Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 15, Problem 45P

A village maintains a large tank with an open top, containing water for emergencies. The water can drain from the tank through a hose of diameter 6.60 cm. The hose ends with a nozzle of diameter 2.20 cm. A rubber stopper is inserted into the nozzle. The water level in the tank is kept 7.50 m above the nozzle. (a) Calculate the friction force exerted on the stopper by the nozzle. (b) The stopper is removed. What mass of water flows from the nozzle in 2.00 h? (c) Calculate the gauge pressure of the flowing water in the hose just behind the nozzle.

(a)

Expert Solution
Check Mark
To determine

The friction force exerted on the stopper by the nozzle.

Answer to Problem 45P

The friction force exerted on the stopper by the nozzle is 27.9 N.

Explanation of Solution

Take point 1 to be at the free surface of the water in the tank and point 2 to be inside the nozzle. Since the top of the tank is open to the atmosphere the pressure at point 1 will be equal to atmospheric pressure. Since the tank has large area compared to the nozzle, the speed of water at point 1 will be very small compared to that at point 2 according to continuity equation and can be considered to be zero. Also assume the position of the nozzle to be y2=0 m.

Write the Bernoulli’s equation.

  P1+12ρv12+ρgy1=P2+12ρv22+ρgy2        (I)

Here, P1 is the pressure at the point 1, ρ is the density of the liquid, v1 is the speed of the liquid at point 1, g is the acceleration due to gravity, y1 is the height of the point 1 above the reference position, P2 is the pressure at the point 2, v2 is the speed of the liquid at point 2 and y2 is the height of the point 2 above the reference position.

Substitute P0 for P1 , 0 m/s for v1 , 0 m/s for v2 and 0 m for y2 in equation (I) and rearrange it.

  P0+12ρ(0 m/s)+ρgy1=P2+12ρ(0 m/s)+ρg(0 m)P0+ρgy1=P2P2P0=ρgy1        (II)

The diagram of the different forces on the system is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 15, Problem 45P

In the diagram f is the frictional force.

The sum of the forces in the horizontal direction must be zero for the stopper.

Write the equilibrium condition for the stopper.

  ΣFx=0        (III)

Here, ΣFx is the sum of the forces in the horizontal direction.

Write the expression for ΣFx.

  ΣFx=FwaterFairf

Here, Fwater is the force exerted by water and Fair is the force exerted by air.

Put the above equation in equation (III) and rewrite it for f.

  FwaterFairf=0f=FwaterFair        (IV)

Write the equation for pressure.

  P=FA

Here, P is the pressure and A is the area over which the force acts.

Rewrite the above equation for F.

  F=PA        (V)

Use equation (V) to write the expression for Fwater.

  Fwater=P2A

Here, A is the area of the nozzle.

Use equation (V) to write the expression for Fair.

  Fair=P0A

Put the above two equations in equation (IV).

  f=P2AP0A=(P2P0)A        (VI)

Write the equation for A.

  A=πdn24        (VII)

Here, dn is the diameter of the nozzle.

Put equations (II) and (VII) in equation (VI).

  f=ρgy1πd24        (VIII)

Conclusion:

The value of P0 is 1.013×105 Pa, The value of g is 9.80 m/s2 and the density of water is 1000 kg/m3.

Substitute 1000 kg/m3 for ρ , 9.80 m/s2 for g , 7.50 m for y1 and 2.20 cm for dn in equation (VIII) to find f.

  f=(1000 kg/m3)(9.80 m/s2)(7.50 m)π(2.20 cm1 m100 cm)24=27.9 N

Therefore, the friction force exerted on the stopper by the nozzle is 27.9 N.

(b)

Expert Solution
Check Mark
To determine

The mass of water that flows from the nozzle in 2.00 h.

Answer to Problem 45P

The mass of water that flows from the nozzle in 2.00 h is 3.32×104 kg.

Explanation of Solution

When the stopper is removed, point 2 is also at the atmospheric pressure.

Substitute P0 for P1 , P0 for P2 , 0 m/s for v1 , and 0 m for y2 in equation (I) and rearrange it for v2.

  P0+12ρ(0 m/s)+ρgy1=P0+12ρv22+ρg(0 m)ρgy1=12ρv22v22=2gy1v2=2gy1        (IX)

Write the equation for density of water.

  ρ=mwaterV        (X)

Here, ρ is the density of water, mwater is the mass of water and V is the volume of water.

Write the equation for volume of water.

  V=Ay        (XI)

Here, y is the distance through which the water falls.

Write the equation for the velocity of water at point 2.

  v2=yt

Here, t is the time.

Rewrite the above equation for y .

  y=v2t

Put equation (IX) in the above equation.

  y=2gy1t        (XII)

Put equations (VII) and (XII) in equation (XI).

  V=πdn242gy1t

Put the above equation in equation (X) and rewrite it for mwater.

  ρ=mwaterπdn242gy1tmwater=πρtdn242gy1        (XIII)

Conclusion:

Substitute 1000 kg/m3 for ρ , 2.00 h for t , 2.20 cm for dn , 9.80 m/s2 for g and 7.50 m for y1 and in equation (XIII) to find mwater.

  mwater=π(1000 kg/m3)(2.00 h60×60 s1.00 h)(2.20 cm1 m100 cm)242(9.80 m/s2)(7.50 m)=3.32×104 kg

Therefore, the mass of water that flows from the nozzle in 2.00 h is 3.32×104 kg.

(c)

Expert Solution
Check Mark
To determine

The gauge pressure of the flowing water in the hose just behind the nozzle.

Answer to Problem 45P

The gauge pressure of the flowing water in the hose just behind the nozzle is 7.26×104 Pa.

Explanation of Solution

Assume point 1 to be in the wide hose and point 2 to be just outside the nozzle.

Write the continuity equation of fluids.

  A1v1=A2v2        (XIV)

Here, A1 is the area through which the liquid moves at point 1 and A2 is the area through which the liquid moves at point 2.

Write the equation for A1.

  A1=πdh24

Here, dh is the diameter of the hose.

Write the equation for A2.

  A2=πdn24

Put the above two equations in equation (XIV) and rewrite it for v1.

  πdh24v1=πdn24v2v1=(dndh)2v2        (XV)

Take the vertical position of point 1 and point 2 to be zero.

Substitute P0 for P2 , 0 m for y1 and 0 m for y2 in equation (I) and rearrange it.

  P1+12ρv12+ρg(0 m)=P0+12ρv22+ρg(0 m)P1+12ρv12=P0+12ρv22P1P0=12ρ(v22v12)        (XVI)

Conclusion:

Substitute 9.80 m/s2 for g and 7.50 m for y1 in equation (IX) to find v2 .

  v2=2(9.80 m/s2)(7.50 m)=12.1 m/s

Substitute 2.20 cm for dn , 6.60 cm for dh and 12.1 m/s for v2 in equation (XV) to find v1 .

  v1=(2.20 cm6.60 cm)2(12.1 m/s)=1.35 m/s

Substitute 1000 kg/m3 for ρ , 12.1 m/s for v2 and 1.35 m/s for v1 in equation (XVI) to find the gauge pressure.

  P1P0=12(1000 kg/m3)[(12.1 m/s)2(1.35 m/s)2]=7.26×104 Pa

Therefore, the gauge pressure of the flowing water in the hose just behind the nozzle is 7.26×104 Pa.

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Chapter 15 Solutions

Principles of Physics: A Calculus-Based Text

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