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2 years ago

5

The motor branch circuit short circuit and ground-fault protective device shall be capable of carrying the ______ current of the

motor.

1 answer:

Varvara68 [4.7K]2 years ago

3 0

Answer:

Starting current

Explanation:

The motor branch circuit short circuit and ground-fault protective device shall be capable of carrying the <u>Starting current</u> of the motor.

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1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per

KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0

3 years ago

Why the velocity potential Φ(x,y,z,t) exists only for irrotational flow

Black_prince [1.1K]

Answer:

$\omega_y,\omega_x,\omega_Z$ all are zero.

Explanation:

We know that if flow is possible then it will satisfy the below equation

$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0$

Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.

And velocity potential function $\phi$ given as follows

$u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}$

Rotationality of fluid is given by $\omega$

$\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z$

$\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x$

$\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y$

So now putting value in the above equations ,we will find

$\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},$

$\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}$

So $\omega_y$ =0

Like this all $\omega_y,\omega_x,\omega_Z$ all are zero.

That is why velocity potential flow is irroational flow.

5 0

4 years ago

Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as a

Goshia [24]

Answer:

a. Wa = 73.14 Btu/lbm

b. Sgen = 0.05042 Btu/lbm °R

c. Isentropic efficiency is 70.76%

d. Minimum specific work for compressor W = -146.2698 Btu/lbm [It is negative because work is being done on the compressor]

Explanation:

Complete question is as follows;

Air initially at 120 psia and 500oF is expanded by an adiabatic turbine to 15 psia and 200oF. Assuming air can be treated as an ideal gas and has variable specific heat.

a) Determine the specific work output of the actual turbine (Btu/lbm).

b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R).

c) Determine the isentropic efficiency of this turbine (%).

d) Suppose the turbine now operates as an ideal compressor (reversible and adiabatic) where the initial pressure is 15 psia, the initial temperature is 200 oF, and the ideal exit state is 120 psia. What is the minimum specific work the compressor will be required to operate (Btu/lbm)?

solution;

Please check attachment for complete solution and step by step explanation

8 0

3 years ago

If a torque of M = 300 N⋅m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD t

Licemer1 [7]

Answer:

<em>866.1 N</em>

Explanation:

The torque on the flywheel = 300 N-m

The force from the hydraulic cylinder will generate a moment on CA about point A.

The part of this moment that will be at point B about A must be proportional to the torque on the cylinder which is 300 N-m

we know that moment = F x d

where F is the force, and

d is the perpendicular distance from the turning point = 1 m

Equating, we have

300 = F x 1

F = 300 N this is the frictional force that stops the flywheel

From F = μN

where F is the frictional force

μ is the coefficient of static friction = 0.4

N is the normal force from the hydraulic cylinder

substituting, we have

300 = 0.4 x N

N = 300/0.4 = 750 N

This normal force calculated is perpendicular to CA. This actual force, is at 30° from the horizontal. To get the force from the hydraulic cylinder R, we use the relationship

N = R sin (90 - 30)

750 = R sin 60°

750 = 0.866R

R = 750/0.866 = <em>866.1 N</em>

3 0

3 years ago

If the reading of mercury manometer was 728 mmHg, what is the reading for another liquid such as water in mH20 units?

vodka [1.7K]

Answer:

mH275 units

Explanation:

that was true

4 0

3 years ago