Applied Fluid Mechanics (7th Edition)
Applied Fluid Mechanics (7th Edition)
7th Edition
ISBN: 9780132558921
Author: Robert L. Mott, Joseph A. Untener
Publisher: PEARSON
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Chapter 12, Problem 12.1PP

Figure 12.7 shows a branched system in which the pressure at A is 700   k P a and the pressure at B is 550   k P a . Each branch is 60   m long. Neglect losses at the junctions, but consider all elbows. If the system carries oil with a specific weight of 8.80   k N / m 3 , calculate the total volume flow rate. The oil has a kinematic viscosity of

   4.8 × 10 m 2 / s .

Expert Solution & Answer
Check Mark
To determine

Total volume flow rate by using the given information.

Answer to Problem 12.1PP

The total volume flow rate is determined below.

Explanation of Solution

First we have to find the head loss at point A and B on applying the energy equation.

   pAγ+zA+v2A2ghL=pBγ+zB+v2B2g

Here pA =pressure

   γ =specific weight

   zA =Head at A

   v2A =velocity at A

And then parallel system head losses for both the parts are same

   ( h L ) a = ( h L ) b

   = p A p B γ

Then substitute pA=700kN/m2,PB=550kN/m2,γ=8.80kN/m3 values into above equation we get

   ( h L ) a = ( h L ) b

   = 700500 8.80

   =17.05m

Then the roughness of steel pipe is ε=(4.6×105)m

Inside diameter of the pipe is 0.1023m

Find the relative roughness of the elbow

   εD=4.6× 10 50.1023=4.49×104

Corresponding to friction factor value is fr=0.017

Find the total head loss of the upper section

   (hL)=hfric+2helb=fa(LD)ua22g+2fr( L e D)ua22g

For 900 elbows curved, effective length ratio is (LeD)=30

Substitute required values we will get

   ( h L ) a = f a ( 60 0.1023 ) u a 2 2g +2×0.017×(30) u a 2 2g

   =[ 587 f a +1.02 ] u a 2 2g

From the table of text book DN 80 pipes the inside diameter of the pipe is 0.0779m

Find the relative rough ness

   εD=4.6× 10 50.0779=5.9×104

Find the total head loss of the lower branch we will get

   (hL)=hfric+2helb=fa(LD)ua22g+fr( L e D)vua22g+2fr( L e D)eua22g

Substitute above values we get

   (hL)=fb( 60 0.0779)ub22g+2×0.018×(30)ub22g+0.018×(240)ub22g=[770fb+5.4]ub22g

Assume fa=fb=0.02

Find the flow velocity at upper branch

Substitute above value we get

   17.05=[587×0.02+1.02]ua22gva=5.08m/s

Find the Reynolds number for the flow

   NR=uaDav

   NR=5.08×0.10234.8× 10 6=1.08×105

Find the relative roughness we get

   εD=4.6× 10 50.1023=4.49×104

Then the value of friction factor is Fa=0.02

Find the flow velocity at B

Substitute above values we get

   17.05=[770×fb+5.4]ub22×9.81334.521=[770×0.02+5.4]ub2vb=4.01m/s

Calculate the Reynolds number for the flow

   NR=uaDav

Substitute above values we get

   NR=4.01×0.00234.8× 10 6=6.51×104

Find the relative roughness value

   εD=4.6× 10 50.0779=5.9×104

Then the value of friction factor is Fa=0.022.

Recomputed the velocity of flow using above equation

   17.05=[770×fb+5.1]ub22×9.81334.521=[770×0.002+5.1]ub2vb=3.87m/s

Find the Reynolds number for the flow

   NR=uaDav

Substitute above values we get

   NR=3.87×0.07794.8× 10 6=6.51*104

Find the value of relative roughness

   εD=4.6× 10 50.0779=5.9×104

From Moody's diagram, the value of relative roughness and Reynolds number Fb=0.022.

Find the flow rate at A,

We get

   Qa=Aaua

   Qa=(8.213×103)×5.08=4.17×102m3/s

Find the flow rate at B

   Q=AbubQb=(4.768×103)×3.87=1.85×102m3/s

Total volume flow rate is

   Q=Qa+Qb=(4.17×102)+(1.85×102)=6.02× 10 2m3/s

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A 3-in Schedule 40 steel pipe is 5000 ft long and carries a lubricating oil between two points A and B such that the Reynolds number is 800. Point B is 20 ft higher than A. The oil has a specific gravity of 0.90 and a dynamic vis- cosity of 4 x 10lb-s/ft. If the pressure at A is 50 psig. calculate the pressure at B.
4. A crude oil of viscosity 0.9 poise and sp. gr. 0.8 is flowing through a horizontal circular pipe of diameter 80 mm and of length 15 m. Calculate the difference of pressure at the two ends of the pipe, if 50 kg of the oil is collected in a tank in 15 seconds.

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