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zloy xaker [14]

3 years ago

12

Conduction is thermal energy transfer by:_______.

1 answer:

adell [148]3 years ago

5 0

Answer:

a. molecular interactions.

Explanation:

Conduction is thermal energy transfer by molecular interactions. Therefore, conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Some examples of conductors include metal, steel, aluminum, copper, graphite, etc.

Hence, conduction is thermal energy transfer as a result of the movement of electrons and collision between the molecules of an object.

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A structural element for a new bridge is designed for a constant load of 1000 psi. Its mean resistance is 1200 psi and the proba

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iii) 3.16 psi

iv) 0.0083

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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