Tanks are a to c from left to right Elevation = 20 mD = 0.08 ml = 200 mf = 0.020Elevation 60 m=D = 0.10 ml = 200 mf = 0.015D = 0.08 ml = 400 mf = 0.020Elevation = 0 MultipleEx. 8.14 Repeat Example 8.14 (Three-Reservoir,Pipe System) with the elevation of Reservoir B moved to 80ft. Determine the flow rates Q₁, Q₂, Q3. Does water flow inor out of Reservoir B?

Given: Elevation of the reservoir A→h1=20 m Elevation of reservoir B→h2=80 m Elevation of reservoir…

Answered: Elevation 20 m = D = 0.08 m l = 200 m f…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Concept explainers

Question

Tanks are a to c from left to right

Elevation = 20 m
D = 0.08 m
l = 200 m
f = 0.020
Elevation 60 m
=
D = 0.10 m
l = 200 m
f = 0.015
D = 0.08 m
l = 400 m
f = 0.020
Elevation = 0

Multiple
Ex. 8.14 Repeat Example 8.14 (Three-Reservoir,
Pipe System) with the elevation of Reservoir B moved to 80
ft. Determine the flow rates Q₁, Q₂, Q3. Does water flow in
or out of Reservoir B?

Expert Solution

Step 1

Given:

Elevation of the reservoir A $\to h_{1} = 20 m$

Elevation of reservoir B $\to h_{2} = 80 m$

Elevation of reservoir C $\to h_{3} = 0 m$

Because the elevation of the reservoir B is highest so the flow out of reservoir B.

Let the head available at the joints of the pipe is at h.

So the head loss can be written as for the pipe A:

$\to h_{1} - h = \frac{f l {Q_{1}}^{2}}{12.1 D^{5}} \to 20 - h = \frac{0.02 \times 200 \times {Q_{1}}^{2}}{12.1 \times 0 . 08^{5}} \to h = 20 - 100884 {Q_{1}}^{2} . . . . . . . . . . . . . . . . . e q u a t i o n 1$

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

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