Answer:
Im guessing this is for CEA for PLTW, if so look up the exact assignment number and look at online examples of the exact same assignment.
Explanation:
Answer:
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Explanation:
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.
Answer:
Option (c) and option (d)
Explanation:
Eutectic system is one in which a solid and homogeneous mixture of two or more substances resulting in the formation of super lattice is formed which can melt or solidify at a temperature lower than the melting point of any individual metal.
Eutectic alloys are those which have its components mixed in a specific ratio.
It is the composition in an alloy system for which both the liquidus and solidus temperatures are equal.
Eutectic alloys have the composition in which the melting point of the metal is lower than the other alloy composition.
Answer:
The required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.
Explanation:
We are given current density of 459 A/cm² and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.
Recall that current density is given by
j = I/A
where I is the current flowing through the wire and A is the area of the wire
A = πr²
but r = d/2 so
A = π(d/2)²
A = πd²/4
so the equation of current density becomes
j = I/πd²/4
j = 4I/πd²
Re-arrange the equation for d
d² = 4I/jπ
d = √4I/jπ
d = √(4*0.53)/(459π)
d = 0.0383 cm
Therefore, the required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.