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Dmitrij [34]
3 years ago
9

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

material has a specific gravity, SG, of 3.1. You may assume that the dam is loosely attached to the ground at its base, though there is significant friction to keep it from sliding.Is the weight of the dam sufficient to prevent it from tipping around its lower right corner?
Engineering
1 answer:
lisabon 2012 [21]3 years ago
6 0

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

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In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at
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To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

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Rearranging,

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Replacing,

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Explanation:

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5 0
3 years ago
How does the engine thermostat regulate the engine temperature?​
Inga [223]

Answer:

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4 years ago
A ship tows a submerged cylinder, 1.5 m in diameter and 22 m long, at U = 5 m/s in fresh water at 20°C. Estimate the towing powe
shusha [124]

Answer:

(a) 120 kW

(b) 800 kW

Explanation:

Given:

diameter: 1.5 m

length = 22 m

U = 5 m/s

temperature = 20°C

For water at 20°C, take ρ = 998 kg/m³  and µ = 0.001 kg/m⋅s

To find:

power in kW

(a) if the cylinder is parallel

Length / Diameter = L / D = 22 / 1.5 = 14.6 = 15

Re(L) = ρ*U*L / µ = 998 * 5 * 22 / 0.001 = 109780000 = 1.1E8

C(D.Frontal) ≈ 1.1

  Force = F = 1.1 * ρ/2 * U² * π / 4 * D

            =  1.1 (998 / 2) (5)²(π / 4)(1.5)²  

            = 1.1 * 499 * 25 * 0.785 * 2.25

            = 24000 N

Power = Force * Displacement / time

           = F * U

           = 24000 * 5

           = 120000

Power = 120 kW

b) if the cylinder is normal to the tow direction.

Re(L) = ρ*U*D / µ = 998 * 5 * 1.5 / 0.001 = 7485000 = 7.5E6

C(D.Frontal) ≈ 0.4

Force = F = 0.4 * ρ/2 * U² * D * L

            =  0.4 (998 / 2) (5)²(1.5)(22)  

            = 164670 ≈ 165000

Power = Force * Displacement / time

           = F * U

           = 165000 * 5 = 825000

           ≈ 800 kW

Power = 800 kW

7 0
4 years ago
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