Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 6, Problem 6.115P
To determine

(a)

The flow rate from reservoir 1 if valve C is closed.

Expert Solution
Check Mark

Answer to Problem 6.115P

Q=0.0152m3/s

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.115P , additional homework tip  1

All pipes are made of cast iron and the diameter is equal to 8cm

If the valve C is closed, the energy equation for the above system can be written as,

z1z2=hfA+hmA+hfB+hmB

Since, the flow is series and the diameter is same for all pipes, the flow rate and the velocity will be equal in both pipe A and pipe B.

Calculation:

Assume, the water at 20°C will have,

ρ=998kg/m3

μ=0.001kg/ms

The loss co efficient will be equal to,

Ksharpentrance=0.5Klinejunction=0.9KBranchjunction=1.3KSubmergedexit=1.0

According to the explanation, the energy equation will be,

z1z2=hfA+h m A +hfB+h m B ΔZ=V22g(f L A D+K sharpentrance+K linejunction+f L B D+K Submergedexit)25m=V22( 9.81m/ s 2 )(f 100m 0.08m+0.5+0.9+f 50m 0.08m+1.0)

Assume,

f=0.02

Therefore,

25m=V22( 9.81m/ s 2 )(( 0.02) 100m 0.08m+0.5+0.9+( 0.02) 50m 0.08m+1.0)V=3.5m/s

Calculate the Reynolds’s number

Red=(ρVd)μ=(998kg/m3)(3.5m/s)(0.08m)0.001kg/ms=279440

Calculate the roughness ratio

εd=0.26mm80mm=3.25×103

Calculate the exact value of friction factor

1f 1/21.8log( 6.9 R e + ( /d 3.7 ) 1.11)1f 1/21.8log( 6.9 279440+ ( 3.25× 10 3 3.7 ) 1.11)f=0.0272

Now, calculate the exact value of the velocity

z1z2=hfA+h m A +hfB+h m B ΔZ=V22g(f L A D+K sharpentrance+K linejunction+f L B D+K Submergedexit)25m=V22( 9.81m/ s 2 )(( 0.0272) 100m 0.08m+0.5+0.9+( 0.0272) 50m 0.08m+1.0)V=3.03m/s

Calculate the flow rate

Q=VA=(3.03m/s)(π( 0.04m)2)=0.0152m3/s

Conclusion:

The flow rate is equal to 0.0152m3/s.

To determine

(b)

The flow arte from reservoir 1 if valve C is open.

Expert Solution
Check Mark

Answer to Problem 6.115P

QA=0.0174m3/sQB=0.00955m3/sQC=0.00784m3/s

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.115P , additional homework tip  2

All pipes are made of cast iron and the diameter is equal to 8cm

If the valve C is opened, the energy equation for the above system can be written as,

z1z2=hfA+hm AB+hfB+hmB=hfA+hm AC+hfB+hmC

Since, the flow is parallel in pipe B and C, and the diameter is same for all pipes,

QA=QB+QC

VA=VB+VC

Calculation:

Assume, the water at 20°C will have,

ρ=998kg/m3

μ=0.001kg/ms

The loss co-efficient will be equal to,

Ksharpentrance=0.5Klinejunction=0.9KBranchjunction=1.3KSubmergedexit=1.0

According to the explanation, the energy equation will be,

z1z2=hfA+h m AB +hfB+h m B =hfA+h m AC +hfB+h m C ΔZ=VA22g(f L A D+K sharpentrance+K linejunction)+VB22g(f L B D+K Submergedexit)25m=VA22( 9.81m/s)(f 100m 0.08m+0.5+0.9)+VB22( 9.81m/s)(f 50m 0.08m+1.0)

Similarly,

ΔZ=VA22g(f L A D+K sharpentrance+K branchjunction)+VC22g(f L C D+K Submergedexit)25m=VA22( 9.81m/s)(f 100m 0.08m+0.5+1.3)+VC22( 9.81m/s)(f 70m 0.08m+1.0)

But, we also know

VA=VB+VC

Assume the friction factor is same for all pipe and it is equal to,

f=0.02

25m=VA22( 9.81m/s)(( 0.027) 100m 0.08m+0.5+0.9)+VB22( 9.81m/s)(( 0.027) 50m 0.08m+1.0)25m=1.791VA2+0.911VB2(1)

Similarly,

25m=VA22( 9.81m/s)(( 0.027) 100m 0.08m+0.5+1.3)+VC22( 9.81m/s)(( 0.027) 70m 0.08m+1.0)25m=1.811VA2+1.255VC2(2)

VA=VB+VC(3)

By solving we get,

VA=3.48m/sVB=1.91m/sVC=1.57m/s

Now find the Reynolds’s number and friction factor separately for each pipe

ReA=( ρVd)μ=( 998kg/ m 3 )( 3.48m/s)( 0.08m)0.001kg/ms=277843.2ReB=( ρVd)μ=( 998kg/ m 3 )( 1.91m/s)( 0.08m)0.001kg/ms=152494.4ReC=( ρVd)μ=( 998kg/ m 3 )( 1.57m/s)( 0.08m)0.001kg/ms=125348.8

1fA 1/21.8log( 6.9 277843.2+ ( 3.25× 10 3 3.7 ) 1.11)fA0.02721fB 1/21.8log( 6.9 152494.4+ ( 3.25× 10 3 3.7 ) 1.11)fB0.02751fC 1/21.8log( 6.9 125348.8+ ( 3.25× 10 3 3.7 ) 1.11)fC0.0277

25m=VA22( 9.81m/s)(( 0.0272) 100m 0.08m+0.5+0.9)+VB22( 9.81m/s)(( 0.0275) 50m 0.08m+1.0)25m=1.804VA2+0.927VB2(1)

25m=VA22( 9.81m/s)(( 0.0272) 100m 0.08m+0.5+1.3)+VC22( 9.81m/s)(( 0.0277) 70m 0.08m+1.0)25m=1.824VA2+1.283VC2(2)

According to the equations,

VA=3.46m/sVB=1.9m/sVC=1.56m/s

The relevant flow rate,

QA=VAA=(3.46m/s)(π ( 0.04m )2)=0.0174m3/sQB=VBA=(1.9m/s)(π ( 0.04m )2)=0.00955m3/sQC=VCA=(1.56m/s)(π ( 0.04m )2)=0.00784m3/s

Conclusion:

The flow rate in pipe A is equal to 0.0174m3/s

The flow arte in pipe B is equal to 0.00955m3/s

The flow arte in pipe C is equal to 0.00784m3/s.

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
(Read Image) (Answer: ω = 1.10 rad/sec CW)
What is the configuration of the control loop if steam must be shut down in case of a problem? (I found this question on the internet and was wondering what the answer is) A.Valve is fail open, PIC is direct-acting, TIC is reverse acting, and controller algorithm is feed-forwarding.B. Valve is fail open, PIC is reverse-acting, TIC is direct acting, and controller algorithm is cascade.C. Valve is fail closed, PIC is direct-acting, TIC is reverse acting, and controller algorithm is feed-forward.D. Valve is fail closed, PIC is reverse-acting, TIC is reverse acting, and controller algorithm is cascade.
What is the procedure to replace the input bellows?(I found this question on the internet and was wondering what the correct answer is out of interest) Remove tubing, old bellows and flapper assembly, install new bellows, connect tubing, install flapper assembly, then calibrate the positioner.Remove tubing, old bellows and cam, install new bellows, connect tubing, install cam, then calibrate the positioner.C. Remove tubing and old bellows, align the quadrant beam, install new bellows and connect tubing, then calibrate the positioner.D. Remove tubing and old bellows, install new bellows and connect tubing, align the quadrant beam, then calibrate the positioner.

Chapter 6 Solutions

Fluid Mechanics

Ch. 6 - Water at 20°C flows upward at 4 m/s in a...Ch. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Prob. 6.14PCh. 6 - Prob. 6.15PCh. 6 - Prob. 6.16PCh. 6 - P6.17 A capillary viscometer measures the time...Ch. 6 - P6.18 SAE 50W oil at 20°C flows from one tank to...Ch. 6 - Prob. 6.19PCh. 6 - The oil tanks in Tinyland are only 160 cm high,...Ch. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - Prob. 6.23PCh. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Let us attack Prob. P6.25 in symbolic fashion,...Ch. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - A laminar flow element (LFE) (Meriam Instrument...Ch. 6 - SAE 30 oil at 20°C flows in the 3-cm.diametcr pipe...Ch. 6 - Prob. 6.33PCh. 6 - Prob. 6.34PCh. 6 - In the overlap layer of Fig. 6.9a, turbulent shear...Ch. 6 - Prob. 6.36PCh. 6 - Prob. 6.37PCh. 6 - Prob. 6.38PCh. 6 - Prob. 6.39PCh. 6 - Prob. 6.40PCh. 6 - P6.41 Two reservoirs, which differ in surface...Ch. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - P6.44 Mercury at 20°C flows through 4 m of...Ch. 6 - P6.45 Oil, SG = 0.88 and v = 4 E-5 m2/s, flows at...Ch. 6 - Prob. 6.46PCh. 6 - Prob. 6.47PCh. 6 - Prob. 6.48PCh. 6 - Prob. 6.49PCh. 6 - Prob. 6.50PCh. 6 - Prob. 6.51PCh. 6 - Prob. 6.52PCh. 6 - Water at 2OC flows by gravity through a smooth...Ch. 6 - A swimming pool W by Y by h deep is to be emptied...Ch. 6 - Prob. 6.55PCh. 6 - Prob. 6.56PCh. 6 - Prob. 6.57PCh. 6 - Prob. 6.58PCh. 6 - P6.59 The following data were obtained for flow of...Ch. 6 - Prob. 6.60PCh. 6 - Prob. 6.61PCh. 6 - Water at 20°C is to be pumped through 2000 ft of...Ch. 6 - Prob. 6.63PCh. 6 - Prob. 6.64PCh. 6 - Prob. 6.65PCh. 6 - Prob. 6.66PCh. 6 - Prob. 6.67PCh. 6 - Prob. 6.68PCh. 6 - P6.69 For Prob. P6.62 suppose the only pump...Ch. 6 - Prob. 6.70PCh. 6 - Prob. 6.71PCh. 6 - Prob. 6.72PCh. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - P6.76 The small turbine in Fig. P6.76 extracts 400...Ch. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - The head-versus-flow-rate characteristics of a...Ch. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Prob. 6.83PCh. 6 - Prob. 6.84PCh. 6 - Prob. 6.85PCh. 6 - SAE 10 oil at 20°C flows at an average velocity of...Ch. 6 - A commercial steel annulus 40 ft long, with a = 1...Ch. 6 - Prob. 6.88PCh. 6 - Prob. 6.89PCh. 6 - Prob. 6.90PCh. 6 - Prob. 6.91PCh. 6 - Prob. 6.92PCh. 6 - Prob. 6.93PCh. 6 - Prob. 6.94PCh. 6 - Prob. 6.95PCh. 6 - Prob. 6.96PCh. 6 - Prob. 6.97PCh. 6 - Prob. 6.98PCh. 6 - Prob. 6.99PCh. 6 - Prob. 6.100PCh. 6 - Prob. 6.101PCh. 6 - *P6.102 A 70 percent efficient pump delivers water...Ch. 6 - Prob. 6.103PCh. 6 - Prob. 6.104PCh. 6 - Prob. 6.105PCh. 6 - Prob. 6.106PCh. 6 - Prob. 6.107PCh. 6 - P6.108 The water pump in Fig. P6.108 maintains a...Ch. 6 - In Fig. P6.109 there are 125 ft of 2-in pipe, 75...Ch. 6 - In Fig. P6.110 the pipe entrance is sharp-edged....Ch. 6 - For the parallel-pipe system of Fig. P6.111, each...Ch. 6 - Prob. 6.112PCh. 6 - Prob. 6.113PCh. 6 - Prob. 6.114PCh. 6 - Prob. 6.115PCh. 6 - Prob. 6.116PCh. 6 - Prob. 6.117PCh. 6 - Prob. 6.118PCh. 6 - Prob. 6.119PCh. 6 - Prob. 6.120PCh. 6 - Prob. 6.121PCh. 6 - Prob. 6.122PCh. 6 - Prob. 6.123PCh. 6 - Prob. 6.124PCh. 6 - Prob. 6.125PCh. 6 - Prob. 6.126PCh. 6 - Prob. 6.127PCh. 6 - In the five-pipe horizontal network of Fig....Ch. 6 - Prob. 6.129PCh. 6 - Prob. 6.130PCh. 6 - Prob. 6.131PCh. 6 - Prob. 6.132PCh. 6 - Prob. 6.133PCh. 6 - Prob. 6.134PCh. 6 - An airplane uses a pitot-static tube as a...Ch. 6 - Prob. 6.136PCh. 6 - Prob. 6.137PCh. 6 - Prob. 6.138PCh. 6 - P6.139 Professor Walter Tunnel needs to measure...Ch. 6 - Prob. 6.140PCh. 6 - Prob. 6.141PCh. 6 - Prob. 6.142PCh. 6 - Prob. 6.143PCh. 6 - Prob. 6.144PCh. 6 - Prob. 6.145PCh. 6 - Prob. 6.146PCh. 6 - Prob. 6.147PCh. 6 - Prob. 6.148PCh. 6 - Prob. 6.149PCh. 6 - Prob. 6.150PCh. 6 - Prob. 6.151PCh. 6 - Prob. 6.152PCh. 6 - Prob. 6.153PCh. 6 - Prob. 6.154PCh. 6 - Prob. 6.155PCh. 6 - Prob. 6.156PCh. 6 - Prob. 6.157PCh. 6 - Prob. 6.158PCh. 6 - Prob. 6.159PCh. 6 - Prob. 6.160PCh. 6 - Prob. 6.161PCh. 6 - Prob. 6.162PCh. 6 - Prob. 6.163PCh. 6 - Prob. 6.1WPCh. 6 - Prob. 6.2WPCh. 6 - Prob. 6.3WPCh. 6 - Prob. 6.4WPCh. 6 - Prob. 6.1FEEPCh. 6 - Prob. 6.2FEEPCh. 6 - Prob. 6.3FEEPCh. 6 - Prob. 6.4FEEPCh. 6 - Prob. 6.5FEEPCh. 6 - Prob. 6.6FEEPCh. 6 - Prob. 6.7FEEPCh. 6 - Prob. 6.8FEEPCh. 6 - Prob. 6.9FEEPCh. 6 - Prob. 6.10FEEPCh. 6 - Prob. 6.11FEEPCh. 6 - Prob. 6.12FEEPCh. 6 - Prob. 6.13FEEPCh. 6 - Prob. 6.14FEEPCh. 6 - Prob. 6.15FEEPCh. 6 - Prob. 6.1CPCh. 6 - Prob. 6.2CPCh. 6 - Prob. 6.3CPCh. 6 - Prob. 6.4CPCh. 6 - Prob. 6.5CPCh. 6 - Prob. 6.6CPCh. 6 - Prob. 6.7CPCh. 6 - Prob. 6.8CPCh. 6 - Prob. 6.9CPCh. 6 - A hydroponic garden uses the 10-m-long...Ch. 6 - It is desired to design a pump-piping system to...
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.
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
Fluid Mechanics - Viscosity and Shear Strain Rate in 9 Minutes!; Author: Less Boring Lectures;https://www.youtube.com/watch?v=_0aaRDAdPTY;License: Standard youtube license