FLUID MECHANICS FUND. (LL)-W/ACCESS
FLUID MECHANICS FUND. (LL)-W/ACCESS
4th Edition
ISBN: 9781266016042
Author: CENGEL
Publisher: MCG CUSTOM
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Chapter 13, Problem 128P
To determine

The velocity before the jump.

The velocity after the jump.

The mechanical power dissipated.

Expert Solution & Answer
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Answer to Problem 128P

The velocity before the jump is 14.132m/s.

The velocity after the jump is 1.978m/s.

The mechanical power dissipated is 559.5787kW.

Explanation of Solution

Given information:

The flow depth before the jump is 0.7m, the flow depth after the jump is 5m, the flow is steady and uniform and the end effect is neglected.

Write the expression for flow depth.

  y2=0.5y1(1+1+8 Fr12)....... (I)

Here, Froude number is Fr1, height before jump is y1, height after the jump is y2.

Write the expression for Froude number.

  Fr=V1gy1...... (II)

Here, acceleration due to gravity is g and velocity before jump is V1.

Write the expression for conservation of mass.

  y1V1=y2V2...... (III)

Here, velocity after jump is V2.

Write the expression for volume flow rate.

  V˙=V1by1...... (IV)

Write the expression for head loss during the jump.

  hL=y1y2+V12V222g...... (V)

Write the expression for mass flow rate.

  m˙=ρV˙...... (VI)

Here, density of water is ρ.

Write the expression for energy dissipated.

  E˙dissipated=m˙ghL...... (VII)

Calculation:

Substitute 0.7m for y1, 5m for y2 in Equation (I).

  5m=0.5(0.7m)(1+ 1+8 Fr 1 2 )(1+ 1+8 Fr 1 2 )=10m0.7m1+8 Fr12=15.2861+8Fr12=233.662Fr12=29.083

  Fr=5.393

Substitute 0.7m for y1, 5.393 for Fr and 9.81m/s2 for g in Equation (II).

  5.393=V1 9.81m/ s 2 ×0.7mV1=(5.393)9.81m/ s 2×0.7mV1=14.132m/s

Substitute 0.7m for y1, 5m for y2 and 14.132m/s for V1 in Equation (III).

  (0.7m)(14.132m/s)=(5m)V2V2=0.7m5m×14.132m/sV2=1.978m/s

Substitute 0.7m for y1, 14.132m/s for V1 and 1m for b in Equation (IV).

  V˙=(14.132m/s)(1m)(0.7m)=14.132m/s×0.7m2=9.892m3/s

Substitute 0.7m for y1, 5m for y2 and 14.132m/s for V1, 1.978m/s for V2 and 9.81m/s2 for g in Equation (V).

  hL=(0.7m)(5m)+ ( 14.132m/s )2 ( 1.978m/s )22( 9.81m/ s 2 )=4.3m+ ( 14.132m/s )2 ( 1.978m/s )219.62m/ s 2=5.778m

Refer to table "Liquid water at 20°C " to obtain density of water as 998kg/m3.

Substitute 998kg/m3 for ρ and 9.892m3/s for V˙ in Equation (VI).

  m˙=(998kg/ m 3)(9.892 m 3/s)=9872.216kg/s

Substitute 9872.216kg/s for m˙, 5.68m for hL and 9.81m/s2 for g in Equation (VII).

  E˙dissipated=(9872.216kg/s)(9.81m/ s 2)(5.778m)=559.5787×103kgm2/s3×1W1 kg m 2 /s 3=559.5787×103W×1kW1000W=559.5787kW

Conclusion:

The velocity before the jump is 14.132m/s.

The velocity after the jump is 1.978m/s.

The mechanical power dissipated is 559.5787kW.

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Chapter 13 Solutions

FLUID MECHANICS FUND. (LL)-W/ACCESS

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