# 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?

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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broliY
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'$$\displaystyle{P}_{{{2}}}$$'