Answer:
r=0.31
Ф=18.03°
Explanation:
Provided:
Original diameter of bar = 75 mm
Diameter post-cutting = 73 mm
Average diameter of the bar d= (75+73)/2=74 mm
Average length of uncut chip = πd
Average length of uncut chip = π x 74 =232.45 mm
Thus, cutting ratio r
r=0.31
Therefore, the cutting ratio equals 0.31.
Now, the shearing angle is given as
Next by substituting the values
\
Ф=18.03°
Concluding, the shearing angle is 18.03°.
Answer:
a) The phase before eutectoid is commonly referred to as cementite, with the chemical formula Fe₃C.
b) The total mass of ferrite obtained is 0.8311 kg.
The total cementite mass equals 0.1689 kg.
c) The total cementite mass accounts for 0.9343 kg.
Explanation:
Provided:
1 kg of austenite
a carbon content of 1.15 wt%
Cooled below 727°C
Questions:
a) Identify the proeutectoid phase.
b) Calculate the mass of total ferrite and cementite, Wf =?, Wc =?
c) Determine the mass of both pearlite and the proeutectoid phase, Wp =?
d) Create a schematic to illustrate the resulting microstructure.
a) The proeutectoid phase is referred to as cementite with the formula Fe₃C.
b) To find the total mass of formed ferrite:
With:
Ccementite = composition of cementite = 6.7 wt%
C₁ = composition of phase 1 = 0.022 wt%
C₂ = overall composition = 1.15 wt%
Inserting the values yields:
For the total mass of cementite:
c) The mass of pearlite:
d) The diagram illustrates the different compositions: (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)
Answer:
Total bandwidth: 8 kHz
Explanation:
Data provided:
Transmitter frequency: 3.9 MHz
Modulation up to: 4 kHz
Solution:
For the upper side frequencies:
Upper side frequencies = 3.9 × + 4 × 10³
Upper side frequencies = 3.904 MHz
For the lower side frequencies:
Lower side frequencies = 3.9 × - 4 × 10³
Lower side frequencies = 3.896 MHz
Consequently, the total bandwidth is computed as:
Total bandwidth = upper side frequencies - lower side frequencies
Total bandwidth = 8 kHz