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podryga [215]
1 year ago
10

What is the value of the thickness of the boundary layer at leading edge of the plate?

Physics
1 answer:
Harman [31]1 year ago
7 0

Along the flow direction, the boundary layer's thickness varies. For ReT values between 260 and 780, the boundary layer thickness—defined as the depth at which the normalized concentration has a value of 1/e—ranges between 800 and 250 m.

<h3>What is the Boundary layer?</h3>
  • A boundary layer is the thin layer of fluid that forms immediately around a bounded surface in physics and fluid mechanics as a result of the fluid flowing along the surface.
  • A no-slip boundary condition is created as a result of the fluid and wall interaction (zero velocity at the wall).
  • After that, the flow velocity above the surface steadily rises until it reaches the bulk flow velocity again.
  • The term "velocity boundary layer" refers to the thin layer of fluid whose velocity has not yet recovered to that of the main flow.
<h3>What purpose does a boundary layer serve?</h3>
  • Because it is on the fluid's boundary, engineers refer to this layer as the boundary layer.
  • Many aerodynamics issues, such as wing stall, skin friction drag on an object, and the heat transfer that takes place in high-speed flight, depend heavily on the specifics of the flow within the boundary layer.

Learn more about Boundary layer here:

brainly.com/question/12974964

#SPJ4

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To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
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Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

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