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3 years ago

How can the direction of rotation of a split-phase motor be changed? *

Engineering

2 answers:

S_A_V [24]3 years ago

8 0

Nejejenshsjsjsjs:

Explanation:rjejsjjejej

Aloiza [94]3 years ago

3 0

Answer:To reverse it, we need to change the direction of the rotating magnetic field produced by the main and starter windings. And this can be accomplished by reversing the polarity of the starter winding. ... If you were to switch the main and starter windings, as one does with a split phase motor, the motor will also reverse.

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What combustion after effects do actuators work to control

Rufina [12.5K]

Answer: I would help you but I don’t know the answer, so sorry

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3 years ago

Read 2 more answers

A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken o

irina [24]

Answer:

the heat transfer from the pipe will decrease when the insulation is taken off for r₂< $r_{cr}$

where;

r₂ = outer radius

$r_{cr}$ = critical radius

Explanation:

Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .

$r_{cr} =\frac{k}{h}$

The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than critical radius (r₂< $r_{cr}$ ) 0, and reaches a maximum when r₂ = $r_{cr}$ , and starts to decrease for r₂< $r_{cr}$ . Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂< $r_{cr}$ .

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3 years ago

Question 1: Final Results = What are the values of the resistances such that the gain = -100, Rin = 1 MI2. Don't use resistances

lidiya [134]

Answer:

Explanation:

In a study of algebra, you will encounter many families of equations, or groups of

equations that share common characteristics. Of interest to us here is the family of

linear equations in one variable, a study that lays the foundation for understanding

more advanced families. In addition to solving linear equations, we’ll use the skills we

develop to solve for a specified variable in a formula, a practice widely used in science,

business, industry, and research.

A. Solving Linear Equations Using Properties of Equality

An equation is a statement that two expressions are

equal. From the expressions and

we can form the equation

which is a linear equation in one variable. To solve

an equation, we attempt to find a specific input or xvalue that will make the equation true, meaning the

left-hand expression will be equal to the right. Using

Table 1.1, we find that is a

true equation when x is replaced by 2, and is a false

equation otherwise. Replacement values that make

the equation true are called solutions or roots of the equation.

4 0

2 years ago

A 0.2-m^3 rigid tank equipped with a pressure regulator contains steam at 2MPa and 320C. The steam in the tank is now heated. Th

prohojiy [21]

Answer:

Q=486.49 KJ/kg

Explanation:

Given that

V= 0.2 m³

At initial condition

P= 2 MPa

T=320 °C

Final condition

P= 2 MPa

T=540°C

From steam table

At P= 2 MPa and T=320 °C

h₁=3070.15 KJ/kg

At P= 2 MPa and T=540°C

h₂=3556.64 KJ/kg

So the heat transfer ,Q=h₂ - h₁

Q= 3556.64 - 3070.15 KJ/kg

Q=486.49 KJ/kg

7 0

3 years ago

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

3 years ago