Compute the discharges for the pipes of the network shown below. Use the Hardy Cross method. Only one iteration cycle is required. The Hazen-Williams coefficient is 130 for all pipes. Begin your computation with the assumed discharges given below and verify that they satisfy the continuity requirement. QAB = 2.4 cfs (from A to B) QBD = 1.5 cfs (from B to D) QBC = 0.3 cfs (from B to C) QCD = 0.5 cfs (from C to D) QAC = 1.8 cfs (from A to C)
Compute the discharges for the pipes of the network shown below. Use the Hardy Cross method. Only one iteration cycle is required. The Hazen-Williams coefficient is 130 for all pipes. Begin your computation with the assumed discharges given below and verify that they satisfy the continuity requirement. QAB = 2.4 cfs (from A to B) QBD = 1.5 cfs (from B to D) QBC = 0.3 cfs (from B to C) QCD = 0.5 cfs (from C to D) QAC = 1.8 cfs (from A to C)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Compute the discharges for the pipes of the network shown below. Use the Hardy Cross
method. Only one iteration cycle is required. The Hazen-Williams coefficient is 130 for
all pipes. Begin your computation with the assumed discharges given below and verify
that they satisfy the continuity requirement.
QAB = 2.4 cfs (from A to B) QBD = 1.5 cfs (from B to D)
QBC = 0.3 cfs (from B to C) QCD = 0.5 cfs (from C to D)
QAC = 1.8 cfs (from A to C)
Transcribed Image Text:0.6 cfs
В
1250 ft
12 in
2 cfs
D
3000 ft
12 in.
2
1000 ft
10 in
1
2000 ft
8 in
4.2 cfs
2500 ft
10 in
C
1.6 cfs
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