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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
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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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-
Transcribed Image Text: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
Transcribed Image Text: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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