91Ó°ÊÓ

• The speed of water at first point is$v_1=3m/s$.
• The gauge pressure at first point is$P_1=5×{10}^{4}Pa$.
• The depth is$∆h=11m$.
• The pipe diameter at second point is$d_2=2d_1$.

In this problem, to evaluate the gauge pressure, firstly estimate the velocity of the water at the second point by using the continuity equation.

The area of pipeline at second point is calculated as:

$A_2=\frac{\mathrm{Ï€}{d}_2^2}{4}$

Substitute$2d_1$ for$d_2$ in the above relation.

$\begin{array}{r}{A}_2=\frac{\mathrm{Ï€}{\left(2d_1\right)}^2}{4}\\ A_2=\frac{4\mathrm{Ï€}{d}_1^2}{4}\\ A_2=4A_1\end{array}$

The relation of speed can be written as:

$\begin{array}{r}{A}_1{V}_1=A_2{V}_2\\ V_2=\left(\frac{A_1{V}_1}{A_2}\right)\end{array}$

Substitute$4A_1$ for$A_2$ and$3m/s$ for$v_1$ in the above relation.

$\begin{array}{r}{v}_2=\left(\frac{A_1(3\mathrm{m}/\mathrm{s})}{4A_1}\right)\\ v_2=0.75\mathrm{m}/\mathrm{s}\end{array}$

The relation of gauge pressure can be written as:

$\begin{array}{r}{P}_1+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2+\mathrm{ÒÏ}g{h}_1=P_2+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2+\mathrm{ÒÏ}g{h}_2\\ P_2=P_1+\frac{1}{2}\mathrm{ÒÏ}\left(v_1^2−v_2^2\right)+\mathrm{ÒÏ}g\left(h_1−h_2\right)\\ P_2=P_1+\frac{1}{2}\mathrm{ÒÏ}\left(v_1^2−v_2^2\right)+\mathrm{ÒÏ}g\left(\mathrm{Δ}h\right)\end{array}$

Here,$\mathrm{ÒÏ}$ is the density of water,$P_2$ is the required pressure and g is the gravitational acceleration.

Substitute$5×{10}^{4}Pa$ for$P_1$ , 1000$kg/m^3$ for$\mathrm{ÒÏ}$ ,$3m/s$ for$v_1$ ,$0.75m/s$ for$v_2$ ,$9.8m/s^2$ for g and 11m for$∆h$ in the above relation.

role="math" localid="1668142301104" $\begin{array}{r}{P}_2=5×{10}^4\mathrm{Pa}+\frac{1}{2}\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left((3\mathrm{m}/\mathrm{s}{)}^2−(0.75\mathrm{m}/\mathrm{s}{)}^2\right)+\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(11\mathrm{m}\right)\\ P_2=162×{10}^3\mathrm{Pa}\end{array}$

Thus, the required gauge pressure is$162×{10}^{3}Pa$ .