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FMA401MAY 2018ASSIGNMENT 1Question 1Determine the flow in each pipe in the network below using the Hardy Cross1.1method. All the pipes have a friction factor f of 0.018 and pipe diameters of 0.1m. The lengths of the pipes and the first guess values are given below. Performonly ONE iteration5 m/s3 m/s2 m/sBCA2 m/s2 m/sPipeLength (m)Flow Rate(m/s)BA41CA31DB31EB4DC4ED521.2 What is the underlying principle behind the Hardy-Cross MethodLO

Question

Hi there, we have handed this in and it has been marked but unfortunately I seem to have a misunderstanding regarding the directions of flow and when you show those in the flow tables. May you please assist by elaborating on the explanation for these questions?

FMA401M
AY 2018
ASSIGNMENT 1
Question 1
Determine the flow in each pipe in the network below using the Hardy Cross
1.1
method. All the pipes have a friction factor f of 0.018 and pipe diameters of 0.1
m. The lengths of the pipes and the first guess values are given below. Perform
only ONE iteration
5 m/s
3 m/s
2 m/s
B
C
A
2 m/s
2 m/s
Pipe
Length (m)
Flow Rate
(m/s)
BA
4
1
CA
3
1
DB
3
1
EB
4
DC
4
ED
5
2
1.2 What is the underlying principle behind the Hardy-Cross Method
LO
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FMA401M AY 2018 ASSIGNMENT 1 Question 1 Determine the flow in each pipe in the network below using the Hardy Cross 1.1 method. All the pipes have a friction factor f of 0.018 and pipe diameters of 0.1 m. The lengths of the pipes and the first guess values are given below. Perform only ONE iteration 5 m/s 3 m/s 2 m/s B C A 2 m/s 2 m/s Pipe Length (m) Flow Rate (m/s) BA 4 1 CA 3 1 DB 3 1 EB 4 DC 4 ED 5 2 1.2 What is the underlying principle behind the Hardy-Cross Method LO

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Step 1

1.1

Draw the balanced pipe network along with the flow and its direction as:

5 m'/5
2 m/s
3mls
2m7s
3m$/s
A
2 mYs
1mirs
2 mi 1s
LL
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5 m'/5 2 m/s 3mls 2m7s 3m$/s A 2 mYs 1mirs 2 mi 1s LL

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Step 2

The Darcy-Weisbach expression for r according to Hardy Cross method is written as:

8f
r = L
gr d
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8f r = L gr d

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Step 3

Substitute the respective values and calculate t...

8(0.018)
T3(4 (9.81 m/s )r2(0.1 m)
- = 594.91 s2/m3
8(0.018)
TCA (3 m)981 m/s? }r°(0.1 m)
446.18 s2/m3
8(0.018)
TDB(3 (9.81 m/s2)r' (0.1 m)
446.18 s2/m3
8(0.018)
(49.81 m/s2)72 (0.1 m)
-= 594.91 s2/m3
5
8(0.018)
Dc (4 m(9.81 m/s* )z°(0.1 m)
594.91 s2/m
8(0.018)
ED(5 m (9.81 m/s?) 72 (0.1 m)
)7
= 743.64 s2/m5
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8(0.018) T3(4 (9.81 m/s )r2(0.1 m) - = 594.91 s2/m3 8(0.018) TCA (3 m)981 m/s? }r°(0.1 m) 446.18 s2/m3 8(0.018) TDB(3 (9.81 m/s2)r' (0.1 m) 446.18 s2/m3 8(0.018) (49.81 m/s2)72 (0.1 m) -= 594.91 s2/m3 5 8(0.018) Dc (4 m(9.81 m/s* )z°(0.1 m) 594.91 s2/m 8(0.018) ED(5 m (9.81 m/s?) 72 (0.1 m) )7 = 743.64 s2/m5

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