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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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\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{   \left{D_2}  }{ {D_1}}   \right )^{3\times n} =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n}

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