Example For the branching system shown in Fig. 11.14, calculate the discharge each pipe. Take f = 0.02 for all pipes. Pipe Dia 1 15 cm 2 10 cm 3 10 cm (Neglect minor losses.) A EL: 126.00 m 1 Length (m) 350 200 250 EL: 100.00 m EL: 109.00 m Connectivity AJ BJ JC B Solution: Putting h₁=rQ², 8fL 7²gD5 = 1.6525 x 10-3 Pipe 1 r = A trial and error solution method is adopted. Flow away from the junction is taken as negative. For first trial assume H; = elevation of hydraulic grade line at the junction J= 114.00 m. First trial r H; = 114.0 m Estimated h (m) = 8 × 0.02 L ²x9.81 DS Q = √h₁|r (m³/s) 7617 126.0114.0 12.0 +0.0397 ted he L DS e 1Q/h₂l (L/s) (Q in L/s) +39.7 3.31 AQ= ElQ/h,

Structural Analysis
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Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Example
For the branching system shown in Fig. 11.14,
calculate the discharge each pipe. Take f = 0.02
for all pipes.
Pipe
Dia
1
15 cm
2
10 cm
3
10 cm
(Neglect minor losses.)
A
EL: 126.00 m
1
Length (m)
350
200
250
EL: 100.00 m
EL: 109.00 m
Connectivity
AJ
BJ
JC
B
Solution: Putting h₁=rQ²,
8fL
7²gD5
= 1.6525 x 10-3
Pipe
1
r =
A trial and error solution method is adopted. Flow away
from the junction is taken as negative.
For first trial assume H; = elevation of hydraulic grade
line at the junction J= 114.00 m.
First trial
r
H; = 114.0 m
Estimated h
(m)
=
8 × 0.02 L
²x9.81 DS
Q = √h₁|r
(m³/s)
7617 126.0114.0 12.0 +0.0397
ted he
L
DS
e 1Q/h₂l
(L/s) (Q in L/s)
+39.7 3.31
AQ=
ElQ/h,
Transcribed Image Text:Example For the branching system shown in Fig. 11.14, calculate the discharge each pipe. Take f = 0.02 for all pipes. Pipe Dia 1 15 cm 2 10 cm 3 10 cm (Neglect minor losses.) A EL: 126.00 m 1 Length (m) 350 200 250 EL: 100.00 m EL: 109.00 m Connectivity AJ BJ JC B Solution: Putting h₁=rQ², 8fL 7²gD5 = 1.6525 x 10-3 Pipe 1 r = A trial and error solution method is adopted. Flow away from the junction is taken as negative. For first trial assume H; = elevation of hydraulic grade line at the junction J= 114.00 m. First trial r H; = 114.0 m Estimated h (m) = 8 × 0.02 L ²x9.81 DS Q = √h₁|r (m³/s) 7617 126.0114.0 12.0 +0.0397 ted he L DS e 1Q/h₂l (L/s) (Q in L/s) +39.7 3.31 AQ= ElQ/h,
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