Question

Models of torpedoes are sometimes tested in a horizontal pipe of flowing water, much as a wind tunnel is used to test model airplanes. Consider a circular pipe of internal diameter $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$and a torpedo model aligned along the long axis of the pipe. The model has a $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$diameter and is to be tested with water flowing past it at $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$.

(a) With what speed must the water flow in the part of the pipe that is unconstricted by the model?

(b) What will the pressure difference be between the constricted and unconstricted parts of the pipe?

Short answer

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

b) The pressure difference between the constricted and unconstricted parts of the pipe is $245\mathrm{Pa}$

Step-by-step solution

Given data

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

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

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.

Determining the concept

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:

$\mathrm{p}\vee -\frac{1}{2}{\mathrm{\rho g}}^2\mathrm{h}+\mathrm{constant}$

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

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 $\mathrm{\rho }$is density.

(a) Determining the speed of the water at the unconstricted section of the pipe

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,

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

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

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

$=2.40\raisebox{1ex}{$\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}$}\right.$

(b) Determining the pressure difference between the constricted and unconstricted parts of the pipe

The water flow obeys Bernoulli’s principle.

So,

$\mathrm{p\vartheta }+\frac{1}{2}{\mathrm{\rho ghh}}_{\mathrm{c}}{\mathrm{p}}_{\mathrm{c}}=\mathrm{\rho \gamma }+\frac{1}{2}\mathrm{\rho gh}+\mathrm{u}$

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

$\mathrm{\Delta p}={\mathrm{p}}_{\mathrm{u}}-{\mathrm{p}}_{\mathrm{c}}$

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

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

$=245\mathrm{Pa}$

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