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2 months ago

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The in situ moist unit weight of a soil is 17.3 kN/m3 and the moisture content is 16%. The specific gravity of soil solids is 2.

72. This soil is to be excavated and transported to a construction site for use in a compacted fill. If the specification calls for the soil to be compacted to a minimum dry unit weight of 18.1 kN/m3 at the same moisture content of 16%, how many cubic meters of soil from the excavation site are needed to produce 2000 m3 of compacted fill?

1 answer:

Mrrafil [318]2 months ago

3 0

Answer:

Explanation:

Given that,

Moisture content w = 16%

The in situ moist unit weight of soil: γ(in situ) = 17.3 kN/m³

Specific gravity of the soil

G(s) = 2.72

Minimum dry unit weight of the soil

γd(compacted) = 18.1 kN/m³

Moisture content remains the same

w = 16%

Question:

How much cubic meters of soil are necessary from the excavation site to create 2000 m³ of compacted fill?

Now, let's work out the in situ dry unit weight γd(in-situ) with the formula:

γd(in-situ) = γ(in-situ) / [1 + (w/100)]

γd(in-situ) = 17.3 / [1 + (16/100)]

γd(in-situ) = 17.3 / (1 + 0.16)

γd(in-situ) = 17.3 / 1.16

γd(in-situ) = 14.89 kN/m³

To find out the Volume of soil to be excavated (Vex):

Let the Volume to be excavated = V

We will use the relationship:

V = V(fill) × γd(compacted) / γd(in situ)

Given that, V(fill) = 2000 m³

V(fill) is the compacted fill volume.

Therefore, we have:

V = V(fill) × γd(compacted) / γd(in situ)

Vex = 2000 × 18.1 / 14.89

Vex = 2423.34 m³

Thus, the soil volume to be excavated equals 2423.34 m³.

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A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

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Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Inertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy =?

(b) rotor acceleration =?

(c) change in torque angle =?

(c) rotor speed =?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Calculate the rotor's stored energy at synchronous speed.

The stored energy is represented as

$E = G \times H$

Where G stands for complex rated power and H signifies the inertia constant of the turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If we suddenly increase the mechanical input to 80 MW against an electrical load of 50 MW, we shall find the rotor's acceleration while ignoring mechanical and electrical losses.

The formula for rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is defined as

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration derived in part (b) persists over 10 cycles, we will calculate both the change in torque angle and the rotor speed in revolutions per minute at the end of this duration.

The change in torque angle is expressed as

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$

Where t is determined from

$1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec$

Consequently,

$\Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2$

$$ \Delta \delta = 6.75 \: elec \: deg$

The change in torque in rpm/s is provided by

$\Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ }$

$\Delta \delta =28.12 \: \: rpm/s$

The rotor speed in rpm at the culmination of this 10-cycle period is calculated as

$Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t$

Where P indicates the number of poles on the turbo-generator.

$Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2$

$Rotor \: speed = 1500 + 5.62$

$$ Rotor \: speed = 1505.62 \:\: rpm$

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2 months ago

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Answer:

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Explanation:

Parameters

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mass flow rate, $\dot{m} = 60.6 kg/s$

flow density, $\rho = 1000 kg/m^3$

efficiency of the pump, $\eta = 0.74$

output gauge pressure, $p_o = 344.75 kPa$

input gauge pressure, $p_i = 68.95 kPa$

cross-sectional area of output pipe, $A_o = 0.069 m^2$

cross-sectional area of input pipe, $A_i = 0.093 m^2$

height of discharge, $z_o = 1.22 m - 0.61 m = 0.61 m$ (evaluated at pump’s maximum height of 1.22 m)

input height, $z_i = 0 m$

hydraulic power of the pump, $P =? kW$

Initially, the volumetric flow (Q) needs to be determined

$Q = \frac{\dot{m}}{\rho}$

$Q = \frac{60.6 kg/s}{1000 kg/m^3}$

$Q = 0.0606 m^3/s$

Next, compute the velocity (v) for both input and output

$v_o = \frac{Q}{A_o}$

$v_o = \frac{0.0606 m^3/s}{0.069 m^2}$

$v_o = 0.88 m/s$

$v_i = \frac{Q}{A_i}$

$v_i = \frac{0.0606 m^3/s}{0.093 m^2}$

$v_i = 0.65 m/s$

Subsequently, the total head (H) can be calculated

$H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}$

$H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}$

$H = 28.74m$

Finally, the computation of pump power is done as follows

$P = \frac{Q \, \rho \, g \, H}{\eta}$

$P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}$

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6 0

2 months ago

A piece of corroded metal alloy plate was found in a submerged ocean vessel. It was estimated that the original area of the plat

Mrrafil [318]

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.

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2 months ago