Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.
Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.
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:
Use GitHub or stackoverflow for this answer
Explanation:
It helps with programming a lot
Answer:
The rate of entropy change of the air is -0.10067kW/K
Explanation:
We'll assume the following
1. It is a steady-flow process;
2. The changes in the kinetic energy and the potential energy are negligible;
3. Lastly, the air is an ideal gas
Energy balance will be required to calculate heat loss;
mh1 + W = mh2 + Q where W = Q.
Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;
Rate of Entropy Change = -Q/T
Where Q = 30Kw
T = Temperature of air = 25°C = 298K
Rate = -30/298
Rate = -0.100671140939597 KW/K
Rate = -0.10067kW/K
Hence, the rate of entropy change of the air is -0.10067kW/K