What is the total thrust in pounds on a 10-inch, 90-degree bend when the water pressure is 100 psi?

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Multiple Choice

What is the total thrust in pounds on a 10-inch, 90-degree bend when the water pressure is 100 psi?

Explanation:
To determine the total thrust on a 10-inch, 90-degree bend at a water pressure of 100 psi, it is essential to understand how thrust is generated in a piping system and how it can be calculated. Thrust is the force exerted by the fluid being directed through a bend, and its calculation involves the fluid pressure, the pipe diameter, and the angle of the bend. In this case, we can use the formula for calculating the thrust on a bend: \[ Thrust = Pressure \times Area \] First, we must calculate the area of the pipe through which the water flows. For a circular pipe, the area can be calculated using the formula: \[ Area = \frac{\pi D^2}{4} \] where D is the diameter of the pipe. For a 10-inch pipe, the diameter is 10 inches. We first need to convert the diameter into feet: \[ D = 10\ \text{inches} \div 12 = 0.8333\ \text{feet} \] Now calculate the area: \[ Area = \frac{\pi \times (0.8333)^2}{4} \approx 0.544 \ \text{square feet

To determine the total thrust on a 10-inch, 90-degree bend at a water pressure of 100 psi, it is essential to understand how thrust is generated in a piping system and how it can be calculated.

Thrust is the force exerted by the fluid being directed through a bend, and its calculation involves the fluid pressure, the pipe diameter, and the angle of the bend. In this case, we can use the formula for calculating the thrust on a bend:

[ Thrust = Pressure \times Area ]

First, we must calculate the area of the pipe through which the water flows. For a circular pipe, the area can be calculated using the formula:

[ Area = \frac{\pi D^2}{4} ]

where D is the diameter of the pipe. For a 10-inch pipe, the diameter is 10 inches. We first need to convert the diameter into feet:

[ D = 10\ \text{inches} \div 12 = 0.8333\ \text{feet} ]

Now calculate the area:

[ Area = \frac{\pi \times (0.8333)^2}{4} \approx 0.544 \ \text{square feet

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