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faust18 [17]
3 years ago
14

Identify the fluid property responsible for the development of the velocity boundary layer and the kinds of fluids. For the idea

lized inviscid fluids, there is no velocity boundary layer. For what kinds of fluids will there be no velocity boundary layer in a pipe?
Engineering
1 answer:
jarptica [38.1K]3 years ago
4 0

Answer:

The fluid property responsible for the development of velocity boundary layer is majorly the fluid's viscosity.

For non-viscous fluids (in theory, because no fluid is entirely non-viscous), there will be no velocity boundary layer.

Explanation:

The velocity boundary layer is the thin layer of viscous fluid that is in direct contact with the pipe surface. The velocity of fluid in this layer is 0 as fluid doesn't move in this layer.

This phenomenon is due to the viscosity of the fluid. Viscosity of the fluid refers to the internal friction that exists between fluid layers, so, the layer of fluid in contact with non-moving, static surface of the pipe experiences friction that causes this layer to not move, causing the fluid velocity to vary from 0 at this surface to the maximum value at the centre of the pipe, before the velocity begins to drop again until it reaches 0 at the other end of the circular pipe.

Since viscosity is the primary cause of this, non-viscous or inviscid fluids are saved from this phenomenon as their flows do not have the velocity boundary layer.

Although, a completely non-viscous or inciscid fluid is an idealized concept because all fluids will experience some sort of viscosity (no matter how small) between their fluid layers. Hence, a velocity boundary layer, no matter how thin (or of minute thickness), will exist in the flow of real fluids.

But, an idealized non-viscous or inviscid fluid will not have a velocity boundary layer.

Hope this Helps!!!

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What is the thermal efficiency of this reheat cycle in terms of enthalpies?
schepotkina [342]

Answer:

   \eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}

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For close gas turbine:

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Now take h_{1},h_{2},,h_{3},h_{4},h_{5},h_{6} represent the enthalpy of point 1,2,3,4,5,6 in the cycle respectively.

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W_{net}=\left (h_5-h_6\right )-\left (h_2-h_1\right )

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4 years ago
Purely resistive loads of 24kW, 18kW and 12kW are connected between the neutral and the red, yellow and the blue respectively of
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Answer:

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Yellow phase = 18kW

Blue phase = 12kW

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For a star connected system, we have;

Phase voltage (V_{p} ) = \frac{Line voltage}{\sqrt{3}}

Phase voltage (V_{p} ) = \frac{415}{\sqrt{3}}

Phase voltage (V_{p} ) = 239.6V

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V_{R} = 239.6

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<em>For the Yellow phase;</em>

I_{Y} = \frac{18000}{239.6

I_{Y} = 75.13< - 120A

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I_{B} = \frac{12000}{239.6

I_{B} = 50.08

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I_{N} =\sqrt{ (I_{R}^{2} +I_{Y}^{2}+I_{B}^{2}-I_{R}I_{Y}-I_{Y}I_{B}-I_{R}I_{B}

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