A resistor, inductor, and capacitor are in parallel in a circuit where the frequency of operation can vary. The R, L, and C valu
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Answer:
A) Ductility = 11% EL
B) Radius after deformation = 4.27 mm
Explanation:
A) From equations in steel test,
Tensile Strength (Ts) = 3.45 x HB
Where HB is brinell hardness;
Thus, Ts = 3.45 x 250 = 862MPa
From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.
Also, from image 2,at CW of 27%,
Ductility is approximately, 11% EL
B) Now we know that formula for %CW is;
%CW = (Ao - Ad)/(Ao)
Where Ao is area with initial radius and Ad is area deformation.
Thus;
%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100
%CW = [1 - (rd)²/(ro)²]
1 - (%CW/100) = (rd)²/(ro)²
So;
(rd)²[1 - (%CW/100)] = (ro)²
So putting the values as gotten initially ;
(ro)² = 5²([1 - (27/100)]
(ro)² = 25 - 6.75
(ro) ² = 18.25
ro = √18.25
So ro = 4.27 mm
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
Answer:
The diffusivity is given as
Explanation:
From the given data as in the attached question found via the search (because the values were not clear in this one)
So the Diffusion coefficient is given as
So the diffusivity is given as