At full load, a commercially available 100hp, three phase induction motor operates at an efficiency of 97% and a power factor of 0.88 lag. The motor is supplied from a three-phase outlet with a line voltage rating of 208V.a. What is the magnitude of the line current drawn from the 208 V outlet? (1 hp = 746 W.) b. Calculate the reactive power supplied to the motor.

Answers

Answer 1
Answer:

Answer:

I = Line Current = 242.58 A

Q = Reactive Power = 41.5 kVAr

Explanation:

Firstly, converting 100 hp to kW.

Since, 1 hp = 0.746 kW,

100 hp = 0.746 kW x 100

100 hp = 74.6 kW

The power of a three phase induction motor can be given as:

P_(in)  = √(3) VI Cos\alpha\n

where,

P in = Input Power required by the motor

V = Line Voltage

I = Line Current

Cosα = Power Factor

Now, calculating Pin:

efficiency = \frac{{P_(out)} }{P_(in) }\n0.97 = (74.6)/(P_(in) ) \nP_(in) = (74.6)/(0.97)\n  P_(in) = 76.9 kW

a) Calculating the line current:

P_(in) = √(3)VICos\alpha   \n76.9 * 1000= √(3)*208*I*0.88\nI = (76.9*1000)/(√(3)*208*0.88 )\nI =   242.58 A

b) Calculating Reactive Power:

The reactive power can be calculated as:

Q = P tanα

where,

Q = Reactive power

P = Active Power

α = power factor angle

Since,

Cos\alpha =0.88\n\alpha =Cos^(-1)(0.88)\n\alpha=28.36

Therefore,

Q = 76.9 * tan (28.36)\nQ = 76.9 * (0.5397)\nQ = 41. 5 kVAr


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When considering free energy change, biochemists usually define a standard state, the biochemical standard state, which is modified from the chemical standard state to fit biochemical applications. Determine which of the phrases describe the biochemical standard state, the chemical standard state, or both.

Answers

Answer:

Maximum work under this condition (∆G) = Maximum work under Standard Condition (∆G°) + Activities defining this condition

Explanation:

In this equation, the term DGo provides us with a value for the maximal work we could obtain from the reaction starting with all reactants and products in their standard states, and going to equilibrium. The term DG' provides us with a value for the maximal work we could obtain under the conditions defined by the activities in the logarithmic term. The logarithmic term can be seen as modifying the value under standard conditions to account for the actual conditions. In describing the work available in metabolic processes, we are concerned with the actual conditions in the reaction medium (whether that is a test-tube, or the cell cytoplasm); the important term is therefore DG'. If we measure the actual activities (in practice, we make do with concentrations), and look up a value for DGo in a reference book, we can calculate DG' from the above equation.

Values for DGo provide a useful indication through which we can compare the relative work potential from different processes, because they refer to a standard set of conditions.

Therefore both phrases describe the Biochemical and Chemical Standard State

Although the viscoelastic response of a polymer can be very complex (time-dependent stress cycling for instance), two special loading scenarios are fairly simple to describe mathematically. _____________refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).
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Answers

Answer:

Viscoelastic stress relaxation

Viscoelastic creep

Explanation:

  • Viscoelastic stress refers to the reduction of tensile stress with the relaxation time that occurs when the body is kept at a certain length under tensile stress. The purpose of this study was to demonstrate viscoelastic stress relaxation
  • Therefore, when the viscous tension stress occurs, the pressure decreases due to the steady-state stress of the phases.
  • When under constant pressure, the viscoelastic material experiences a time-dependent increase in pressure. This phenomenon is called viscoelastic creep.
  • Therefore the phase constant pressure decreases when there is relaxation in the viscoelastic stress

Air is contained in a cylinder device fitted with a piston-cylinder. The piston initially rests on a set of stops, and a pressure of 300 kPa is required to move the piston. Initially, the air is at 100 kPa and 27°C and occupies a volume of 0.4 m^3. A) Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K. Assume air has constant specific heats evaluated at 300 K.

Answers

Answer:

The amount of heat transferred to the air is 340.24 kJ

Explanation:

From P-V diagram,

Initial temperature T1 = 27°C

Initial pressure P1 = 100 kPa

final pressure P3 = P2 = 300 kPa

volume at point 2, V2 = V1 = 0.4 m³

final temperature T2 = T3 = 1200 K

To determine the final pressure V3, use ideal gas equation

PV = mRT

Where R is the specific gas constant = 0.2870 KPa m³ kg K

But,

from initial condition, mass m = PV/RT

m = (P1*V1)/R*T1

T1 = 27+273 = 300K

m = (100*0.4)/(0.2870*300) = 0.4646 kg

Then;

Final volume V3 = mRT3/P3

V3 = (0.4646*0.2870*1200)/300

V3 = 0.5334 m³

Total work done W is determined where there is volume change which from point 2 to 3.

W = P3*(V3-V2)

W = 300*(0.5334-0.4) = 40.02 kJ

To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K

∆U = m*Cv*(T2-T1)

∆U = 0.4646*0.718(1200-300)

∆U = 300.22 kJ

The heat transfer Q = W + ∆U

Q = 40.02 + 300.22 = 340.24 kJ

Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ

The attached file shows the Pressure - Volume relationship (P -V graph)

A 15. μF capacitor is connected to a 50. V battery and becomes fully charged. The battery is removed and a slab of dielectric that completely fills the space between the plates is inserted. If the dielectric has a dielectric constant of 5.0, what is the voltage across the capacitor's plates after the slab is inserted?

Answers

Answer:

voltage across the capacitor's is 75 μF

Explanation:

given data

capacitor = 15 μF

voltage = 50V

dielectric constant k = 5

to find out

voltage across the capacitor's

solution

we will find here voltage across the capacitor's  by this formula

voltage across the capacitor's  = kC_(o)

here k is 5 and C_(o) = 15

put these value we get

voltage across the capacitor's  = kC_(o)

voltage across the capacitor's  = 5 ( 15 )  = 75 μF

so voltage across the capacitor's is 75 μF

Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are measured to be 300 kPa, 1 L; 290 kPa, 1.1 L; 270 kPa, 1.2 L; 250 kPa, 1.4 L; 220 kPa, 1.7 L; and 200 kPa, 2 L

Answers

By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.

How to determine the boundary work done by a gas during an expansion process

A process is a consecution of states of a system. The boundary work (W), in kilojoules, is the work done by the system on surroundings and in a P-Vdiagram this kind of work is equal to the area below the curve, which can be approximated by Riemann sums:

W = \sum\limits_(i=1)^(n-1) p_(i)\cdot (V_(i+1)-V_(i)) + (1)/(2)\sum\limits_(i=1)^(n-1) (p_(i+1)-p_(i))\cdot (V_(i+1)-V_(i))     (1)

Where:

  • p - Pressure, in kilopascals.
  • V - Volume, in cubic meters.

W = (1)/(2) \sum\limits_(i=1)^(n-1) (p_(i+1)+p_(i))\cdot (V_(i+1)-V_(i))

Now we proceed to calculate the boundary work:

W = 0.5 · [(300 kPa + 290 kPa) · (1.1 × 10⁻³ m³ - 1 × 10⁻³ m³) + (270 kPa + 290 kPa) · (1.2 × 10⁻³ m³ - 1.1 × 10⁻³ m³) + (250 kPa + 270 kPa) · (1.4 × 10⁻³ m³ - 1.2 × 10⁻³ m³) + (220 kPa + 250 kPa) · (1.7 × 10⁻³ m³ - 1.4 × 10⁻³ m³) + (200 kPa + 220 kPa) · (2 × 10⁻³ m³ - 1.7 × 10⁻³ m³)]

W = 0.243 kJ

By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.

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Answer:

attached below

Explanation:

Consider the products you use and the activities you perform on a daily basis. Describe three examples that use both SI units and customary units for measurement.

Answers

Three activities that I can do on a daily basis that involve both metric units (SI units) and customary units are: measuring the length of a door with a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that calls for one teaspoon (customary unit) of baking soda, which can also be converted to four grams (SI unit); and weighing myself on a weighing scale, which can be measured in pound and kilogram (metric unit).

Answer: Three examples of activities that I can perform on a daily basis that involves both metric units (SI units) and customary units include: measuring the length of a door using a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that requires one teaspoon (customary unit) of baking soda, which could also be converted into four grams (SI unit); weighing myself on a weighing scale, which can be measured by pounds (customary unit) or kilograms (metric unit).

Explanation:I big brain :) (Not Really I Just Wanted To Help) I hope this helped! ;)