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3 years ago

Help Please!!!!!!! Whatever3443 Please help!

Engineering

2 answers:

KIM [24]3 years ago

8 0

Answer:

Thanks for the pointssssssssssssss

Explanation:

Olin [163]3 years ago

7 0

Answer:

I need some help understanding this question what is it you need to know about doc. whatever 3443?

744 with 3040 1230 0 215 141 184 247 353 whatever 3443 1230 0 215 140 246 337 400 492 581 673 737 for 4232 1230 0 215 50 154 218 which 4499 1230

Explanation:

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A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-v

adelina 88 [10]

Answer:

The power produced by the turbine is 23309.1856 kW

Explanation:

h₁ = 3755.39

s₁ = 7.0955

s₂ = sf + x₂sfg =

Interpolating fot the pressure at 3.25 bar gives;

570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175

2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345

h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg

Power output of the turbine formula =

$Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ]$

Which gives;

$560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ]$

= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856

$- \dot{W }$ = -22749.1856 - 560 = -23309.1856 kJ

$\dot{W }$ = 23309.1856 kJ

Power produced by the turbine = Work done per second = 23309.1856 kW.

5 0

3 years ago

30POINTS

garri49 [273]

Concentrating solar power (CSP) plants use mirrors to concentrate the sun's energy to drive traditional steam turbines or engines that create electricity. The thermal energy concentrated in a CSP plant can be stored and used to produce electricity when it is needed, day or night. Today, roughly 1,815 megawatts (MWac) of CSP plants are in operation in the United States.

Parabolic Trough
Parabolic trough systems use curved mirrors to focus the sun’s energy onto a receiver tube that runs down the center of a trough. In the receiver tube, a high-temperature heat transfer fluid (such as a synthetic oil) absorbs the sun’s energy, reaching temperatures of 750°F or higher, and passes through a heat exchanger to heat water and produce steam. The steam drives a conventional steam turbine power system to generate electricity. A typical solar collector field contains hundreds of parallel rows of troughs connected as a series of loops, which are placed on a north-south axis so the troughs can track the sun from east to west. Individual collector modules are typically 15-20 feet tall and 300-450 feet long.

Compact Linear Fresnel Reflector
CLFR uses the principles of curved-mirror trough systems, but with long parallel rows of lower-cost flat mirrors. These modular reflectors focus the sun's energy onto elevated receivers, which consist of a system of tubes through which water flows. The concentrated sunlight boils the water, generating high-pressure steam for direct use in power generation and industrial steam applications.

3 0

3 years ago

Read 2 more answers

What is required when setting up a smart phone as a WIFI hotspot?

emmainna [20.7K]

How to create a personal hot spot on an iPhone?

Go to Settings | Cellular | Personal Hotspot.

Tap the slider next to Allow Others to Join. ...

Your Wi-Fi Password will be shown right underneath the Allow Others to Join option. ...

Now, on another device, such as a laptop, go to the Wi-Fi section and search for nearby networks.

5 0

3 years ago

Storm sewer backup causes your basement to flood at the steady rate of 1 in. of depth per hour. The basement floor area is 2600

Wittaler [7]

Answer:

attached below

Explanation:

4 0

3 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.

In this case, for [1 2 1]:

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

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

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

for $[1 2 \overline{4}]$ :

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

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

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

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.

In this case, for [1 2 1]:

$d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a$

for $[1 2 \overline{4}]$ :

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

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