2 years ago

Problem 2

Engineering

1 answer:

mamaluj [8]2 years ago

8 0

Answer:

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Explanation:

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Two kilograms of air within a piston-cylinder assembly execute a Carnot power cycle with maximum and minimum temperatures of 750

dezoksy [38]

Answer:

because it is not over the world please me 25 kilometre

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2 years ago

In order to produce a certain semiconductor, a process called doping is performed in which phosphorus is diffused into germanium

Nimfa-mama [501]

Answer:

The diffusivity is given as $8.064\times 10^{-16} m^2/s$

Explanation:

From the given data as in the attached question found via the search (because the values were not clear in this one)

$D_o=2.0\times10^{-4}m^2/s\\Q_d=240.6 kJ/mol=2.406 \times 10^5 J/mol\\Gas Constant=R=8.314 J/mol K\\Temperature =830 C =830+273 =1103 K$

So the Diffusion coefficient is given as

$D=D_oe^{\frac{-Q_d}{RT}}\\D=(2.0\times 10^{-4})e^{\frac{-2.406\times 10^5}{8.314\times 1103}}\\D=8.0640\times 10^{-16} m^2/s$

So the diffusivity is given as $8.064\times 10^{-16} m^2/s$

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3 years ago

Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5060 kJ/h for each kW of power it c

Olin [163]

Answer:

$COP_{R} = 1.406$ , $\dot Q_{H} = 2.406 W$

Explanation:

The COP of a refrigerator is modelled after this expression:

$COP_{R} = \frac{\dot Q_{L}}{\dot W} \\COP_{R} = \frac{(5060 \frac{kJ}{h})(\frac{1 h}{3600 s})}{1 kW}\\COP_{R} = 1.406$

The rate of heat rejection to the outside air is:

$\dot Q_{H} = \dot Q_{L} + \dot W\\\dot Q_{H} = (5060 \frac{kJ}{h} )\cdot (\frac{1}{3600 s} ) + 1 kW\\\dot Q_{H} = 2.406 kW$

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2 years ago

Why is engineering so important

Gekata [30.6K]

Explanation:

because it helps in our daily life

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2 years ago

Ground effect is felt at a height that is equal to the plane's wingspan.<br> True or False

Oliga [24]

Answer:

True

Explanation:

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2 years ago