Which of the following positions would be responsible for attaching I-beams to a crane?

The line voltage of a balanced three-phase transmission line is 4200 V rms. The transmission line has an impedance of 4 6 Z j l

balu736 [363]

Answer:

1.23MVA, 226.74kW, 5.16kV

Explanation:

Parameters Given

line impedance, Zl = (4 + j6) ohms per phase

load voltage, Vl = 4200V

received complex power, S = 1 × 10⁶VA

power factor, cosФ = 0.75

Ф = 41.41°

sinФ = 0.66

Solution:

S = √3 * Vl * I (that is √3 × line voltage × line current)

1 × 10⁶ = √3 × 4200 × I

I = 137.46A

Vl = 4200∠0

I = 137.46∠- 41.41° lagging

source voltage, Vs = load voltage, Vl + voltage drop along the line, Vd

Vd = Zl * I where ( Zl = 4 + j6, = √(4² + 6²)∠tan⁻¹(6/4), = 7.21∠56.31° )

Vd = 7.21∠56.31° × 137.46∠- 41.41°

= 991.22∠14.9°

Vs = Vl + Vd

= 4200∠0° + 991.22∠14.9°

= 4200(cos 0° + j sin 0°) + 991.22(cos 14.9° + jsin 14.9°)

= 4200 + 957.68 + j254.88

= 5157.68 + j254.88

or

= 5163.97∠2.83° V (line voltage at the sending end of the transmission line)

Sending end current, I = 137.46∠-41.41 A

(a) Complex power = √3 × Vs × I

= √3 × 5163.97∠2.83° ×137.46∠-41.41

= 1229477.76∠-38.58°VA

= 1.23∠-38.58MVA

complex power = 1.23MVA

(b) power loss in the three phase line, Pl = 3 × square of line current, I × line impedance, Rl

Pl =3 × I² × Rl where Zl = R + j X = 4 + j6 hence R = 4

= 3 × 137.46² × 4

= 226743.02W

= 226.74kW

(c) from the above, the line voltage at the sending end of the transmission line is = 5163.97V

= 5.16kW

4 0

3 years ago

PLEASE ANSWER THIS DIAL CALIPER

Molodets [167]

Answer:

1) 2.365cm

2)4.443cm

3)2.515cm

4)0.271cm

8 0

3 years ago

A pump is used to transport water from a reservoir at one elevation to another reservoir at a higher elevation. If the elevation

erastova [34]

Answer:b

Explanation:

We know power delivered by Pump is

$P=\rho \times Q\times g\times \Delta H$

where $P\rho$ =Density of fluid

$Q$ =Flow rate

$g$ =acceleration due to gravity

$\Delta H$ =Change in Elevation

If $\Delta H$ is increased by 4 time then

$P'=\rho \times Q\times g\times (4\Delta H)$

$P'=4 P$

So power increases by four times.

4 0

3 years ago

When you are configuring data deduplication, you must choose a usage type for the volume you are configuring. Which of the follo

Finger [1]

Answer:

1. General purpose file server.

2. Virtual Desktop Infrastructure Server.

3. Virtualized Backup Server.

All except : Database server

Explanation:

As a simple definition, we can tell, data deduplication is an elimination of redundant data in data set and storing only one copy of the same data. It is done by identifying double byte patterns through data analysis, removing double data and replacing it with reference pointed to stored, single piece of data.

6 0

3 years ago

A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem

lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is $\Delta h=9.975\,m$

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

$\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2} \Delta v^2 =0$

( $\rho$ stands here for density, $h$ for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

$\frac{\Delta P}{\rho}+g\, \Delta h =0$

$\Delta P= -g\, \rho\, \Delta h$

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

$\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\$

We insert that into our last equation and get:

$\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m$

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

8 0

4 years ago