Fluid Mechanics?? Water is pumped through the steel (e = 0.046 mm) pipeline syst

Question

Fluid Mechanics??

Water is pumped through the steel (e = 0.046 mm) pipeline system. Pipes 1 and 2 are the same kind of pipe, but pipe 2 is three times as long as pipe 1. They arc the same diameter (30 cm). The volume flow rate (discharge) of water in pipe 2 is measured to be 0.025 m2/sec. 1000 kg/m3 Ignore all minor losses and assume the same value of f in both pipes, what will be the discharge in pipe I (m3/sec)? Based on the discharge obtained from part (a), calculate the friction coefficient/in pipe I and 2. With the new/factor, recalculate the discharge in pipe 1. What is the % error for assuming same value of f? If the length of pipe 1 is 100 m. recalculate the discharge in pipe I considering all minor losses (flanged tee and pipe bend: r/d = 10). What is the % error for ignoring all minor losses? (Use f factors obtained in part (b). The % error can be calculated between results for part (b) and (c).)

Explanation / Answer

a)

Pipe cross-section area A = pi/4 *d^2 = 3.14 / 4 * 0.3^2 = 0.07065 m^2

Velocity in pipe 2 is V2 = Q2 / A2

= 0.025 / 0.07065

= 0.35386 m/s

Drop in head between A and B based on pipe 2 is = f*(3L)/d *V2 ^2 / (2g)

Drop in head based on pipe 1 = f*L/d * V1 ^2 / (2g)

Equating both,

3*V2 ^2 = V1 ^2

3*0.35386 ^2 = V1 ^2

V1 = 0.6129 m/s

Q1 = A*V1

= 0.07065*0.6129

= 0.0433 m^3/s

b)

Re in pipe 2 is = rho*V*d / mu

= 1000*0.35386*0.3 / (1.51*10^-3)

= 70303

Since Re > 2300, flow is turbulent.

Relative roughness e = eps / d

= 0.046 / 300

= 0.0001533

From Moody chart for Re = 70303 and e = 0.0001533 we get f = 0.019

Drop in head between A and B = f*(3L)/d *V2 ^2 / (2g)

= 0.019*3*L / 0.3 *0.35386^2 / (2*9.81)

= 0.0012126*L m

Re1 = 1000*0.6129*0.3 / (1.51*10^-3) = 1.21768*10^5

From Moody chart for Re = 1.21768*10^5 and e = 0.0001533 we get f = 0.018

Drop in head based on pipe 1 = f*L/d * V1 ^2 / (2g)

= 0.018*L / 0.3 *V1 ^2 / (2*9.81)

= 0.003058*L*V1 ^2

Equating both,

0.0012126*L = 0.003058*L* V1^2

V1 = 0.6297 m/s

Q1 = A*V1

= 0.07065*0.6297

= 0.04449 m^3/s

%error = (0.04449 - 0.0433)*100 / 0.04449

= 2.67%

c)

There are 2 90 deg bends. From table, for r/d = 10, we have K = 0.32

There are 2 flanged tees. For each K = 0.5

Total minor losses = (2*0.32 + 2*0.5)*V2 ^2 / (2g)

= 1.64*0.35386^2 / (2*9.81)

= 0.0104666 m

Drop in head based on pipe 2 = 0.0012126*(100) + 0.0104666 = 0.1317266 m

0.1317266 = 0.003058*100* V1^2

V1 = 0.6563 m/s

Q1 = A*V1

= 0.07065*0.6563

= 0.04637 m^3/s

% error = (0.04637 - 0.04449)*100 / 0.04449

= 4.22 %

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