Cem Hayes

2021-02-18

In an industrial cooling process, water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of 400 torr from the first floor through a 6.0-cm diameter pipe, what will be the pressure on the next floor 4.0 m above in a pipe with a diameter of 2.0 cm?

broliY

In an industrial cooling process, water is circulated through a system.
If the water is pumped with a speed of(${v}_{1}=0.45\frac{m}{s}$)
under a pressure of (${P}_{1}$ = 400 torr) from the first floor througha(${D}_{1}$ = 6.0cm) diameter pipe,
Let '${P}_{2}$' be the pressure on the next floor ( h = 4.0 m ) above in a pipe with adiameter of (${D}_{2}$ = 2.0 cm)
Density of water = $1000k\frac{g}{{m}^{3}}$
From the equation of continuity we have
The rate is ${A}_{1}{v}_{1}={A}_{2}{v}_{2}$
Or, ${v}_{2}={A}_{1}\frac{{v}_{1}}{{A}_{2}}$
Or, ${v}_{2}={\left(\frac{{D}_{1}}{{D}_{2}}\right)}^{2}{v}_{1}$
Apply; Bernoulli's Theorem
${P}_{1}+\left(\frac{1}{2}\right){v}_{1}^{2}+0={P}_{2}+\left(\frac{1}{2}\right)\cdot {v}_{2}^{2}+h\cdot g$
Put the above given values in the above expression , to get'${P}_{2}$'

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