Answer:
Step-by-step explanation:
To determine the required size of standard schedule 40 steel pipe to carry 192 m³/hour of water with a minimum velocity of 6.0 m/sec, we can use the following formula:
Q = A × v
where Q is the volumetric flow rate of water, A is the cross-sectional area of the pipe, and v is the velocity of water.
First, we need to convert the volumetric flow rate from m³/hour to m³/sec.
192 m³/hour = 0.0533 m³/sec
Next, we can rearrange the formula to solve for the cross-sectional area:
A = Q / v
A = 0.0533 m³/sec / 6.0 m/sec
A = 0.0089 m²
The cross-sectional area of the pipe is 0.0089 m².
Standard schedule 40 steel pipe has a nominal inside diameter (ID) of 1.049 inches, which is approximately 0.0266 meters. The cross-sectional area of the pipe can be calculated using the formula for the area of a circle:
A = π × (ID/2)²
A = 3.14 × (0.0266/2)²
A = 5.58×10^-4 m²
To determine the required size of the pipe, we can rearrange the formula for the area of a circle to solve for the diameter:
ID = 2 × √(A/π)
ID = 2 × √(0.0089/π)
ID = 0.106 meters
Therefore, the required size of standard schedule 40 steel pipe to carry 192 m³/hour of water with a minimum velocity of 6.0 m/sec is a nominal size of 4 inches, with an inside diameter of 0.102 meters (or 102 millimeters).