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3 years ago

Use the drop-down menus to choose the correct term or words to complete the statements.

Engineering

1 answer:

aleksklad [387]3 years ago

7 0

Explanation:

Search PubMed v use drop down menu and choose MeSH MeSH term: select MeSH ... Tree) Use Links (right side of screen) to return to PubMed and complete the search (Default): ...

Iven Klineberg, ‎Diana Kingston - 2012 -

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At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

ludmilkaskok [199]

Answer:

a) $COP_{R} = 25.014$ , b) $T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

$COP_{R} = 5$

But:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

$T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5} \right)$

$T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

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3 years ago

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A 3-oz serving of roasted, skinless chicken breast contains 140 Cal, 27 g of protein, 3 g of fat, 13 mg of calcium, and 64 mg of

Art [367]

Answer:

Explanation:

a) Given C [ cal pro fat calc sod]

[140 27 3 13 64]

P = [cal pro fat calc sod]

[180 4 11 24 662]

B = [cal pro fat calc sod]

[50 5 1 82 20]

To find C+2P+3B = [140 27 3 13 64] + 2[180 4 11 24 662] + 3[50 5 1 82 20]

= [650 54 28 307 1448]

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Answer:

A single- shear pin connection.

Explanation:

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Tema [17]

Answer: True

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