Respuesta :
Answer:
The pressure difference across the 60 m long section of the pipe is 1106.28 kPa
The pumping power required to maintain the flow is 4.171 kW
Explanation:
For a fully developed laminar flow in a pipe we have
[tex]u(r) = u_{max} (1-\frac{r^2}{R^2} )[/tex]
Plugging in the values gives
[tex]u(r) = 6\hspace {0.09cm} m/s(1-\frac{r^2}{(0.02 \hspace {0.09cm} m)^2} )[/tex] = 6·(1-2500·r²)
The average velocity =
[tex]V = V_{avg} = \frac{u_{max}}{2} =\frac{6 \hspace{0.09cm} m/s}{2}[/tex] = 3 m/s
[tex]\dot{\nu }[/tex] = [tex]V_{avg}\cdot A_c[/tex] = v(πD²/4) = 3 m/s(π·(0.04 m)²/4) =3.7699× 10⁻³ m³/s
Reynolds' number is given by;
[tex]Re =\frac{\rho VD}{\mu}= \frac{1252\times3\times0.04}{0.3072}[/tex]
Re = 488.903
Since Re < 2300 we have a laminar flow
[tex]f = \frac{64}{Re}= \frac{64}{488.903}[/tex]
f = 0.131
Head loss
[tex]h_L = f\frac{LV^2}{2Dg} = 0.131\frac{60\times3^2}{2\times0.04\times9.81} = 90.07 \hspace{0.09cm}m[/tex]
The energy equation is given by
[tex]\frac{P_1}{\rho g} +\alpha_1\frac{V^2_1}{2g} + z_1+h_{pump} = \frac{P_2}{\rho g} +\alpha_2\frac{V^2_2}{2g} + z_2+h_{turbne}[/tex]
Where, V₁ =V₂, and z₁ = z₂
[tex]h_{pump} =h_{turbne} = 0[/tex], Therefore
P₁ - P₂ = ΔP = ρg([tex]h_L[/tex]) =1252×9.81×90.07 = 1106280.796 Pa = 1106.28 kPa
The power;
[tex]\dot{W}_{pump,u}=\dot{\nu}\Delta P = 3.7699\times 10^{-3} \times 1106.28 = 4.171\hspace{0.09cm} kW[/tex].
