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Anastaziya [24]
3 years ago
12

Describing Tasks for Stationary Engineers Click this link to view O*NET’s Tasks section for Stationary Engineers. Note that comm

on tasks are listed toward the top, and less common tasks are listed toward the bottom. According to O*NET, what are common tasks performed by Stationary Engineers? Check all that apply. observing and interpreting readings on gauges, meters, and charts managing financial resources to order supplies and equipment writing computer programs to control equipment testing boiler water quality or arranging for testing creating safety regulations after experimentation operating or tending stationary engines, boilers, and auxiliary equipment
Engineering
2 answers:
Natali5045456 [20]3 years ago
8 0

Answer:

A, D, F

Explanation:

took on edge.

11111nata11111 [884]3 years ago
8 0

Answer:

The answer is A,D,F

A:observing and interpreting readings on gauges, meters, and charts

D:testing boiler water quality or arranging for testing

F:operating or tending stationary engines, boilers, and auxiliary equipment

Explanation:

I got all right on edge.

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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
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th
xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }

\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }

if Gl≈El(l,5800)

\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt= 2*5800=11600 um-K, at this value, F=0.941

\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77

The hemispherical emissivity is equal to:

E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt=2*333=666 K, at this value, F=0

E=0+(1-0.7)(1)=0.3

The hemispherical absorptivity is equal to:

qe=EoTs^{4}+h(Ts-T\alpha  )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}

3 0
3 years ago
The electrical panel schedules are located on EWR Plan number ___.
Stells [14]
A8 is the answer because yea and because I am a teacher
5 0
2 years ago
The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100),
skelet666 [1.2K]

Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = \frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}

a)(100)

a=5.28 Å

Distance = \frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}=5.28 Å

b)(110)

Distance = \frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}} = 3.73 Å

c)(111)

Distance= \frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}= 3.048 Å

6 0
3 years ago
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