A single phase inductive load draws 10 MW at 0.6 power factor lagging. Draw the power triangle and determine the reactive power of a capacitor to be connected in parallel with the load to raise the power factor to 0.85.

Answers

Answer 1

Answer: attached below is the power triangles

7.13589 MVAR

Explanation:

Power ( P1 ) = 10 MW

power factor ( cos ∅ ) = 0.6 lagging

New power factor = 0.85

Calculate the reactive power of a capacitor to be connected in parallel

Cos ∅ = 0.6

therefore ∅ = 53.13°

S = P1 / cos ∅ = 16.67 MVA

Q1 = S ( sin ∅ ) = 13.33 MVAR  ( reactive power before capacitor was connected in parallel )

note : the connection of a capacitor in parallel will cause a change in power factor and reactive power while the active power will be unchanged i.e. p1 = p2

cos ∅2 = 0.85 ( new power factor )

hence ∅2 =  31.78°

Qsh ( reactive power when power factor is raised to 0.85 )

= P1 ( tan∅1 - tan∅2 )

= 10 ( 1.333 - 0.6197 )

= 7.13589 MVAR

A Single Phase Inductive Load Draws 10 MW At 0.6 Power Factor Lagging. Draw The Power Triangle And Determine

Related Questions

A rectangular channel 3-m-wide carries 12 m^3/s at a depth of 90cm. Is the flow subcritical or supercritical? For the same flowrate, what depth will five critical flow?

Answers

Answer:

Super critical

1.2 m

Explanation:

Q = Flow rate = [tex]12\ \text{m}^3/\text{s}[/tex]

w = Width = 3 m

d = Depth = 90 cm = 0.9 m

A = Area = wd

v = Velocity

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

[tex]Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}[/tex]

Froude number is given by

[tex]Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5[/tex]

Since [tex]F_r>1[/tex] the flow is super critical.

Flow is critical when [tex]Fr=1[/tex]

Depth is given by

[tex]d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}[/tex]

The depth of the channel will be 1.2 m for critical flow.

Indicate similarities between a nucleus and a liquid droplet; why small droplets are stable and very big droplets are not?

Answers

Answer:

There are several similarities between the nucleus and a liquid droplet.

Explanation:

A droplet of liquid simply is is very small or tiny drop of liquid. It is also considered as a tiny column of liquid that is surrounded by surfaces that have zero shear stress.

A nucleus on the other hand is an assembly between protons and neutrons. The latter is electrically charged whilst the former is positively charged. The number of protons present in an element is very crucial to the qualities of an element.

The main similarities between a nucleus and a liquid droplet are:

1. a nucleus consists of a large amount of neutrons and protons in the same volume as would a liquid which contains large numbers of molecules in the same volume;

2. both the nucleus and the droplet are similar for their homogeneity in electric charge and density;

3. the molecules exert the same amount for forces towards one another as would the nuclear forces in the nucleons.

4. both of them cannot be compressed

5. both molecules and nucleus are can be subject to nuclear fission which simply mean the breaking apart into smaller units (in the case of the nucleus) or the breaking apart into smaller droplets in the case of the liquid molecule.

6. There are two types of phenomena which occurs in both the liquid droplet and the nucleus which are similar to one another. They are:

Evaporation (in the case of the liquid molecule) and reaction emission (in the case of the nucleus). In evaporation, particles are lost, in Atomic transmutation, particles are lost as well.

B)  the forces which determine the stability of droplets are surface tension and gravitation. The smaller the area, the stronger the surface tension available to keep the drops from going out of shape.

Cheers

A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 318 g/mol
b. 204 g/mol
c. 175 g/mol
d. 298 g/mol

Answers

Answer: Option D) 298 g/mol  is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

so

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol  ≈ 298 g/mol

Therefore Option D) 298 g/mol  is the correct answer

A settling tank has an influent rate of 0.6 mgd. It is 12 ft deep and has a surface area of 8000 ft². What is the hydraulic retention time?

Answers

Answer: hydraulic retention time,τ=28.67 hours

Explanation:

The hydraulic retention time  τ (tau),  is given as  The volume of the settling tank(V) divided by the influent flowrate(Q)

τ =V/Q

But Volume is not known  and is given as

Volume =  surface area  x depth of the tank

= 8000 ft² X 12 ft

= 96,000 ft³

Also, the influent flow rate is in mgd ( million gallons per day), we change  it to  ft³/sec so as to be in same unit with the volume in ft³

1 million gallons/day = 1.5472286365101 cubic feet/second

0.6mgd =  1.5472286365101 cubic feet/second  x 0.6

=0.93cubic feet/second

τ =V/Q

96,000 ft³/0.93 ft³/sec

τ=103,225.8 secs

changing to hours

103,225.8 /3600 =28.67 hours

The hydraulic retention time =28.67 hours

An insulated 40 ft^3 rigid tank contains air at 50 psia and 580°R. A valve connected to the tank is now opened, and air is allowed to escape until the pressure inside drops to 25 psia. The air temperature during this process is kept constant by an electric resistance heater placed in the tank. Determine the electrical work done during this process in Btu assuming variable specific heats.

Answers

Answer:

The answer is "[tex]\bold{W_{in} = 645.434573 \ Btu}[/tex]".

Explanation:

Its air enthalpy is obtained only at a given temperature by A-17E.  

The solution from the carbon cycle is acquired:

[tex]\to \Delta U= W_{in}-m_{out}h_{out}=0\\\\\to W_{in} = (m_1 - m_2)h[/tex]

           [tex]=\frac{Vh}{RT}(P_1- P_2)\\\\= \frac{40 \times 138.66}{0.3704\times 580}(50-25) Btu\\\\= \frac{5,546.4}{214.832}(25) Btu\\\\= 25.8173829(25) Btu\\\\ =645.434573 Btu[/tex]

[tex]W_{in} = 645.434573 \ Btu[/tex]

Which of following are coding languages used in controlling a robot? *
A. Scratch
B. B/B--
C. C/C++
D. Robot Z

Answers

Answer:

C/C++

Explanation:

C/C++

the differences between building technology vs architectural technology​

Answers

Architectural technologists are concerned more with the technical and functional elements of the building, including building code, drafting and CAD, conceptual drawings and 3D models They also generally cost less than an architect.

Products exit a combustor at a rate of 100 kg/sec, and the air-fuel ratio is 9. Determine the air flow rate. a. 9 kg/sec b. 90 kg/sec c. 100 kg/sec d. 10 kg/sec

Answers

Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer

Explanation:

Given that;

product of combustor flow rate m = 100 kg/s

air-fuel = 9

Airflow rate = ?

⇒We know that in the combustor, air fuel are mixed and then ignited,

⇒air fuel products are exited at the combustor

let air and fuel be a and b respectively

⇒ a + b = 100 kg/sec ----- let this be equation 1

now

⇒ air / fuel = 9

a / b = 9

a = 9b -----------let this be equation 2

now input a = 9b in equation 1

9b + b = 100 kg/sec

10b = 100 kg/sec

b = 10 kg/sec

we know that

a = 9b

so a = 9 × 10 = 90 kg/sec

Therefore the air flow rate a is 90 kg/sec

A three-phase motor rated 25 hp, 480 V, operates with a power factor of 0.74 lagging and supplies the rated load. The motor efficiency is 96%. Calculate the motor input power, reactive power and current.

Answers

Answer:

the motor input power is 19.42 KW

the Reactive power is 17.65 KVAR

Current is 31.56 A

Explanation:

Given that;

V = 480V

h.p = 25 hp

p.f = 0.74 lagging

n_motor = 96%

so output = 25hp

and we know that;

1hp = 746 watt

watt = hp × 1hp

so output in watt = 25 × 746 = 18650 Watt = 18.65 KW

n_motor = (output / input) × 100

96 = 1865 / Input

96Input = 1865

Input = 1865 / 96

Input = 19.42 KW

Therefore the motor input power is 19.42 KW

P = √( 3 × V × I × cos∅)

19.42 = √( 3 ×480 × I × 0.74)

I = 31.56 A

Therefore Current is 31.56 A

Q = √( 3 × V × I × sin∅)

we know that

cos∅ = 0.74

so ∅ = cos⁻¹(0.74) = 42.26

so we substitute

Q = √( 3 × 480 × 31.56 × sin(42.26))

 = 17.65 KVAR

Therefore the Reactive power is 17.65 KVAR

A vortex tube receives 0.3 m^3 /min of air at 600 kPa and 300 K. The discharge from the cold end of the tube is 0.6 kg/min at 245 K and 100 kPa. The discharge from the hot end is at 325 K and 100 kPa. Determine the irreversibility.

Answers

Answer:

Irreversibility = 5.361 kW

Explanation:

From the given information:

By applying ideal gas equation at entry:

PV =  mRT

600 × 0.3 = m × 0.287 × 300      (where R = 0.287 kJ/kg)

180 = m × 86.1

m = 180/86.1

m = 2.0905 kg/min

At the hot end, using the same ideal gas equation:

PV = mRT

100 × V = 1.4905 × 0.287 × 325

V = 139.026/100

V = 1.3903 m³/ min

This implies that: The total entropy change = Entropy of the universe

So,

[tex]m\bigg [ c_p \ In \dfrac{T_2}{T_o}-R \ In \dfrac{P_2}{P_o} \bigg] + m_2 \bigg [ c_p \ In \dfrac{T_2}{T_o}- R \ In\dfrac{P_2}{P_o} \bigg][/tex]

[tex]= 0.6\bigg [ 1.004 \ In \dfrac{245}{300}-0.287 \ In \dfrac{100}{600} \bigg] +1.4905\bigg [1.004 \ In \dfrac{325}{300}- 0.287\ In\dfrac{100}{600} \bigg][/tex]

= 0.6[-0.2033 + 0.5142] + 1.4905 [0.08036 + 0.5142]

= 1.0727 kJ/min.K

= 0.01787 kw/K

Irreversibility = [tex]T_o [ \Delta S][/tex]

Irreversibility = 300 × 0.01787

Irreversibility = 5.361 kW

How many snaps points does an object have?

Answers

Answer:

what do you mean by that ? snap points ?

A stream leaving a sewage pond (containing 80 mg/L of sewage) moves as a plug with a velocity of 40 m/hr. A concentration of 50 mg/L is measured 5,000 m downstream. What is the 1st order decay rate constant in the stream?

Answers

Answer:Decay rate constant,k  = 0.00376/hr

Explanation:

IsT Order  Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time  t  and  [A]o  is the inital concentration at time 0, and  k  is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

              5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k  = 0.00376/hr

Calculate the LER for the rectangular wing from the previous question if the weight of the glider is 0.0500 Newton’s.

Answers

Answer:

0.2

Explanation:

Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.

Let the span of the rectangular wing be 0.225 m

Let the chord of the rectangular wing be 0.045 m.

Then, the area of any rectangular chord is

A = chord * span

A = 0.045 * 0.225

A = 0.010 m²

And using the weight of the glider given to us from the question, we can find the LER for the wing.

LER = Area / weight.

LER = 0.010 / 0.05

LER = 0.2.

Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2

Please mark brainliest...

Answer: 0.2025

Explanation: I got it correct

People tend to self-disclose to others that are in age, social status, religion, and personality.

Answers

Answer:people tend to do this when they are in a different environment they lose something or just have something going on in their life

Explanation:

Series aiding is a term sometimes used to describe voltage sources of the same polarity in series. If a 5 V and a 9 V source are connected in this manner, what is the total voltage?

Answers

Answer:Total Voltage = 14V

Explanation: it is possible that a circuit  can contain more than one source of electromotive force which can cause flow of current in the same or opposite direction . When the  connection to  voltage sources  allows for current  from the voltage sources to flow in  same direction,it is termed  Series aiding  Thus, the  Total/effective voltage in a series aiding circuit is  computed as the sum of series aiding voltages .

 Here we have the series aiding voltages to be 5V and 9V ,

therefore,

Total Voltage = 5V + 9V

= 14V

Copy bits 3..0 in $s1 to 6..3 in $s2. Bits 6..3 in $s2 are already set to 0. Registers$s0 0..01111$s1 0..0101$s3 0

Answers

Answer:

Following are the solution to this question:

Explanation:

To copy 3.0 bits in 50 dollars or run at 50 dollars, it takes just 3.0 bits as well as other bits but masks, and 50 dollars.  

Instead of shifting the $ 50 by 3 bits to 6...3 bits of [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex] This procedure instead took place at $53 and $50  

AND [tex]\$ \ 50,\$ \ 50,0*0000 000f[/tex], take 3..0  bits

SLL [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex]Shifts the bits to 6..3

O R [tex]\$ \ 53,\$ \ 53,\$ \ 50 ,[/tex] coping to [tex]\$ \ 53[/tex]

4. The instant the ignition switch is turned to the start position,

A. The starter motor starts to rotate before energizing the starter p

O B. Only the pull-in winding is energized.

C. Only the hold-in winding is energized.

D. Both pull-in and hold-in windings are energized.

Answers

Answer:

D. Both pull-in and hold-in windings are energized.

Explanation:

The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.

The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.

A roadway with a rough-asphalt pavement has a cross slope of 2%, a longitudinal slope of 2.5%, a curb height of 8 cm, and a 90-cm-wide concrete gutter. If the flow rate in the gutter is 0.07 m/s, determine the size (W XL, in mm) and interception capacity (m/s) of a reticuline grate that should be used to intercept as much of the flow as possible.
a. Reticuline grate size?
b. Interception capacity?

Answers

Answer:

b

Explanation:

please help me make a lesson plan. the topic is Zigzag line. and heres the format.
A. Objective
B. Subject matter
C. Learning activities.
D. Assessment.
E. Reinforcement​

Answers

Explanation:

D. B. C. A. E. Is this a good idea

Air is compressed by a 30-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as a result of heat transfer to the surrounding medium at 20°C. Determine the rate of entropy change of the air.

Answers

Answer:

-0.1006Kw/K

Explanation:

The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,

ΔS = Q/T

Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

S(air) = - Q/T(air) .......1

Where S.air =

Q = 30-kW

T.air = 298k

Substitute the values into equation 1

S(air) = - 30/298

= -0.1006Kw/K

A fluid has a mass of 5 kg and occupies a volume of 1 m3 at a pressure of 150 kPa. If the internal energy is 25000 kJ/kg, what is the total enthalpy?

Answers

Answer:

155 KJ

Explanation:

The total enthalpy is given by

ΔH=ΔU + PV

Where;

ΔH = enthalpy

ΔU = internal energy = 25000 kJ/kg/ 5 kg = 5000 KJ

P = 150 kPa = 150,000 Pa

V =  1 m3

ΔH=  5000 + (150,000 * 1)

ΔH=  155 KJ

A battery with a nominal voltage of 200-V with a resistance of 10 milliohms to be charged at a constant current of 20 amps from a 3-phase semi-converter with a 220-V (line-to-line) Y-connected 60 Hs supply. Determine:

a. The firing angle of the thyristors for the charging process.
b. The displacement power factor and the supply power factor.

Answers

Answer:

a)  ( ∝ ) = 69.6548

b) supply power factor = 0.6709

  displacement power factor = 0.8208

Explanation:

Given data:

Nominal voltage ( E ) = 200-V

resistance (r) = 10 milliohms

constant current ( I )  = 20 amps

Phase ; 3-phase

semi-converter  with 220-v ( line-to-line ) ,  220√2  ( phase voltage )

frequency ; 60 Hz

a) determine the firing angle of thyristors

Vo = E + I*r

    = 200 + 20*10*10^-3

    = 200.2 v

attached below is the remaining part of the solution

firing angle of thyristors for charging process ( ∝ ) = 69.6548

b) determine displacement power factor and supply power factor

attached below is the detailed solution

Displacement power factor ( Dpf ) = cos ( ∝ /2 ) = 0.8208

displacement power factor = g * Dpf

                                             = 0.81747 * 0.8208 = 0.6709

In a CS amplifier, the resistance of the signal source Rsig = 100 kQ, amplifier input resistance (which is due to the biasing network) Rin = 100kQ, Cgs = 1 pF, Cgd = 0.2 pF, gm = 5 mA/V, ro = 25 kΩ, and RL = 20 kΩ. Determine the expected 3-dB cutoff frequency.

Answers

Answer:

406.140 KHz

Explanation:

Given data:

Rsig = 100 kΩ

Rin = 100kΩ

Cgs = 1 pF,

Cgd = 0.2 pF,  and   etc.

Determine the expected 3-dB cutoff frequency

first find the CM miller capacitance

CM = ( 1 + gm*ro || RL )( Cgd )

     = ( 1 + 5*10^-3 * 25 || 20 ) ( 0.2 )

     = ( 11.311 ) pF

now we apply open time constant method to determine the cutoff frequency

Th = 1 / Fh

hence : Fh = 1 / Th = [tex]\frac{1}{(Rsig +Rin) (Cm + Cgs )}[/tex]

                               = [tex]\frac{1}{( 200*10^3 ) ( 12.311 * 10^{-12} )}[/tex] =  406.140 KHz

What is the amount of pearlite formed during the equilibrium cooling of a 1055 steel from 1000°C to room temperature?

Answers

Answer: 98.5% of pearlite was formed during the equilibrium cooling

Explanation:

First we calculate the fraction of pro-eutectoid phase which forms for equilibrium cooling of the 1085 steel from 1000°C at room temperature;

we know that in 1085 steel, last two digits denotes the carbon percentage

so 1085 steel contains 0.85% carbon.

Now from the diagram, carbon percentage is greater than the eutectoid com[psition

i.e 0.85 > 0.76

it is a hyper eutectoid steel

so

fraction of pro eutectoid phase W_Fe₃C = (0.85 - 0.76) / ( 6.7 - 0.76)

= 0.09 / 5.94 = 0.015 = 1.5%

Now, the amount of pearlite formed during the equilibrium cooling of the 1055 steel from 1000°C to room temperature will be;

pearlite (C') = (1 - W_Fe₃C)

= 1 - 0.015

= 0.985 = 98.5%

Therefore 98.5% of pearlite was formed during the equilibrium cooling

Help this is very hard and I don't get it

Answers

Answer:

yes it is very hard you should find a reccomended doctor to aid in your situation. But in the meantime how about you give me that lil brainliest thingy :p

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