Problem #4If the pressure difference between points 1 and 2 is 25 psi, whatwill be the flow rate? The pipes are galvanized iron with0.0005 ft. Take v =1.06 × 10−5 ft²/s and neglectKis=minor losses.10 in. dia1.2000 ft, 8 in. diaAB1600 ft, 6 in. diaс2.800 ft, 10 in. dia- Step 1: Determine the given variableGiven:Pressure difference between points 1 and 2 is 25 psiKs =0.0005 ftv=1.06*10-5ft²/sStep 2: Calculate the mass flow rateFriction losses:10.68L*Q1.852C1.852D4.87losses reamins the same in parallel pipes(hp) A = (h₂) Bh₂=10.68L AC1.852 + |*DAremove the common things2000*12 1.852=*QA84.874.871.852=QA*QA1600*1264.870.959*Q₁1.852= 3.116*Qg1.8521.852=1.8523.1160.959Q1.852=3.249*Q¹1.8*QB10.68L B1.852 4.87'DB1.852.*QB1.852mass flow rate remains constantQ₁ = QA+QB = Qc = Q₂1.852Applying bernoullis equation between 1 and 2P₂ V₂²pgP₁ v₁²+ +2₁=pg2gZ₁ = 2₂V₁ = V₂Q₁ = Q₂P₁-P₂pg= (h₂)₂ + (1₂) g + (h₂) c25*6894.757*39.371000*9.81691.759 =2g- +2₂+(1₂) ₁-210.68LAC1.852 DA10.68 2000*12*3.25 Qg¹C1.85284.87691.759 *1291.852=3.119 Qg"10.68QB-73373.209 1.852Q=423.818 in ³/secflow rate:QA+Q8=2.89 QB4.87=1.852 +3.116 QQ1.8521.8521.852525044.0109-Q1.852 (3.119+3.116+0.9208)525044.0109(3.119+3.116+0.9208)Q1.852=73373.209flow rate=2.89*423.818 (0.000578)flow rate=0.707 cfs.10.68LcC1.852 DC1600*12*Q1.852 800*12*(Q+Q)1.85264.87104.8710.68L BC1.852 D4.871.852+"QB1.852 +0.129 (1.89+1)1.852 Qg¹.1.8524.87*Qc

The objective of the problem is to determine the flow rate in a pipe system with known pressure…

Problem #4 If the pressure difference between points 1 and 2 is 25 psi, what will be the flow rate? The pipes are galvanized iron with k, = 0.0005 ft. Take v = 1.06 × 10-5 ft²/s and neglect minor losses. 10 in. dia 2000 ft, 8 in. dia A B 1600 ft, 6 in. dia C 800 ft, 10 in. dia- 2.

Problem #4 If the pressure difference between points 1 and 2 is 25 psi, what will be the flow rate? The pipes are galvanized iron with k, = 0.0005 ft. Take v = 1.06 × 10-5 ft²/s and neglect minor losses. 10 in. dia 2000 ft, 8 in. dia A B 1600 ft, 6 in. dia C 800 ft, 10 in. dia- 2.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Pressurized pipe flow

A 15-kilometer-long pipe with a pipe diameter of one meter transports water at a speed of one meter per second. The pipe has a friction coefficient of 0.005. Calculate the frictional head loss?

Question

Problem #4
If the pressure difference between points 1 and 2 is 25 psi, what
will be the flow rate? The pipes are galvanized iron with
0.0005 ft. Take v =
1.06 × 10−5 ft²/s and neglect
Kis
=
minor losses.
10 in. dia
1.
2000 ft, 8 in. dia
A
B
1600 ft, 6 in. dia
с
2.
800 ft, 10 in. dia-

Step 1: Determine the given variable
Given:
Pressure difference between points 1 and 2 is 25 psi
Ks =0.0005 ft
v=1.06*10-5ft²/s
Step 2: Calculate the mass flow rate
Friction losses:
10.68L
*Q1.852
C1.852D4.87
losses reamins the same in parallel pipes
(hp) A = (h₂) B
h₂
=
10.68L A
C1.852 + |
*DA
remove the common things
2000*12 1.852=
*QA
84.87
4.87
1.852=
QA
*QA
1600*12
64.87
0.959*Q₁1.852= 3.116*Qg1.852
1.852=
1.852
3.116
0.959
Q1.852=3.249*Q¹1.8
*QB
10.68L B
1.852 4.87
'DB
1.852
.*QB
1.852
mass flow rate remains constant
Q₁ = QA+QB = Qc = Q₂
1.852
Applying bernoullis equation between 1 and 2
P₂ V₂²
pg
P₁ v₁²
+ +2₁=
pg
2g
Z₁ = 2₂
V₁ = V₂
Q₁ = Q₂
P₁-P₂
pg
= (h₂)₂ + (1₂) g + (h₂) c
25*6894.757*39.37
1000*9.81
691.759 =
2g
- +2₂+(1₂) ₁-2
10.68LA
C1.852 DA
10.68 2000*12*3.25 Qg¹
C1.852
84.87
691.759 *1291.852=3.119 Qg"
10.68
QB-73373.209 1.852
Q=423.818 in ³/sec
flow rate:
QA+Q8=2.89 QB
4.87
=
1.852 +3.116 Q
Q1.852
1.852
1.852
525044.0109-Q1.852 (3.119+3.116+0.9208)
525044.0109
(3.119+3.116+0.9208)
Q1.852=73373.209
flow rate=2.89*423.818 (0.000578)
flow rate=0.707 cfs.
10.68Lc
C1.852 DC
1600*12*Q1.852 800*12*(Q+Q)1.852
64.87
104.87
10.68L B
C1.852 D
4.87
1.852+
"QB
1.852 +0.129 (1.89+1)1.852 Qg¹.
1.852
4.87
*Qc

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