The loss coefficients for the pipe shown is: up to A it is 0.8, from A to B it is 1.2, from B to C it is 0.8, from C to D it is 2.2.Estimate the flow rate and the pressures at A, B, C, and D. The elevations are shown.al 10mal. 3mOP₁BUSolution:1. The energy equation can be simplified to02₁-22=1el. 12 mP2D3 cm dia.4. Pressure at A, PA=m;5. Pressure at B, PB=m;10282go The total energy loss occurred through this system is calculated as: h=o Relationship between Upipe and Vexit2. Thus, velocity in the pipe is Upipe3. Calculate the flow rate Q=-m;=is:-Vexit=m³/s;m/s;Upipe ikPa; (energy equation from surface to A)kPa; (energy equation from surface to B)Upipe2g

Given Data: KA=0.8KA-B=1.2KB-C=0.8KC-D=2.2 To Find:

The loss coefficients for the pipe shown is: up to A it is 0.8, from A to B it is 1.2, from B to C it is 0.8, from C to D it is 2.2. Estimate the flow rate and the pressures at A, B, C, and D. The elevations are shown. al. 10m O Solution: 1. The energy equation can be simplified to 02₁-22= P O B Y C P₂ Y 1 441 m; v² 28 28 o The total energy loss occurred through this system is calculated as: h= Relationship between Upipe and Vexit is:-Vexit= m; 4. Pressure at A, PA= 5. Pressure at B, PB= -m; 2. Thus, velocity in the pipe is Upipe 3. Calculate the flow rate Q= = m³/s; m/s; Upipe i kPa; (energy equation from surface to A) kPa; (energy equation from surface to B) Upipe 28

The loss coefficients for the pipe shown is: up to A it is 0.8, from A to B it is 1.2, from B to C it is 0.8, from C to D it is 2.2. Estimate the flow rate and the pressures at A, B, C, and D. The elevations are shown. al. 10m O Solution: 1. The energy equation can be simplified to 02₁-22= P O B Y C P₂ Y 1 441 m; v² 28 28 o The total energy loss occurred through this system is calculated as: h= Relationship between Upipe and Vexit is:-Vexit= m; 4. Pressure at A, PA= 5. Pressure at B, PB= -m; 2. Thus, velocity in the pipe is Upipe 3. Calculate the flow rate Q= = m³/s; m/s; Upipe i kPa; (energy equation from surface to A) kPa; (energy equation from surface to B) Upipe 28

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

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Concept explainers

Properties of Fluids

The term fluid represents a type of substance that can easily change its shape under any external pressure/force and can flow. Fluids can acquire any shape in which it is stored and can resist normal stresses but cannot resist shear stress in rest condition. Whenever external shear stress is applied to fluids, it starts to flow. The fluids may be liquid or gas. Example- water, air, etc.

Question

The loss coefficients for the pipe shown is: up to A it is 0.8, from A to B it is 1.2, from B to C it is 0.8, from C to D it is 2.2.
Estimate the flow rate and the pressures at A, B, C, and D. The elevations are shown.
al 10m
al. 3m
O
P₁
B
U
Solution:
1. The energy equation can be simplified to
02₁-22=
1
el. 12 m
P2
D
3 cm dia.
4. Pressure at A, PA=
m;
5. Pressure at B, PB=
m;
10
28
2g
o The total energy loss occurred through this system is calculated as: h=
o Relationship between Upipe and Vexit
2. Thus, velocity in the pipe is Upipe
3. Calculate the flow rate Q=
-m;
=
is:-Vexit=
m³/s;
m/s;
Upipe i
kPa; (energy equation from surface to A)
kPa; (energy equation from surface to B)
Upipe
2g

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Step 2: Write the energy equation, determine the total energy loss.

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Step 3: Determine the volumetric flow rate.

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