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:
Temperature T = 394.38 K
Explanation:
The full solution and detailed explanation regarding the above question and its specified conditions can be found below in the accompanying document. I trust my explanation will assist you in grasping this particular topic.
Answer:
The power of the pump is 23.09 kW.
Explanation:
Parameters
gravitational constant,
mass flow rate,
flow density,
efficiency of the pump,
output gauge pressure,
input gauge pressure,
cross-sectional area of output pipe,
cross-sectional area of input pipe,
height of discharge, (evaluated at pump’s maximum height of 1.22 m)
input height,
hydraulic power of the pump,
Initially, the volumetric flow (Q) needs to be determined
Next, compute the velocity (v) for both input and output
Subsequently, the total head (H) can be calculated
Finally, the computation of pump power is done as follows
Answer:
Refer to the attached document
1512 ft
Explanation:
Because the acceleration is constant or zero, the acceleration-time graph consists of horizontal segments. The values for t2 and a4 are derived as follows:
Acceleration - Time
0 < t < 6: Velocity change = area beneath the a–t graph
V_6 - 0 = (6 s)(4 ft/s²) = 24 ft/s
6 < t < t2: The velocity rises from 24 to 48 ft/s,
Velocity change = area beneath the a–t graph
48 - 24 = (t2 - 6) * 6
t2 = 10 s
t2 < t < 34: The velocity remains constant, meaning acceleration is zero.
34 < t < 40: Velocity change = area beneath the a–t graph
0 - 42 = 6*a4
a4 = - 8 ft / s²
A negative acceleration shows the area lies below the t-axis, indicating a decrease in speed.
Velocity - Time
Since acceleration remains constant or zero, the v−t graph is made up of linear segments connecting the calculated points.
Position change = area beneath the v−t graph
0 < t < 6: x6 - 0 = 0.5*6*24 = 72 ft
6 < t < 10: x10 - x6 = 0.5*4*(24 + 48) = 144 ft
10 < t < 34: x34 - x10 = 48*24 = 1152 ft
34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft
Summing these position changes yields the distance from A to B:
d = x40 - 0 = 1512 ft
