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3 years ago

Introduction to fluid

Engineering

1 answer:

otez555 [7]3 years ago

8 0

Answer:

A fluid is a substance [ liquid and gas state ] in which motion of another substance in it is opposed due to viscous drag [ viscosity ]

Explanation:

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Assume the transistor is biased in the saturation region at VGS 4 V. (a) Calculate the ideal cutoff frequency. (b) Assume that t

insens350 [35]

Answer:

hello your question is incomplete attached below is the complete question and the detailed solution

Answer: A) 5.17 GHz

B) 1.01 GHz

Explanation:

Assuming the transistor is biased and considering the two conditions as given in A and B attached below is a detailed solution to the given problem

4 0

3 years ago

Write the definitions for engineering stress, true stress, engineering strain, and true strain for loading along a single axis.

shusha [124]

Answer and Explanation:

Engineering Stress of a material is defined as the applied load or force divided by the original cross sectional Area of the material.

σ(engineering) = F/(Ao)

True Stress is defined as the applied load or force divided by the actual cross-sectional area (the changing area with respect to time) of the material at that point in time. It's an instantaneous stress.

σ(true) = F/A

Engineering strain is a measure of how much a material deforms under a particular load. It is the amount of deformation in the direction of the applied force divided by the initial length of the material.

ε(engineering) = Δl/lo

True Strain measures instantaneous deformation. It is obtained mathematically by integrating strain over small time periods and summing them up. Hence,

ε(true) = In (lf/lo)

The calculations,

First step, 10m to 10.1m, Δl = 0.1m, lf = 10.1m, lo = 10m

ε(engineering) = 0.1/10 = 0.01

ε(true) = In (10.1/10) = 0.00995

Second step, 10.1m to 10.2m, Δl = 0.1m, lf = 10.2m, lo = 10.1m

ε(engineering) = 0.1/10.1 = 0.0099

ε(true) = In (10.2/10.1) = 0.00985

Overall, 10m to 10.2m, Δl = 0.2m, lf = 10.2m, lo = 10m

ε(engineering) = 0.2/10 = 0.02

ε(true) = In (10.2/10) = 0.0198

QED!

5 0

3 years ago

Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the

ra1l [238]

The evaporation rate of the n-Hexane is $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

Explanation:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

$N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}$

$\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}$

$P_{t}=1 a t m=101325 P a\\$

$\text { so, } P_{A 1}=$ the vapor pressure at hexane $25 \mathrm{C}$ $=20158.2 \mathrm{Pa}$

For wind, assume negligible hexane is present, hence $P_{A 2}=0$

Now,

$\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}$

$\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}$

$P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\$

$=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}$

$=90873.57 \mathrm{Pa}$

$R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}$

$z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\$

where T = 298 K

substituting all in the equation, we get

$\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}$

$=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\$

Now,Flux $\times$ area = Molar rate of evaporation

Evaporation rate = $0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)$

Evaporation rate = $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

6 0

3 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 5 × 10-4

White raven [17]

Answer:

The question is a problem that requires the principles of fracture mechanics.

and we will need this equation below to get the Max. Stress that exist at the tip of an internal crack.

Explanation:

Max Stress, σ = 2σ₀√(α/ρ)

where,

σ₀ = Tensile stress = 190MPa = 1.9x10⁸Pa

α = Length of the cracked surface = (4.5x10⁻²mm)/2 = 2.25x10⁻⁵m

ρ = Radius of curvature of the cracked surface = 5x10⁻⁴mm = 5x10⁻⁷m

Max Stress, σ = 2 x 1.9x10⁸ x (2.25x10⁻⁵/5x10⁻⁷)⁰°⁵

Max Stress, σ = 2 x 1.9x10⁸ x 6.708 Pa

Max Stress, σ = 2549MPa

Hence, the magnitude of the maximum stress that exists at the tip of an internal crack = 2549MPa

7 0

4 years ago

Discuss how, as a safety professional, you would respond to the overlap in Occupational Safety and Health Administration (OSHA)

katrin2010 [14]

Answer:

Explanation:

As a security professional, I will respond positively to the OSHA requirements overlap. OSHA guidelines are meant to provide general guidance to all members of various entities throughout the country, while local or state codes also ensure compliance with laws unique to their areas, taking into account workplace safety and security.

OSHA accepts the security codes of the state building To the degree that such codes comply with OSHA regulations, such as BOCA. All the codes and regulations for local, state-owned construction, electrical and life protection are under the same umbrella. Generally, all security protocols and specifications are in accordance with OSHA guidelines. Nonetheless, certain points will overlap, while localized codes will also be addressed to a particular community or state that may

3 0

3 years ago

Introduction to fluid​

Introduction to fluid