Question

Consider the fully developed flow of glycerin at 40°C through a 60-m-long, 4-cm-diameter, horizontal, circular pipe. If the flow velocity at the centerline is measured to be 6 m/s, determine the pressure difference across this 60-m-long section of the pipe and the useful pumping power required to maintain this flow. The density and dynamic viscosity of glycerin at 40°C are rho = 1252 kg/m3 and µ = 0.3073 kg/m·s, respectively.

Answer 1

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

$u(r) = u_{max} (1-\frac{r^2}{R^2} )$

Plugging in the values gives

$u(r) = 6\hspace {0.09cm} m/s(1-\frac{r^2}{(0.02 \hspace {0.09cm} m)^2} )$ = 6·(1-2500·r²)

The average velocity =

$V = V_{avg} = \frac{u_{max}}{2} =\frac{6 \hspace{0.09cm} m/s}{2}$ = 3 m/s

$\dot{\nu }$ = $V_{avg}\cdot A_c$ = v(πD²/4) = 3 m/s(π·(0.04 m)²/4) =3.7699× 10⁻³ m³/s

Reynolds' number is given by;

$Re =\frac{\rho VD}{\mu}= \frac{1252\times3\times0.04}{0.3072}$

Re = 488.903

Since Re < 2300 we have a laminar flow

$f = \frac{64}{Re}= \frac{64}{488.903}$

f = 0.131

Head loss

$h_L = f\frac{LV^2}{2Dg} = 0.131\frac{60\times3^2}{2\times0.04\times9.81} = 90.07 \hspace{0.09cm}m$

The energy equation is given by

$\frac{P_1}{\rho g} +\alpha_1\frac{V^2_1}{2g} + z_1+h_{pump} = \frac{P_2}{\rho g} +\alpha_2\frac{V^2_2}{2g} + z_2+h_{turbne}$

Where, V₁ =V₂, and z₁ = z₂

$h_{pump} =h_{turbne} = 0$, Therefore

P₁ - P₂ = ΔP = ρg($h_L$) =1252×9.81×90.07 = 1106280.796 Pa = 1106.28 kPa

The power;

$\dot{W}_{pump,u}=\dot{\nu}\Delta P = 3.7699\times 10^{-3} \times 1106.28 = 4.171\hspace{0.09cm} kW$.