Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 13, Problem 64P

(a)

To determine

To Show:The speed of the root beer leaving the spigot is approximately 2gh .

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Height of the root beer is h

Calculation:

Let us apply the Bernoulli’s equation to the beer at the top of the keg and at the spigot,

  P1+ρbeergh1+12ρbeerv12=P2+ρbeergh2+12ρbeerv22(1)

Where, P1 and P2 are the pressure at the top of the keg and at the spigot respectively,

  ρbeer is density of the beer,

  g is acceleration due to gravity,

  h1 and h2 are height of the keg and spigot respectively

  v1 is speed of the root beer at the top of the keg,

  v2 is speed of the root beer leaving the spigot,

Since, v10 , h2=0 , P1=P2=Patm and h1=h , equation (1) becomes,

  gh=12v22v22=2ghv2=2gh

Conclusion:

Thus, the speed of the root beer leaving the spigot is approximately 2gh .

(b)

To determine

To Show:If

  A2A1 , the rate of change of the height h of the root beer is given by dhdt=(A2A1)(2gh)12 .

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Height of keg is H

Cross sectional area of keg is A1

Area of spigot opening is A2

Calculation:

Let, v1 be the speed of the root beer at the top of the keg, and v2 be the speed of the root beer leaving the spigot.

Let us relate the speeds v1 and v2 using the continuity equation, that is

  A1v1=A2v2

Substituting dhdt for v1 and 2gh For

  v2 ,

  dhdtA1=A22gh

  dhdt=A22ghA1

Conclusion:

Thus, the speed of the root beer leaving the spigot is approximately 2gh .

(c)

To determine

To Find: h as a function of time.

(c)

Expert Solution
Check Mark

Answer to Problem 64P

  h=(H+tA22g2A1)2

Explanation of Solution

Given:

  h=H at t=0

Calculation:

Consider the equation,

  dhdt=A22ghA1

Separating the variables,

  A1A22gdhh=dt

Taking the integral from the limits, H to h and 0 to t ,

  A1A22gHhdhh=0tdt

  2A1A22g[Hh]=t

  Hh=tA22g2A1

  h=H+tA22g2A1

  h=(H+tA22g2A1)2

Conclusion:

  h as a function of time is h=(H+tA22g2A1)2

(d)

To determine

To Find:The total time needed to drain the keg.

(d)

Expert Solution
Check Mark

Answer to Problem 64P

  1h 46 min

Explanation of Solution

Given:

  H=2.0 m

  A1=0.8 m2

  A2=1×104A1

Formula used:

Consider the time dependent expression for h ,

  h=(H+tA22g2A1)2

Solving this expression for the time to drain (t') ,

  t'=A1A22Hg(2)

Calculation:

Substituting the numerical values in equation (2)

  t'=A11×104A12(2.00m)9.8m/s2=6.39×103s= 1h 46 min

Conclusion:

The total time needed to drain the keg is 1h 46 min .

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Chapter 13 Solutions

Physics for Scientists and Engineers

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