Question

Oil enters the bend of a pipe in the horizontal plane with velocity $4{\text{ms}}^{-1}$ and pressure $280\times {10}^3{\text{Nm}}^{-2}$ as shown in the figure. The pressure of oil at the point Q is $n\times {10}^2{\text{kNm}}^{-2}$. Then $n=$
(Take specific gravity of oil as 0.9 and $\sin {37}^{\circ }=0.6$)

Answer 1

Let $A_1$ and $A_2$ be the cross-sectional area of the pipe at points P and Q respectively. Let $v_1$ and $v_2$ be the velocities of oil at the points P and Q respctively.
By conservation of mass,
$Q=A_1{v}_1=A_2{v}_2$
$\Rightarrow v_2=\left(\frac{A_1}{A_2}\right)\left(v_1\right)=4v_1$
Applying Bernoullis's equation between points P and Q, we have
$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$
$P_2=P_1+\frac{1}{2}\rho (v_1^2-v_2^2)$
$=P_1+\frac{1}{2}\times 0.9[16-256]\times {10}^3$
$=280\times {10}^3+(-108)\times {10}^3$
$P_2=172\times {10}^3{\text{Nm}}^{-2}$
$P_2=1.72\times {10}^2\left({\text{kNm}}^{-2}\right)$.