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Gelneren [198K]
3 years ago
5

On a piece of paper, sketch the x-y stress state and the properly oriented principal stress state. Use the resulting sketch to a

nswer the questions in the subsequent parts of this GO exercise.(a) The angle -4.45° locates a principal stress plane. Give the stress value including sign if any.(b) The angle 85.6° locates the other in-plane principal stress plane. Give the stress value including sign if any.(c) What stress or stresses occur on a plane at 40.6°? Give the stress value or values including sign if any.

Engineering
1 answer:
stealth61 [152]3 years ago
6 0

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step and very detailed solution of the given problem

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A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle
Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

X_{s} =1.23(\frac{V_{line}^{2}  }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms

the network voltage phase is

V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV

the power transmitted is equal to:

|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV

the line induced voltage is

|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV

7 0
3 years ago
A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The
AleksandrR [38]

Answer:

The work furnished by the compressor is 69.77kJ/s

The minimum work required for the state to change is 55.26kW

Explanation:

The explanation to these solution is on the first, second , third and fourth uploaded image respectively

8 0
3 years ago
Which system of linear inequalities is represented by the graph? y > x – 2 and y x + 1 y x + 1 y > x – 2 and y < x + 1
KatRina [158]

Answer:

The graph representing the linear inequalities is attached below.

Explanation:

The inequalities given are :

y>x-2   and y<x+1

For tables for values of x and y and get coordinates to plot for both equation.

In the first equation;

y>x-2

y=x-2

y-x = -2

The table will be :

x    y

-2  -4

-1    -3

0     -2

1      -1

2      0

The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)

Use a dotted line and shade the part right hand side of the line.

Do the same for the second inequality equation and plot then shade the part satisfying the inequality.

The graph attached shows results.

5 0
3 years ago
Read 2 more answers
A circuit contains a resistor, an inductor, and a capacitor. When an AC voltage is applied across the circuit, the impedance of
katen-ka-za [31]

Answer:

The impedance of  the circuit depends on the angular frequency of the voltage source.

Explanation:

  • In a electric circuit, the magnitude of  the impedance, is given by the following expression:

       Z = \sqrt{R^{2} + (Xl-Xc)^{2} (1)

        where R = Resistance

                   Xl = Inductive reactance = ω*L

                   Xc = Capacitive Reactance = 1/ωC

        and  ω = angular frequency of the voltage source.

  • So, it can be seen that the impedance depends on the value of the constants R,L and C, and on the angular frequency ω.

3 0
3 years ago
A community plans to build a facility to convert solar radiation to electrical power. The community requires 2.80 MW of power, a
Greeley [361]

Given Information:  

Output power required = Pout = 2.80 MW

Efficiency = η = 30%

Intensity = I = 1180 W/m²

Required Information:  

Effective area = A = ?

Answer:  

Effective area = A = 7.907x10³ m²

Step-by-step explanation:  

A community plans to build a facility to convert solar power into electrical power and this facility has an efficiency of 30%

As we know efficiency is given by

η = Pout/Pin

Where Pout is the output power and Pin is the input power.

Pin = Pout/η

Pin = 2.80x10⁶/0.30

Pin = 9.33x10⁶ W

The effective area of a perfectly absorbing surface used in such an installation can be found using

A = Pin/I

Where I is the in Intensity of the sunlight in W/m²

A = 9.33x10⁶/1180

A = 7.907x10³ m²

Therefore, the effective area of the absorbing surface would be 7.907x10³ m².

6 0
3 years ago
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