The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diam
eter of 50 mu m has a strength of 140 MPa. (a) Determine the constants for the Hall-Patch equation for this alloy, (b) How much more should you reduce the grain size if you desired a strength of 220 MPa?
1 answer:
Answer:
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Explanation:
Given that; The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diameter of 50 mu m has a strength of 140 MPa. (a) Determine the constants for the Hall-Patch equation for this alloy, (b) How much more should you reduce the grain size if you desired a strength of 220 MPa
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Answer:
153.2 J
Explanation:
Let's first list our given parameters;
mass (m) of the block = 10 kg
which slides down ( i.e displacement) = 2 m
kinetic coefficient of friction (μk) = 0.2
In the diagram shown below; if we take an integral look at the component of force in the direction of the displacement; we have
Fcos 40°
100 (cos 40°)
76.60 N
Workdone by the friction force can now be determined as:
W = × displacement
W = 76.60 × 2
W = 153.2 J
∴ the work done by the friction force = 153.2 J
Answer:
Yes, the flow is turbulent.
Explanation:
Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.
Given:
Diameter of pipe is 10mm.
Velocity of the pipe is 1m/s.
Temperature of water is 200°C.
The kinematic viscosity at temperature 200°C is m2/s.
Calculation:
Step1
Expression for Reynolds number is given as follows:
Here, v is velocity, is kinematic viscosity, d is diameter and Re is Reynolds number.
Substitute the values in the above equation as follows:
Re=64226.07579
Thus, the Reynolds number is 64226.07579. This is greater than 2000.
Hence, the given flow is turbulent flow.