Select the most accurate statement. A diffuser converts some of a fluid's _(i)____ to ___(ii)___. Select one: a. (i) pressure, (
1 answer:
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Answer:
The maximum theoretical height that the pump can be placed above liquid level is
Explanation:
To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:
( stands here for density,
for height)
Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:
This means that pressure drop is proportional to the suction lift's height.
We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.
That means:
We insert that into our last equation and get:
And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.
Answer:
The head loss in Psi is 0.390625 psi.
Explanation:
Fluid looses energy in the form of head loss. Fluid looses energy in the form of head loss when passes through the valve as well.
Given:
Factor cv is 48.
Flow rate of water is 30 GPM.
GPM means gallon per minute.
Calculation:
Step1
Expression for head loss for the water is given as follows:
Here, cv is valve coefficient, Q is flow rate in GPM and h is head loss is psi.
Step2
Substitute 48 for cv and 30 for Q in above equation as follows:
h = 0.390625 psi.
Thus, the head loss in Psi is 0.390625 psi.
Answer:
The speed of the sound for the adiabatic gas is 313 m/s
Explanation:
For adiabatic state gas, the speed of the sound c is calculated by the following expression:
Where R is the gas's particular constant defined in terms of Cp and Cv:
For particular values given:
The gamma undimensional constant is also expressed as a function of Cv and Cp:
And the variable T is the temperature in Kelvin. Thus for the known temperature:
The Jules unit can expressing by:
Replacing the new units for the speed of the sound:
Answer:
A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].
B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with lattice constant a.
C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.
Compleated question:
1. Miller Indices:
a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.
b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?
c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.
Explanation:
A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:
1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]
2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).
3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.
4- Join the points.
<u>In this case, for [1 2 1]:</u>
<u>for </u><u>:</u>
B) The closest distance between planes with the same Miller indices can be calculated as:
With ,the distance is
with lattice constant a.
<u>In this case, for [1 2 1]:</u>
<u /><u />
<u>for </u><u>:</u>
C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:
dir₁=|0 0 1|
dir₂=|0 0.5 1.5|≡|0 1 3|
dir₃=|0 1 1|
dir₄=|0 1.5 0.5|≡|0 3 1|
dir₅=|0 0 1|
