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3 years ago

A 1000 KVA three phase transformer has a secondary voltage of 208/120. What is the secondary full load amperage?

Engineering

1 answer:

IceJOKER [234]3 years ago

4 0

Answer:

The three phase full load secondary amperage is 2775.7 A

Explanation:

Following data is given,

S = Apparent Power = 1000 kVA

No. of phases = 3

Secondary Voltage: 208 V/120 V (Here 208 V is three phase voltage and 120 V is single phase voltage)

Since,

$V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\$

$V_{1ph} = 120 V$

The formula for apparent power in three phase system is given as:

$S = \sqrt{3} VI$

Where:

S = Apparent Power

V = Line Voltage

I = Line Current

In order to calculate the Current on Secondary Side, substituting values in above formula,

$1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A$

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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

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

3 years ago

assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2

amm1812

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

4 0

3 years ago

Intravenous infusions are usually driven by gravity by hanging the bottle at a sufficient height to counteract the blood pressur

Bingel [31]

Answer:

(a) BP = 11.99 KPa

(b) h = 2 m

Explanation:

(a)

Since, the fluid pressure and blood pressure balance each other. Therefore:

BP = ρgh

where,

BP = Blood Pressure

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = 1.2 m

Therefore,

BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)

BP = 11995.2 Pa = 11.99 KPa

(b)

Again using the equation:

P = ρgh

with data:

P = Gauge Pressure = 20 KPa = 20000 Pa

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = ?

Therefore,

20000 Pa = (1020 kg/m³)(9.8 m/s²)h

h = 2 m

7 0

3 years ago

Harlin is designing a new car engine that does not create pollution. Which technological design factor is probably the most

kari74 [83]

Answer: environmental impact

Explanation:

From the question, we are informed that Harlin is designing a new car engine that does not create pollution.

The technological design factor which is probably the most important for the design is the impact on the environment.

Pollution has a negative effect on the environment a d doing this shows that emphasis is placed on how the environmental issue of pollution can be tackled.

8 0

3 years ago