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stiv31 [10]
2 years ago
9

Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is

0.9 cm. The water density is 1.0 kg/L. The empty weight of the metal pipe is 2500 N/m. In kN, what is the total weight (pipe plus water)?

Engineering
1 answer:
Yuki888 [10]2 years ago
7 0

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

60.1cm*\frac{1m}{100cm}=0.601m

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

V_{water}=\pi  r^{2}h

V_{water}=\pi (\frac{0.601m}{2})^{2}*120m

V_{water}=113.28m^{3}

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}

d_{water}=1000\frac{Kg}{m^{3}}

Now, water density is given by the equation d=\frac{m}{V}, where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

m_{water}=d_{water}.V_{water}

m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}

m_{water}=113280Kg

With the water mass we can find the weight of water:

w_{water}=m_{water} *g

w_{water}=113280kg*9.8\frac{m}{s^{2}}

w_{water}=1110144N

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

w_{total}=w_{water}+w_{pipe}

w_{total}=1110144N+2500N

w_{total}=1112644N

Converting this total weight to kN, we have:

1112644N*\frac{0.001kN}{1N}=1113kN

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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

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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

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3 years ago
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