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?

Cem Hayes 2021-02-18 Answered
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?

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Expert Answer

broliY
Answered 2021-02-20 Author has 7970 answers
In an industrial cooling process, water is circulated through a system.
If the water is pumped with a speed of(\(\displaystyle{v}_{{{1}}}={0.45}\frac{{m}}{{s}}\))
under a pressure of (\(\displaystyle{P}_{{{1}}}\) = 400 torr) from the first floor througha(\(\displaystyle{D}_{{{1}}}\) = 6.0cm) diameter pipe,
Let '\(\displaystyle{P}_{{{2}}}\)' be the pressure on the next floor ( h = 4.0 m ) above in a pipe with adiameter of (\(\displaystyle{D}_{{{2}}}\) = 2.0 cm)
Density of water = \(\displaystyle{1000}{k}\frac{{g}}{{m}^{{{3}}}}\)
From the equation of continuity we have
The rate is \(\displaystyle{A}_{{{1}}}{v}_{{{1}}}={A}_{{{2}}}{v}_{{{2}}}\)
Or, \(\displaystyle{v}_{{{2}}}={A}_{{{1}}}\frac{{v}_{{{1}}}}{{A}_{{{2}}}}\)
Or, \(\displaystyle{v}_{{{2}}}={\left(\frac{{D}_{{{1}}}}{{D}_{{{2}}}}\right)}^{{{2}}}{v}_{{{1}}}\)
Apply; Bernoulli's Theorem
\(\displaystyle{P}_{{{1}}}+{\left(\frac{{1}}{{2}}\right)}{{v}_{{{1}}}^{{{2}}}}+{0}={P}_{{{2}}}+{\left(\frac{{1}}{{2}}\right)}\cdot{{v}_{{{2}}}^{{{2}}}}+{h}\cdot{g}\)
Put the above given values in the above expression , to get'\(\displaystyle{P}_{{{2}}}\)'
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