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kaheart [24]
3 years ago
11

A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the g

as pressure at the inlet is 5.5 psig, what is the absolute pressure, in psia, of the gas at the exit? Atmospheric pressure is 14.5 lbf/in.2
Engineering
1 answer:
olga55 [171]3 years ago
7 0

Answer:

P_2_{abs}=160\ psia (absolute).

Explanation:

Given that

Pressure ratio r

r=8

r=\dfrac{P_2_{abs}}{P_1_{abs}}

  8=\dfrac{P_2_{abs}}{P_1_{abs}}                                  -----1

P₁(gauge) = 5.5 psig

We know that

Absolute pressure = Atmospheric pressure  + Gauge  pressure

Given that

Atmospheric pressure = 14.5 lbf/in²

P₁(abs) = 14.5 + 5.5  psia

P₁(abs) =20 psia

Now by putting the values in the above equation 1

8=\dfrac{P_2_{abs}}{20}

P_2_{abs}=8\times 20\ psia

P_2_{abs}=160\ psia

Therefore the exit gas pressure will be 160 psia (absolute).

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In the planning process of the product development life cycle what is it important to inventory
Verizon [17]

Your Answer would be A I believe.

6 0
3 years ago
You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you
telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

7 0
2 years ago
Miller Indices:
svetlana [45]

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>

l=1=1/a_0 \rightarrow a_0=1

m=2=2/b_0 \rightarrow b_0=0.5

n=1=1/c_0 \rightarrow c_0=1

<u>for </u>[1 2 \overline{4}]<u>:</u>

l=1=1/a_0 \rightarrow a_0=1

m=2=2/b_0 \rightarrow b_0=0.5

n=\overline{4}=-4/c_0 \rightarrow c_0=-0.25

B) The closest distance between planes with the same Miller indices can be calculated as:

With \pi:[l m n] ,the distance is d_{lmn}= \displaystyle \frac{a}{\sqrt{l^2+m^2+n^2}} with lattice constant a.

<u>In this case, for [1 2 1]:</u>

<u />d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a<u />

<u>for </u>[1 2 \overline{4}]<u>:</u>

d_{12\overline{4}}= \displaystyle \frac{a}{\sqrt{1^2+2^2+(-4)^2}}=\frac{a}{\sqrt{21}}=0.22a

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|

5 0
3 years ago
The section of the area to be examined is shown circumscribed by broken lines with circles at
Aloiza [94]

This question is about Circle Geometry. it evaluates connected and broken lines with respect to circles.

<h3>What is Circle Geometry?</h3>

This refers to the body of knowledge in mathematics that has to do with the various problems associated with the Circle.

In real-world scenarios, circle geometry is used in technologies involving:

  • Camera lenses
  • Circular Architectural structures
  • Steering Wheels
  • Buttons etc.

Learn more about Circle Geometry at:
brainly.com/question/24375372

3 0
2 years ago
Is the COP of a heat pump always larger than 1?
Liono4ka [1.6K]

Answer:

Yes

Explanation:

Yes it is true that COP of heat pump always greater than 1.But the COP of refrigeration can be greater or less than 1.

We know that

COP of heat pump=  1 + COP of refrigeration

It is clear that COP can not be negative .So from the above expression we can say that COP of heat pump is always greater than one.  

3 0
3 years ago
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