can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and angle. thanks
can anyone help me with this please.i have the current - 1

Answers

Answer 1
Answer:

Answer:

  • branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
  • branch 2: i = 21.466∠63.435°; pf = 0.447 leading
  • total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

  X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915*10^(-3)\approx j4.99984\,\Omega

The capacitive reactance is ...

  X_C=(1)/(j\omega C)=(-j)/(100\pi\cdot 318.31*10^(-6)F)\approx -j10.00000\,\Omega

Branch 1

The impedance of branch 1 is ...

  Z1 = 8 +j4.99984 Ω

so the current is ...

  I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

Branch 2

The impedance of branch 2 is ...

  Z2 = 5 -j10 Ω

so the current is ...

  I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

Total current

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

  It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

  It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

__

The phasor diagram of the currents is attached.

_____

Additional comment

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.


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B. Either vertically or horizontally polarized antennas may be used for transmission or reception

C. FM voice is unusable

D. Both the transmitting and receiving antennas must be of the same polarization

Answers

Answer:

B

Explanation:

The fact that skip signals refracted from the ionosphere are elliptically polarized is a result of either vertically or horizontally polarized antennas may be used for transmission or reception.

The Rappahannock River near Warrenton, VA, has a flow rate of 3.00 m3/s. Tin Pot Run (a pristine stream) discharges into the Rappahannock at a flow rate of 0.05 m3/s. To study mixing of the stream and river, a conserva- tive tracer is to be added to Tin Pot Run. If the instruments that can mea- sure the tracer can detect a concentration of 1.0 mg/L, what minimum concentration must be achieved in Tin Pot Run so that 1.0 mg/L of tracer can be measured after the river and stream mix? Assume that the 1.0 mg/L of tracer is to be measured after complete mixing of the stream and Rappa- hannock has been achieved and that no tracer is in Tin Pot Run or the Rap- pahannock above the point where the two streams mix. What mass rate (kg/d) of tracer must be added to Tin Pot Run?

Answers

Find the given attachments for complete explanation

A power plant burns natural gas to supply heat to a heat engine which rejects heat to the adjacent river. The power plant produces 800 MW of electrical power and has a thermal efficiency of 38%. Determine the heat transfer rates from the natural gas and to the river, in MW.

Answers

A. The heat transfer rate from natural gas is 2105.26 MW

B. The heat transfer rate to river is 1305.26 MW

Efficiency formula

Efficiency = (power output / power input) × 100

A. How to determine the heat transfer from natural gas

  • Efficiency = 38%
  • Power output = 800 MW
  • Power input =?

Power input = Power input / efficiency

Power input = 800 / 38%

Power input = 800 / 0.38

Power input = 2105.26 MW

Thus, the heat transfer from natural gas is 2105.26 MW

B. How to determine the heat transfer to the river

  • Total heat = 2105.26 MW
  • Heat used by plant = 800 MW
  • Heat to the river =?

Heat to the river = 2105.26 – 800

Heat to the river = 1305.26 MW

Learn more about efficiency:

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Answer:

heat transfer from natural gas is 2105.26 MW

heat transfer to river is 1305.26 MW

Explanation:

given data

power output Wn = 800 MW

efficiency = 38%

solution

we know that efficiency is express as

\eta = (Wn)/(Qin)    ......................1

put here value we get

38% = (800)/(Qin)  

Qin  = 2105.26 MW

so heat supply is 2105.26

so we can say

Wn = Qin - Qout

800 = 2105.26 - Qout

Qout = 2105.26 - 800

Qout = 1305.26 MW

so heat transfer from natural gas is 2105.26 MW

and heat transfer to river is 1305.26 MW

A 24.3-foot long pipe use to carry solvent through a chemical plant is made of two layers. The inner layer is a one-inch schedule 30 stainless steel (AISI 304) pipe. An outer coating is made of plain carbon steel and is 0.075 inches thick. The hot solvent stream can be assumed to have a constant temperature of 180.0oF as it passes through the pipe. On the outside of the pipe is air at 85.0oF. Given the solvent has a convective heat transfer coefficient of 275 Btu/h-ft^2-oF, while the air has a convective heat transfer coefficient of 140 Btu/h-ft^2-oF.A. Determine UA and q for heat transfer in this system. (Since the problem is given in English units, your answer should also be in English units)
B. Determine the overall heat transfer coefficients, U; and U., for the pipe.
C. Would your answer change if the two materials were swapped so that the inside material were carbon steel and the outside material of the pipe were made of AISI 304 stainless steel? If so, calculate new values for UA and q. If not, explain why your answer would not change. (Here, assume the dimensions of the inner material (now plain carbon steel) match those of the AISI 304 schedule 30 stainless steel from part A, and the stainless steel is 0.075 inches thick on the outside.)

Answers

Sorry man i dont know the answer to this one

When looking at a chain of processes with a low yield (high defective rate), what is a good place to start investigating the source of variation?

Answers

The most upstream process with issues would be a good location to start exploring the cause of the variance.

Chain processes

A manufacturing technique would be a specific procedure for generating a commodity.

Throughout manufacturing, a six sigma process has been utilized just to generate a product throughout which 99.99966 percent among all possibilities to produce certain aspects of a part seem to be likely toward being defect-free.

Thus the response above is correct.

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brainly.com/question/25646504

Answer: The furthest upstream process that has problems.

A process in manufacturing is a particular method used for producing a product.

A six sigma process is used in processing to produce a product that is 99.99966% of all opportunities to produce some feature of a part are statistically expected to be free of defects.

According to the rules of the six sigma process, when there's a defect, the best thing to do is investigate the furthest upstream process that has problems.

An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity kA while the other is fabricated from the material whose thermal conductivity kB is desired. Both rods are attached at one end to a heat source of fixed temperature Tb, are exposed to a fluid of temperature [infinity] T[infinity], and are instrumented with thermocouples to measure the temperature at a fixed distance x1 from the heat source. If the standard material is aluminum, with kA= 200 W/m·K, and measurements reveal values of TA= 75°C and TB= 70°C at x1 for Tb= 100°C and [infinity] T[infinity]= 25°C, what is the thermal conductivity kB of the test material?

Answers

Answer: the thermal conductivity of the second material is 125.9 W/m.k

Explanation:

Given that;

The two rods could be approximated as a fins of infinite length.

TA = 75°C,    θA = (TA - T∞) = 75 - 25 = 50°C

TB = 70°C     θB = (TB - T∞) = 70 - 25 = 45°C

Tb = 100°C    θb = (Tb - T∞) = (100 - 25) = 75°C

T∞ = 25°C

KA = 200 W/m · K,   KB = ?

Now

The temperature distribution for the infinite fins are given by

θ/θb = e^(-mx)

θA/θb= e^-√(hp/A.kA) x 1  --------------1

θB/θb = e^-√(hp/A.kB) x 1---------------2

next we  take the natural logof both sides,  

ln(θA/θb) = -√(hp/A.kA) x 1 ------------3

In(θB/θb) = -√(hp/A.kB) x 1 ------------4

now we divide 3 by 4

[ ln(θA/θb) /in(θB/θb)] = √(KB/KA)

we substitute

 [ In(50/75) /In(45/75)] = √(KB/200)

In(0.6666) / In(0.6) = √KB / √200

-0.4055/-0.5108 = √KB / √200

0.7938 = √KB / 14.14

√KB = 11.22

KB = 125.9 W/m.k

So the thermal conductivity of the second material is 125.9 W/m.k