Is an ideal way for a high school student to see what an engineer does on a typical day but does not provide a hands-on experien
2 answers:
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Answer:
the graph and the answer can be found in the explanation section
Explanation:
Given:
Network rated voltage = 24 kV
Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi
Rn = 0.07 * 8 = 0.56 Ω
Xn = 0.5 * 8 = 4 Ω
If the alternator terminal voltage is equal to network rated voltage will have
Vt = 24 kV/√3 = 13.85 kV/phase
The alternative current is
The impedance Zn is
The voltage drop is
rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω
The effective armature resistance is
The induced voltage for leading power factor is
if cosθ = 0.5
if cosθ= 0.6
EF = 12790.8 V
if cosθ = 0.7
EF = 13731.05 V
if cosθ = 0.8
EF = 14741.6 V
if cosθ = 0.9
EF = 15809.02 V
if cosθ = 1
EF = 13975.6 V
The voltage regulation is
For each value:
if cosθ = 0.5
voltage regulation = -13.8%
if cosθ = 0.6
voltage regulation = -7.6%
if cosθ = 0.7
voltage regulation = -0.85%
if cosθ = 0.8
voltage regulation = 6.4%
if cosθ = 0.9
voltage regulation = 14%
if cosθ = 1
voltage regulation = 0.9%
the graph is shown in the attached image
for 10% of regulation the power factor is 0.81
Answer:
a) Mechanical efficiency ()=63.15% b) Temperature rise= 0.028ºC
Explanation:
For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).
The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, .
Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump . So
.
The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is
For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and the density of the water:
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Finally, the temperature rise is: