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Oksi-84 [34.3K]
4 years ago
6

Discuss the trends in reaction forces versus jet velocity. Is the trend consistent with the theory? Does it make sense?

Engineering
1 answer:
Snowcat [4.5K]4 years ago
3 0

Answer:

The rate of fluid motion(Jet Velocity) exert a force on the object in contact with it. This force is also knowns as reactions forces.

In theory, this is related to Newton Second of motion which States that:

The rate of change of momentum is directly proportional to impressed force.

This makes sense and it is consistent with theory. Detailed explanation below:

Explanation:

A jet which can be illustrated as a moving fluid, in natural or artificial systems, may exert forces on objects in contact with it.

To analyze fluid motion, a finite region of the fluid (control volume) is usually selected, and the gross effects of the flow, such as its force or torque on an object, is determined by calculating the net mass rate that flows into and out of the control volume.

These forces can be determined, as in solid mechanics, by the use of Newton’s second law, or by the momentum equation(Consistent with the theory). The force exerted by a jet of fluid on a flat or curve surface can be resolved by applying the momentum equation. The study of these forces is essential to the study of fluid mechanics and hydraulic machinery.

In practice, Engineers and designers use the momentum equation to accurately calculate the force that moving fluid may exert on a solid body. For example, in hydropower plants, turbines are utilized to generate electricity. Turbines rotate due to force exerted by one or more water jets that are directed tangentially onto the turbine’s vanes or buckets. The impact of the water on the vanes generates a torque on the wheel, causing it to rotate and to generate electricity.

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Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
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Answer:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

h = 20 \frac{W}{m^2 K} represent the heat transfer coefficient.

p represent the perimeter for this case and would be given by:

p = 2*0.05m +2*0.001m= 0.102m

k = 200 \frac{W}{m C} represent the thermal conductivity

w = 5cm =0.05 m represent the width

h = 1mm =0.001m represent the thickness

A= wh= 0.05m *0.001m = 0.00005 m^2

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}

Where m is a coefficient given by:

m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}

The value of x for this case represent the distance x =5 cm =0.05m

T_b =130 C represent the base temperature

T_{\infty}= 20 represent the temperature of the sorroundings or the ambient.

If we replace we have this:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

3 0
3 years ago
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