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Dmitry [639]
3 years ago
13

What is the physical mechanism that causes the friction factor to be higher in turbulent flow?

Engineering
1 answer:
Montano1993 [528]3 years ago
3 0

Answer:

The standard enthalpy of combustion of ethene gas,  , is  at  . Given the following enthalpies of formation, calculate  for  .

Explanation:

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As a general rule of thumb, the ratio of the rate of etch-product formation to the flow rate of etch gas should be greater than
g100num [7]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

8 0
3 years ago
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa (50 ksi ) is exposed to a stress of 2023 MPa
Neko [114]

Answer:

Explanation:

The formula for critical stress is

\sigma_c=\frac{K}{Y\sqrt{\pi a} }

\sigma_c =\texttt{critical stress}

K is the plane strain fracture toughness

Y is dimensionless parameters

We are to Determine the Critical stress

Now replacing the critical stress with 54.8

a with 0.2mm = 0.2 x 10⁻³

Y with 1

\sigma_c=\frac{54.8}{1\sqrt{\pi  \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa

The fracture will not occur because this material can handle a stress of 2186.20Mpa  before fracture. it is obvious that is greater than 2023Mpa

Therefore, the specimen does not failure for surface crack of 0.2mm

4 0
3 years ago
A nanometer connected to a pipe indicates a negative gauge pressure of 70mm of mercury .what is the pressure in the pipe in N/m^
prohojiy [21]

Based on the calculations, the absolute pressure in the pipe is equal to 90,670.4‬ N/m².

<h3>How to calculate the absolute pressure?</h3>

Mathematically, absolute pressure can be calculated by using this formula:

P_{abs} = P_{atm} + P_{guage}

<u>Scientific data:</u>

Atmospheric pressure (Patm) = 1 bar = 1 × 10⁵ N/m²

Head (h) = -70mmhg = -0.07 m hg

Acceleration due to gravity = 9.8 m/s²

Density of mercury = 13,600 kg/m³.

Next, we would determine the gauge pressure:

P_{guage} = \rho gh\\\\P_{guage} =  -13600 \times 9.8 \times 0.07\\\\P_{guage} = -9,329.6\;N/m^2

Now, we can calculate the absolute pressure:

P_{abs} = P_{atm} + P_{guage}\\\\P_{abs} = 1 \times 10^5 - 9329.6

Absolute pressure = ‭90,670.4‬ N/m².

Read more on absolute pressure here: brainly.com/question/10013312

#SPJ1

6 0
2 years ago
A rigid bar ABCD is pinned at A and supported by two steel rods connected at B and C, as shown. There is no strain in the vertic
mylen [45]

Answer:

See attached picture.

Explanation:

4 0
3 years ago
H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation f s m o o t h ≈ 0.3164 Re − 1 / 4 fsmooth≈
tiny-mole [99]

Answer:

Therefore the angle  the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

Explanation:

The first step to take is to calculate the the velocity of flow through a pipe

Q =Av

Where Q = is the discharge through pipe

A = Area of the pipe

v = the flow of velocity

We substitute 0.001 m^3/s for Q and 0.03 m for D

Q= Av

0.001=Av

Substitute π/4 D² for A

0.001 = π/4 D² (v)

v = 0.004/πD²

D = he diameter of the pipe

substitute 3 cm  for D

v=  0.004/π * [3 cm * 1 m/100 cm]²

v =1.414 m/s

Obtain fluid properties from the table Kinematic viscosity and Dynamic of water

p =1000 kg /m³

μ= 1.002 * 10^ ⁻³ N.s/m³

Thus,

we write the expression to determine  the Reynolds number of flow

Re = pvD/μ

Re = is the Reynolds number

p =density

μ = dynamic viscosity at 20⁰C

We then substitute 1000 kg /m³ in place of p, 1.002 * 10^ ⁻³ N.s/m³ for μ,

1.414 m/s for v and 0.03 m for D

Thus,

Re = 1000 * 1.414 * 0.03/ 1.002 * 10^ ⁻³ = 42335

The next step is to calculate the friction factor form the Blasius equation

f = 0.3164 (Re)^1/4

f = friction factor

We substitute 42335 for Re

f = 0.3164 (42335)1/4

=0.022

The next step is to write the expression to determine the friction head loss

hl = flv²/2gD

hl = head loss

l = length of pipe

g=  acceleration due to gravity

We then again substitute 0.022 for f, 1.414 m/s for v, 0.03 m for D, and 9.8 m/s² for g.

so,

hl = flv²/2gD

hl/L = 0.022 * 1.414²/2 * 9.81 * 0.03

sinθ = 0.07473

θ = 4° 16'

Therefore the angle  the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

3 0
3 years ago
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