Kirchoff's Law Kirchoff's Law states that, by the time current has returned to its source is explained in the following.
Explanation:
- Kirchhoff's Current Law (KCL) is Kirchhoff's first law that deals with the conservation of charge entering and leaving a junction. ... In other words the algebraic sum of ALL the currents entering and leaving a junction must be equal to zero.
- Kirchoff's laws apply for a given instant in time. So the voltages at a given moment around a loop will all sum to zero, or currents in a node sum to zero if you look at the instantaneous voltage and current. But they will be out of phase.
- Kirchhoff Voltage Law states that ''The algebraic sum of all voltages (source voltage and voltage drops) is equal to zero around a close path''. This is called KVL ( Kirchhoff Voltage Law) equation. The source voltage is equal to the sum of all voltage drops.
- Kirchhoff's Voltage Law (KVL) is Kirchhoff's second law that deals with the conservation of energy around a closed circuit path.
- Kirchhoff's laws can be used to determine the values of unknown values like current, Voltage in the circuit. These laws can be applied on any circuit (with some limitation), and useful to find the unknown values in complex circuits and networks.
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Answer:</h3><h3><em>1. Ask questions</em></h3><h3><em>2. Thank the interviewer for their time </em></h3><h3>
Explanation:</h3>
1<em>. When the interviewer asked if you have any questions at the end of the interview don't say no. You should always say yes your interviewer is expecting you to ask a few good questions before ending the interview. </em>
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<em>2. Always thank the interviewer for their time and effort to interview you. This would look very good for you and its a nice way to help wrap up the interview. </em>
Answer:
0.0297M^3/s
W=68.48kW
Explanation:
Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
state 1
X=quality=1
T=-26C
density 1=α1=5.27kg/m^3
entalpy1=h1=234.7KJ/kg
state 2
T2=70
P2=8bar=800kPa
density 2=α2=31.91kg/m^3
entalpy2=h2=306.9KJ/kg
Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.
m1=m2
(Q1)(α1)=(Q2)(α2)
the volumetric flow rate at the exit is 0.0297M^3/s
To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation
W=m(h2-h1)
m=Qα
W=(0.18)(5.27)(306.9-234.7)
W=68.48kW
the compressor power is 68.48kW
