A manometer measures a pressure difference as 40 inches of water. Take the density of water to be 62.4 lbm/ft3.What is this pres

Answer:

t = 5.27 years

Explanation:

Firstly, the corrosion penetration rate is defined by the formula;

CPR = (KW)/(ρAt)

Where;

K = constant based on exposed area A.

W - mass lost over time

t- duration

ρ - density

A - area exposed

From the problem, we have;

W = 7.6kg or 7.6 x 10^(6) mg

CPR = 4 mm/yr

ρ = 4.5 g/cm³

Area = 800 cm²

K is a constant valued at 87.6cm

Rearranging the CPR formula to isolate t, we derive;

t = KW/(ρA(CPR))

t = (87.6 x 7.6 x 10^(6))/(4.5 x 800 x 4) = 46233.3 hours

The duration in question needs to be expressed in years.

Thus, converting hours to years;

There are 8760 hours in a year.

Therefore;

t = 46233.3/8760 = 5.27 years.

Answer:

1.015 ha.

Explanation:

To calculate the landfill area required for 12,000 people producing waste over 10 years, follow these steps:[STEP ONE: Calculate the average solid waste generated per person per year (kg p^-1 ^y(kg/py)).

According to the problem, the average solid waste produced is 2.78 kg per person daily (kg/pd), hence converting to kg/py involves:

2.78 × 365 days = 1014.7 kg/py.

STEP TWO: Determine yearly volume of refuse per person.

Thus, volume = 1014.7 kg/py ÷ 500 kg/m^3 = 2.03 m^3 per person per year.

STEP THREE: Calculate total solid waste volume over 10 years for 12,000 individuals.

Total waste volume over 10 years = 10 × 12,000 × 2.03 = 243,600 m^3.

STEP FOUR: Find the required area for the landfill.

Note: The total height for the landfill should be 20 + 4 = 24m.

Thus, the area for the landfill = 243,600 m^3 / 24m = 10,150 m^2.

If 10,000 m^2 equals 1 ha, then 10,150 m^2 ÷ 10,000 m^2 = 1.015 ha.

(f). Ensure to expand the landfill area for enhancements.

By securing stakeholder needs throughout the project planning stage via thorough process documentation that is revised after each work cycle.

Answer:

Ps=19.62N

Explanation:

A thorough explanation of the answer can be found in the attached files.

In the scenario of a metal ingot cooling slowly, the microstructure tends to be coarse. The surface, exposed to higher temperatures for extended periods during cooling, features smaller grain sizes as they have less time to form. However, as we delve deeper into the ingot, the grains gradually extend, leading to equiaxed grain formation at the center.