Answer:
overcurrent
Explanation:
It is any electrical current in excess of the nominal value indicated in the protection device, in the electrical equipment or in the current carrying capacity of a conductor. The overcurrent can be caused by an overload, a short circuit or a ground fault.
The overcurrent raises the operating temperature in the different elements of the electrical installation where this presentation is.
An overcurrent can be an overload or electric shock current.
<u>Overload</u>: the overload current is an excessive current in relation to the nominal operating current. It occurs in drivers and other components of a distribution system. Overloads are in most cases, more frequent between a range of one to six times the nominal current level. They are caused by temporary increases in current and occur when the motors start or when the transformers are energized.
Electric shock: as the name implies, a electric shock current is one that flows out of the normal conduction pathways. Electric shock or fault currents can be hundreds of times greater than the nominal operating current.
Answer:
Answer to this question is option D i.e. unit price.
Explanation:
The unit price of the item can be understood as the price of a single product or one single commodity which forms a part of a group of items. When only one unit is to be sold then here comes the importance of 'unit price.' This is generally helpful in the retail sector where the products are bought in bulk after calculating the per-unit price of each commodity in that particular bulk.
Answer:
heat transfer for the process is - 643.3 kJ
Explanation:
given data
mass m = 2 kg
pressure p1 = 500 kPa
temperature t1 = 400°C = 673.15 K
temperature t2 = 40°C = 313.15 K
pressure p2 = 300 kPa
to find out
heat transfer for the process
solution
we know here mass is constant so
m1 = m2
so by energy equation
m ( u2 - u1 ) = Q - W
Q is heat transfer
and in process P = A+ N that is linear spring
so
W = ∫PdV
= 0.5 ( P1+P2) ( V1 - V2)
so for case 1
P1V1 = mRT
put here value
500 V1 = 2 (0.18892) (673.15)
V1 = 0.5087 m³
and
for case 2
P2V2 = nRT
300 V2 = 2 (0.18892) (313.15)
V2 = 0.3944 m³
and
here W will be
W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )
W = -45.72 kJ
and
Q is here for Cv = 0.83 from ideal gas table
Q = mCv ( T2-T1 ) + W
Q = 2 × 0.83 ( 40 - 400 ) - 45.72
Q = - 643.3 kJ
heat transfer for the process is - 643.3 kJ
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