1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Reika [66]
3 years ago
14

1. Design a circuit, utilizing set/reset coils where PB 1 starts Motor 1 and PB2 stops Motor 1. Pressing and releasing either pu

sh button should latch the motor either on or off depending on the function

Engineering
1 answer:
lianna [129]3 years ago
5 0

Answer:

Circuit attached with explanation

Explanation:

Hi Dear,

A circuit is attached for your reference.

When you press "start" PB, the supply reaches the motor starter relay coil "M" that is also in parallel with the "start" PB which allows the motor to remain ON even when you release "start" PB as supply to relay coil is directly from supply "L" through "M".

To stop motor just press "stop" PB and the circuit breaks which de-energize the relay coil and the motor stops.

Hope this finds easy to you.

You might be interested in
“In a List of Positive Integers, Set MINIMUM to 1. For each number X in the list L, compare it to MINIMUM. If X is smaller, set
artcher [175]

Answer:

no i cant answer it sorry

7 0
2 years ago
Using the characteristics equation, determine the dynamic behavior of a PI controller with τI = 4 applied to a second order proc
Sladkaya [172]

Answer:

The values of Kc that render this closed-loop process unstable are in the interval

(Kc < 0)

Explanation:

The transfer function of a PI controller is given as

Gc = Kc {1 + (1/sτI)}

τI = 4

Gc = Kc {1 + (1/4s)}

Gc = Kc {(4s+1)/(4s)}

Divide numerator and denominator by 4

Gc = Kc {(s+0.25)/(s)}

For a second order process, the general transfer function is given by

Gp = Kp {1/(τn²s² + 2ζτns + 1)}

Kp = 2, τn = 5 and ζ = 1.5

Gp = 2/(25s² + 15s + 1)

Divide numerator and denominator by 25

Gp = 0.08/(s² + 0.6s + 0.04)

Ga = 1

Gs = 1

We need to find the value(s) of Kc that makes the closed loop transfer function unstable. Gp*Ga*Gc*Gs + 1 = 0

The closed loop transfer function is unstable when the solution(s) of the characteristic equation obtained is positive.

Gp*Ga*Gc*Gs + 1 = 0

Becomes

[0.08/(s² + 0.6s + 0.04)] × [Kc (s+0.25)/(s)] + 1 = 0

[0.08Kc (s + 0.25)/(s³ + 0.6s² + 0.04s)] = - 1

0.08Kc (s + 0.25) = -s³ - 0.6s² - 0.04s

0.08Kc s + 0.02Kc = -s³ - 0.6s² - 0.04s

s³ + 0.6s² + 0.04s + 0.08Kc s + 0.02Kc = 0

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

We will use the direct substitution method to evaluate the values of Kc that matter. The values of Kc at the turning points of the closed loop transfer function.

For the substitution,

We put s = jw into the equation. (frequency analysis)

Note that j = √(-1)

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

(jw)³ + 0.6(jw)² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

-jw³ - 0.6w² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

we then collect terms with j and terms without.

(0.08Kcw + 0.04w - w³)j + (0.02Kc - 0.6w²) = 0

Meaning,

0.08Kcw + 0.04w - w³ = 0 (eqn 1)

0.02Kc - 0.6w² = 0 (eqn 2)

0.02 Kc = 0.6 w²

Kc = 15w²

Substituting this into eqn 1

0.08Kcw + 0.04w - w³ = 0

Kc = 15w²

0.08(15w²)w + 0.04w - w³ = 0

1.2w³ + 0.04w - w³ = 0

0.2w³ + 0.04w = 0

w = 0 or 0.2w² + 0.04 = 0

0.2w² = -0.04

w² = -0.2

w = ± √(-0.2)

w = ± 0.4472j or w = 0

Recall, Kc = 15w² = 15(-0.2) = -3 or Kc = 0

The turning points for the curve of the closed loop transfer function occur when

Kc = 0 or Kc = -3

To investigate, we pick values around these turning points to investigate the behaviour of the closed loop transfer function at those points.

Kc < -3, Kc = -3, (-3 < Kc < 0), Kc = 0 and Kc > 0

Note that, one positive characteristic root or pole is enough to make the system unstable.

We pick a value for Kc in that interval and evaluate the closed loop transfer function.

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

- First of, let Kc = - 4 (Kc < -3)

s³ + 0.6s² - 0.28s - 0.08 = 0

Solving the polynomial

s = (-0.22002), 0.44223, (-0.82221)

One positive pole means the closed loop transfer function is unstable in this region

Let Kc = -3

s³ + 0.6s² - 0.20s - 0.06 = 0

s = 0.37183, (-0.21251) or (-0.75933)

One positive pole still means that the closed loop transfer function is still unstable.

Then the next interval

Let Kc = -1

s³ + 0.6s² - 0.04s - 0.02 = 0

Solving this polynomial,

s = 0.18686, (-0.1749) or (-0.61196)

The function is unstable in the region being investigated.

Let Kc = 0

s³ + 0.6s² + 0.04s = 0

s = 0, -0.0769, -0.5236

One zero, all negative roots, indicate that the closed loop transfer function is marginally stable at this point.

Let Kc = 1, Kc > 0

s³ + 0.6s² + 0.12s + 0.02 = 0

s = (-0.42894), (-0.08553 + 0.1983j) or (-0.08553 - 0.1983j)

All the real negative parts of the poles are all negative, this indicates stability.

Hence, after examining the turning points of the closed loop transfer function, it is evident that, the region's of Kc where the closed loop transfer function is unstable is (Kc < 0)

Hope this Helps!!!

8 0
3 years ago
Which of the following can easily reverse motion and are better at varying speeds than electrical motors?
satela [25.4K]
Fluid power system

Example
4 0
2 years ago
Read 2 more answers
An adiabatic air compressor compresses 10 L/s of air at 120 kPa and 20 degree C to 1000 kPa and 300 degree C.
Oksana_A [137]

Answer:

work=281.4KJ/kg

Power=4Kw

Explanation:

Hi!

To solve follow the steps below!

1. Find the density of the air at the entrance using the equation for ideal gases

density=\frac{P}{RT}

where

P=pressure=120kPa

T=20C=293k

R= 0.287 kJ/(kg*K)= gas constant ideal for air

density=\frac{120}{(0.287)(293)}=1.43kg/m^3

2.find the mass flow by finding the product between the flow rate and the density

m=(density)(flow rate)

flow rate=10L/s=0.01m^3/s

m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s

3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow

Work

w=Cp(T1-T2)

Where

Cp= specific heat for air=1.005KJ/kgK

w=work

T1=inlet temperature=20C

T2=outlet temperature=300C

w=1.005(300-20)=281.4KJ/kg

Power

W=mw

W=(0.0143)(281.4KJ/kg)=4Kw

5 0
3 years ago
Mathew is manufacturing overhead transmission wires for the working of electric trains. On what does the insulation of these wir
elena-14-01-66 [18.8K]

Answer:

B

Explanation:

4 0
3 years ago
Other questions:
  • Create a program named PaintingDemo that instantiates an array of eight Room objects and demonstrates the Room methods. The Room
    10·1 answer
  • A group of n Ghostbusters is battling n ghosts. Each Ghostbuster carries a proton pack, which shoots a stream at a ghost, eradic
    9·1 answer
  • provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki
    12·1 answer
  • What are the four categories of engineering materials used in manufacturing?
    9·1 answer
  • How to go about the designing of a multirange voltmeter​
    8·1 answer
  • Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th
    8·1 answer
  • Please help, Artificial Intelligence class test
    12·1 answer
  • During her soccer game, Brittany hears her coach tell her team to look for better shot selection. The offensive strategy her coa
    7·2 answers
  • Which type of software is LibreOffice an example? open source freeware O beta O firmware​
    15·2 answers
  • The volume of microbial culture is observed to increase according to the formula
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!