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Leno4ka [110]
3 years ago
15

What is the pressure at the bottom of a 32m high water tower in gage pressure and absolute pressure (answers in kPa, and bar)

Engineering
1 answer:
creativ13 [48]3 years ago
5 0

Answer:

P_{g}=313.6kPa, P_{g}=3.13bar, P_{a}=414.93kPa, P_{a}=4.15bar

Explanation:

The gauge pressure is the one made by the water over the bottom of the tank without taking into account the pressure of the atmosphere. It can be calculated by using the following equation:

P_{g}=\gamma _{H_{2}O} h

in this case:

\gamma _{H_{2}O}=9.8kN/m^{3}

and h=32m

so the equation would then solve to:

P_{g}=(9.8kN(m^{3})(32m)

Which yields

P_{g}=313.6kPa

now, in order to find the bars, we must remember that:

1bar=100kPa, so

313.6kPa*\frac{1bar}{100kPa}=3.136bar

Now, in order to find the absolute pressure we need to add the atmospheric pressur to the pressures previously found.

P_{a}=313.6kPa+101.325kPa

P_{a}=414.93kPa

and likewise we can do the conversion of kPa to bars, so we get:

414.93kPa*\frac{1bar}{100kPa}=4.15bar

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3 years ago
What is an advantage of a nuclear-fission reactor?.
Kobotan [32]

Answer:

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advantages:
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Disadvantages:
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4 0
2 years ago
How fast is a 2012 nissan sentra<br>speed and acceleration ​
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3 years ago
A long corridor has a single light bulb and two doors with light switch at each door.
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8 0
2 years ago
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
3 years ago
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