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NISA [10]
3 years ago
6

What is the friction factor for fully developed flow in a circular pipe where Reynolds number is 1000

Engineering
1 answer:
Pie3 years ago
7 0
Shear stress decreases along the flow direction. That is why the pressure drop is highest in the entrance region of a pipe, which increases the average friction factor for the whole pipe. ... In fully developed region the pressure gradient and the shear stress in flow are in balance.
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The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its s
Gnoma [55]

Answer:

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Explanation:

The length of the spring when the collar is in point A is equal to:

lA=\sqrt{0.2^{2}+0.2^{2}  }=0.2\sqrt{2}m

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B} (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2

Replacing in eq. 1:

\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+      \frac{1}{2}k(l_{A}-l_{ul})  ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul})  ^{2}

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

\frac{m_{C}v_{B}^{2}   }{R}-N_{B}-k(l_{B}-l_{ul})=0

Replacing values and clearing NB:

NB = 694 N

6 0
4 years ago
Read 2 more answers
A 993-kg car starts from rest at the bottom of a drive way and has a speed of 3.00 m/s at a point where the drive way has risen
tino4ka555 [31]

Answer:

13177.34 J

Explanation:

Work done = force × distance

work done by the engine = kinetic energy + potential energy + work done friction

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work done by friction = 2870 J

work done by engine = 5838.84 J + 2870 J + 4468.5 J = 13177.34 J

7 0
3 years ago
A turbojet aircraft is flying with a velocity of 280 m/s at an altitude of 9150 m, where the ambient conditions are 32 kPa and -
artcher [175]

Answer:

(a) The velocity of the exhaust gases. is 832.7 m/s

(b) The rate of fuel consumption is 0.6243 kg/s

Explanation:

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Use "ideal-gas specific heats of various common gases" to find the properties of air at room temperature.

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Specific heat ratio, k = 1.4

3 0
4 years ago
What is the least common denominator for the fractions 16and34?
mihalych1998 [28]

Answer:

272

Explanation:

3 0
3 years ago
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a horizontal jet of water (at 100c) that is 6 cm in diameter and has a velocity of 20 m/s is deflected by the vane as shown. if
Reptile [31]

The net resultant direct force and angle on the vane is created when the water jet exits the vane at position 2 with 92% of its initial velocity.

<h3>What is mean by velocity?</h3>
  • The speed at which a body or object is moving determines its direction of motion. A scalar quantity, speed is primarily. As a matter of fact, velocity is a vector quantity.
  • The rate at which distance changes is what it is. It measures the displacement's rate of change. A body's velocity is defined as its speed in a particular direction.
  • Velocity is a measure of how quickly a distance changes in relation to time. Having both magnitude and direction, velocity is a vector quantity.
  • The rate of change in a body's displacement with respect to time is referred to as velocity. In the SI, m/s is its unit.

Given,

External angle of Curved Vane = 158°

mean velocity at 1 = 12 m/s

Volumetric flow rate = 55 \mathrm{~m}^3 / \mathrm{h}=\frac{55}{3600} \mathrm{~m}^3 / \mathrm{s}$.

mean velocity at $2=12 \times 0.92=11.04 \mathrm{~m} / \mathrm{s}$

i) Force exerted in x - friction A C 1 =  Volume

F_{S_x} &=\rho A C_1\left[C_2 \cos \theta-C_1\right] \\

&=1000 \times \frac{55}{3600}\left[\left(11.04 \cos 158^{\circ}\right)-12\right]

i\rangle F_{\text {sc }}=\supseteq A c_1\left[C_2 \sin \theta\right] \\

&=1000 \times \frac{55}{3600} \times \text { TI. 04 } \sin (1589 \\

&F_{\text {syn }}=63.18 \mathrm{~N} \\

&\text { Angle } \Rightarrow \frac{F_{s y}}{F_{3 x}}-\tan \theta \\

&\tan \theta=\frac{63.18}{339.18}, \theta=160-10-5.3 \\

&\theta=\tan ^{-}\left(\frac{-63 \cdot 18}{339728}\right) \\

&\theta=-10.540^{\circ} \\

The complete question is:

A horizontal jet of water strikes a curved vane as shown in Figure C.1. The external angle of the curved vane is 158°.The mean velocity and volumetric flow rate of the water jet at position 1 are 12 m/s and 55 m³/h respectively. Due to friction, the water jet leaves the vane at position 2 with 92 % its original velocity.

(i) Direct force exerted by the water jet on the vane in the x - direction.

(ii) Direct force exerted by the water jet on the vane in the y - direction.

(ii) Net resultant direct force and angle on the vane.

To learn more about velocity, refer to:

brainly.com/question/24681896

#SPJ4

4 0
2 years ago
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