An ac generator provides emf to a resistive load in a remote factory over a two-cable transmission line. At the factory a step-
1 answer:
Answer:
a) 1.875 volts
b) 5.86 W
Explanation:
Given data:
transmission line resistance = 0.30/cable
power of the generator = 250 kW
if Vt = 80 kV
<u>a) Determine the decrease V along the transmission line </u>
Rms current in cable = P / Vt = 250 / 80 = 3.125 amps
hence the rms voltage drop ( Δrmsvoltage )
Δrmsvoltage = Irms* R = 3.125 * 2 * 0.30 =<em> 1.875 volts </em>
<u>b) Determine the rate Pd at which energy is dissipated in line as thermal energy </u>
Pd =
= 3.125^2 * 2* 0.30 = <em>5.86 W</em>
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Answer: see the graph attached (straight line, passing through the origin and positive slope).
Justification:
1) Kinetic energy and temperature are in direct proportion. That means:
i) Being kinetic energy y and temperature x: y α x
ii) That implies: y = kx,where k is the constant of proportionality.
iii) The graph is a line that passes through the origin and has positive slope k (k = y / x).
2) The proportional relationship between kinetic energy (KE) and temperature (T) is shown by the Boltzman law, which states:
Average KE = [3 / 2] KT, where K is Boltzman's constant, whose graph is of the form shown in the figure attached.
Justification:
1) Kinetic energy and temperature are in direct proportion. That means:
i) Being kinetic energy y and temperature x: y α x
ii) That implies: y = kx,where k is the constant of proportionality.
iii) The graph is a line that passes through the origin and has positive slope k (k = y / x).
2) The proportional relationship between kinetic energy (KE) and temperature (T) is shown by the Boltzman law, which states:
Average KE = [3 / 2] KT, where K is Boltzman's constant, whose graph is of the form shown in the figure attached.