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2 years ago

Why the power factor is Low in no load test in induction motor ?

Engineering

1 answer:

marshall27 [118]2 years ago

5 0

Answer:

At no load, the Cosine of the phase angle between the stator current(I1) and the stator applied voltage(V1) is the power factor of an induction motor. The reason of the low power factor is high rotor reactance at lower loading. The rotor reactance value gets decreased with increased loading on induction motor..','.

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A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar

jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading 114.06 m^3/d/m

Explanation:

calculation for single clarifier

$sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d$

$surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2$

$surface area = 314.16 m^2$

volume of tank $V = A\times side\ water\ depth$

$=314.16\times 2 = 628.32m^3$

$Length\ of\ weir = \pi \times diameter of weir$

$= \pi \times 18 = 56.549 m$

overflow rate = $v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2$

Detention time $t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr$

weir loading $= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m$

6 0

3 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago

The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re

Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 % , Irreversible

b) W_cycle = 600 KW , n_th = 100 % , Impossible

c) W_cycle = 400 KW , n_th = 66.67 % , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

TL = 400 K

TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

n_th < n_max ...... irreversible

n_th = n_max ...... reversible

n_th > n_max ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And, n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 400 = 200 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 200 / 600 = 33.33 %

- The type of process according to second Law of thermodynamics:

n_th = 33.333 % n_max = 66.67 %

n_th < n_max

Hence, Irreversible Process

b) QH = 600 kW, QC = 0 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 0 = 600 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 600 / 600 = 100 %

- The type of process according to second Law of thermodynamics:

n_th = 100 % n_max = 66.67 %

n_th > n_max

Hence, Impossible Process

c) QH = 600 kW, QC = 200 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 200 = 400 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 400 / 600 = 66.67 %

- The type of process according to second Law of thermodynamics:

n_th = 66.67 % n_max = 66.67 %

n_th = n_max

Hence, Reversible Process

7 0

3 years ago

Consider the following statement concerned with the collection of data, and determine the best selection of terms to complete th

snow_lady [41]

Answer:

a. population, units, sample

Explanation:

In a survey or in a research, population is defined as the total number of people or total number of items in the group that we want to study in a research. It is the entire pool from where a sample is drawn.

An unit is defined as the individual members for which the information or data is collected.

A sample is defined is defined as the group or part of the selection from where the information or data is to be obtained.

8 0

3 years ago

An ant starts at one edge of a long strip of paper that is 34.2 cm wide. She travels at 1.3 cm/s at an angle of 61◦ with the lon

ankoles [38]

Answer:

t = 30.1 sec

Explanation:

If the ant is moving at a constant speed, the velocity vector will have the same magnitude at any point, and can be decomposed in two vectors, along directions perpendicular each other.

If we choose these directions coincident with the long edge of the paper, and the other perpendicular to it, the components of the velocity vector, along these axes, can be calculated as the projections of this vector along these axes.

We are only interested in the component of the velocity across the paper, that can be calculated as follows:

vₓ = v* sin θ, where v is the magnitude of the velocity, and θ the angle that forms v with the long edge.

We know that v= 1.3 cm/s, and θ = 61º, so we can find vₓ as follows:

vₓ = 1.3 cm/s * sin 61º = 1.3 cm/s * 0.875 = 1.14 cm/s

Applying the definition of average velocity, we can solve for t:

t = $\frac{x}{vx}$ = $\frac{34.2 cm}{1.14 cm/s} =30.1 sec$

⇒ t = 30.1 sec

7 0

3 years ago

Why the power factor is Low in no load test in induction motor ?​

Why the power factor is Low in no load test in induction motor ?