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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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Atmospheric pressure is measured to be 14.769 psia. a. What would be the equivalent reading of a water barometer (inches of H20)

Fofino [41]

Answer:

(a) water height =408.66 in.

(b) mercury height=30.04 in.

Explanation:

Given: P=14.769 psi ( 1 psi= 6894.76 $\frac{N}{m^2}$ )

we know that $P=\rho\times g\timesh$

where $\rho =Density,g=9.81\frac{m}{s^2}$

h=height.

Given that P=14.769 psi ⇒P= 101828.6 7 $\dfrac{N}{m^2}$

(a) $P=\rho_{w}\times g\times h_{w}$

$\rho_{w}=1000\frac{Kg}{m^3}$

⇒101828.67= $1000\times 9.81\times h_{w}$

$h_{w}$ =10.38 m

So water barometer will read 408.66 in. (1 m=39.37 in)

(b) $P=\rho_{hg}\times g\times h_{hg}$

$\rho_{hg}$ =13600

So 101828.67= $13600\times 9.81\times h_{hg}$

$h_{hg}$ =0.763 m

So mercury barometer will read 30.04 in.

6 0

3 years ago

(a) Consider a germanium semiconductor at T 300 K. Calculate the thermal equilibrium electron and hole concentrations for (i) Nd

padilas [110]

Answer:

i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2 × 10¹¹ cm⁻³

ii. electron concentration, n₀ = 1.33 × 10¹¹ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³

i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2.205 × 10⁻³ cm⁻³

ii. electron concentration, n₀ = 1.47 × 10⁻³ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³

c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.

Explanation:

a. For Germanium, intrinsic concentration n₁ = 2 × 10¹³ cm⁻³.

i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 0,

n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ] since N₁ > N₂

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ - 0) +√[(2 × 10¹⁵ cm⁻³ - 0)² + 4(2 × 10¹³ cm⁻³)²] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4.0016 × 10³⁰ cm⁻⁶] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ + 2.0004 × 10¹⁵ cm⁻³ ]

n₀ = 1/2[4.0004 × 10¹⁵ cm⁻³ ]

n₀ = 2.0002 × 10¹⁵ cm⁻³ ≅ 2 × 10¹⁵ cm⁻³

The hole concentration p₀ is gotten from

n₀p₀ = n₁²

p₀ = n₁²/n₀ = (2 × 10¹³ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/2 × 10¹⁵ cm⁻³

p₀ = 2 × 10¹¹ cm⁻³

ii. For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 10¹⁶ cm⁻³,

p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ] since N₂ > N₁

p₀ = 1/2[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³) +√[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³)² + 4(2 × 10¹³ cm⁻³)²] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9.0016 × 10³⁰ cm⁻⁶] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ + 3.0003 × 10¹⁵ cm⁻³ ]

p₀ = 1/2[6.0003 × 10¹⁵ cm⁻³ ]

p₀ = 3.00015 × 10¹⁵ cm⁻³ ≅ 3 × 10¹⁵ cm⁻³

Te electron concentration n₀ is gotten from

n₀p₀ = n₁²

n₀ = n₁²/p₀ = (2 × 10¹³ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/3 × 10¹⁵ cm⁻³

n₀ = 1.33 × 10¹¹ cm⁻³

b. For GaAs, intrinsic concentration n₁ = 2 × 10⁶ cm⁻³.

i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 0,

n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ] since N₁ > N₂ and N₁ - N₂ = 2 × 10¹⁵ cm⁻³ >> n₁ = 2 × 10⁶ cm⁻³

n₀ = (N₁ - N₂) = 2 × 10¹⁵ cm⁻³ - 0 = 2 × 10¹⁵ cm⁻³

The hole concentration p₀ is gotten from

n₀p₀ = n₁²

p₀ = n₁²/n₀ = (2.1 × 10⁶ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/2 × 10¹⁵ cm⁻³

p₀ = 2.205 × 10⁻³ cm⁻³

ii. For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 10¹⁶ cm⁻³,

p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ] since N₂ > N₁ and N₂ - N₁ = 10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³ = 3 × 10¹⁵ cm⁻³ >> n₁ = 2.1 × 10⁶ cm⁻³

p₀ ≅ N₂ - N₁ = 3 × 10¹⁵ cm⁻³

The electron concentration n₀ is gotten from

n₀p₀ = n₁²

n₀ = n₁²/p₀ = (2.1 × 10⁶ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/3 × 10¹⁵ cm⁻³

n₀ = 1.47 × 10⁻³ cm⁻³

c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.

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3 years ago

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Naddik [55]

Answer:

There are actually multiple types of processes a manufacturer uses, and those can be grouped into four main categories: casting and molding, machining, joining, and shearing and forming.

Explanation:

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Explanation:

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s2008m [1.1K]

Answer:

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Explanation:

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3 years ago