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3 years ago

In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?

Engineering

1 answer:

taurus [48]3 years ago

3 0

Answer:

<em>No, the velocity profile does not change in the flow direction.</em>

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, <em>then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.</em>

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A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr

Hoochie [10]

Answer:

$\Delta m = 102.25\,kg$

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

$m_{o} = \rho_{air}\cdot V_{tank}$

$m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{o} = 31.25\,kg$

The final mass inside the rigid tank is:

$m_{f} = \rho \cdot V_{tank}$

$m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{f}= 133.5\,kg$

The supplied air mass is:

$\Delta m = m_{f}-m_{o}$

$\Delta m = 133.5\,kg-31.25\,kg$

$\Delta m = 102.25\,kg$

4 0

3 years ago

How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?

Murljashka [212]

Answer:

hello your question is incomplete attached below is the missing part of the question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = $\frac{q^2Nd^2*Xn^2}{6Eo} * f$

also note that ; Pactive ∝ Nd2 (

tD = K . $\frac{Vdd}{(Vdd - Vt )^2}$ since K = constant

Hence : Nd ∝ rt

5 0

3 years ago

A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft an

Zigmanuir [339]

Answer:

3.03 INCHES

Explanation:

According to ASTM D198 ;

Modulus of rupture = ( M / I ) * y ----- ( 1 )

M ( bending moment ) = R * length of span / 2

= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in

I ( moment of inertia ) = bd^3 / 12

= ( 2 )*( d )^3 / 12 = 2d^3 / 12

b = 2 in , d = ?

length of span = 4 * 12 = 48 inches

R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib

y ( centroid distance ) = d / 2 inches

back to equation ( 1 )

( M / I ) * y

940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2

= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2

940300 = 34560000* d / 4d^3

4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )

4d^2 = 34560000 / 940300

d^2 = 9.188 ∴ Value of d ≈ 3.03 in

8 0

3 years ago

Technician A says high resistance causes an increase in current flow. Technician B says a higher than

Kryger [21]

The correct statement is: a higher than a normal voltage drop could indicate high resistance. Technician B is correct.

Ohm's law states that the current flowing through a metallic conductor is directly proportional to the voltage provided all physical conditions are constant. Mathematically, it is expressed as

V = IR

Where

V is the potential difference

I is the current

R is the resistance

<h3>Technician A</h3>

High resistance causes an increase in current flow

V = IR

Divide both side by I

R = V / I

Thus, technician A is wrong as high resistance suggest low current flow

<h3>Technician B</h3>

Higher than normal voltage drop could indicate high resistance

V = IR

Thus, technician B is correct as high voltage indicates high resistance

<h3>Conclusion </h3>

From the above illustration, we can see that technician B is correct

Learn more about Ohm's law:

brainly.com/question/796939

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3 years ago

A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r

velikii [3]

Answer:

Explanation:

4140-40 I’d pick wood

I hope this helps! :)

4 0

3 years ago

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