Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 13, Problem 48RQ
To determine

The force acting on the flanges of the elbow.

Expert Solution & Answer
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Explanation of Solution

Given:

The flow rate (Q) of the water is 0.16m3/s.

The Diameter (D) of the elbow is 30cm.

The exit Diameter (d) of the elbow is 10cm.

The height (h) of the elbow is 10cm.

The internal volume (V) of the elbow is 0.03m3.

Calculation:

Assume the density of water to be 1000kg/m3.

Calculate the velocity (V1) using the relation.

    V1=qA1V1=(0.16m3/sπ4(30cm×1m100cm))=2.26m/s

Calculate the velocity (V2) using the relation.

  V2=qA2V2=(0.16m3/sπ4(10cm×1m100cm))=20.37m/s

Calculate the constant (z1) using the relation.

  z1=0.5

Calculate the pressure (P) using the Bernoulli’s relation.

    P1+12ρV12+yz1=P2+12ρV22+yz2P1=12ρ(V22V12)yz1=12(1000kg/m3)((20.37m/s)2(2.26m/s)2)(9810)(0.5)=20010Pa

Calculate the linear momentum equation (Fx) using the relation.

    Fx=cvVρVndA+tcvρVdV

Calculate the component (Fx) using the relation.

    V1ρ(V1)A1+(V2cosθ)ρ(V2)A2=P1A1+FxρQV12ρQV22cosθP1A1=FxρQ(V12V22cosθ)P1A1=Fx[(1000kg/m3)(0.16m3/s)((2.26m/s)2(20.37m/s)2cos(60°))(20010Pa×0.03m3)]=FxFx=48150N

Calculate the component (Fy) using the relation.

  0+(V2sinθ)ρ(V2)A2=Wwater,cv+FyρV22sinθ+Wcv=Fy[(1000kg/m3)(20.37m/s)2sin(60)+9810×π4(0.03m3)]=FyFy=359052N

Calculate the resultant force (R) using the relation.

    R=Fx2+Fy2=(48150N)2+(359052N)2=362266N

Calculate the direction of resultant R.

  tanβ=FyFx=(359052N48150N)β=82.4°

Thus, the resultant required force is R=362266N.

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Chapter 13 Solutions

Fundamentals of Thermal-Fluid Sciences

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