A three-phase voltage source with a terminal voltage of 22kV is connected to a three-phase transformer rated 5MVA 22kV/220V. The
transformer has a per-phase series impedance of 0.05 pu and is connected to a distribution line with series impedance of 0.01+0.05Ω per phase. The line supplies a balanced three-phase load rated 1MVA, 220V with 0.8 lagging power factor.
a. Draw the one-line diagram of the system described.
b. Draw the equivalent per phase circuit of the system indicating all impedances in per unit. Use 1MVA, 220V as the base at the load.
c. Calculate the real losses in the transmission line in kW.
a. Draw the one-line diagram of the system described.
b. Draw the equivalent per phase circuit of the system indicating all impedances in per unit. Use 1MVA, 220V as the base at the load.
c. Calculate the real losses in the transmission line in kW.
1 answer:
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Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
Answer:
The code is attached.
Explanation:
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Answer:
The temperature T= 648.07k
Explanation:
T1=input temperature of the first heat engine =1400k
T=output temperature of the first heat engine and input temperature of the second heat engine= unknown
T3=output temperature of the second heat engine=300k
but carnot efficiency of heat engine =
where Th =temperature at which the heat enters the engine
Tl is the temperature of the environment
since both engines have the same thermal capacities <em> </em> therefore
We have now that
multiplying through by T
multiplying through by 300
-
The temperature T= 648.07k