Which of the following is an essential component of reinforced concrete?

At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t

nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

$PGP = \rho *\frac{\pi r^2}{2} *V*e \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}$

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0

2 years ago

A water reservoir contains 108 metric tons of water at an average elevation of 84 m. The maximum amount of electric energy that

zavuch27 [327]

Answer:

24.72 kwh

Explanation:

Electric energy=potential energy=mgz where m is mass, g is acceleration due to gravity and z is the elevation.

Substituting the given values while taking g as 9.81 and dividing by 3600 to convert to per hour we obtain

PE=(108*9.81*84)/3600=24.72 kWh

8 0

3 years ago

Based on experimental observations, the acceleration of a particle is defined by the relationa = -( 0.1 + sin(x/b) ),where a and

yKpoI14uk [10]

Answer:

a) v = +/- 0.323 m/s

b) x = -0.080134 m

c) v = +/- 1.004 m/s

Explanation:

Given:

a = - (0.1 + sin(x/b))

b = 0.8

v = 1 m/s @ x = 0

Find:

(a) the velocity of the particle when x = -1 m

(b) the position where the velocity is maximum

(c) the maximum velocity.

Solution:

- We will compute the velocity by integrating a by dt.

a = v*dv / dx = - (0.1 + sin(x/0.8))

- Separate variables:

v*dv = - (0.1 + sin(x/0.8)) . dx

-Integrate from v = 1 m/s @ x = 0:

0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5

0.5v^2 = 0.8cos(x/0.8) - 0.1x - 0.3

- Evaluate @ x = -1

0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3

v = sqrt (0.104516)

v = +/- 0.323 m/s

- v = v_max when a = 0:

-0.1 = sin(x/0.8)

x = -0.8*0.1002

x = -0.080134 m

- Hence,

v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134

v = sqrt (0.504)

v = +/- 1.004 m/s

4 0

3 years ago

The cantilevered W530 x 150 beam shown is subjected to a 9.8-kN force F applied by means of a welded plate at A. Determine the e

snow_lady [41]

The equivalent force-couple system at O is the force and couple experienced when at point O due to the applied force at point A

The equivalent force couple system at O due to force F are;

Force, F = (8.65·i - 4.6·j) KN

Couple, M₀ ≈ 40.9 k kN·m

The reason the above values are correct is as follows:

The known values for the cantilever are;

The height of the beam = 0.65 m

The magnitude of the applied force, F = 9.8 kN

The length of the beam = 4.9 m

The angle away from the vertical the force is applied = 26°

The required parameter:

The equivalent force-couple system at the centroid of the beam cross-section of the cantilever

Solution:

The equivalent force-couple system is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;

The equivalent force $\overset \longrightarrow F$ = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j