Question

A nuclear power plant draws $3.0\times {10}^6\mathrm{L}/\min$ of cooling water from the ocean. If the water is drawn in through two parallel,$3.0-\mathrm{m}$-diameter pipes, what is the water speed in each pipe?

Short answer

Therefore, the speed of the water through each pipe is$3.53\mathrm{m}/\mathrm{s}$

Step-by-step solution

Step :1 Introduction 

The volume from one pipe is given follow

$V_t=\frac{3.0\times {10}^6\mathrm{L}/\min }{2}$

The conversion is given as follows:

$V_t=1.5\times {10}^6\frac{\mathrm{L}}{min}\left(\frac{{10}^{-3}{\mathrm{m}}^3}{60\mathrm{s}}\right)$

$=25\frac{{\mathrm{m}}^3}{\mathrm{s}}$

Step :2 Explanation 

The area of the pipe is given as follows:

$A=\pi {\left(\frac{d}{2}\right)}^2$

Here, $d$is the diameter of the pipe.

Substitute $3m$for $d$.

$A=\pi {\left(\frac{3\mathrm{m}}{2}\right)}^2$

$=7.065{\mathrm{m}}^2$

Step :3  Expression of speed

The expression for the speed of the water flowing through the pipe related to the volume flow rate is given as follows:

$v=\frac{V_i}{A}$

Substitute $25\frac{{\mathrm{m}}^3}{\mathrm{s}}\text{for}{V}_1\text{and}7.065{\mathrm{m}}^2\text{for}A$

$v=\frac{\left(25\frac{{\mathrm{m}}^3}{\mathrm{s}}\right)}{\left(7.065{\mathrm{m}}^2\right)}$

$=3.53\mathrm{m}/\mathrm{s}$