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ra1l [238]
3 years ago
14

The value 100 MW is equivalent to (a) 100×10^6 w (b) 100 x 10^-6 w (c) 100 x 10^-3 w (d) 100 x 10^3 w

Engineering
1 answer:
insens350 [35]3 years ago
5 0

Answer:

(a) 100×10⁶ W

Explanation:

Power is defined as the rate of doing work. Thus,

Power=\frac{Work\ done}{Time\ taken}

<u>The S.I. unit of Power is watt which is J/s.</u>

Mega is a unit prefix which denotes a factor of 1 million (10⁶). It is denoted by 'M'.

Thus,

<u>1 MW=10⁶ W</u>

The Conversion of 100 MW into W is shown below:

<u>100 MW = 100×10⁶ W</u>

The option that correct resembles is option (a)

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A

Explanation:

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2 years ago
It is true about Metals and alloys: a)-They are good electrical and thermal conductors b)-They can be used as semi-conductors c)
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Answer:

(d) a and c are correct

Explanation:

METALS : Metal are those materials which has very high ductility, high modulus of elasticity, good thermal and electrical conductivity

for example : iron, gold ,silver, copper

ALLOYS: Alloys are those materials which are made up of combining of two or more than two metals these also have good thermal and electrical conductivity and me liable property

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so from above discussion it is clear that option (d) will be the correct option

8 0
3 years ago
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Entropy change is evaluated using Eq. 6.2a based on an internally reversible process. Can the entropy change between two states
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Answer:

YES

Explanation:

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4 0
3 years ago
The state of plane strain on an element is:
balu736 [363]

Answer:

a. ε₁=-0.000317

   ε₂=0.000017

θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2}  \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2}  \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6}  \pm 1.67 \times 10^{-4}

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5

θ= -13.28° and  76.72°

To determine the direction of ε₁ and ε₂

\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2}  + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta  + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2}  + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56)  + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain

\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}

=3.335 *10^-4

\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )

ε(avg) =150 *10^-6

orientation of γmax

tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}

θ = 31.71 or -58.29

To determine the direction of γmax

\gamma _{x'y' }=  - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta  + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }=  - \frac{-300*10^{-6} - \ 0}{2} sin(63.42)  + \frac{150*10^{-6}}{2}cos(63.42)

= 1.67 *10^-4

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Answer:

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Explanation:

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These type of pressure valve are used to pour fluid into the container at very low or no pressure.

b. Safety valve:

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c. Pressure Reducing Valve:

These are basically used for the control of the pressure in downstream not exceeding the design limits.

d. Pressure Relief Valves:

These are basically used to limit and regulate the pressure of any system.

e. Counter Balance Valve:

These are used to develop pressure in the reverse direction at the actuator's return line in order to keep the load under control.

f. Sequence Valve:

These are used to maintain sequence or order in the operations of two parts or branches.

8 0
3 years ago
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