91Ó°ÊÓ

i) The internal diameter of the pipe, $d_p=25.0\mathrm{cm}−25.0×{10}^{−2}\mathrm{m}$.

ii) The diameter of the torpedoes model,${\mathrm{Î\pm }}_m=5.00\mathrm{cm}=5.00×{10}^2\mathrm{m}$ .

iii) The speed of the water at a constricted section of pipe, role="math" localid="1657633745547" $v_a=2.50\mathrm{m}/\mathrm{s}$.

iv) The pipe is horizontal.

Determine the speed of the water at the unconstricted section of the pipe using the equation of continuity. Then, using Bernoulli’s equation, find the pressure difference between the constricted and unconstricted parts of the pipe. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

Formulae are as follows:

$\mathrm{p}∨−\frac{1}{2}{\mathrm{Òϲ\mu }}^2\mathrm{h}+\mathrm{constant}$

$\mathrm{Av}=\mathrm{constant}$

Where, $\mathrm{p}$is pressure, $\mathrm{v}$is velocity, $\mathrm{h}$is height, $\mathrm{g}$is an acceleration due to gravity, $\mathrm{h}$is height, $\mathrm{A}$is the area, and $\mathrm{ÒÏ}$is density.

The effective diameter of the constricted section of the pipe is,

${\mathrm{d}}_{\mathrm{c}}={\mathrm{d}}_{\mathrm{p}}−{\mathrm{d}}_{\mathrm{m}}$

The water flowing through the pipe obeys the equation of continuity.

${\mathrm{A}}_{\mathrm{c}}{\mathrm{V}}_{\mathrm{c}}={\mathrm{A}}_{\mathrm{u}}{\mathrm{V}}_{\mathrm{u}}$

Hence,

${\mathrm{v}}_{\mathrm{u}}=\frac{{\mathrm{A}}_{\mathrm{c}}{\mathrm{v}}_{\mathrm{c}}}{{\mathrm{A}}_{\mathrm{u}}}$

$=\frac{\left({\mathrm{d}}_{\mathrm{p}}^2−{\mathrm{d}}_{\mathrm{m}}^2\right){\mathrm{v}}_{\mathrm{c}}}{\left({\mathrm{d}}_{\mathrm{p}}^2\right)}$

$=\frac{\left({\left(25.0×{10}^{−2}\mathrm{m}\right)}^2−{\left(5.00×{10}^{−2}\mathrm{m}\right)}^2\right)×2.50\mathrm{m}/\mathrm{s}}{{\left(25×{10}^{−2}\mathrm{m}\right)}^2}$

$=2.40\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

Hence, the speed of the water at the unconstricted section of the pipe is$2.40\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

The water flow obeys Bernoulli’s principle.

So,

$\mathrm{\pm èÏ‘}+\frac{1}{2}{\mathrm{Òϲ\mu hh}}_{\mathrm{c}}{\mathrm{p}}_{\mathrm{c}}=\mathrm{ÒÏγ}+\frac{1}{2}\mathrm{Òϲ\mu h}+\mathrm{u}$

Given that the pipe is horizontal, i.e., ${\mathrm{h}}_{\mathrm{c}}={\mathrm{h}}_{\mathrm{u}}$and $\mathrm{ÒÏ}=\mathrm{density}\mathrm{of}\mathrm{water}=1000\mathrm{kg}/\mathrm{m}3.$

$\mathrm{Δ\pm è}={\mathrm{p}}_{\mathrm{u}}−{\mathrm{p}}_{\mathrm{c}}$

$=\frac{1}{2}\mathrm{ÒÏ}\left({\mathrm{v}}_{\mathrm{c}}^2−{\mathrm{v}}_{\mathrm{u}}^2\right)$

$=\frac{1}{2}×1000×\left((2.50{)}^2−(2.40{)}^2\right)$

$=245\mathrm{Pa}$

Hence, the pressure difference between the constricted and unconstricted parts of the pipe is$245\mathrm{Pa}$.