.    Here is the step-by-step explanation:   Given: Displacement of the ship (W) = 6500 tonnes = 6500×1000=6,500,000kg Mass moved transversely (w) = 30 tonnes=30×1000=30,000kg The transverse shift of mass (d) = 6.0 meters Pendulum length (L) = 6.0 meters Pendulum deflection (x) = 30 cm = 0.30 meters Step 1: Formula for Metacentric Height…
Answer the assignment question, expert only
A 1 inch rod diameter   B 3/4 inch rod diameter   C 1/2 inch rod diameter   D 3/8 inch rod diameter

Chapter 10 Solutions

Fluid Mechanics

Ch. 10 - Prob. 10.11PCh. 10 - Prob. 10.12PCh. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Prob. 10.15PCh. 10 - Prob. 10.16PCh. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - An unfinished concrete sewer pipe, of diameter 4...Ch. 10 - Prob. 10.21PCh. 10 - Prob. 10.22PCh. 10 - Prob. 10.23PCh. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Prob. 10.26PCh. 10 - Prob. 10.27PCh. 10 - Prob. 10.28PCh. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Prob. 10.31PCh. 10 - Prob. 10.32PCh. 10 - Prob. 10.33PCh. 10 - Prob. 10.34PCh. 10 - Prob. 10.35PCh. 10 - Prob. 10.36PCh. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Pl0.39 A trapezoidal channel has n = 0.022 and Sn...Ch. 10 - Prob. 10.40PCh. 10 - Prob. 10.41PCh. 10 - Prob. 10.42PCh. 10 - Prob. 10.43PCh. 10 - Prob. 10.44PCh. 10 - Prob. 10.45PCh. 10 - Prob. 10.46PCh. 10 - Prob. 10.47PCh. 10 - Prob. 10.48PCh. 10 - Prob. 10.49PCh. 10 - Prob. 10.50PCh. 10 - Prob. 10.51PCh. 10 - Prob. 10.52PCh. 10 - Prob. 10.53PCh. 10 - A clay tile V-shaped channel has an included angle...Ch. 10 - Prob. 10.55PCh. 10 - Prob. 10.56PCh. 10 - Prob. 10.57PCh. 10 - Prob. 10.58PCh. 10 - Prob. 10.59PCh. 10 - Prob. 10.60PCh. 10 - P10.59 Uniform water flow in a wide brick channel...Ch. 10 - P10.62 Consider the flow in a wide channel over a...Ch. 10 - Prob. 10.63PCh. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Prob. 10.66PCh. 10 - Prob. 10.67PCh. 10 - Prob. 10.68PCh. 10 - Given is the flow of a channel of large width b...Ch. 10 - Prob. 10.70PCh. 10 - Prob. 10.71PCh. 10 - Prob. 10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. 10.74PCh. 10 - Prob. 10.75PCh. 10 - Prob. 10.76PCh. 10 - Prob. 10.77PCh. 10 - Prob. 10.78PCh. 10 - Prob. 10.79PCh. 10 - Prob. 10.80PCh. 10 - Prob. 10.81PCh. 10 - Prob. 10.82PCh. 10 - Prob. 10.83PCh. 10 - Prob. 10.84PCh. 10 - Pl0.85 The analogy between a hydraulic jump and a...Ch. 10 - Prob. 10.86PCh. 10 - Prob. 10.87PCh. 10 - Prob. 10.88PCh. 10 - Prob. 10.89PCh. 10 - Prob. 10.90PCh. 10 - Prob. 10.91PCh. 10 - Prob. 10.92PCh. 10 - Prob. 10.93PCh. 10 - Prob. 10.94PCh. 10 - Prob. 10.95PCh. 10 - Prob. 10.96PCh. 10 - Prob. 10.97PCh. 10 - Prob. 10.98PCh. 10 - Prob. 10.99PCh. 10 - Prob. 10.100PCh. 10 - Prob. 10.101PCh. 10 - Prob. 10.102PCh. 10 - Prob. 10.103PCh. 10 - Prob. 10.104PCh. 10 - Prob. 10.105PCh. 10 - Prob. 10.106PCh. 10 - Prob. 10.107PCh. 10 - Prob. 10.108PCh. 10 - Prob. 10.109PCh. 10 - Prob. 10.110PCh. 10 - Prob. 10.111PCh. 10 - Prob. 10.112PCh. 10 - Prob. 10.113PCh. 10 - Prob. 10.114PCh. 10 - Prob. 10.115PCh. 10 - Prob. 10.116PCh. 10 - Prob. 10.117PCh. 10 - Prob. 10.118PCh. 10 - Prob. 10.119PCh. 10 - The rectangular channel in Fig. P10.120 contains a...Ch. 10 - Prob. 10.121PCh. 10 - Prob. 10.122PCh. 10 - Prob. 10.123PCh. 10 - Prob. 10.124PCh. 10 - Prob. 10.125PCh. 10 - Prob. 10.126PCh. 10 - Prob. 10.127PCh. 10 - Prob. 10.128PCh. 10 - Prob. 10.1WPCh. 10 - Prob. 10.2WPCh. 10 - Prob. 10.3WPCh. 10 - Prob. 10.4WPCh. 10 - Prob. 10.5WPCh. 10 - Prob. 10.6WPCh. 10 - Prob. 10.7WPCh. 10 - Prob. 10.8WPCh. 10 - Prob. 10.9WPCh. 10 - Prob. 10.10WPCh. 10 - Prob. 10.11WPCh. 10 - Prob. 10.12WPCh. 10 - Prob. 10.13WPCh. 10 - Prob. 10.1FEEPCh. 10 - Prob. 10.2FEEPCh. 10 - Prob. 10.3FEEPCh. 10 - Prob. 10.4FEEPCh. 10 - Prob. 10.5FEEPCh. 10 - Prob. 10.6FEEPCh. 10 - Prob. 10.7FEEPCh. 10 - February 1998 saw the failure of the earthen dam...Ch. 10 - Prob. 10.2CPCh. 10 - Prob. 10.3CPCh. 10 - Prob. 10.4CPCh. 10 - Prob. 10.5CPCh. 10 - Prob. 10.6CPCh. 10 - Prob. 10.7CPCh. 10 - Prob. 10.1DPCh. 10 - Prob. 10.2DP
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Intro to Compressible Flows — Lesson 1; Author: Ansys Learning;https://www.youtube.com/watch?v=OgR6j8TzA5Y;License: Standard Youtube License