Question

An upward flow of oil (mass density 800 kg/m${}^3,$ dynamic viscosity 0.8 kg/m-s) takes place under laminar conditions in an inclined pipe of 0.1 m diameter as shown in the figure. The pressures at sections 1 and 2 are measured as p${}_1$=435 kN/m${}^2$ and p${}_2$=200kN/m${}^2$.

The discharge in the pipe is equal to

• A. $0.100m^3/s$
• B. $0.127m^3/s$
• C. $0.144m^3/s$
• D. $0.161m^3/s$

Answer 1

The correct option is B $0.127m^3/s$


Apply energy equation between (1) and (2)
$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+h_f$

Where

$h_f=\frac{32\mu VL}{\rho g{D}^2}$

$\frac{435\times {10}^3}{800\times 9.81}+\frac{V^2}{2g}+0=\frac{200\times {10}^3}{800\times 9.81}+\frac{V^2}{2g}+\frac{5}{\sqrt{2}}+\frac{32\left(0.8\right)V\left(5\right)}{\left(800\right)\left(9.81\right){\left(0.1\right)}^2}$

$\Rightarrow \frac{435\times {10}^3-200\times {10}^3}{800\times 9.81}-\frac{5}{\sqrt{2}}=\frac{32\times 0.8\times V\times 5}{800\times 9.81\times {\left(0.1\right)}^2}$

$\Rightarrow V=16.19m/s$

$\therefore Q=AV$

$=\frac{\pi \times {\left(0.1\right)}^2}{4}\times 16.19$

$=0.127m^3/s$