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Vanyuwa [196]
3 years ago
6

What does a peak flow meter allow you to assess?

Engineering
1 answer:
Alex Ar [27]3 years ago
3 0

Answer:

  peak flow and any engineering considerations related thereto

Explanation:

It should be no surprise that a peak flow meter will report peak flow, sometimes with important maximum-value, time-constant, or bandwidth limitations. There are many engineering issues related to flow rates. A peak flow meter can allow you to assess those issues with respect to the flows actually encountered.

Peak flow can allow you to assess adequacy of flow and whether there may be blockages or impediments to flow that reduce peak levels below expected values. An appropriate peak flow meter can help you assess the length of time that peak flow can be maintained, and whether that delivers sufficient volume.

It can also allow you to assess whether appropriate accommodation is made for unexpectedly high flow rates. (Are buffers or overflow tanks of sufficient size? Is there adequate protection against possible erosion? Is there adequate support where flow changes direction?)

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D Reflection

Explanation:

It reflects the light from the sun. Then when the earth gets in the way, it casts a shadow causing crescent moons.

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2 years ago
To do the right thing or to have moral principles is called _____ . a code
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Answer:

Ethics

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- Loyalty

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3 years ago
A right triangle has a height of 20 centimeters and a base of 10 centimeters. What is the área of the triangle
zmey [24]

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100 cm

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8 0
2 years ago
6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe
Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}

Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t

Therefore, we have;

Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

7 0
2 years ago
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