Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 13, Problem 59RQ
To determine

The total x and z forces at the two flanges connecting the pipe.

Expert Solution & Answer
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Explanation of Solution

Given:

The total absolute pressure (P1) in flange (1) is 200KPa.

Mass flow rate (m˙1) in flange (1) is 55kg/s.

The total absolute pressure (P2) in flange (2) is 150KPa.

Mass flow rate (m˙2) in flange (2) is 40kg/s.

The total absolute pressure (P3) in flange (3) is 100KPa.

Mass flow rate (m˙3) in flange (3) is 15kg/s.

The diameter (d1) of flange (1) is 5cm.

The diameter (d2) of flange (2) is 10cm.

The diameter (d3) of flange (3) is 3cm.

The density (ρ) of fluid is 1000kg/m3.

The momentum-flux correction factor (β) is 1.03.

Calculation:

Calculate the flange (1) velocity by using the relation.

  v1=m˙1ρ(πd124)=55kg/s1000kg/m3(π×(5cm×(1×102m1cm))24)=28.01m/s

Calculate the flange (2) velocity by using the relation.

  v2=m˙2ρ(πd224)=40kg/s1000kg/m3(π×(10cm×(1×102m1cm))24)=5.093m/s

Calculate the flange (3) velocity by using the relation.

  v3=m˙3ρ(πd324)=15kg/s1000kg/m3(π×(3cm×(1×102m1cm))24)=21.22m/s

Calculate the force in x axis by using the relation.

  FRx=Px1A1Px2A2β(m˙2v2+m˙1v1)=(P1P3)A1(P2P3)A2β(m˙2v2+m˙1v1)={(200100)kPa×1kN/m21kPa[π×(5cm×1m100cm)24](150100)kPa×1kN/m21kPa[π×(10cm×1m100cm)24]1.03[40kg/s×5.093m/s×1kN1000kgm/s2+55kg/s×28.01m/s×1kN1000kgm/s2]}

  FRx=2.386kN=2386N2390N

Calculate the force for y axis by using the relation.

  FRz=βm˙3v3=1.03×15kg/s×21.22m/s=327.8kgm/s2×(1Nkgm/s2)=327.8N

Thus the force in x axis is 2390N and the force along y axis is 327.8N.

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Chapter 13 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

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