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Levart [38]
3 years ago
9

Identify the correct statements in the context of friction factors of laminar and turbulent flows

Engineering
1 answer:
soldi70 [24.7K]3 years ago
5 0

Answer:

a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces.

Explanation:

Turbulent flow is a type of fluid flow in which fluid will undergo irregular fluctuations. The tubes with rough surfaces have higher friction factors than the tubes with smooth surfaces. In laminar flow the effect of effect of surface roughness is negligible on friction factors.

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Find the compressibility factor Z for oxygen at 3 MPa and 160 K.
saveliy_v [14]

Answer:

Z= 0.868

Explanation:

Given that

P= 3 MPa

T = 160 K

We know that

P v= Z R T

P= Pressure

v = specific volume

R= gas constant

T = Absolute temperature

Z=  Compressibility factor

Here specific volume of gas is not given so we assume that specific volume gas

v=0.012\ m^3/kg

We know that for oxygen gas constant

R = 0.259 KJ/kg.K

Now by putting the values

P v = Z R T

3000 x 0.012 = Z x 0.259 x 160

Z= 0.868

So  Compressibility factor is 0.868.

5 0
3 years ago
When its 100-hp engine is generating full power, a small airplane with mass 700 kggains altitude at a rate of 2.5 m/s. What frac
snow_lady [41]

For the given problem it is necessary to recap the concepts about Power, that is, is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time.

The equation for power can be written as

P = FV

Where,

F= Net Force

V =Velocity

By the second newton law, force can be:

F = mg

Where m means the mass and g the gravity acceleration.

We can also write the equation as,

P = mgv

Replacing the values

P = 700*9.8*25

P = 17.15kW

Tenemos unidades en dos sistemas diferentes, por lo que convertimos los HP a Sistema internacional y tenemos que

1hp = 0.746kW

Then the fraction \etais,

\eta = \frac{17.15kW}{0.746kW}

\eta = 22.989

Therefore the fraction of the engine power is being used to make the airplane climb is 22.984%

5 0
3 years ago
Compression is maintained during combustion because on top of the motor is a _____
stealth61 [152]

Answer:

fan

Explanation:

7 0
3 years ago
A three-story school has interior column bays that are spaced 25 ft apart in both directions. If the loading on the flat roof is
Lady_Fox [76]

Answer:

Explanation:

Floor Load:

Lo= 50psf

At= 25x25 = 625 square feet

L= Lo(0.25 +15/\sqrt{KuAt)}

L=50(0.25+15/\sqrt{(4)(625)}= 13.1psf

%reduction= 13.1/50 = 26%

Fr= 3[(13.1psf)(25ft)(25ft)+(20psf)(25ft)(25ft)]= 62k

7 0
3 years ago
Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one e
marshall27 [118]

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

\Delta x=\frac{PL^3}{3EI}

Hence we can write

P=\frac{3EI\cdot \Delta x}{L^3}

Comparing with the standard spring equation F=kx we find the cantilever analogous to spring with k=\frac{3EI}{L^3}

Now the angular frequency of a spring is given by

\omega =\sqrt{\frac{k}{m}}

where

'm' is the mass of the load

Thus applying values we get

\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}

\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec

8 0
3 years ago
Read 2 more answers
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