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2 months ago

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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)?

1 answer:

Viktor [391]2 months ago

7 0

Answer:

1113kN

Explanation:

The outer diameter of the pipe is 61cm, and its thickness is 0.9cm, leading to an inner diameter of:

Inner Diameter = Outer Diameter - Thickness

Inner Diameter = 61cm - 0.9cm = 60.1cm

Converting this measurement into meters gives us:

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

This inner diameter is relevant for calculating the volume V of water contained within the pipe, which can be defined as 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 water density provided is d = 1.0kg/L, but we must adjust it to appropriate units:

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

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

Substituting into the mass of the water yields:

$d=\frac{m}{V}$

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

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

This mass allows us to find the water's weight:

$m_{water}=113280Kg$

$w_{water}=m_{water} *g$

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

We then add this weight to the weight of the empty pipe to calculate the total weight:

$w_{water}=1110144N$

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

$w_{total}=1110144N+2500N$

Finally, converting this total weight into kN results in:

$w_{total}=1112644N$

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

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