Question

Water at 20 C flows through a 5-cm-diameter pipe that has a 180 vertical bend, as in Fig. P3.43. The total length of pipe between flanges 1 and 2 is 75 cm. When the weight flow rate is 230 N/s, p1

Answer 1

Answer:

F = 749 [N]

Explanation:

We must give full information on this problem, as well as the question that needs to be resolved.

Water at 20°C flows through a 5-cm-diameter pipe that has a 180° vertical bend. The total length of pipe between flanges 1 and 2 is 75 cm. When the weight flow rate is 230 N/s,

P1=165kPa

and

P2=134kPa

Neglecting pipe weight, determine the total force that the flanges must withstand for this flow.

We must make a U-shaped body diagram of the pipe in order to visualize the forces acting according to the pressures and the area of the pipe.

Then by means of the second law of motion of Newton, which says that the sum of the forces must be equal to the product of mass times acceleration, we can find the force Fp

Let us remember that pressure is defined as the divided force over the area, therefore:

F = P *A

where:

P = pressure

A = area

The product of the mass by acceleration, is equal to the product of the speed of the fluid by the mass flow, since we know the weight of the fluid we can find its mass flow.

$W_{flow}=230[N/s]\\W_{flow} =g*m_{flow}\\m_{flow} = W_{flow} / g\\m_{flow} = 230/9.81\\m_{flow}= 23.45[kg/s]$

In the function of the mass flow, we can find the velocity of the fluid, as we also know the diameter of the pipe

$m_{flow} = density*v*A\\where\\density = 1000[kg/m^{3}]\\ v= velocity[m/s]\\A = area [m^{2}]\\v=\frac{m_{flow}}{density*A} \\v=\frac{23.45}{1000*\frac{\pi}{4}*(5*10^{-2})^{2} } \\v= 11.94 [m/s]$

We know that the atmospheric pressure is equal to:

$P_{atm} = 101.325[kpa]$

The value of the pipe area is calculated for a circular section

$A = \frac{\pi}{4} * (0.05)^{2}\\ A = 0.00196[m^{2} ]$

The resultant force is 749 [N]

The solution of the equations and the free body diagram can be seen in the attached picture.