The following is a series of questions pertaining to the NSPE Code of Ethics. Please indicate whether the statement are true or
1 answer:
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Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;
where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²
According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Answer:
2074.2 KW
Explanation:
<u>Determine power developed at steady state </u>
First step : Determine mass flow rate ( m )
m / Mmax = ( AV )₁ P₁ / RT₁ -------------------- ( 1 )
<em> where : ( AV )₁ = 8.2 kg/s, P₁ = 0.35 * 10^6 N/m^2, R = 8.314 N.M / kmol , </em>
<em> T₁ = 720 K . </em>
insert values into equation 1
m = 0.1871 kmol/s ( mix )
Next : calculate power developed at steady state ( using ideal gas tables to get the h values of the gases )
W( power developed at steady state )
W = m [ Yco2 ( h1 - h2 )co2
Attached below is the remaining part of the detailed solution
