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3 years ago

A fluidis flowing through a capillary

Engineering

1 answer:

Illusion [34]3 years ago

6 0

Answer:

The velocity of the fluid is 1.1012 m/s

Solution:

As per the question, for the fluid:

Diameter of the capillary tube, d = 1.0 mm = $1.0\times 10^{- 3} m$

Reynolds No., R = 1000

Kinematic viscosity, $\mu_{k} = 1.1012\times 10^{- 6} m^{2}/s$

Now, for the fluid velocity, we use the relation:

$R = \frac{v_{f}\times d}{\mu_{k}}$

where

$v_{f}$ = velocity of fluid

$v_{f} = \frac{R\times \mu_{k}}{d}$

$v_{f} = \frac{1000\times 1.1012\times 10^{- 6}}{1.0\times 10^{- 3}} = 1.1012 m/s$

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Consider the following set of processes, with the length of the CPU burst given in milliseconds:Process Burst Time PriorityP1 2

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Answer and Explanation:

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3 years ago

A projectile is launched horizontally 1m above the ground. If it lands 300m away from the initial launch position, find: a)-the

PIT_PIT [208]

Answer:

(a): The launch velocity is Vx= 666.66 m/s.

(b): The angle wich the projectile contacts the ground is α= 0.38°

Explanation:

h= 1m

g= 9.8 m/s²

h= g*t²/2

t= 0.45 s

Vy= g*t

Vy= 4.42 m/s

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3 years ago

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Answer:

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Explanation:

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3 years ago

Read 2 more answers

Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$

$h_{pump} = 7.705\,m$

The power that pump adds to the fluid is:

$\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}$

$\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)$

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

4 0

3 years ago

In successive an object moves from start position, then moves 1ft, 4ft and 8ft. This is an example of a. non-uniform motion b. u

Nikolay [14]

Answer:Non-uniform motion

Explanation:

This is an example of Non-uniform motion because unequal distance traveled by an object in equal interval of time is termed as Non-Uniform motion and here also the distance traveled by object is 1 ft ,4 ft and 8 ft which is different from each other.

In Uniform motion distance traveled is equal in equal interval of time .

6 0

3 years ago