A pipeline, 300mm in diameter and 3200 M long is used to pump a 50 kg per second of oil whose density is 950kg/m3950kg/m3and whose kinematics viscosity is 2.1 Stokes. The centre of pipeline at the upper end is 40 m above than that at the lower end. The discharge at the upper and is atmospheric. find the pressure at the lower end and draw the hydraulic gradient line and Total energy line diagram.

 A pipeline, 300mm in diameter and 3200 M long is used to pump a 50 kg per second of oil whose density is 

950kg/m3$950kg/m^3$and whose kinematics viscosity is 2.1 Stokes. The centre of pipeline at the upper end is 40 m above than that at the lower end. The discharge at the upper and is atmospheric. find the pressure at the lower end and draw the hydraulic gradient line and Total energy line diagram.

Solution:

 Given:

Diameter of pipe, d=300mm=0.3m$d=300mm=0.3m$

Length of pipe, L=3200m$L=3200m$

Mass, m=50kg/s=ρ×Q$m=50kg/s=\rho \times Q$

(ρ=950kg/m3$\rho =950kg/m^3$)

∴$\therefore$ Discharge, Q=50ρ$Q={\displaystyle \frac{50}{\rho }}$

Q=50950=0.0526m3/s$Q={\displaystyle \frac{50}{950}}=0.0526m^3/s$

• kinematic Viscosity, v=2.1 stokes

v=2.1cm2/s=2.1×10−4m2/s$v=2.1c{m}^2/s=2.1\times {10}^{-4}{m}^2/s$

• Height of upper end= 40m

• Pressure at the upper end = atmospheric = 0

-Reynolds number, Re=V×dv$R_e={\displaystyle \frac{V\times d}{v}}$

where V=QA=0.0526π4×(0.3)2=0.744m/s$V={\displaystyle \frac{Q}{A}}={\displaystyle \frac{0.0526}{{\displaystyle \frac{\pi }{4}}\times (0.3{)}^2}}=0.744m/s$

∴,Re=0.744×0.302.1×10−4=1062.8$\therefore ,R_e={\displaystyle \frac{0.744\times 0.30}{2.1\times {10}^{-4}}}=1062.8$

As the flow of oil through the pipe is laminated,

f=16Re$f={\displaystyle \frac{16}{R_e}}$

∴f=161062.8=0.015$\therefore f={\displaystyle \frac{16}{1062.8}}=0.015$

$\mathrm{<i><br></i>}$

Step 1: Head lost due to friction,

hf=4fLV2d×2g$hf={\displaystyle \frac{4fL{V}^2}{d\times 2g}}$

hf=4×0.015×3200×(0.744)20.3×2×9.81$hf={\displaystyle \frac{4\times 0.015\times 3200\times (0.744{)}^2}{0.3\times 2\times 9.81}}$

hf=18.05m of oil

Step 2: Applying Bernoulli's equation at the lower end of pipe and taking datum line passing through the lower end

P1ρg+V212g+z1=P2ρg+V222g+z2+hf${\displaystyle \frac{P_1}{\rho g}}+{\displaystyle \frac{V_1^2}{2g}}+z_1={\displaystyle \frac{P_2}{\rho g}}+{\displaystyle \frac{V_2^2}{2g}}+z_2+hf$

But z1=0,z2=40m,V1=V2$z_1=0,z_2=40m,V_1=V_2$ (as diagram is same)

P2=0,hf=18.05m$P_2=0,hf=18.05m$

Therefore, substituting the values, we get

P1ρg=40+18.05=58.05m${\displaystyle \frac{P_1}{\rho g}}=40+18.05=58.05m$ of oil

∴P1=58.05×ρg$\therefore P_1=58.05\times \rho g$

P1=58.05×950×9.81=540997N/m2$P_1=58.05\times 950\times 9.81=540997N/m^2$

${\mathrm{<i><br></i>}}^{}$

∴P1=54.099N/cm2$\therefore P_1=54.099N/c{m}^2$

Step 3: Hydraulic gradient Line and Total energy line

V22g=(0.744)22×9.81=0.0282m${\displaystyle \frac{V^2}{2g}}={\displaystyle \frac{(0.744{)}^2}{2\times 9.81}}=0.0282m$

P1ρg=58.05m${\displaystyle \frac{P_1}{\rho g}}=58.05m$ of oil

P2ρg=0${\displaystyle \frac{P_2}{\rho g}}=0$