Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 14, Problem 62P

Repeat Prob. 14-60, but with the pipe diameter increased by a factor of 2 (all else being equal). Does the volume flow rate at which cavitation occurs in the pump increase or decrease with the larger pipe? Discuss.

Expert Solution & Answer
Check Mark
To determine

Whether the volume flow rate at which cavitations occurs in the pump increase or decrease of the larger pipe.

Answer to Problem 62P

The diameter of the pipe is doubled then the flow rate increase by 7.69%. But the average speed is decreased by the four.

Explanation of Solution

Given information:

The pipe diameter is increased by the factor two.

Write the expression for the net positive suction head.

  NPSH=(P atmPvρg)+z1z2hL   ....... (I)

Here, the net positive suction head is NPSH, the atmospheric pressure is Patm, the vapor pressure is Pv, the acceleration due to gravity is g, the density of flowing fluids is ρ, the datum head difference between point 1 and point 2 is z1z2 and the total head loss is hL.

Write the expression for the total head loss.

  hL=(fLD+KL1+KL2)V22g   ....... (II)

Here, the friction factor is f, the minor loss due to sharp edged reentrant inlet is KL1, the minor loss due to flanged smooth at right angle regular elbow, the velocity is V, the length of the pipe is L and the diameter of the pipe is D.

Write the expression for the average speed.

  V=4V˙πD2     ....... (III)

Here, the diameter of the pipe is D and the volume flow rate is V˙.

Write the expression for the Reynolds number.

  Re=ρVDμ   ....... (IV)

Here, the Reynolds number is Re and the kinematic viscosity is μ.

Write the expression for the friction factor.

  1f=1.8log[6.9Re]   ....... (V)

Substitute (fLD+KL1+KL2)V22g for hL in Equation (I).

  NPSH=(P atmPvρg)+z1z2(fLD+KL1+KL2)V22g   ....... (VI)

Write the expression for the required net positive suction head.

  NPSHrequired=2.2m+[0.0013m/Lpm2]V˙2.......(VII)

Here, the required net positive head is NPSHrequired.

Write the expression for the percentage increase.

  E=( ( V ) 60 ( V ) 20 ( V ) 60)×100   ......... (VIII)

Here, the percentage increase is E.

Calculation:

Refer the Table B-1, "The physical properties of saturated liquid" to obtain the value of the density is 997kg/m3 corresponding to the temperature 25°C.

Refer the Table B-1, "The physical properties of saturated liquid" to obtain the value of the kinematic viscosity is 8.91×104kg/ms corresponding to the temperature 25°C.

Refer the Table B-1, "The physical properties of saturated liquid" to obtain the value of the vapor pressure is 3.169kPa corresponding to the temperature 25°C.

Substitute 997kg/m3 for ρ, 101.3kPa for Patm, 3.169kPa for Pv, 2.8m for L, 24mm for D, 9.81m/s2 for g, 0.85 for KL1, 0.3 for KL2 and 2.2m for (z2z1) in Equation (VI).

  NPSH=[( 101.3kPa3.169kPa 997 kg/ m 3 ×9.81m/ s 2 )2.2m( f 2.8m 24mm +0.85+0.3) V 2 2×9.81m/ s 2 ]NPSH=[( 98.131kPa( 1000Pa 1kPa ) 9780.57 kg/ m 2 s 2 ( 1Pa 1 kg/ m 1 s 2 ) )2.2m( f 2.8m 24mm( 1m 1000mm ) +1.15) V 2 2×9.81m/ s 2 ]NPSH=7.833m5.946fV20.0586V2m/s2.......(IX)

Substitute 20Lpm for V˙ and 24mm for D in Equation (III).

  V=4×20Lpmπ ( 24mm )2=80Lpmπ ( 24mm( 1m 1000mm ) )2=80Lpm( 1 m 3 /s 60000Lpm )0.001809m2=0.736828m/s

Substitute 0.736828m/s for V, 997kg/m3 for ρ, 24mm for D, 9.81m/s2 for g and 8.91×104kg/ms for μ in Equation (IV).

  Re=( 997 kg/ m 3 )( 0.736828m/s )( 24mm)8.91× 10 4kg/ms=( 997 kg/ m 3 )( 0.736828m/s )( 24mm( 1m 1000mm ))8.91× 10 4kg/ms=17.63082kg/ms8.91× 10 4kg/ms=19787.6772

Substitute 19787.6772 for Re in Equation (V).

  1f=1.8log[6.919787.6772]1f=1.8log(0.000348)f=0.160679f=0.02581

Substitute 0.02581 for f and 0.736828m/s for V in Equation (IX).

  NPSH=[7.8335.946( 0.02581) ( 0.736828m/s ) 2m/ s 2 0.0586 ( 0.736828m/s ) 2m/ s 2 ]=7.833m0.0837m0.03181m=7.7174m

Substitute 20Lpm for V˙ in Equation (VII).

  NPSHrequired=2.2m+[0.0013m/ Lpm2](20Lpm)2=2.2m+[0.0013m/ Lpm2](400 Lpm2)=2.2m+0.52m=2.72m

The following table represents the net positive suction head, the required net positive suction head, and discharge.

    V˙NPSH

      (m)

    NPSHrequired

      (m)

    07.8332.2
    107.752.33
    207.71742.72
    507.16665.46
    607.8244m6.88

Plot the graph by using tabulated values.

  Fluid Mechanics: Fundamentals and Applications, Chapter 14, Problem 62P , additional homework tip  1

Figure-(1)

From Figure-(1), the cavitations occurs at the flow rate 60Lpm.

Substitute 997kg/m3 for ρ, 101.3kPa for Patm, 3.169kPa for Pv, 2.8m for L, 24mm for D, 9.81m/s2 for g, 0.85 for KL1, 0.3 for KL2 and 2.2m for (z2z1) in Equation (VI).

  NPSH=[( 101.3kPa3.169kPa 997 kg/ m 3 ×9.81m/ s 2 )2.2m( f 2.8m 48mm +0.85+0.3) V 2 2×9.81m/ s 2 ]NPSH=[( 98.131kPa( 1000Pa 1kPa ) 9780.57 kg/ m 2 s 2 ( 1Pa 1 kg/ m 1 s 2 ) )2.2m( f 2.8m 48mm( 1m 1000mm ) +1.15) V 2 2×9.81m/ s 2 ]NPSH=7.833m2.973fV20.0586V2m/s2   ....... (X)

Substitute 60Lpm for V˙ and 48mm for D in Equation (III).

  V=4×20Lpmπ ( 24mm )2=80Lpmπ ( 48mm( 1m 1000mm ) )2=80Lpm( 1 m 3 /s 60000Lpm )0.007238m2=0.18420m/s

Substitute 0.18420m/s for V, 997kg/m3 for ρ, 48mm for D, 9.81m/s2 for g and 8.91×104kg/ms for μ in Equation (IV).

  Re=( 997 kg/ m 3 )( 0.18420m/s )( 48mm)8.91× 10 4kg/ms=( 997 kg/ m 3 )( 0.18420m/s )( 48mm( 1m 1000mm ))8.91× 10 4kg/ms=17.63082kg/ms8.91× 10 4kg/ms=9893.46

Substitute 9893.46 for Re in Equation (V).

  1f=1.8log[6.99893.46]1f=1.8log(0.000697)f=0.1750f=0.03097

Substitute 0.03097 for f and 0.18420m/s for V in Equation (X).

  NPSH=[7.8335.946( 0.03097) ( 0.18420m/s ) 2m/ s 2 0.0586 ( 0.18420m/s ) 2m/ s 2 ]=7.833m0.0062339m0.00.1988m=7.8244m

Substitute 60Lpm for V˙ in Equation (VII).

  NPSHrequired=2.2m+[0.0013m/ Lpm2](60Lpm)2=2.2m+[0.0013m/ Lpm2](3600 Lpm2)=2.2m+4.68m=6.88m

The following table represents the net positive suction head, the required net positive suction head, and discharge.

    V˙NPSH

      (m)

    NPSHrequired

      (m)

    07.8332.2
    107.752.33
    207.71742.72
    507.16665.46
    607.8244m6.88

Plot the graph by using tabulated values.

  Fluid Mechanics: Fundamentals and Applications, Chapter 14, Problem 62P , additional homework tip  2

Figure-(2)

From Figure-(2), the cavitations occurs at the flow rate 65Lpm.

Substitute 65Lpm for (V)60 and 20Lpm for (V)20 in Equation (VIII).

  E=( 65Lpm60Lpm 65Lpm)×100=( 5Lpm 65Lpm)×100=0.07692×100=7.69%

The diameter of the pipe is doubled then the flow rate increase by 7.69%. But the average speed is decrease by the four.

Conclusion:

The diameter of the pipe is doubled then the flow rate increase by 7.69%. But the average speed is decrease by the four.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A centrifugal pump is used to pump the water at its temperature in a reservoir with the free water surface located 2 m above the center of the pump inlet to the atmosphere. Piping system in 5ng from tank to pump; It consists of 3 galvanized iron pipes with an internal diameter of 20 cm and an average internal roughness of 0.15 mm. Local ancillary elements above the piping system are years; Sharp edged inlet (Kg-0,85), flanged smooth 3 pieces 90 ° bend -0,3 each thousand and one flanged full open globe valve (KK-8). Find the net positive charge on the usable suction since the minimum volumetric flow is 30 L / s that can be mpaed without cavitation. Evaporation pressure for water at 25 ° temperature: 3.169 kPa) (Standard atmospheric pressure: 101.3 kPa) -997 kg / m3 (-1.05) (dynamic viscosity of water at 25 ° temperature: 8.91.10-kg / ms)
Answer with complete solutions.
plzzzz help me with this one plzzz  Water (viscosity of 0.001 kg/m.s  and density of 1000 kg/m3 ) is to be pumped through 50 m of pipe from lower reservoir  to a higher reservoir at a rate of 0.2 m3/s. If the pipe is cast iron of diameter 12 cm and the pump efficient is =0.44%, the major head loss inside the pipe is ¼ of the velocity head what horsepower pump is needed? Hint: consider the abrupt expansion only as a minor loss.

Chapter 14 Solutions

Fluid Mechanics: Fundamentals and Applications

Ch. 14 - There are three main categories of dynamic pumps....Ch. 14 - For each statement about cow cetrifugal the...Ch. 14 - Prob. 13CPCh. 14 - Consider flow through a water pump. For each...Ch. 14 - Write the equation that defines actual (available)...Ch. 14 - Consider a typical centrifugal liquid pump. For...Ch. 14 - Prob. 17CPCh. 14 - Consider steady, incompressible flow through two...Ch. 14 - Prob. 19CPCh. 14 - Prob. 20PCh. 14 - Suppose the pump of Fig. P1 4-19C is situated...Ch. 14 - Prob. 22PCh. 14 - Prob. 23EPCh. 14 - Consider the flow system sketched in Fig. PI 4-24....Ch. 14 - Prob. 25PCh. 14 - Repeat Prob. 14-25, but with a rough pipe-pipe...Ch. 14 - Consider the piping system of Fig. P14—24. with...Ch. 14 - The performance data for a centrifugal water pump...Ch. 14 - For the centrifugal water pump of Prob. 14-29,...Ch. 14 - Suppose the pump of Probs. 14-29 and 14-30 is used...Ch. 14 - Suppose you are looking into purchasing a water...Ch. 14 - The performance data of a water pump follow the...Ch. 14 - For the application at hand, the flow rate of...Ch. 14 - A water pump is used to pump water from one large...Ch. 14 - For the pump and piping system of Prob. 14-35E,...Ch. 14 - A water pump is used to pump water from one large...Ch. 14 - Suppose that the free surface of the inlet...Ch. 14 - Calculate the volume flow rate between the...Ch. 14 - Comparing the results of Probs. 14-39 and 14-43,...Ch. 14 - Prob. 45PCh. 14 - The performance data for a centrifugal water pump...Ch. 14 - Transform each column of the pump performance data...Ch. 14 - 14-51 A local ventilation system (a hood and duct...Ch. 14 - Prob. 52PCh. 14 - Repeat Prob. 14-51, ignoring all minor losses. How...Ch. 14 - Suppose the one- way of Fig. P14-51 malfunctions...Ch. 14 - A local ventilation system (a hood and duct...Ch. 14 - For the duct system and fan of Prob. 14-55E,...Ch. 14 - Repeat Prob. 14-55E, ignoring all minor losses....Ch. 14 - A self-priming centrifugal pump is used to pump...Ch. 14 - Repeat Prob. 14-60. but at a water temperature of...Ch. 14 - Repeat Prob. 14-60, but with the pipe diameter...Ch. 14 - Prob. 63EPCh. 14 - Prob. 64EPCh. 14 - Prob. 66PCh. 14 - Prob. 67PCh. 14 - Prob. 68PCh. 14 - Prob. 69PCh. 14 - Two water pumps are arranged in Series. The...Ch. 14 - The same two water pumps of Prob. 14-70 are...Ch. 14 - Prob. 72CPCh. 14 - Name and briefly describe the differences between...Ch. 14 - Discuss the meaning of reverse swirl in reaction...Ch. 14 - Prob. 75CPCh. 14 - Prob. 76CPCh. 14 - Prob. 77PCh. 14 - Prob. 78PCh. 14 - Prob. 79PCh. 14 - Prob. 80PCh. 14 - Wind ( =1.204kg/m3 ) blows through a HAWT wind...Ch. 14 - Prob. 82PCh. 14 - Prob. 84CPCh. 14 - A Francis radial-flow hydroturbine has the...Ch. 14 - Prob. 87PCh. 14 - Prob. 88PCh. 14 - Prob. 89PCh. 14 - Prob. 90CPCh. 14 - Prob. 91CPCh. 14 - Discuss which dimensionless pump performance...Ch. 14 - Prob. 93CPCh. 14 - Prob. 94PCh. 14 - Prob. 95PCh. 14 - Prob. 96PCh. 14 - Prob. 97PCh. 14 - Prob. 98PCh. 14 - Prob. 99PCh. 14 - Prob. 100EPCh. 14 - Prob. 101PCh. 14 - Calculate the pump specific speed of the pump of...Ch. 14 - Prob. 103PCh. 14 - Prob. 104PCh. 14 - Prob. 105PCh. 14 - Prob. 106PCh. 14 - Prob. 107EPCh. 14 - Prob. 108PCh. 14 - Prob. 109PCh. 14 - Prob. 110PCh. 14 - Prove that the model turbine (Prob. 14-109) and...Ch. 14 - Prob. 112PCh. 14 - Prob. 113PCh. 14 - Prob. 114PCh. 14 - Prob. 115CPCh. 14 - Prob. 116CPCh. 14 - Prob. 117CPCh. 14 - Prob. 118PCh. 14 - For two dynamically similar pumps, manipulate the...Ch. 14 - Prob. 120PCh. 14 - Prob. 121PCh. 14 - Prob. 122PCh. 14 - Calculate and compare the turbine specific speed...Ch. 14 - Prob. 124PCh. 14 - Prob. 125PCh. 14 - Prob. 126PCh. 14 - Prob. 127PCh. 14 - Prob. 128PCh. 14 - Prob. 129PCh. 14 - Prob. 130PCh. 14 - Prob. 131PCh. 14 - Prob. 132PCh. 14 - Prob. 133PCh. 14 - Prob. 134PCh. 14 - Prob. 135PCh. 14 - A two-lobe rotary positive-displacement pump moves...Ch. 14 - Prob. 137PCh. 14 - Prob. 138PCh. 14 - Prob. 139PCh. 14 - Prob. 140PCh. 14 - Which choice is correct for the comparison of the...Ch. 14 - Prob. 142PCh. 14 - In a hydroelectric power plant, water flows...Ch. 14 - Prob. 144PCh. 14 - Prob. 145PCh. 14 - Prob. 146PCh. 14 - Prob. 147PCh. 14 - Prob. 148PCh. 14 - Prob. 149PCh. 14 - Prob. 150PCh. 14 - Prob. 151P
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
8.01x - Lect 27 - Fluid Mechanics, Hydrostatics, Pascal's Principle, Atmosph. Pressure; Author: Lectures by Walter Lewin. They will make you ♥ Physics.;https://www.youtube.com/watch?v=O_HQklhIlwQ;License: Standard YouTube License, CC-BY
Dynamics of Fluid Flow - Introduction; Author: Tutorials Point (India) Ltd.;https://www.youtube.com/watch?v=djx9jlkYAt4;License: Standard Youtube License