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matrenka [14]
3 years ago
10

A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The

compressor has a cooling jacket where water flows at 1500 kg/hr and undergoes a 20K temperature rise. Assuming air is an ideal gas, calculate the work furnished by the compressor, and also determine the minimum work required for the same state change of air.

Engineering
1 answer:
AleksandrR [38]3 years ago
8 0

Answer:

The work furnished by the compressor is 69.77kJ/s

The minimum work required for the state to change is 55.26kW

Explanation:

The explanation to these solution is on the first, second , third and fourth uploaded image respectively

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In order to defend against side channel power analysis, we should: ______________
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3 years ago
A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co
NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx  32355 liters.

Explanation:

The tank is shown in the attached figure below

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V_{hem}=\frac{2}{3}\pi r^3

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

V_{cyl}=\pi r^2\cdot h

3) Conical bottom

Volume of conical bottom of radius'r' and angle \theta is

V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}

Applying the given values we obtain the volume of the container up to cylinder is

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3 0
3 years ago
You buy a 75-W lightbulb in Europe, where electricity is delivered to homes at 240V. If you use the lightbulb in the United Stat
givi [52]

Answer:

The bulb will be \frac{1}{4} times as bright as it is in Europe.

Explanation:

Data provided in the question:

Power of bulb in Europe = 75 W

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Now,

We know,

Power = Voltage² ÷ Resistance

Therefore,

75 = 240² ÷ Resistance

or

Resistance = 240² ÷ 75

or

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Therefore,

Power in United States = Voltage² ÷ Resistance

= 120² ÷ 768

= 18.75 W

Therefore,

Ratio of powers

\frac{\text{Power in united states}}{\text{Power in Europe}}=\frac{18.75}{75}

= \frac{1}{4}

Hence,

The bulb will be \frac{1}{4} times as bright as it is in Europe.

7 0
2 years ago
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