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3 years ago

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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Answer:

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Explanation:

https://opendsa-server.cs.vt.edu/embed/mergesortAV

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3 years ago

A freshwater jet boat takes in water through side vents and ejects it through a nozzle of diameter D = 75 mm; the jet speed is V

Radda [10]

Answer:

a) 0.0663 m³/s

b) 3.312 N/(m/s)²

c) 16.665 m/s

d) 0.1105 m³/s

Explanation:

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3 years ago

Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r

My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0

3 years ago

Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th

Elodia [21]

Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0

3 years ago

An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in

Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, $T_{a} = 140^{\circ}C$ = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, $T_{i} = 60^{\circ}C$ = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, $C_{p} = 0.949\ kJ/kgK$

At room temperature

Specific heat of iron, $C_{p} = 0.45\ kJ/kgK$

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

$MC_{p}\Delta T = mC_{p}\Delta T$

$28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)$

Thus

$T_{e} = 109.71^{\circ}C$ = 273 + 109.71 = 382.71 K

where

$T_{e}$ = Equilibrium temperature

Now,

To calculate the changer in entropy:

$\Delta s = \Delta s_{a} + \Delta s_{i}$

Now,

For Aluminium:

$\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K$

For Iron:

$\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K$

Thus

$\Delta s =-2.025 + 2.253 = 0.228\ kJ/K$

6 0

3 years ago