1: Determine the dynamic pressure of water received at the site described below.

Water Tower holds water at an elevation of 265 feet

Site is at an elevation of 145 feet

The water supply system uses cast iron pipes.

The water travels through 3.2 miles of pipes before reaching the site

The pipe has a diameter of 8 inches

The water travels through 9 90-degree bends, 4 Branch tees, and 1 Swing Check Valve

The water has a flow rate of 105 gpm

Round to the hundredths place (2 places after the decimal)

2:
What is the static head of a water supply system if the water tower holds water at and elevation of 462 feet and the site that uses the water is at an elevation of 294 feet?
Answer value

20.87 Pa

Explanation:

The formula for dynamic pressure is given as;

q= 1/2*ρ*v²

where ;

q=dynamic pressure

ρ = density of fluid

v = velocity of fluid

First find v by applying the formula for flow rate as;

Q = v*A   where ;

Q= fluid flow rate

v = flow velocity

A= cross-sectional area.

A= cross-sectional vector area of the pipe given by the formula;

A= πr² = 3.14 * 4² = 50.27 in²   where r=radius of pipe obtained from the diameter given divided by 2.

Q = fluid flow rate = 105 gpmchange to m³/s as

1 gpm = 0.00006309

105 gpm = 105 * 0.00006309 = 0.006624 m³/s

A= cross-sectional vector area = 50.27 in² change to m² as:

1 in² = 0.0006452 m²

50.27 in² = 50.27 * 0.0006452 = 0.03243 m²

Now calculate flow velocity as;

Q =v * A

Q/A = v

0.006624 m³/s / 0.03243 m² =v

0.2043 m/s = v

Now find the dynamic pressure q given as;

q= 1/2 * ρ*v²

q= 1/2 * 1000 * 0.2043² = 20.87 Pa

answer: make you irritable

answer:

the average distance would be 20ly

Explanation:

100,000ly times 5,000ly=20ly