Question

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)?

Answer 1

1113kN

Step-by-step 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*(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 ((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(Kg)/(L)*(1L)/(1000cm^(3))*((100cm)/(1m))^(3)`

`d_(water)=1000(Kg)/(m^(3))`

Now, water density is given by the equation
`d=(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(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(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*(0.001kN)/(1N)=1113kN`