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 $p_1=435kN/m^2$ and $p{}_2=200kN/m^2$.

If the flow is reversed, keeping the same discharge, and the pressure at section 1 is maintained as $435kN/m^2$, the pressure at section 2 is equal to

• A. $488kN/m^2$
• B. $549kN/m^2$
• C. $586kN/m^2$
• D. $614kN/m^2$

Answer 1

The correct option is D $614kN/m^2$


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

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

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

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

$\Rightarrow V=16.19m/s$

For reversal of flow
Total energy at section (2) - loses = Total energy at section (1)
When the flow is reversed, then,
$\frac{p_1}{\rho g}+\frac{32\mu VL}{\rho g{D}^2}=\frac{p_2}{\rho g}+5sin{45}^{\circ }$
$\Rightarrow \frac{435\times {10}^3}{800\times 9.81}+\frac{32\times 0.8\times 16.19\times 5}{800\times 9.81\times (0.1{)}^2}$
$=\frac{p_2}{800\times 9.81}+\frac{5}{\sqrt{2}}$
$\Rightarrow p_2=614.48kN/m^2=614kN/m^2$