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il63 [147K]
3 years ago
7

Can you determine the critical distance along a flat surface?

Engineering
1 answer:
Keith_Richards [23]3 years ago
6 0

Explanation:

Consider a fluid of density, ρ moving with a velocity, U over a flat plate of length, L.

Let the Kinematic viscosity of the fluid be ν.

Let the flow over the fluid be laminar for a distance x from the leading edge.

Now this distance is called the critical distance.

Therefore, for a laminar flow, the critical distance can be defined as the distance from the leading edge of the plate where the Reynolds number is equal to 5 x 10^{5}

And Reynolds number is a dimensionless number which determines whether a flow is laminar or turbulent.  

Mathematically, we can write,

    Re = \frac{\rho .U.x}{\mu }

or 5 x 10^{5} =  \frac{\rho .U.x}{\mu } ( for a laminar flow )

Therefore, critical distance

x=\frac{5\times 10^{5}\times \mu }{\rho \times U}

So x is defined as the critical distance upto which the flow is laminar.

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A 3.7 g mass is released from rest at C which has a height of 1.1 m above the base of a loop-the-loop and a radius of 0.2 m . Th
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Answer:

Normal force = 0.326N

Explanation:

Given that:

mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg

height of the mass = 1.1 m

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acceleration due to gravity = 9.8 m/s²

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Let consider the conservation of energy relation; which says:

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However; the normal force will result to a centripetal force; as such, using the relation

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Normal force = 0.326N

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Air enters an adiabatic compressor through 0:5m2 opening and exhausts through a 0:2m2 opening. The inlet conditions of air are 2
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Answer:

i) 43.55 kg/s

ii) 40 m/s

iii)  -199.32 KW

Explanation:

To resolve the above question we have to make some assumptions :

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  • The only interactions that are between the system and the surrounding are work and heat
  • The fluid is uniform

i) first we have to determine the mass flow rate of the air

M = (\frac{P1}{RT1} )A1v1

    = ( \frac{200}{0.287*325} ) 0.5 * V1 ---------- (1) hence  M = 43.55 kg/s

ii) using this relationship : A1V1 = A2V2 hence V1 = (0.2/0.5) * 100 = 40m/s (    inlet velocity )

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attached below

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