# An air-operated isolation valve requires 2,400 pounds-force applied to the top of the actuator diaphragm to open against spring pressure ___

An air-operated isolation valve requires 2,400 pounds-force applied to the top of the actuator diaphragm to open against spring pressure.

The actuator diaphragm has a diameter of 12 inches.

If control air pressure to the valve actuator begins to decrease from 100 psig, which one of the following is the approximate air pressure at which the valve will begin to close?

A. 5 psig
B. 17 psig
C. 21 psig
D. 66 psig

Share your understanding of this question with the correct explanation.

Explanation:

To find the approximate air pressure at which the valve will begin to close, we can use the formula for pressure:

Pressure = Force / Area

First, let’s find the area of the actuator diaphragm:

Area = π * (6 inches)^2
Area = 36π square inches

Now, let’s find the pressure at which the valve will begin to close. The force required to close the valve is the same as the force required to open it, which is 2,400 pounds-force.

Pressure to close = Force / Area
Pressure to close = 2,400 pounds-force / (36π square inches)

To convert pounds-force to psi (pounds per square inch), we divide by the conversion factor of 144 (1 square inch = 0.00694 square feet, and 1 pound-force per square foot = 1 psi):

Pressure to close ≈ (2,400 pounds-force / (36π square inches)) / 144
Pressure to close ≈ (66.67 / π) psi

Now, let’s approximate the value of π (pi) to 3.14:

Pressure to close ≈ (66.67 / 3.14) psi
Pressure to close ≈ 21.23 psi

So, the approximate air pressure at which the valve will begin to close is 21.23 psig.

The closest option to this value is:

C. 21 psig