Respuesta :
Answer:
pipe is old one with increased roughness
Explanation:
discharge is given as
[tex]V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}[/tex]
V = 7.07 m/s
from bernou;ii's theorem we have
[tex]\frac{p_1}{\gamma} +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} +\frac{V_2^2}{2g} + z_2 + h_l[/tex]
as we know pipe is horizontal and with constant velocity so we have
[tex]\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}[/tex]
[tex]P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma[/tex]
[tex]135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81[/tex]
solving for friction factor f
f = 0.0324
fro galvanized iron pipe we have [tex]\epsilon = 0.15 mm[/tex]
[tex]\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025[/tex]
reynold number is
[tex]Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}[/tex]
Re = 378750
from moody chart
[tex]For Re = 378750 and \frac{\epsilon}{d} = 0.0025[/tex]
[tex]f_{new} = 0.025[/tex]
therefore new friction factor is less than old friction factoer hence pipe is not new one
now for Re = 378750 and f = 0.0324
from moody chart
we have [tex]\frac{\epsilon}{d} =0.006[/tex]
[tex]\epsilon = 0.006 \times 60[/tex]
[tex]\epsilon = 0.36 mm[/tex]
thus pipe is old one with increased roughness
