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2 years ago

The maximum voltage in a cooling system should not exceed ______ volts.

Engineering

1 answer:

Natali [406]2 years ago

7 0

Answer:2%

When the voltages between the three phases are not equal, the current increases dramatically in the motor winding's, and if allowed to continue, the motor will be damaged.

Explanation:

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A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para

mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c) 54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

$V_p=\frac{3810.5}{\sqrt{3} }=2200 V$

The complex power S is given as:

$S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA$

$where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA$

The line current I is given as:

$I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A$

The phase voltage at the sending end is:

$V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV$

The magnitude of the line voltage at the source end of the line ( $V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V$

b) The Total real and reactive power loss in the line is:

$S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000$

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

$S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000$

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0

4 years ago

Which of the following has special properties that allow forces and pressure to be distributed evenly?

Thepotemich [5.8K]

Answer:

Fluids

Explanation:

Fluids has special properties that allow forces and pressure to be distributed evenly within them.

Fluids are gases and liquids whose intermolecular forces of attraction are generally weak or non-existence.
Therefore, when pressure is applied to them, it permeates evenly on all parts.
Their ability to tend to randomness makes liquids and gases very viable for distributing pressure.

7 0

3 years ago

A chemical process stream enters a shell-and-tube exchanger at a temperature of 200.0°Fand does two passes on the shell side, ex

ikadub [295]

Answer:

a) Decrease

b) Decrease

c) Decrease

d) Decrease

Explanation:

Ti= 200°F ,

Te = 170°F

Area of heat exchanger = $\pi *(2 )* 10$ = 20π

A) when The flow rate of the cooling fluid is increased

Temperature of process stream will decrease and this is because the tube side heat transfer coefficient will increase and this will increase the rate of heat transfer thereby decreasing the temperature of the process stream.

B) when There are 200 tubes that are 1.0-in. OD and 20.0ft long

The temperature of the process stream will decrease and this is because the heat transfer coefficient will increase likewise the heat transfer rate

C) When The number of shell passes is doubled

This will cause an increase in the overall length of the shell, an increase in velocity of constant volumetric flowrate, hence the Temperature of the process steam will decrease as well

D) When The tube material is changed to copper.

Due to the high thermal conductivity of copper when compared to steel , switching to copper will cause a decrease in the temperature of the process steam

3 0

3 years ago

A pipe leads from a storage tank on the roof of a building to the ground floor. The absolute pressure of the water in the storag

Fudgin [204]

Answer:

6.4 m/s

Explanation:

From the equation of continuity

A1V1=A2V2 where A1 and V1 are area and velocity of inlet respectively while A2 and V2 are the area and velocity of outlet respectively

$A1=\pi (r1)^{2}$

$A2=\pi (r2)^{2}$

where r1 and r2 are radius of inlet and outlet respectively

v1 is given as 1.6 m/s

Therefore

$\pi (0.01)^{2}\times 1.6 = \pi (0.005)^{2}v2$

$V2=\frac {\pi (0.01)^{2}\times 1.6}{\pi (0.005)^{2}}=6.4 m/s$

3 0

4 years ago

The high-pressure air system at OSU's Aerospace Research Center is fed by a set of two cylindrical tanks. Each tank has an outer

lana66690 [7]

Answer:

179000 lb

Explanation:

The supports must be able to hold the weight of the tank and the contents. Since tanks are pressure tested with water, and the supports cannot fail during testing, we disregard the air and will consider the weight of water.

The specific weight of water is ρw = 62.4 lbf/ft^3

These tanks are thin walled because

D / t = 4.6 / 0.1 = 46 > 10

To calculate the volume of steel we can approximate it by multiplying the total surface area by the thickness:

A = 2 * π/4 * D^2 + π * D * h

The steel volume is:

V = A * t

The specific weight is

ρ = δ * g

ρs = 499 lbm/ft^3 * 1 lbf/lbm = 499 lbf/ft^3

The weight of the steel tank is:

Ws = ρs * V

Ws = ρs * A * t

Ws = ρs * (2 * π/4 * D^2 + π * D * h) * t

Ws = 499 * (π/2 * 4.6^2 + π * 4.6 * 50) * 0.1 = 37700 lb

The weight of water can be approximated with the volume of the tank:

Vw = π/4 * D^2 * h

Ww = ρw * π/4 * D^2 * h

Ww = 62.4 * π/4 * 4.6^2 * 50 = 51800 lb

Wt = Ws + Ww = 37700 + 51800 = 89500 lb

Assuming the support holds both tanks

2 * 89500 = 179000 lb

The support must be able to carry 179000 lb

3 0

4 years ago