ENGINEERING FUNDAMENTALS
ENGINEERING FUNDAMENTALS
6th Edition
ISBN: 9781337705011
Author: MOAVENI
Publisher: CENGAGE L
Question
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Chapter 8, Problem 36P
To determine

Find the average velocity of the water leaving the tank.

Expert Solution & Answer
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Answer to Problem 36P

The average velocity of the leaving water in the tank is 2.79fts_.

Explanation of Solution

Given data:

Refer to the Figure Problem 8.36 in the textbook, which shows the tank is filled by water using pipes 1 and 2.

The water level increases in the rate of 0.1 in/s.

The diameter of the tank is (dtank) 6 in.

Formula used:

Write the expression for the volume flow rate.

Q=VaverageAc (1)

Here,

Ac is cross-sectional area, and

Vaverage is average velocity.

Calculation:

Consider that the constant water density.

From the Figure Problem 8.35 in the textbook, write the volume flow rate expression for inlet and outlet volumetric flow rate expressions.

Q1+Q2Q3=Atankchangeintheheightofwatertime

Re-arrange the equation,

Q1+Q2Atankchangeintheheightofwatertime=Q3V1A1+V2A2Atankchangeintheheightofwatertime=V3A3[V1(π4)(d1)2+V2(π4)(d2)2(π4)(dtank)2(0.1ins)=V3(π4)(d3)2=Q3]{A=π4d2,and[changeintheheightofwater]time=0.1ins}

Refer to the Figure Problem 8.35 in the textbook, substitute 2fts for V1, 1.5fts for V2, 1 in. for d1, 1.75 in. for d2, 1.5 in. for d3, and 6 in. for dtank,

(2fts)(π4)(1in.)2+(1.5fts)(π4)(1.75in.)2(π4)(6in.)2(0.1in.s)=(V3)(π4)(1.5in.)2=Q3{[(2fts)(π4)(1ft12)2+(1.5fts)(π4)(1.75 ft12)2(π4)(6ft12)2(0.112fts)=(V3)(π4)(1.5ft12)2=Q3]}[1in.=112ft] 

Reduce the equation as follows,

0.0109ft3s+0.02505ft3s0.001636ft3s=0.01227ft2(V3)=Q30.0343ft3s=0.01227ft2(V3)=Q3

The values of Q3 and V3 are,

Q3=0.0343ft3s,

And

V3=0.0343ft3s0.01227ft2=2.79fts

Conclusion:

Hence, the average velocity of the leaving water in the tank is 2.79fts_.

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