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Nutka1998 [239]
3 years ago
14

What is the relationship between compressor work and COPR?

Engineering
1 answer:
FrozenT [24]3 years ago
5 0
Oh that’s a question you can just look up on safari the right answer is there thank me later tho
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Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di
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Answer:

The solution is attached in the attachment.

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3 years ago
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An escalator handles a steady load of 26 people per minute in elevating them from the first to the second floor through a vertic
photoshop1234 [79]

Answer:

\eta = 70.711\,\%

Explanation:

The power needed to make the escalator working is obtained by means of the Work-Energy Theorem:

\dot W  = \dot U_{g}

\dot W = \dot n \cdot m_{p}\cdot g \cdot \Delta y

\dot W = \left(26\,\frac{persons}{min}\right)\cdot (124\,lbm)\cdot \left(32.174\,\frac{ft}{s^{2}}\right)\cdot \left(\frac{1\,lbf}{32.174\,\frac{lbm\cdot ft}{s^{2}} } \right)\cdot (27.5\,ft)

\dot W = 88660\,\frac{lbf\cdot ft}{min}\,\left(2.687\,hp\right)

The mechanical efficiency of the escalator is:

\eta = \frac{2.687\,hp}{3.8\,hp}\times 100\,\%

\eta = 70.711\,\%

3 0
3 years ago
You can help build a safe work environment by using your knowledge of violence prevention strategies to spot what?
Ostrovityanka [42]

Answer:

warning signs

Explanation:

give directions on your surroundings

4 0
3 years ago
Which of the following answer options is a hazard in using scaffolding?​
kenny6666 [7]

Answer:

One of the most common injuries that result from working on a scaffold is electrocution. If workers make contact with those wires while working, it can lead to electric shock, especially if that contact damages the wire or if the scaffold is made of metal.

6 0
3 years ago
A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055
Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}

So the mean conductivity

K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )

T_m=\dfrac{160+273+40+273}{2}

T_m=373 K

K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )

K_m=0.112 \frac{W}{mK}

So heat conduction through cylinder

Q=kA\dfrac{\Delta T}{L}

Q=0.112\times \pi \times 0.15^2\times 120

Q=0.95 W/m

4 0
3 years ago
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