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3 years ago

A 60-Hz 3-phase induction motor is required to drive a load at approximately 850 rpm. How many poles should the motor have

Engineering

1 answer:

Anastaziya [24]3 years ago

7 0

Answer:

8 poles

Explanation:

Given

$Frequency, f = 60Hz$

$Speed, n_s = 850rpm$

Required

Calculate the number of poles the motor should have?

To solve this question, we make use of synchronous angular velocity formula

$n_s = \frac{120f}{P}$

Where

$P = Number\ of\ Poles$

Substitute values for $n_s$ and f in $n_s = \frac{120f}{P}$

$850 = \frac{120 * 60}{P}$

$850 = \frac{7200}{P}$

Multiply both sides by P

$P * 850 = \frac{7200}{P} * P$

$P * 850 = 7200$

Divide both sides by 850

$\frac{P * 850}{850} = \frac{7200}{850}$

$P = \frac{7200}{850}$

$P = 8.47058823529$

However, the number of poles must be an integer.

So:

$P = 8$

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Derive the expression ε=ln(1+e), where ε is the true strain and e is the engineering strain. Note that this expression is not va

Rudiy27

The formula for true strain after derivation from basic terms is; ε_t = In(1 + ε_e)

<h3>How to derive the expression for True Strain?</h3>

Formula for Engineering Stress is;

σ_e = Load/Area

Formula for true stress is;

σ_t = Force/Instantaneous Area

Formula for Engineering Strain is;

ε_e = ΔL/L₀

Formula for true strain is;

dε_t = dL/L

Total true strain is gotten from;

ε_t = ∫(dL/L) between boundaries of L_f and L_o

When we integrate between those boundaries, we have;

ε_t = In[(L₀ + ΔL)/L₀

⇒ ε_t = In[(1+ ΔL/L₀)

⇒ ε_t = In(1 + ε_e)

Read more about True Strain at; brainly.com/question/20717759

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2 years ago

A two-dimensional flow field described by

Oduvanchick [21]

Answer:

the answer is

Explanation:

<h2> We now focus on purely two-dimensional flows, in which the velocity takes the form </h2><h2>u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) </h2><h2>With the velocity given by (2.1), the vorticity takes the form </h2><h2>ω = ∇ × u = </h2><h2> </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y </h2><h2>k. (2.2) </h2><h2>We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y = 0. (2.3) </h2><h2>We have already shown in Section 1 that this condition implies the existence of a velocity </h2><h2>potential φ such that u ≡ ∇φ, that is </h2><h2>u = </h2><h2>∂φ </h2><h2>∂x, v = </h2><h2>∂φ </h2><h2>∂y . (2.4) </h2><h2>We also recall the definition of φ as </h2><h2>φ(x, y, t) = φ0(t) + Z x </h2><h2>0 </h2><h2>u · dx = φ0(t) + Z x </h2><h2>0 </h2><h2>(u dx + v dy), (2.5) </h2><h2>where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent </h2><h2>of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is </h2><h2>even easier to establish when we restrict our attention to two dimensions. If we consider </h2><h2>two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, </h2><h2>Green’s Theorem implies that </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2></h2><h2></h2>

5 0

3 years ago

A car is moving at 68 miles per hour. The kinetic energy of that car is 5 × 10 5 J.How much energy does the same car have when i

Blababa [14]

Answer:

The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

Explanation:

Given the data in the question;

Initial velocity v₁ = 68 miles per hour = 30.398 meter per seconds

let mass of the car be m

kinetic energy of that car is 5 × 10⁵ J

E₁ = $\frac{1}{2}$ mv²

we substitute

5 × 10⁵ = $\frac{1}{2}$ × m × ( 30.398 )²

5 × 10⁵ = m × 462.019

m = 5 × 10⁵ / 462.019

m = 1082.2065 kg

Now, Also given that; v₂ = 97 miles per hour = 43.362 meter per seconds

E₂ = $\frac{1}{2}$ mv₂²

we substitute

E₂ = $\frac{1}{2}$ × 1082.2065 × ( 43.362 )²

E₂ = $\frac{1}{2}$ × 1082.2065 × 1880.263

E₂ = 1.017 × 10⁵ J

Therefore, The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

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3 years ago

The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate

Sonja [21]

Answer:

a) Mechanical efficiency ( $\varepsilon$ )=63.15% b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, $Wmec=0.95*6kW=5.7 kW$ .

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump $\Delta P$ . So $P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W$ .

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is $\varepsilon=3.6kW/5.7kW=0.6315=63.15\%$

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and $\rho$ the density of the water:

$[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T$ [/tex]

Finally, the temperature rise is:

$\Delta T=2100/75348 \ºC=0.028 \ºC$

7 0

3 years ago

Saul is testing an installation and discovers a short circuit. What's causing this?

sergey [27]

A high voltage!! Hope this helps

6 0

3 years ago