Answer:
the estimated time of submersion is 17.7 years
Explanation:
Given the data in the question;
estimate the time of submersion in years.
we write down the relation between time of submersion and corrosion penetration as follows;
CPR(mpy) = K × W(mg) / [ A(in²) × p(g/cm³) × t(hr) ]
we solve for t
t = (K × W) / ( AP × CPR )
given that;
Area A = 5 in²
W = 2.3 kg = 2.3 × 10⁶ mg
density of steel p = 7.9 g/cm³
CPR = 200
we know that K is 534
so we substitute
t = (534 × 2.3 × 10⁶ mg) / ( 5 in² × 7.9 g/cm³ × 200 mpy )
t = 1,228,200,000 / 7900
t = 155468.3544 hr
t = 155468.3544 hr × ( 1 yrs / ( 365 × 24 hrs )
t = 17.7 years
Therefore, the estimated time of submersion is 17.7 years
Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.
Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.
Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:
- <u><em>Decrease the length of the pipes</em></u>: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
- <u><em>Reduce the surface roughness of the pipes</em></u>: By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
- <u><em>Increase the pipe diameter: </em></u>By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.
You can learn more about friction losses at
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Answer:
17.658 kPa
Explanation:
The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.
Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:
Meanwhile, the volume of a column is the area of the base multiplied by the height:
Replacing:
The base cancels out, so:
The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.
If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:
We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:
Answer:
115 Ib/ft^3
Explanation:
To determine the maximum dry density in Ib/ft3 we have to calculate :
Bulk unit weight ( yb ) ; W / v
Dry unit weight ( yd. ) : yb / ( 1 + w )
For every set of data given
assuming v = 1/30 ft^3
calculating for the 3 data set ( maximum dry density )
weight of soil (W) = 4.40
moisture content (%) (w) = 15.0 = 0.15
Bulk unit weight (yb) = 4.40 / (1/30) = 132 Ib/ft^3
Dry unit weight ( yd. ) = 132 / ( 1 + 0.15 ) = 114.702 Ib/ft^3
therefore after calculations the maximum dry density in Ib/ft^3 ≈ 115 Ib/ft^3